\nWe know that if we divide or multiply a polynomial by any constant (real number), then the zeroes of polynomial remains same.
\nHere, \u03b1 = \u20142 and \u03b2 = +5
\n\u03b1 + \u03b2 = \u20142 + 5 = 3 and \u03b1\u03b2 = \u2014 2 x 5 =- 10
\nSo, required polynomial is
\nx2 \u2014 (\u03b1 + \u03b2)x + \u03b1\u03b2 = x2 \u2014 3x \u201410
\nIf we multiply this polynomial by any real number, let 5 and 2, we get 5x2<\/sup> \u2014 15x \u2014 50 and 2x2<\/sup> \u2014 6x \u2014 20 which are different polynomials but having same zeroes -2 and 5. So, we can obtain so many (infinite polynomials) from two given zeroes.<\/p>\n
![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions.png?resize=159%2C13&ssl=1\")
<\/p>\n
Question 2.<\/p>\n
Given that one of the zeroes of the cubic polynomial ax3<\/sup>+ bx2<\/sup> + cx + d is zero, the product of the other two zeroes is:<\/h2>\n(A) \\(-\\frac{c}{a}\\)
\n(B) \\(\\frac{c}{a}\\)
\n(C) 0
\n(D)\\(-\\frac{b}{a}\\)<\/p>\n
Answer:
\n(B) \\(\\frac{c}{a}\\)<\/p>\n
Explanation:
\nLet\/(x) = ax3<\/sup>+ bx2<\/sup> + cx + d If a, p, y are the zeroes of f(x), then
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 =\\(=\\frac{c}{a}\\)
\nOne root is zero (given) so, \u03b1 = 0. \u03b2\u03b3 =\\(=\\frac{c}{a}\\)<\/p>\nQuestion 3.<\/p>\n
If one of the zeroes of the cubic polynomial x3<\/sup> + ax2<\/sup> + bx + c is \u2014 1, then the product of the other two zeroes is:<\/h2>\n(A) b – a +1
\n(B) b – a – 1
\n(C) a – b + 1
\n(D) a – b – 1
\nAnswer:
\n(A) b – a +1<\/p>\n
Explanation:
\nLet f(x) = x3<\/sup> + ax2<\/sup> + bx + c
\n\u2234 One of the zeroes of f(x) is – I so
\nf(-1) = 0
\n(-1)3<\/sup> + a(-1)2<\/sup> + b(-1) + c = 0
\n-1 + a – b + c = 0
\na – b + c = 1
\nc = 1 + b – a
\nNow, \u03b1\u03b2\u03b3 = \\(\\frac{-d}{a}\\) [\u2234a = 1, d = c]
\n\\(-1 \\beta y=\\frac{-c}{1}\\)
\n\u03b2\u03b3 = c
\n\u03b2\u03b3 = 1 + b – a<\/p>\n<\/p>\n
Question 4.<\/p>\n
The zeroes of the quadratic polynomial x2<\/sup> + 99x 127 are:<\/h2>\n(A) both positive
\n(B) both negative
\n(C) one positive and one negative
\n(D) both equal
\nAnswer:
\n(B) both negative<\/p>\n
Explanation:
\nLet given quadratic polynomial
\np(x) = x2<\/sup>+ 99x + 127
\nOn comparing p(x) with ax2<\/sup>+ bx + c. we get a = 1,b = 99 and c = 127
\nWe know that,
\nx = \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\)
\n= \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\)
\n= \\(\\frac{-99 \\pm \\sqrt{9801-508}}{2}\\)
\n= \\(\\frac{-99 \\pm \\sqrt{9293}}{2}=\\frac{-99 \\pm 96.4}{2}\\)
\n= \\(\\frac{-99+96.4}{2}, \\frac{-99-96.4}{2}\\)
\n= \\(\\frac{-2.6}{2}, \\frac{-195.4}{2}\\)
\n= -1.3, – 97.7
\nHence, both zeroes of the given quadratic polynomial p(x) are negative.<\/p>\nQuestion 5.<\/p>\n
The zeroes of the quadratic polynomial x2<\/sup> + kx + k, k \u2260 0,<\/h2>\n(A) cannot both be positive
\n(B) cannot both be negative
\n(C) are always unequal
\n(D) are always equal
\nAnswer:
\n(A) cannot both be positive<\/p>\n
Explanation:
\nLet f(x) = x 2<\/sup> + kx + k, k 0.
\nOn comparing the given polynomial with ax2<\/sup> + bx + c, we get a = 1, b – k,c = k
\nIf \u03b1 and \u03b2 be the zeroes of the polynomial (x).
\nWe know thaL,
\nSum of zeroes, \u03b1 + \u03b2 = \\(-\\frac{b}{a}\\)
\n\u03b1 + \u03b2 = \\(-\\frac{k}{1}\\) = -k ……(i)
\nAnd ofzeroes, \u03b1\u03b2 = \\(\\frac{c}{a}\\)
\n\u03b1\u03b2 = \\(\\frac{k}{1}\\)=k
\nCase I: If k is negative, \u03b1\u03b2[from equation (ii)] is negative. It means \u03b1 and \u03b2 are of opposite sign.
\nCase II: If k is positive, then \u03b1\u03b2 [from equation (ii)] is positive but \u03b1 + \u03b2 is negative. If, the product of two numbers is positive, then either both are negative or both are positive. But the sum of these numbers is negative, so numbers must be negative. Hence, in any case zeroes of the given quadratic polynomial cannoL both be positive.<\/p>\nQuestion 6.<\/p>\n
If the zeroes of the quadratic polynomial ax2<\/sup> + bx + c, a \u2260 0 are equal, then :<\/h2>\n(A) c and a have opposite signs
\n(B) c and b have opposite signs
\n(C) c and a have the same sign
\n(D) c and b have the same sign
\nAnswer:
\n(C) c and a have the same sign<\/p>\n
Explanation:
\nL Tor equal roots b2<\/sup> – 4ac = 0 b2<\/sup> = 4ac
\nb2<\/sup> is always positive so 4ac must be positive, i.e., product of a and r must be positive, i.e., a and r must have same sign either positive or negative.<\/p>\nQuestion 7.<\/p>\n
If one of the zeroes of a quadratic polynomial of the form x2<\/sup> + ax + b is the negative of the other, then it<\/h2>\n(A) has no linear term and the constant term is negative.
\n(B) has no linear term and the constant term is positive.
\n(C) can have a linear term but the constant term is negative.
\n(D) can have a linear term but the constant term is positive. [U]
\nAnswer:
\n(A) has no linear term and the constant term is negative.<\/p>\n
Explanation:
\nLet f(x) = x2<\/sup> + ax + b and \u03b1, \u03b2 are tha roots of it.
\nthen, \u03b2 = – \u03b1(given)
\n\u03b1 + \u03b2 = \\(-\\frac{b}{a}\\) and \u03b1\u03b2 = \\(\\frac{c}{a}\\)
\n\u03b1 – \u03b1 = \\(-\\frac{a}{1}\\) and \u03b1(-\u03b1) = \\(\\frac{b}{1}\\)
\n– a = 0 – \u03b12<\/sup> = b
\na = 0 b<0 or b is negative
\nso, f(x) = x2<\/sup> + b shows that it has no linear term<\/p>\n<\/p>\n
Question 8.<\/p>\n
Which of the following is not the graph of a quadratic polynomial?<\/h2>\n
![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-22.png?resize=316%2C225&ssl=1\")
<\/p>\n
Answer:
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-20-1.png?resize=230%2C117&ssl=1\")
<\/p>\n
Explanation:
\nGraph (D) intersect at three points cm .v-axis so the roots ot polynomial of graph is three, so it is cubic polynomial. Other graphs are of quadratic polynomial. Graph a have no real zeroes and Graph b has coincident zeroes.<\/p>\n
Assertion and Reason Based MCQs<\/span><\/p>\nDirections: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\n
Question 1.<\/p>\n
Assertion (A): If the zeroes of a quadratic polynomial ax2<\/sup> + bx + c are both positive, then a, b and c all have the same sign.
\nReason (R): If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.<\/h2>\nAnswer:
\n(D) A is false and R is True<\/p>\n
Explanation:
\nIn case of assertion: Let \u03b1 and \u03b2 be the roots of the quadratic polynomial. If \u03b1 and \u03b2 are positive then
\n\u03b1 + \u03b2 = \\(\\frac{-b}{a}\\) it shows that \\(\\frac{-b}{a}\\) is negetive a sum of two positive numbers (\u03b1 , \u03b2) must be Tve i.c., either b or a must be negative. So, a, b and c will have different signs.
\n\u2234 Given statement is incorrect.
\nIn case of reason:
\nLet \u03b2 = 0, \u03b3 =0
\nf(x) = (x – \u03b1) (x – \u03b2) (x – \u03b3)
\n= (x – \u03b1) x.<\/sup>x
\nf(x) = x3<\/sup>– ax2<\/sup>
\nwhich has no linear (coefficient of x) and constant terms. Given statement is correct. Thus, assertion is incorrect but reason is correct.<\/p>\nQuestion 2.<\/p>\n
Assertion (A): The value of k for which the quadratic polynomial kx2<\/sup> + x + k has equal zeroes are \u00b1\\(\\frac{1}{2}\\)
\nReason (R): If all the three zeroes of a cubic polynomial x 3<\/sup> + ax2<\/sup> – bx + c are positive, then at least one of a, b and c is non-negative.<\/h2>\nAnswer:
\n(C) A is true but R is false<\/p>\n
Explanation:
\nIn case of assertion:
\nf(x) = kx2<\/sup> + x + k (a = k, b = 1, c = k)
\nLor equal roots b2<\/sup>– 4ac = 0
\n(1)2<\/sup> \u2014 4(k) (k) = 0
\n4k2<\/sup>=1
\nk2<\/sup> = \\(\\frac{1}{4}\\)
\nk = \\(\\pm \\frac{1}{2}\\)
\nSo, there are \\(+\\frac{1}{2}\\) and \\(-\\frac{1}{2}\\) values of k so that the given equation has equal roots.
\n\u2234 Given statement is correct.
\nIn case of reason:
\nAll the zeroes’of cubic polynomial are positive only when all the constants a, b, and c are i negative.
\n\u2234Given statement is incorrect.
\nThus, assertion is correct but reason is incorrect.
\nQuestion 3.<\/p>\nAssertion (A): The graph of y = p(x), where p(x) is a polynomial in variable x, is as follows:
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-21.png?resize=237%2C146&ssl=1\")
\nThe number of zeroes of p(x) is 5.
\nReason (R): If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.<\/h2>\n
Answer:
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
Explanation:
\nIn case of assertion: Since the graph touches the r-axis 5 times. So, the number of zeroes of p{x) is 5.
\n\u2234Given statement is correct.
\nIn case of reason:
\nIf a polynomial of degree more than two hastwo . real zeroes aqd other zeroes are not real or are \u2022 imaginary, and then graph of the polynomial jj will intersect at two points on x-axis.
\n\u2234Given statement is correct:
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\n
<\/p>\n
Question 4.<\/p>\n
Assertion (A): If the zeroes of the quadratic polynomial (k – 1) x2<\/sup> + kx + 1 is – 3, then the value of k is \\(\\frac{4}{3}\\)
\nReason (R): If – 1 is a zero of the polynomial p(x) = kx2<\/sup> -4x + k, then the value of k is -2.<\/h2>\nAnswer:
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
Explanation:
\nIn Case of assertion:
\nLet p(x) = (k – 1)x2<\/sup> + kx + 1
\nAs -3 is a zero of p(x) then p(-3) = 0
\n(k-1)(-3)2 + k(-3) + 1 = 0
\n9k- 9-3k + 1 = 0
\n9k – 3k = + 9 – 1
\n6k = 8
\nk = \\(\\frac{4}{3}\\)<\/p>\n\u2234 Given statement is correct.
\nIn case of reason:
\nSince,-1is a zero of the polynomial and
\np (x) = kx2<\/sup>– 4x + k,
\nP (-1) = 0
\nCk(-1) 2<\/sup> -4(-1) + k = 0
\nk + 4 + k = 0
\n2k + 4 = 0
\n2k = -4
\nk = -2
\n\u2234 Given statement is correct.
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\nCase -Based MCQs<\/span><\/p>\nAttempt any four sub-parts from each question. Each sub-part carries 1 mark.
\nI. Read the following text and answer the following questions on the basis of the same:
\nThe below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-1.png?resize=383%2C189&ssl=1\")
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-2.png?resize=317%2C348&ssl=1\")
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-3.png?resize=359%2C245&ssl=1\")
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-4.png?resize=357%2C266&ssl=1\")
<\/p>\n
Question 1.<\/p>\n
In the standard form of quadratic polynomial, ax2<\/sup> + bx, c, a, b and c are<\/h2>\n(A) All are real numbers.
\n(B) All are rational numbers.
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.
\n(D) All are integers.
\nAnswer:
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.<\/p>\n
Question 2.<\/p>\n
If the roots of the quadratic polynomial are equal, where the discriminant D = b2<\/sup>– 4ac, then<\/h2>\n(A) D > 0
\n(B) D < 0
\n(C) D
\n(D) D = 0
\nAnswer:
\n(D) D = 0<\/p>\n
Explanation:
\nIf the roots of the quadratic polynomial are equal, then discriminant is equal to zero
\nD = b2<\/sup> – 4ac = 0<\/p>\nQuestion 3.<\/p>\n
If a are \\(\\frac{1}{\\alpha}\\) the zeroes of the quadratic polynomial a 2x2<\/sup>– a + 8k, then k is<\/h2>\n(A) 4
\n(B) \\(\\frac{1}{4}\\)
\n(C)\\(\\frac{-1}{4}\\)
\n(D) 2
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
Explanation:
\nGiven equation, 2x2<\/sup> – x + 8k
\nSum of zeroes = \u03b1 +\\(+\\frac{1}{\\alpha}\\)
\nProduct of zeroes = \u03b1. \\(\\frac{1}{\\alpha}\\)=1
\nProduct of zeroes = \\(\\frac{c}{a}=\\frac{8 k}{2}\\)
\n\\(\\begin{aligned}
\n\\frac{8 k}{2} &=1 \\\\
\nk &=\\frac{2}{8} \\\\
\nk &=\\frac{1}{4}
\n\\end{aligned}\\)<\/p>\n<\/p>\n
Question 4.<\/p>\n
The graph of x2<\/sup> + 1 = 0<\/h2>\n(A) Intersects a-axis at two distinct points.
\n(B) Touches a-axis at a point.
\n(C) Neither touches nor intersects a-axis.
\n(D) Either touches or intersects a-axis.
\nAnswer:
\n(C) Neither touches nor intersects a-axis.<\/p>\n
Question 5.<\/p>\n
If the sum of the roots is -p and product of the roots is \\(-\\frac{1}{p}\\) , then the quadratic polynomial is<\/h2>\n
\\(\\text { (A) } k\\left(-p x^{2}+\\frac{x}{p}+1\\right)\\)
\n\\(\\text { (B) } k\\left(p x^{2}-\\frac{x}{p}-1\\right)\\)
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)
\n\\(\\text { (D) } k\\left(x^{2}+p x+\\frac{1}{p}\\right)\\)<\/p>\n
Answer:
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)<\/p>\n
II. Read the following text and answer the following questions on the basis of the same:
\nAn asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-5.png?resize=319%2C224&ssl=1\")
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-6.png?resize=238%2C272&ssl=1\")
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-7.png?resize=312%2C188&ssl=1\")
<\/p>\n
Question 1.<\/p>\n
The shape of the poses shown is<\/h2>\n
(A) Spiral
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\n
Question 2.<\/p>\n
The graph of parabola opens downwards, if<\/h2>\n
(A) a \u2265 0
\n(B) a = 0
\n(C) a < 0
\n(D) a > 0
\nAnswer:
\n(C) a < 0<\/p>\n
Question 3.<\/p>\n
In the graph, how many zeroes are there for the polynomial?<\/h2>\n
![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-8.png?resize=282%2C288&ssl=1\")
\n(A) 0
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(C) 2<\/p>\n
Question 4.<\/p>\n
The two zeroes in the above shown graph are<\/h2>\n
(A) 2,4
\n(B) -2,4
\n(C) -8,4
\n(D) 2,-8
\nAnswer:
\n(B) -2,4<\/p>\n
Question 5.<\/p>\n
The zeroes of the quadratic polynomial \\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)are<\/h2>\n
\\(
\n[latex]\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (C) } \\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (D) }-\\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
Explanation:
\n\\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)
\n\\(\\begin{aligned}
\n&=4 \\sqrt{3} x^{2}+(8-3) x-2 \\sqrt{3} \\\\
\n&=4 \\sqrt{3} x^{2}+8 x-3 x-2 \\sqrt{3} \\\\
\n&=4 x(\\sqrt{3} x+2)-\\sqrt{3}(\\sqrt{3} x+2) \\\\
\n&=(\\sqrt{3} x+2)(4 x-\\sqrt{3})
\n\\end{aligned}\\)
\nHence, zeroes of polynomial\\(\\frac{-2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
III. Read the following text and answer the following questions on the basis of the same:
\nBasketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoor on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing quadratic polynomial.
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-9.png?resize=342%2C239&ssl=1\")
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-10.png?resize=359%2C207&ssl=1\")
<\/p>\n
Question 1.<\/p>\n
The shape of the path traced shown is<\/h2>\n
(A) Spiral
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\n
<\/p>\n
Question 2.<\/p>\n
The graph of parabola opens upwards, if ………..<\/h2>\n
(A) a = 0
\n(B) a < 0 (C) a>0
\n(D) o \u2265 0
\nAnswer:
\n(C) a>0<\/p>\n
Question 3.<\/p>\n
Observe the following graph and answer
\n![\"MCQ](\"https:\/\/i2.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-11.png?resize=400%2C290&ssl=1\")
\nIn the above graph, how many zeroes are there for the polynomial?<\/h2>\n
(A) 0
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(D) 3<\/p>\n
Explanation:
\nThe number of zeroes of polynomial is the number of times the curve intersects the x-axis, i.e. attains the value 0. Here, the polynomial meets the x-axis at 3 points. So, number of zeroes = 3.<\/p>\n
Question 4.<\/p>\n
The three zeroes in the above shown graph are<\/h2>\n
(A) 2,3, -1
\n(B) -2, 3,1
\n(C) -3, -1,2
\n(D) -2, -3, -1
\nAnswer:
\n(C) -3, -1,2<\/p>\n
Question 5.<\/p>\n
What will be the expression of the polynomial?<\/h2>\n
(A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6
\n(B) x3<\/sup> + 2 x2<\/sup> – 5 x – 6
\n(C) x3<\/sup> + 2 x2<\/sup>+ 5 x-6
\n(D) x3<\/sup> + 2 x2<\/sup> + 5 x + 6
\nAnswer:
\n(A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6<\/p>\nExplanation:
\nSince, the three zeroes = – 3, -1,2 Hence, the expression is (x + 3)(x + l)(x – 2)
\n= [x2<\/sup> + x + 3 x + 3] (x – 2)
\n= x3<\/sup>+ 4x2<\/sup> + 3x-2x2<\/sup>-8x-6
\n= x3<\/sup>+ 2x2<\/sup> – 5x – 6<\/p>\nlV. Read the following text and answer the following questions on the basis of the same: Applications of Parabolas: Highway Overpasses\/ Underpasses
\nA highway underpass is parabolic in shape.
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-12.png?resize=245%2C172&ssl=1\")
\n![\"MCQ](\"https:\/\/i0.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-13.png?resize=387%2C151&ssl=1\")
\nParabolic camber equation y = 2x2<\/sup>\/nw
\nParabola
\nA parabola is the graph that results from p(x) = ax2<\/sup> + bx + c.
\nParabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the vertex.
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-14.png?resize=430%2C213&ssl=1\")
<\/p>\nQuestion 1.<\/p>\n
If the highway overpass is represented by x2<\/sup>– 2x – 8. Then its zeroes are<\/h2>\n(A) (2,-4)
\n(B) (4,-2)
\n(C) (-2,-2)
\n(D) (- 4, – 4)
\nAnswer:
\n(C) (-2,-2)<\/p>\n
Explanation:
\nx2<\/sup>– 2 x – 8 = 0
\nor, x2 <\/sup>-4x + 2x- 8 = 0
\nor, x(x – 4) + 2(x – 4) = 0
\nor, (x- 4) (x + 2) = 0
\nor, x = 4, x = – 2<\/p>\nQuestion 2.<\/p>\n
The highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial:<\/h2>\n
(A) Intersects X-axis
\n(B) Intersects Y-axis
\n(C) Intersects Y-axis or X-axis
\n(D) None of the above
\nAnswer:
\n(A) Intersects X-axis<\/p>\n
Explanation:
\nWe know that the number of zeroes of polynomial is equal to number of points where the graph of polynomial intersects X-axis.<\/p>\n
Question 3.<\/p>\n
Graph of a quadratic polynomial is a:<\/h2>\n
(A) straight line
\n(B) circle
\n(C) parabola
\n(D) ellipse
\nAnswer:
\n(C) parabola<\/p>\n
Explanation:
\nHere, the given graph of a quadratic polynomial is a parabola.<\/p>\n
<\/p>\n
Question 4.<\/p>\n
The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is:<\/h2>\n
(A) x2<\/sup> – 6x + 2
\n(B) x2<\/sup>– 36
\n(C) x2 <\/sup>– 6
\n(D) x2 <\/sup>– 3
\nAnswer:
\n(B) x2 <\/sup>– 36<\/p>\nExplanation:
\nx2<\/sup>-36 =0 => x2<\/sup> = 36
\n=>x 6, – 6. => x = \\(\\pm \\sqrt{36}\\)<\/p>\nQuestion 5.<\/p>\n
The number of zeroes that polynomial f(x) = (x – 2)2<\/sup>+ 4 can have is:<\/h2>\n(A) 1
\n(B) 2
\n(C) 0
\n(D) 3
\nAnswer:
\n(C) 0<\/p>\n
Explanation:
\nWe have,
\nf(x) = (x – 2)2<\/sup> + 4
\n= x2<\/sup> + 4 – 4 x + 4
\n=x2<\/sup>-4x + 8. i.e., It has no factorisation:
\nHence no real value of x is possible, i.e., no zero.<\/p>\nV. Read the following text and answer the following questions on the basis of the same:
\nFor a linear polynomial kx + c, k \u2260 0, the graph of y = kx + c is a straight line which intersects the x-axis at exactly one point, namely, \\(\\left(\\frac{-c}{k}, 0\\right),\\) the linear polynomial kx + c, k \u2260 0, has exactly one zero, namely, the X-coordinate of the point where the graph of y = kx + c intersects the x-axis.
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-15.png?resize=281%2C169&ssl=1\")
<\/p>\n
Question 1.<\/p>\n
If a linear polynomial is 2x + 3, then the zero of 2x + 3 is:<\/h2>\n
\\(\\text { (A) } \\frac{3}{2}\\)
\n\\(\\text { (B) }-\\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{3}{2}\\)<\/p>\n
Explanation:
\nGiven, polynomial = 2+ 3
\nLet p(x) = 2x + 3
\nFora zero of p(x),
\n2x + 3 =0
\n2x = -3
\nx = \\(-\\frac{3}{2}\\)<\/p>\n
Question 2.<\/p>\n
The graph of y = p(x) is given in figure below for some polynomial p(x). The number of zero\/zeroes of p(x) is\/are:
\n![\"MCQ](\"https:\/\/i1.wp.com\/mcq-questions.com\/wp-content\/uploads\/2022\/01\/MCQ-Questions-for-Class-10-Maths-Chapter-2-Polynomials-16.png?resize=276%2C167&ssl=1\")
<\/h2>\n
(A) 1
\n(B) 2
\n(C) 3
\n(D) 0
\nimg-10
\nAnswer:
\n(D) 0<\/p>\n
Explanation:
\nSince the graph does not intersect the x-axis therefore it has no zero.<\/p>\n
Question 3.<\/p>\n
If a and p are the zeroes of the quadratic polynomial x2<\/sup> – 5x + k such that \u03b1 – \u03b2 = 1, then the value of k is:<\/h2>\n(A) 4
\n(B) 5
\n(C) 6
\n(D) 3
\nAnswer:
\n(C) 6<\/p>\n
Explanation:
\nS \u2234p (x) = x2<\/sup>-5x + k
\n\u03b1 – \u03b2 = \\(\\frac{-(-5)}{1}\\) = 5
\n\u03b1\u03b2 = \\(\\frac{k}{1}\\) = k
\nAlso given, \u03b1 – \u03b2 = 1
\n\\(\\sqrt{(\\alpha+\\beta)^{2}-4 \\alpha \\beta}=1\\)<\/p>\nQuestion 4.<\/p>\n
If \u03b1 and \u03b2 are the zeroes of the quadratic polynomial p(x) = 4x2<\/sup> + 5x + 1, then the product of zeroes is:<\/h2>\n(A) -1
\n(B) \\(\\frac{1}{4}\\)
\n(C) -2
\n(D) \\(-\\frac{5}{4}\\)
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
Explanation:
\nWe have, p(x) = 4x2<\/sup> + 5x + 1
\n\u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{1}{4}\\)<\/p>\n<\/p>\n
Question 5.<\/p>\n
If the product of the zeroes of the quadratic polynomial p(x) = ax2<\/sup>– 6x – 6 is 4, then the value of a is: –<\/h2>\n\\(\\text { (A) }-\\frac{3}{2}\\)
\n\\(\\text { (B) } \\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (A) }-\\frac{3}{2}\\)<\/p>\n
Explanation:
\nWWe have,
\np (x) = ax2<\/sup> – 6x – 6
\nLet a and p be the zeroes of the given polynomial, then
\n\\(\\begin{aligned}
\n\\alpha \\beta &=\\frac{c}{a} \\\\
\n4 &=\\frac{-6}{a} \\\\
\n4 a &=-6 \\\\
\na &=-\\frac{6}{4}=-\\frac{3}{2}
\n\\end{aligned}\\)<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"Polynomials Class 10 MCQ Questions with Answers Question 1. The number of polynomials having zeroes as -2 and 5 is: (A) 1 (B) 2 (C) 3 (D) more than 3 Answer: (D) more than 3 Explanation: We know that if we divide or multiply a polynomial by any constant (real number), then the zeroes of …<\/p>\n
\n(B) \\(\\frac{c}{a}\\)
\n(C) 0
\n(D)\\(-\\frac{b}{a}\\)<\/p>\n
\n(B) \\(\\frac{c}{a}\\)<\/p>\n
\nLet\/(x) = ax3<\/sup>+ bx2<\/sup> + cx + d If a, p, y are the zeroes of f(x), then
\n\u03b1\u03b2 + \u03b2\u03b3 + \u03b3\u03b1 =\\(=\\frac{c}{a}\\)
\nOne root is zero (given) so, \u03b1 = 0. \u03b2\u03b3 =\\(=\\frac{c}{a}\\)<\/p>\n
Question 3.<\/p>\n
If one of the zeroes of the cubic polynomial x3<\/sup> + ax2<\/sup> + bx + c is \u2014 1, then the product of the other two zeroes is:<\/h2>\n(A) b – a +1
\n(B) b – a – 1
\n(C) a – b + 1
\n(D) a – b – 1
\nAnswer:
\n(A) b – a +1<\/p>\n
Explanation:
\nLet f(x) = x3<\/sup> + ax2<\/sup> + bx + c
\n\u2234 One of the zeroes of f(x) is – I so
\nf(-1) = 0
\n(-1)3<\/sup> + a(-1)2<\/sup> + b(-1) + c = 0
\n-1 + a – b + c = 0
\na – b + c = 1
\nc = 1 + b – a
\nNow, \u03b1\u03b2\u03b3 = \\(\\frac{-d}{a}\\) [\u2234a = 1, d = c]
\n\\(-1 \\beta y=\\frac{-c}{1}\\)
\n\u03b2\u03b3 = c
\n\u03b2\u03b3 = 1 + b – a<\/p>\n<\/p>\n
Question 4.<\/p>\n
The zeroes of the quadratic polynomial x2<\/sup> + 99x 127 are:<\/h2>\n(A) both positive
\n(B) both negative
\n(C) one positive and one negative
\n(D) both equal
\nAnswer:
\n(B) both negative<\/p>\n
Explanation:
\nLet given quadratic polynomial
\np(x) = x2<\/sup>+ 99x + 127
\nOn comparing p(x) with ax2<\/sup>+ bx + c. we get a = 1,b = 99 and c = 127
\nWe know that,
\nx = \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\)
\n= \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\)
\n= \\(\\frac{-99 \\pm \\sqrt{9801-508}}{2}\\)
\n= \\(\\frac{-99 \\pm \\sqrt{9293}}{2}=\\frac{-99 \\pm 96.4}{2}\\)
\n= \\(\\frac{-99+96.4}{2}, \\frac{-99-96.4}{2}\\)
\n= \\(\\frac{-2.6}{2}, \\frac{-195.4}{2}\\)
\n= -1.3, – 97.7
\nHence, both zeroes of the given quadratic polynomial p(x) are negative.<\/p>\nQuestion 5.<\/p>\n
The zeroes of the quadratic polynomial x2<\/sup> + kx + k, k \u2260 0,<\/h2>\n(A) cannot both be positive
\n(B) cannot both be negative
\n(C) are always unequal
\n(D) are always equal
\nAnswer:
\n(A) cannot both be positive<\/p>\n
Explanation:
\nLet f(x) = x 2<\/sup> + kx + k, k 0.
\nOn comparing the given polynomial with ax2<\/sup> + bx + c, we get a = 1, b – k,c = k
\nIf \u03b1 and \u03b2 be the zeroes of the polynomial (x).
\nWe know thaL,
\nSum of zeroes, \u03b1 + \u03b2 = \\(-\\frac{b}{a}\\)
\n\u03b1 + \u03b2 = \\(-\\frac{k}{1}\\) = -k ……(i)
\nAnd ofzeroes, \u03b1\u03b2 = \\(\\frac{c}{a}\\)
\n\u03b1\u03b2 = \\(\\frac{k}{1}\\)=k
\nCase I: If k is negative, \u03b1\u03b2[from equation (ii)] is negative. It means \u03b1 and \u03b2 are of opposite sign.
\nCase II: If k is positive, then \u03b1\u03b2 [from equation (ii)] is positive but \u03b1 + \u03b2 is negative. If, the product of two numbers is positive, then either both are negative or both are positive. But the sum of these numbers is negative, so numbers must be negative. Hence, in any case zeroes of the given quadratic polynomial cannoL both be positive.<\/p>\nQuestion 6.<\/p>\n
If the zeroes of the quadratic polynomial ax2<\/sup> + bx + c, a \u2260 0 are equal, then :<\/h2>\n(A) c and a have opposite signs
\n(B) c and b have opposite signs
\n(C) c and a have the same sign
\n(D) c and b have the same sign
\nAnswer:
\n(C) c and a have the same sign<\/p>\n
Explanation:
\nL Tor equal roots b2<\/sup> – 4ac = 0 b2<\/sup> = 4ac
\nb2<\/sup> is always positive so 4ac must be positive, i.e., product of a and r must be positive, i.e., a and r must have same sign either positive or negative.<\/p>\nQuestion 7.<\/p>\n
If one of the zeroes of a quadratic polynomial of the form x2<\/sup> + ax + b is the negative of the other, then it<\/h2>\n(A) has no linear term and the constant term is negative.
\n(B) has no linear term and the constant term is positive.
\n(C) can have a linear term but the constant term is negative.
\n(D) can have a linear term but the constant term is positive. [U]
\nAnswer:
\n(A) has no linear term and the constant term is negative.<\/p>\n
Explanation:
\nLet f(x) = x2<\/sup> + ax + b and \u03b1, \u03b2 are tha roots of it.
\nthen, \u03b2 = – \u03b1(given)
\n\u03b1 + \u03b2 = \\(-\\frac{b}{a}\\) and \u03b1\u03b2 = \\(\\frac{c}{a}\\)
\n\u03b1 – \u03b1 = \\(-\\frac{a}{1}\\) and \u03b1(-\u03b1) = \\(\\frac{b}{1}\\)
\n– a = 0 – \u03b12<\/sup> = b
\na = 0 b<0 or b is negative
\nso, f(x) = x2<\/sup> + b shows that it has no linear term<\/p>\n<\/p>\n
Question 8.<\/p>\n
Which of the following is not the graph of a quadratic polynomial?<\/h2>\n
<\/p>\n
Answer:
\n<\/p>\n
Explanation:
\nGraph (D) intersect at three points cm .v-axis so the roots ot polynomial of graph is three, so it is cubic polynomial. Other graphs are of quadratic polynomial. Graph a have no real zeroes and Graph b has coincident zeroes.<\/p>\n
Assertion and Reason Based MCQs<\/span><\/p>\nDirections: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\n
Question 1.<\/p>\n
Assertion (A): If the zeroes of a quadratic polynomial ax2<\/sup> + bx + c are both positive, then a, b and c all have the same sign.
\nReason (R): If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.<\/h2>\nAnswer:
\n(D) A is false and R is True<\/p>\n
Explanation:
\nIn case of assertion: Let \u03b1 and \u03b2 be the roots of the quadratic polynomial. If \u03b1 and \u03b2 are positive then
\n\u03b1 + \u03b2 = \\(\\frac{-b}{a}\\) it shows that \\(\\frac{-b}{a}\\) is negetive a sum of two positive numbers (\u03b1 , \u03b2) must be Tve i.c., either b or a must be negative. So, a, b and c will have different signs.
\n\u2234 Given statement is incorrect.
\nIn case of reason:
\nLet \u03b2 = 0, \u03b3 =0
\nf(x) = (x – \u03b1) (x – \u03b2) (x – \u03b3)
\n= (x – \u03b1) x.<\/sup>x
\nf(x) = x3<\/sup>– ax2<\/sup>
\nwhich has no linear (coefficient of x) and constant terms. Given statement is correct. Thus, assertion is incorrect but reason is correct.<\/p>\nQuestion 2.<\/p>\n
Assertion (A): The value of k for which the quadratic polynomial kx2<\/sup> + x + k has equal zeroes are \u00b1\\(\\frac{1}{2}\\)
\nReason (R): If all the three zeroes of a cubic polynomial x 3<\/sup> + ax2<\/sup> – bx + c are positive, then at least one of a, b and c is non-negative.<\/h2>\nAnswer:
\n(C) A is true but R is false<\/p>\n
Explanation:
\nIn case of assertion:
\nf(x) = kx2<\/sup> + x + k (a = k, b = 1, c = k)
\nLor equal roots b2<\/sup>– 4ac = 0
\n(1)2<\/sup> \u2014 4(k) (k) = 0
\n4k2<\/sup>=1
\nk2<\/sup> = \\(\\frac{1}{4}\\)
\nk = \\(\\pm \\frac{1}{2}\\)
\nSo, there are \\(+\\frac{1}{2}\\) and \\(-\\frac{1}{2}\\) values of k so that the given equation has equal roots.
\n\u2234 Given statement is correct.
\nIn case of reason:
\nAll the zeroes’of cubic polynomial are positive only when all the constants a, b, and c are i negative.
\n\u2234Given statement is incorrect.
\nThus, assertion is correct but reason is incorrect.
\nQuestion 3.<\/p>\nAssertion (A): The graph of y = p(x), where p(x) is a polynomial in variable x, is as follows:
\n
\nThe number of zeroes of p(x) is 5.
\nReason (R): If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.<\/h2>\n
Answer:
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
Explanation:
\nIn case of assertion: Since the graph touches the r-axis 5 times. So, the number of zeroes of p{x) is 5.
\n\u2234Given statement is correct.
\nIn case of reason:
\nIf a polynomial of degree more than two hastwo . real zeroes aqd other zeroes are not real or are \u2022 imaginary, and then graph of the polynomial jj will intersect at two points on x-axis.
\n\u2234Given statement is correct:
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\n
<\/p>\n
Question 4.<\/p>\n
Assertion (A): If the zeroes of the quadratic polynomial (k – 1) x2<\/sup> + kx + 1 is – 3, then the value of k is \\(\\frac{4}{3}\\)
\nReason (R): If – 1 is a zero of the polynomial p(x) = kx2<\/sup> -4x + k, then the value of k is -2.<\/h2>\nAnswer:
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
Explanation:
\nIn Case of assertion:
\nLet p(x) = (k – 1)x2<\/sup> + kx + 1
\nAs -3 is a zero of p(x) then p(-3) = 0
\n(k-1)(-3)2 + k(-3) + 1 = 0
\n9k- 9-3k + 1 = 0
\n9k – 3k = + 9 – 1
\n6k = 8
\nk = \\(\\frac{4}{3}\\)<\/p>\n\u2234 Given statement is correct.
\nIn case of reason:
\nSince,-1is a zero of the polynomial and
\np (x) = kx2<\/sup>– 4x + k,
\nP (-1) = 0
\nCk(-1) 2<\/sup> -4(-1) + k = 0
\nk + 4 + k = 0
\n2k + 4 = 0
\n2k = -4
\nk = -2
\n\u2234 Given statement is correct.
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\nCase -Based MCQs<\/span><\/p>\nAttempt any four sub-parts from each question. Each sub-part carries 1 mark.
\nI. Read the following text and answer the following questions on the basis of the same:
\nThe below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.
\n
\n
\n
\n<\/p>\n
Question 1.<\/p>\n
In the standard form of quadratic polynomial, ax2<\/sup> + bx, c, a, b and c are<\/h2>\n(A) All are real numbers.
\n(B) All are rational numbers.
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.
\n(D) All are integers.
\nAnswer:
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.<\/p>\n
Question 2.<\/p>\n
If the roots of the quadratic polynomial are equal, where the discriminant D = b2<\/sup>– 4ac, then<\/h2>\n(A) D > 0
\n(B) D < 0
\n(C) D
\n(D) D = 0
\nAnswer:
\n(D) D = 0<\/p>\n
Explanation:
\nIf the roots of the quadratic polynomial are equal, then discriminant is equal to zero
\nD = b2<\/sup> – 4ac = 0<\/p>\nQuestion 3.<\/p>\n
If a are \\(\\frac{1}{\\alpha}\\) the zeroes of the quadratic polynomial a 2x2<\/sup>– a + 8k, then k is<\/h2>\n(A) 4
\n(B) \\(\\frac{1}{4}\\)
\n(C)\\(\\frac{-1}{4}\\)
\n(D) 2
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
Explanation:
\nGiven equation, 2x2<\/sup> – x + 8k
\nSum of zeroes = \u03b1 +\\(+\\frac{1}{\\alpha}\\)
\nProduct of zeroes = \u03b1. \\(\\frac{1}{\\alpha}\\)=1
\nProduct of zeroes = \\(\\frac{c}{a}=\\frac{8 k}{2}\\)
\n\\(\\begin{aligned}
\n\\frac{8 k}{2} &=1 \\\\
\nk &=\\frac{2}{8} \\\\
\nk &=\\frac{1}{4}
\n\\end{aligned}\\)<\/p>\n<\/p>\n
Question 4.<\/p>\n
The graph of x2<\/sup> + 1 = 0<\/h2>\n(A) Intersects a-axis at two distinct points.
\n(B) Touches a-axis at a point.
\n(C) Neither touches nor intersects a-axis.
\n(D) Either touches or intersects a-axis.
\nAnswer:
\n(C) Neither touches nor intersects a-axis.<\/p>\n
Question 5.<\/p>\n
If the sum of the roots is -p and product of the roots is \\(-\\frac{1}{p}\\) , then the quadratic polynomial is<\/h2>\n
\\(\\text { (A) } k\\left(-p x^{2}+\\frac{x}{p}+1\\right)\\)
\n\\(\\text { (B) } k\\left(p x^{2}-\\frac{x}{p}-1\\right)\\)
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)
\n\\(\\text { (D) } k\\left(x^{2}+p x+\\frac{1}{p}\\right)\\)<\/p>\n
Answer:
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)<\/p>\n
II. Read the following text and answer the following questions on the basis of the same:
\nAn asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.
\n
\n
\n<\/p>\n
Question 1.<\/p>\n
The shape of the poses shown is<\/h2>\n
(A) Spiral
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\n
Question 2.<\/p>\n
The graph of parabola opens downwards, if<\/h2>\n
(A) a \u2265 0
\n(B) a = 0
\n(C) a < 0
\n(D) a > 0
\nAnswer:
\n(C) a < 0<\/p>\n
Question 3.<\/p>\n
In the graph, how many zeroes are there for the polynomial?<\/h2>\n
\n(A) 0
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(C) 2<\/p>\n
Question 4.<\/p>\n
The two zeroes in the above shown graph are<\/h2>\n
(A) 2,4
\n(B) -2,4
\n(C) -8,4
\n(D) 2,-8
\nAnswer:
\n(B) -2,4<\/p>\n
Question 5.<\/p>\n
The zeroes of the quadratic polynomial \\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)are<\/h2>\n
\\(
\n[latex]\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (C) } \\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (D) }-\\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
Explanation:
\n\\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)
\n\\(\\begin{aligned}
\n&=4 \\sqrt{3} x^{2}+(8-3) x-2 \\sqrt{3} \\\\
\n&=4 \\sqrt{3} x^{2}+8 x-3 x-2 \\sqrt{3} \\\\
\n&=4 x(\\sqrt{3} x+2)-\\sqrt{3}(\\sqrt{3} x+2) \\\\
\n&=(\\sqrt{3} x+2)(4 x-\\sqrt{3})
\n\\end{aligned}\\)
\nHence, zeroes of polynomial\\(\\frac{-2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
III. Read the following text and answer the following questions on the basis of the same:
\nBasketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoor on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing quadratic polynomial.
\n
\n<\/p>\n
Question 1.<\/p>\n
The shape of the path traced shown is<\/h2>\n
(A) Spiral
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\n
<\/p>\n
Question 2.<\/p>\n
The graph of parabola opens upwards, if ………..<\/h2>\n
(A) a = 0
\n(B) a < 0 (C) a>0
\n(D) o \u2265 0
\nAnswer:
\n(C) a>0<\/p>\n
Question 3.<\/p>\n
Observe the following graph and answer
\n
\nIn the above graph, how many zeroes are there for the polynomial?<\/h2>\n
(A) 0
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(D) 3<\/p>\n
Explanation:
\nThe number of zeroes of polynomial is the number of times the curve intersects the x-axis, i.e. attains the value 0. Here, the polynomial meets the x-axis at 3 points. So, number of zeroes = 3.<\/p>\n
Question 4.<\/p>\n
The three zeroes in the above shown graph are<\/h2>\n
(A) 2,3, -1
\n(B) -2, 3,1
\n(C) -3, -1,2
\n(D) -2, -3, -1
\nAnswer:
\n(C) -3, -1,2<\/p>\n
Question 5.<\/p>\n
What will be the expression of the polynomial?<\/h2>\n
(A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6
\n(B) x3<\/sup> + 2 x2<\/sup> – 5 x – 6
\n(C) x3<\/sup> + 2 x2<\/sup>+ 5 x-6
\n(D) x3<\/sup> + 2 x2<\/sup> + 5 x + 6
\nAnswer:
\n(A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6<\/p>\nExplanation:
\nSince, the three zeroes = – 3, -1,2 Hence, the expression is (x + 3)(x + l)(x – 2)
\n= [x2<\/sup> + x + 3 x + 3] (x – 2)
\n= x3<\/sup>+ 4x2<\/sup> + 3x-2x2<\/sup>-8x-6
\n= x3<\/sup>+ 2x2<\/sup> – 5x – 6<\/p>\nlV. Read the following text and answer the following questions on the basis of the same: Applications of Parabolas: Highway Overpasses\/ Underpasses
\nA highway underpass is parabolic in shape.
\n
\n
\nParabolic camber equation y = 2x2<\/sup>\/nw
\nParabola
\nA parabola is the graph that results from p(x) = ax2<\/sup> + bx + c.
\nParabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the vertex.
\n<\/p>\nQuestion 1.<\/p>\n
If the highway overpass is represented by x2<\/sup>– 2x – 8. Then its zeroes are<\/h2>\n(A) (2,-4)
\n(B) (4,-2)
\n(C) (-2,-2)
\n(D) (- 4, – 4)
\nAnswer:
\n(C) (-2,-2)<\/p>\n
Explanation:
\nx2<\/sup>– 2 x – 8 = 0
\nor, x2 <\/sup>-4x + 2x- 8 = 0
\nor, x(x – 4) + 2(x – 4) = 0
\nor, (x- 4) (x + 2) = 0
\nor, x = 4, x = – 2<\/p>\nQuestion 2.<\/p>\n
The highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial:<\/h2>\n
(A) Intersects X-axis
\n(B) Intersects Y-axis
\n(C) Intersects Y-axis or X-axis
\n(D) None of the above
\nAnswer:
\n(A) Intersects X-axis<\/p>\n
Explanation:
\nWe know that the number of zeroes of polynomial is equal to number of points where the graph of polynomial intersects X-axis.<\/p>\n
Question 3.<\/p>\n
Graph of a quadratic polynomial is a:<\/h2>\n
(A) straight line
\n(B) circle
\n(C) parabola
\n(D) ellipse
\nAnswer:
\n(C) parabola<\/p>\n
Explanation:
\nHere, the given graph of a quadratic polynomial is a parabola.<\/p>\n
<\/p>\n
Question 4.<\/p>\n
The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is:<\/h2>\n
(A) x2<\/sup> – 6x + 2
\n(B) x2<\/sup>– 36
\n(C) x2 <\/sup>– 6
\n(D) x2 <\/sup>– 3
\nAnswer:
\n(B) x2 <\/sup>– 36<\/p>\nExplanation:
\nx2<\/sup>-36 =0 => x2<\/sup> = 36
\n=>x 6, – 6. => x = \\(\\pm \\sqrt{36}\\)<\/p>\nQuestion 5.<\/p>\n
The number of zeroes that polynomial f(x) = (x – 2)2<\/sup>+ 4 can have is:<\/h2>\n(A) 1
\n(B) 2
\n(C) 0
\n(D) 3
\nAnswer:
\n(C) 0<\/p>\n
Explanation:
\nWe have,
\nf(x) = (x – 2)2<\/sup> + 4
\n= x2<\/sup> + 4 – 4 x + 4
\n=x2<\/sup>-4x + 8. i.e., It has no factorisation:
\nHence no real value of x is possible, i.e., no zero.<\/p>\nV. Read the following text and answer the following questions on the basis of the same:
\nFor a linear polynomial kx + c, k \u2260 0, the graph of y = kx + c is a straight line which intersects the x-axis at exactly one point, namely, \\(\\left(\\frac{-c}{k}, 0\\right),\\) the linear polynomial kx + c, k \u2260 0, has exactly one zero, namely, the X-coordinate of the point where the graph of y = kx + c intersects the x-axis.
\n<\/p>\n
Question 1.<\/p>\n
If a linear polynomial is 2x + 3, then the zero of 2x + 3 is:<\/h2>\n
\\(\\text { (A) } \\frac{3}{2}\\)
\n\\(\\text { (B) }-\\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{3}{2}\\)<\/p>\n
Explanation:
\nGiven, polynomial = 2+ 3
\nLet p(x) = 2x + 3
\nFora zero of p(x),
\n2x + 3 =0
\n2x = -3
\nx = \\(-\\frac{3}{2}\\)<\/p>\n
Question 2.<\/p>\n
The graph of y = p(x) is given in figure below for some polynomial p(x). The number of zero\/zeroes of p(x) is\/are:
\n<\/h2>\n
(A) 1
\n(B) 2
\n(C) 3
\n(D) 0
\nimg-10
\nAnswer:
\n(D) 0<\/p>\n
Explanation:
\nSince the graph does not intersect the x-axis therefore it has no zero.<\/p>\n
Question 3.<\/p>\n
If a and p are the zeroes of the quadratic polynomial x2<\/sup> – 5x + k such that \u03b1 – \u03b2 = 1, then the value of k is:<\/h2>\n(A) 4
\n(B) 5
\n(C) 6
\n(D) 3
\nAnswer:
\n(C) 6<\/p>\n
Explanation:
\nS \u2234p (x) = x2<\/sup>-5x + k
\n\u03b1 – \u03b2 = \\(\\frac{-(-5)}{1}\\) = 5
\n\u03b1\u03b2 = \\(\\frac{k}{1}\\) = k
\nAlso given, \u03b1 – \u03b2 = 1
\n\\(\\sqrt{(\\alpha+\\beta)^{2}-4 \\alpha \\beta}=1\\)<\/p>\nQuestion 4.<\/p>\n
If \u03b1 and \u03b2 are the zeroes of the quadratic polynomial p(x) = 4x2<\/sup> + 5x + 1, then the product of zeroes is:<\/h2>\n(A) -1
\n(B) \\(\\frac{1}{4}\\)
\n(C) -2
\n(D) \\(-\\frac{5}{4}\\)
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
Explanation:
\nWe have, p(x) = 4x2<\/sup> + 5x + 1
\n\u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{1}{4}\\)<\/p>\n<\/p>\n
Question 5.<\/p>\n
If the product of the zeroes of the quadratic polynomial p(x) = ax2<\/sup>– 6x – 6 is 4, then the value of a is: –<\/h2>\n\\(\\text { (A) }-\\frac{3}{2}\\)
\n\\(\\text { (B) } \\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (A) }-\\frac{3}{2}\\)<\/p>\n
Explanation:
\nWWe have,
\np (x) = ax2<\/sup> – 6x – 6
\nLet a and p be the zeroes of the given polynomial, then
\n\\(\\begin{aligned}
\n\\alpha \\beta &=\\frac{c}{a} \\\\
\n4 &=\\frac{-6}{a} \\\\
\n4 a &=-6 \\\\
\na &=-\\frac{6}{4}=-\\frac{3}{2}
\n\\end{aligned}\\)<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"Polynomials Class 10 MCQ Questions with Answers Question 1. The number of polynomials having zeroes as -2 and 5 is: (A) 1 (B) 2 (C) 3 (D) more than 3 Answer: (D) more than 3 Explanation: We know that if we divide or multiply a polynomial by any constant (real number), then the zeroes of …<\/p>\n
\n(B) b – a – 1
\n(C) a – b + 1
\n(D) a – b – 1
\nAnswer:
\n(A) b – a +1<\/p>\n
\nLet f(x) = x3<\/sup> + ax2<\/sup> + bx + c
\n\u2234 One of the zeroes of f(x) is – I so
\nf(-1) = 0
\n(-1)3<\/sup> + a(-1)2<\/sup> + b(-1) + c = 0
\n-1 + a – b + c = 0
\na – b + c = 1
\nc = 1 + b – a
\nNow, \u03b1\u03b2\u03b3 = \\(\\frac{-d}{a}\\) [\u2234a = 1, d = c]
\n\\(-1 \\beta y=\\frac{-c}{1}\\)
\n\u03b2\u03b3 = c
\n\u03b2\u03b3 = 1 + b – a<\/p>\n
<\/p>\n
Question 4.<\/p>\n
The zeroes of the quadratic polynomial x2<\/sup> + 99x 127 are:<\/h2>\n(A) both positive
\n(B) both negative
\n(C) one positive and one negative
\n(D) both equal
\nAnswer:
\n(B) both negative<\/p>\n
Explanation:
\nLet given quadratic polynomial
\np(x) = x2<\/sup>+ 99x + 127
\nOn comparing p(x) with ax2<\/sup>+ bx + c. we get a = 1,b = 99 and c = 127
\nWe know that,
\nx = \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\)
\n= \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\)
\n= \\(\\frac{-99 \\pm \\sqrt{9801-508}}{2}\\)
\n= \\(\\frac{-99 \\pm \\sqrt{9293}}{2}=\\frac{-99 \\pm 96.4}{2}\\)
\n= \\(\\frac{-99+96.4}{2}, \\frac{-99-96.4}{2}\\)
\n= \\(\\frac{-2.6}{2}, \\frac{-195.4}{2}\\)
\n= -1.3, – 97.7
\nHence, both zeroes of the given quadratic polynomial p(x) are negative.<\/p>\nQuestion 5.<\/p>\n
The zeroes of the quadratic polynomial x2<\/sup> + kx + k, k \u2260 0,<\/h2>\n(A) cannot both be positive
\n(B) cannot both be negative
\n(C) are always unequal
\n(D) are always equal
\nAnswer:
\n(A) cannot both be positive<\/p>\n
Explanation:
\nLet f(x) = x 2<\/sup> + kx + k, k 0.
\nOn comparing the given polynomial with ax2<\/sup> + bx + c, we get a = 1, b – k,c = k
\nIf \u03b1 and \u03b2 be the zeroes of the polynomial (x).
\nWe know thaL,
\nSum of zeroes, \u03b1 + \u03b2 = \\(-\\frac{b}{a}\\)
\n\u03b1 + \u03b2 = \\(-\\frac{k}{1}\\) = -k ……(i)
\nAnd ofzeroes, \u03b1\u03b2 = \\(\\frac{c}{a}\\)
\n\u03b1\u03b2 = \\(\\frac{k}{1}\\)=k
\nCase I: If k is negative, \u03b1\u03b2[from equation (ii)] is negative. It means \u03b1 and \u03b2 are of opposite sign.
\nCase II: If k is positive, then \u03b1\u03b2 [from equation (ii)] is positive but \u03b1 + \u03b2 is negative. If, the product of two numbers is positive, then either both are negative or both are positive. But the sum of these numbers is negative, so numbers must be negative. Hence, in any case zeroes of the given quadratic polynomial cannoL both be positive.<\/p>\nQuestion 6.<\/p>\n
If the zeroes of the quadratic polynomial ax2<\/sup> + bx + c, a \u2260 0 are equal, then :<\/h2>\n(A) c and a have opposite signs
\n(B) c and b have opposite signs
\n(C) c and a have the same sign
\n(D) c and b have the same sign
\nAnswer:
\n(C) c and a have the same sign<\/p>\n
Explanation:
\nL Tor equal roots b2<\/sup> – 4ac = 0 b2<\/sup> = 4ac
\nb2<\/sup> is always positive so 4ac must be positive, i.e., product of a and r must be positive, i.e., a and r must have same sign either positive or negative.<\/p>\nQuestion 7.<\/p>\n
If one of the zeroes of a quadratic polynomial of the form x2<\/sup> + ax + b is the negative of the other, then it<\/h2>\n(A) has no linear term and the constant term is negative.
\n(B) has no linear term and the constant term is positive.
\n(C) can have a linear term but the constant term is negative.
\n(D) can have a linear term but the constant term is positive. [U]
\nAnswer:
\n(A) has no linear term and the constant term is negative.<\/p>\n
Explanation:
\nLet f(x) = x2<\/sup> + ax + b and \u03b1, \u03b2 are tha roots of it.
\nthen, \u03b2 = – \u03b1(given)
\n\u03b1 + \u03b2 = \\(-\\frac{b}{a}\\) and \u03b1\u03b2 = \\(\\frac{c}{a}\\)
\n\u03b1 – \u03b1 = \\(-\\frac{a}{1}\\) and \u03b1(-\u03b1) = \\(\\frac{b}{1}\\)
\n– a = 0 – \u03b12<\/sup> = b
\na = 0 b<0 or b is negative
\nso, f(x) = x2<\/sup> + b shows that it has no linear term<\/p>\n<\/p>\n
Question 8.<\/p>\n
Which of the following is not the graph of a quadratic polynomial?<\/h2>\n
<\/p>\n
Answer:
\n<\/p>\n
Explanation:
\nGraph (D) intersect at three points cm .v-axis so the roots ot polynomial of graph is three, so it is cubic polynomial. Other graphs are of quadratic polynomial. Graph a have no real zeroes and Graph b has coincident zeroes.<\/p>\n
Assertion and Reason Based MCQs<\/span><\/p>\nDirections: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\n
Question 1.<\/p>\n
Assertion (A): If the zeroes of a quadratic polynomial ax2<\/sup> + bx + c are both positive, then a, b and c all have the same sign.
\nReason (R): If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.<\/h2>\nAnswer:
\n(D) A is false and R is True<\/p>\n
Explanation:
\nIn case of assertion: Let \u03b1 and \u03b2 be the roots of the quadratic polynomial. If \u03b1 and \u03b2 are positive then
\n\u03b1 + \u03b2 = \\(\\frac{-b}{a}\\) it shows that \\(\\frac{-b}{a}\\) is negetive a sum of two positive numbers (\u03b1 , \u03b2) must be Tve i.c., either b or a must be negative. So, a, b and c will have different signs.
\n\u2234 Given statement is incorrect.
\nIn case of reason:
\nLet \u03b2 = 0, \u03b3 =0
\nf(x) = (x – \u03b1) (x – \u03b2) (x – \u03b3)
\n= (x – \u03b1) x.<\/sup>x
\nf(x) = x3<\/sup>– ax2<\/sup>
\nwhich has no linear (coefficient of x) and constant terms. Given statement is correct. Thus, assertion is incorrect but reason is correct.<\/p>\nQuestion 2.<\/p>\n
Assertion (A): The value of k for which the quadratic polynomial kx2<\/sup> + x + k has equal zeroes are \u00b1\\(\\frac{1}{2}\\)
\nReason (R): If all the three zeroes of a cubic polynomial x 3<\/sup> + ax2<\/sup> – bx + c are positive, then at least one of a, b and c is non-negative.<\/h2>\nAnswer:
\n(C) A is true but R is false<\/p>\n
Explanation:
\nIn case of assertion:
\nf(x) = kx2<\/sup> + x + k (a = k, b = 1, c = k)
\nLor equal roots b2<\/sup>– 4ac = 0
\n(1)2<\/sup> \u2014 4(k) (k) = 0
\n4k2<\/sup>=1
\nk2<\/sup> = \\(\\frac{1}{4}\\)
\nk = \\(\\pm \\frac{1}{2}\\)
\nSo, there are \\(+\\frac{1}{2}\\) and \\(-\\frac{1}{2}\\) values of k so that the given equation has equal roots.
\n\u2234 Given statement is correct.
\nIn case of reason:
\nAll the zeroes’of cubic polynomial are positive only when all the constants a, b, and c are i negative.
\n\u2234Given statement is incorrect.
\nThus, assertion is correct but reason is incorrect.
\nQuestion 3.<\/p>\nAssertion (A): The graph of y = p(x), where p(x) is a polynomial in variable x, is as follows:
\n
\nThe number of zeroes of p(x) is 5.
\nReason (R): If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.<\/h2>\n
Answer:
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
Explanation:
\nIn case of assertion: Since the graph touches the r-axis 5 times. So, the number of zeroes of p{x) is 5.
\n\u2234Given statement is correct.
\nIn case of reason:
\nIf a polynomial of degree more than two hastwo . real zeroes aqd other zeroes are not real or are \u2022 imaginary, and then graph of the polynomial jj will intersect at two points on x-axis.
\n\u2234Given statement is correct:
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\n
<\/p>\n
Question 4.<\/p>\n
Assertion (A): If the zeroes of the quadratic polynomial (k – 1) x2<\/sup> + kx + 1 is – 3, then the value of k is \\(\\frac{4}{3}\\)
\nReason (R): If – 1 is a zero of the polynomial p(x) = kx2<\/sup> -4x + k, then the value of k is -2.<\/h2>\nAnswer:
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
Explanation:
\nIn Case of assertion:
\nLet p(x) = (k – 1)x2<\/sup> + kx + 1
\nAs -3 is a zero of p(x) then p(-3) = 0
\n(k-1)(-3)2 + k(-3) + 1 = 0
\n9k- 9-3k + 1 = 0
\n9k – 3k = + 9 – 1
\n6k = 8
\nk = \\(\\frac{4}{3}\\)<\/p>\n\u2234 Given statement is correct.
\nIn case of reason:
\nSince,-1is a zero of the polynomial and
\np (x) = kx2<\/sup>– 4x + k,
\nP (-1) = 0
\nCk(-1) 2<\/sup> -4(-1) + k = 0
\nk + 4 + k = 0
\n2k + 4 = 0
\n2k = -4
\nk = -2
\n\u2234 Given statement is correct.
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\nCase -Based MCQs<\/span><\/p>\nAttempt any four sub-parts from each question. Each sub-part carries 1 mark.
\nI. Read the following text and answer the following questions on the basis of the same:
\nThe below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.
\n
\n
\n
\n<\/p>\n
Question 1.<\/p>\n
In the standard form of quadratic polynomial, ax2<\/sup> + bx, c, a, b and c are<\/h2>\n(A) All are real numbers.
\n(B) All are rational numbers.
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.
\n(D) All are integers.
\nAnswer:
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.<\/p>\n
Question 2.<\/p>\n
If the roots of the quadratic polynomial are equal, where the discriminant D = b2<\/sup>– 4ac, then<\/h2>\n(A) D > 0
\n(B) D < 0
\n(C) D
\n(D) D = 0
\nAnswer:
\n(D) D = 0<\/p>\n
Explanation:
\nIf the roots of the quadratic polynomial are equal, then discriminant is equal to zero
\nD = b2<\/sup> – 4ac = 0<\/p>\nQuestion 3.<\/p>\n
If a are \\(\\frac{1}{\\alpha}\\) the zeroes of the quadratic polynomial a 2x2<\/sup>– a + 8k, then k is<\/h2>\n(A) 4
\n(B) \\(\\frac{1}{4}\\)
\n(C)\\(\\frac{-1}{4}\\)
\n(D) 2
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
Explanation:
\nGiven equation, 2x2<\/sup> – x + 8k
\nSum of zeroes = \u03b1 +\\(+\\frac{1}{\\alpha}\\)
\nProduct of zeroes = \u03b1. \\(\\frac{1}{\\alpha}\\)=1
\nProduct of zeroes = \\(\\frac{c}{a}=\\frac{8 k}{2}\\)
\n\\(\\begin{aligned}
\n\\frac{8 k}{2} &=1 \\\\
\nk &=\\frac{2}{8} \\\\
\nk &=\\frac{1}{4}
\n\\end{aligned}\\)<\/p>\n<\/p>\n
Question 4.<\/p>\n
The graph of x2<\/sup> + 1 = 0<\/h2>\n(A) Intersects a-axis at two distinct points.
\n(B) Touches a-axis at a point.
\n(C) Neither touches nor intersects a-axis.
\n(D) Either touches or intersects a-axis.
\nAnswer:
\n(C) Neither touches nor intersects a-axis.<\/p>\n
Question 5.<\/p>\n
If the sum of the roots is -p and product of the roots is \\(-\\frac{1}{p}\\) , then the quadratic polynomial is<\/h2>\n
\\(\\text { (A) } k\\left(-p x^{2}+\\frac{x}{p}+1\\right)\\)
\n\\(\\text { (B) } k\\left(p x^{2}-\\frac{x}{p}-1\\right)\\)
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)
\n\\(\\text { (D) } k\\left(x^{2}+p x+\\frac{1}{p}\\right)\\)<\/p>\n
Answer:
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)<\/p>\n
II. Read the following text and answer the following questions on the basis of the same:
\nAn asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.
\n
\n
\n<\/p>\n
Question 1.<\/p>\n
The shape of the poses shown is<\/h2>\n
(A) Spiral
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\n
Question 2.<\/p>\n
The graph of parabola opens downwards, if<\/h2>\n
(A) a \u2265 0
\n(B) a = 0
\n(C) a < 0
\n(D) a > 0
\nAnswer:
\n(C) a < 0<\/p>\n
Question 3.<\/p>\n
In the graph, how many zeroes are there for the polynomial?<\/h2>\n
\n(A) 0
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(C) 2<\/p>\n
Question 4.<\/p>\n
The two zeroes in the above shown graph are<\/h2>\n
(A) 2,4
\n(B) -2,4
\n(C) -8,4
\n(D) 2,-8
\nAnswer:
\n(B) -2,4<\/p>\n
Question 5.<\/p>\n
The zeroes of the quadratic polynomial \\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)are<\/h2>\n
\\(
\n[latex]\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (C) } \\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (D) }-\\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
Explanation:
\n\\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)
\n\\(\\begin{aligned}
\n&=4 \\sqrt{3} x^{2}+(8-3) x-2 \\sqrt{3} \\\\
\n&=4 \\sqrt{3} x^{2}+8 x-3 x-2 \\sqrt{3} \\\\
\n&=4 x(\\sqrt{3} x+2)-\\sqrt{3}(\\sqrt{3} x+2) \\\\
\n&=(\\sqrt{3} x+2)(4 x-\\sqrt{3})
\n\\end{aligned}\\)
\nHence, zeroes of polynomial\\(\\frac{-2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
III. Read the following text and answer the following questions on the basis of the same:
\nBasketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoor on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing quadratic polynomial.
\n
\n<\/p>\n
Question 1.<\/p>\n
The shape of the path traced shown is<\/h2>\n
(A) Spiral
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\n
<\/p>\n
Question 2.<\/p>\n
The graph of parabola opens upwards, if ………..<\/h2>\n
(A) a = 0
\n(B) a < 0 (C) a>0
\n(D) o \u2265 0
\nAnswer:
\n(C) a>0<\/p>\n
Question 3.<\/p>\n
Observe the following graph and answer
\n
\nIn the above graph, how many zeroes are there for the polynomial?<\/h2>\n
(A) 0
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(D) 3<\/p>\n
Explanation:
\nThe number of zeroes of polynomial is the number of times the curve intersects the x-axis, i.e. attains the value 0. Here, the polynomial meets the x-axis at 3 points. So, number of zeroes = 3.<\/p>\n
Question 4.<\/p>\n
The three zeroes in the above shown graph are<\/h2>\n
(A) 2,3, -1
\n(B) -2, 3,1
\n(C) -3, -1,2
\n(D) -2, -3, -1
\nAnswer:
\n(C) -3, -1,2<\/p>\n
Question 5.<\/p>\n
What will be the expression of the polynomial?<\/h2>\n
(A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6
\n(B) x3<\/sup> + 2 x2<\/sup> – 5 x – 6
\n(C) x3<\/sup> + 2 x2<\/sup>+ 5 x-6
\n(D) x3<\/sup> + 2 x2<\/sup> + 5 x + 6
\nAnswer:
\n(A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6<\/p>\nExplanation:
\nSince, the three zeroes = – 3, -1,2 Hence, the expression is (x + 3)(x + l)(x – 2)
\n= [x2<\/sup> + x + 3 x + 3] (x – 2)
\n= x3<\/sup>+ 4x2<\/sup> + 3x-2x2<\/sup>-8x-6
\n= x3<\/sup>+ 2x2<\/sup> – 5x – 6<\/p>\nlV. Read the following text and answer the following questions on the basis of the same: Applications of Parabolas: Highway Overpasses\/ Underpasses
\nA highway underpass is parabolic in shape.
\n
\n
\nParabolic camber equation y = 2x2<\/sup>\/nw
\nParabola
\nA parabola is the graph that results from p(x) = ax2<\/sup> + bx + c.
\nParabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the vertex.
\n<\/p>\nQuestion 1.<\/p>\n
If the highway overpass is represented by x2<\/sup>– 2x – 8. Then its zeroes are<\/h2>\n(A) (2,-4)
\n(B) (4,-2)
\n(C) (-2,-2)
\n(D) (- 4, – 4)
\nAnswer:
\n(C) (-2,-2)<\/p>\n
Explanation:
\nx2<\/sup>– 2 x – 8 = 0
\nor, x2 <\/sup>-4x + 2x- 8 = 0
\nor, x(x – 4) + 2(x – 4) = 0
\nor, (x- 4) (x + 2) = 0
\nor, x = 4, x = – 2<\/p>\nQuestion 2.<\/p>\n
The highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial:<\/h2>\n
(A) Intersects X-axis
\n(B) Intersects Y-axis
\n(C) Intersects Y-axis or X-axis
\n(D) None of the above
\nAnswer:
\n(A) Intersects X-axis<\/p>\n
Explanation:
\nWe know that the number of zeroes of polynomial is equal to number of points where the graph of polynomial intersects X-axis.<\/p>\n
Question 3.<\/p>\n
Graph of a quadratic polynomial is a:<\/h2>\n
(A) straight line
\n(B) circle
\n(C) parabola
\n(D) ellipse
\nAnswer:
\n(C) parabola<\/p>\n
Explanation:
\nHere, the given graph of a quadratic polynomial is a parabola.<\/p>\n
<\/p>\n
Question 4.<\/p>\n
The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is:<\/h2>\n
(A) x2<\/sup> – 6x + 2
\n(B) x2<\/sup>– 36
\n(C) x2 <\/sup>– 6
\n(D) x2 <\/sup>– 3
\nAnswer:
\n(B) x2 <\/sup>– 36<\/p>\nExplanation:
\nx2<\/sup>-36 =0 => x2<\/sup> = 36
\n=>x 6, – 6. => x = \\(\\pm \\sqrt{36}\\)<\/p>\nQuestion 5.<\/p>\n
The number of zeroes that polynomial f(x) = (x – 2)2<\/sup>+ 4 can have is:<\/h2>\n(A) 1
\n(B) 2
\n(C) 0
\n(D) 3
\nAnswer:
\n(C) 0<\/p>\n
Explanation:
\nWe have,
\nf(x) = (x – 2)2<\/sup> + 4
\n= x2<\/sup> + 4 – 4 x + 4
\n=x2<\/sup>-4x + 8. i.e., It has no factorisation:
\nHence no real value of x is possible, i.e., no zero.<\/p>\nV. Read the following text and answer the following questions on the basis of the same:
\nFor a linear polynomial kx + c, k \u2260 0, the graph of y = kx + c is a straight line which intersects the x-axis at exactly one point, namely, \\(\\left(\\frac{-c}{k}, 0\\right),\\) the linear polynomial kx + c, k \u2260 0, has exactly one zero, namely, the X-coordinate of the point where the graph of y = kx + c intersects the x-axis.
\n<\/p>\n
Question 1.<\/p>\n
If a linear polynomial is 2x + 3, then the zero of 2x + 3 is:<\/h2>\n
\\(\\text { (A) } \\frac{3}{2}\\)
\n\\(\\text { (B) }-\\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{3}{2}\\)<\/p>\n
Explanation:
\nGiven, polynomial = 2+ 3
\nLet p(x) = 2x + 3
\nFora zero of p(x),
\n2x + 3 =0
\n2x = -3
\nx = \\(-\\frac{3}{2}\\)<\/p>\n
Question 2.<\/p>\n
The graph of y = p(x) is given in figure below for some polynomial p(x). The number of zero\/zeroes of p(x) is\/are:
\n<\/h2>\n
(A) 1
\n(B) 2
\n(C) 3
\n(D) 0
\nimg-10
\nAnswer:
\n(D) 0<\/p>\n
Explanation:
\nSince the graph does not intersect the x-axis therefore it has no zero.<\/p>\n
Question 3.<\/p>\n
If a and p are the zeroes of the quadratic polynomial x2<\/sup> – 5x + k such that \u03b1 – \u03b2 = 1, then the value of k is:<\/h2>\n(A) 4
\n(B) 5
\n(C) 6
\n(D) 3
\nAnswer:
\n(C) 6<\/p>\n
Explanation:
\nS \u2234p (x) = x2<\/sup>-5x + k
\n\u03b1 – \u03b2 = \\(\\frac{-(-5)}{1}\\) = 5
\n\u03b1\u03b2 = \\(\\frac{k}{1}\\) = k
\nAlso given, \u03b1 – \u03b2 = 1
\n\\(\\sqrt{(\\alpha+\\beta)^{2}-4 \\alpha \\beta}=1\\)<\/p>\nQuestion 4.<\/p>\n
If \u03b1 and \u03b2 are the zeroes of the quadratic polynomial p(x) = 4x2<\/sup> + 5x + 1, then the product of zeroes is:<\/h2>\n(A) -1
\n(B) \\(\\frac{1}{4}\\)
\n(C) -2
\n(D) \\(-\\frac{5}{4}\\)
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
Explanation:
\nWe have, p(x) = 4x2<\/sup> + 5x + 1
\n\u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{1}{4}\\)<\/p>\n<\/p>\n
Question 5.<\/p>\n
If the product of the zeroes of the quadratic polynomial p(x) = ax2<\/sup>– 6x – 6 is 4, then the value of a is: –<\/h2>\n\\(\\text { (A) }-\\frac{3}{2}\\)
\n\\(\\text { (B) } \\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (A) }-\\frac{3}{2}\\)<\/p>\n
Explanation:
\nWWe have,
\np (x) = ax2<\/sup> – 6x – 6
\nLet a and p be the zeroes of the given polynomial, then
\n\\(\\begin{aligned}
\n\\alpha \\beta &=\\frac{c}{a} \\\\
\n4 &=\\frac{-6}{a} \\\\
\n4 a &=-6 \\\\
\na &=-\\frac{6}{4}=-\\frac{3}{2}
\n\\end{aligned}\\)<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"Polynomials Class 10 MCQ Questions with Answers Question 1. The number of polynomials having zeroes as -2 and 5 is: (A) 1 (B) 2 (C) 3 (D) more than 3 Answer: (D) more than 3 Explanation: We know that if we divide or multiply a polynomial by any constant (real number), then the zeroes of …<\/p>\n
\n(B) both negative
\n(C) one positive and one negative
\n(D) both equal
\nAnswer:
\n(B) both negative<\/p>\n
\nLet given quadratic polynomial
\np(x) = x2<\/sup>+ 99x + 127
\nOn comparing p(x) with ax2<\/sup>+ bx + c. we get a = 1,b = 99 and c = 127
\nWe know that,
\nx = \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\)
\n= \\(\\frac{-b \\pm \\sqrt{b^{2}-4 a c}}{2 a}\\)
\n= \\(\\frac{-99 \\pm \\sqrt{9801-508}}{2}\\)
\n= \\(\\frac{-99 \\pm \\sqrt{9293}}{2}=\\frac{-99 \\pm 96.4}{2}\\)
\n= \\(\\frac{-99+96.4}{2}, \\frac{-99-96.4}{2}\\)
\n= \\(\\frac{-2.6}{2}, \\frac{-195.4}{2}\\)
\n= -1.3, – 97.7
\nHence, both zeroes of the given quadratic polynomial p(x) are negative.<\/p>\n
Question 5.<\/p>\n
The zeroes of the quadratic polynomial x2<\/sup> + kx + k, k \u2260 0,<\/h2>\n(A) cannot both be positive
\n(B) cannot both be negative
\n(C) are always unequal
\n(D) are always equal
\nAnswer:
\n(A) cannot both be positive<\/p>\n
Explanation:
\nLet f(x) = x 2<\/sup> + kx + k, k 0.
\nOn comparing the given polynomial with ax2<\/sup> + bx + c, we get a = 1, b – k,c = k
\nIf \u03b1 and \u03b2 be the zeroes of the polynomial (x).
\nWe know thaL,
\nSum of zeroes, \u03b1 + \u03b2 = \\(-\\frac{b}{a}\\)
\n\u03b1 + \u03b2 = \\(-\\frac{k}{1}\\) = -k ……(i)
\nAnd ofzeroes, \u03b1\u03b2 = \\(\\frac{c}{a}\\)
\n\u03b1\u03b2 = \\(\\frac{k}{1}\\)=k
\nCase I: If k is negative, \u03b1\u03b2[from equation (ii)] is negative. It means \u03b1 and \u03b2 are of opposite sign.
\nCase II: If k is positive, then \u03b1\u03b2 [from equation (ii)] is positive but \u03b1 + \u03b2 is negative. If, the product of two numbers is positive, then either both are negative or both are positive. But the sum of these numbers is negative, so numbers must be negative. Hence, in any case zeroes of the given quadratic polynomial cannoL both be positive.<\/p>\nQuestion 6.<\/p>\n
If the zeroes of the quadratic polynomial ax2<\/sup> + bx + c, a \u2260 0 are equal, then :<\/h2>\n(A) c and a have opposite signs
\n(B) c and b have opposite signs
\n(C) c and a have the same sign
\n(D) c and b have the same sign
\nAnswer:
\n(C) c and a have the same sign<\/p>\n
Explanation:
\nL Tor equal roots b2<\/sup> – 4ac = 0 b2<\/sup> = 4ac
\nb2<\/sup> is always positive so 4ac must be positive, i.e., product of a and r must be positive, i.e., a and r must have same sign either positive or negative.<\/p>\nQuestion 7.<\/p>\n
If one of the zeroes of a quadratic polynomial of the form x2<\/sup> + ax + b is the negative of the other, then it<\/h2>\n(A) has no linear term and the constant term is negative.
\n(B) has no linear term and the constant term is positive.
\n(C) can have a linear term but the constant term is negative.
\n(D) can have a linear term but the constant term is positive. [U]
\nAnswer:
\n(A) has no linear term and the constant term is negative.<\/p>\n
Explanation:
\nLet f(x) = x2<\/sup> + ax + b and \u03b1, \u03b2 are tha roots of it.
\nthen, \u03b2 = – \u03b1(given)
\n\u03b1 + \u03b2 = \\(-\\frac{b}{a}\\) and \u03b1\u03b2 = \\(\\frac{c}{a}\\)
\n\u03b1 – \u03b1 = \\(-\\frac{a}{1}\\) and \u03b1(-\u03b1) = \\(\\frac{b}{1}\\)
\n– a = 0 – \u03b12<\/sup> = b
\na = 0 b<0 or b is negative
\nso, f(x) = x2<\/sup> + b shows that it has no linear term<\/p>\n<\/p>\n
Question 8.<\/p>\n
Which of the following is not the graph of a quadratic polynomial?<\/h2>\n
<\/p>\n
Answer:
\n<\/p>\n
Explanation:
\nGraph (D) intersect at three points cm .v-axis so the roots ot polynomial of graph is three, so it is cubic polynomial. Other graphs are of quadratic polynomial. Graph a have no real zeroes and Graph b has coincident zeroes.<\/p>\n
Assertion and Reason Based MCQs<\/span><\/p>\nDirections: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\n
Question 1.<\/p>\n
Assertion (A): If the zeroes of a quadratic polynomial ax2<\/sup> + bx + c are both positive, then a, b and c all have the same sign.
\nReason (R): If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.<\/h2>\nAnswer:
\n(D) A is false and R is True<\/p>\n
Explanation:
\nIn case of assertion: Let \u03b1 and \u03b2 be the roots of the quadratic polynomial. If \u03b1 and \u03b2 are positive then
\n\u03b1 + \u03b2 = \\(\\frac{-b}{a}\\) it shows that \\(\\frac{-b}{a}\\) is negetive a sum of two positive numbers (\u03b1 , \u03b2) must be Tve i.c., either b or a must be negative. So, a, b and c will have different signs.
\n\u2234 Given statement is incorrect.
\nIn case of reason:
\nLet \u03b2 = 0, \u03b3 =0
\nf(x) = (x – \u03b1) (x – \u03b2) (x – \u03b3)
\n= (x – \u03b1) x.<\/sup>x
\nf(x) = x3<\/sup>– ax2<\/sup>
\nwhich has no linear (coefficient of x) and constant terms. Given statement is correct. Thus, assertion is incorrect but reason is correct.<\/p>\nQuestion 2.<\/p>\n
Assertion (A): The value of k for which the quadratic polynomial kx2<\/sup> + x + k has equal zeroes are \u00b1\\(\\frac{1}{2}\\)
\nReason (R): If all the three zeroes of a cubic polynomial x 3<\/sup> + ax2<\/sup> – bx + c are positive, then at least one of a, b and c is non-negative.<\/h2>\nAnswer:
\n(C) A is true but R is false<\/p>\n
Explanation:
\nIn case of assertion:
\nf(x) = kx2<\/sup> + x + k (a = k, b = 1, c = k)
\nLor equal roots b2<\/sup>– 4ac = 0
\n(1)2<\/sup> \u2014 4(k) (k) = 0
\n4k2<\/sup>=1
\nk2<\/sup> = \\(\\frac{1}{4}\\)
\nk = \\(\\pm \\frac{1}{2}\\)
\nSo, there are \\(+\\frac{1}{2}\\) and \\(-\\frac{1}{2}\\) values of k so that the given equation has equal roots.
\n\u2234 Given statement is correct.
\nIn case of reason:
\nAll the zeroes’of cubic polynomial are positive only when all the constants a, b, and c are i negative.
\n\u2234Given statement is incorrect.
\nThus, assertion is correct but reason is incorrect.
\nQuestion 3.<\/p>\nAssertion (A): The graph of y = p(x), where p(x) is a polynomial in variable x, is as follows:
\n
\nThe number of zeroes of p(x) is 5.
\nReason (R): If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.<\/h2>\n
Answer:
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
Explanation:
\nIn case of assertion: Since the graph touches the r-axis 5 times. So, the number of zeroes of p{x) is 5.
\n\u2234Given statement is correct.
\nIn case of reason:
\nIf a polynomial of degree more than two hastwo . real zeroes aqd other zeroes are not real or are \u2022 imaginary, and then graph of the polynomial jj will intersect at two points on x-axis.
\n\u2234Given statement is correct:
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\n
<\/p>\n
Question 4.<\/p>\n
Assertion (A): If the zeroes of the quadratic polynomial (k – 1) x2<\/sup> + kx + 1 is – 3, then the value of k is \\(\\frac{4}{3}\\)
\nReason (R): If – 1 is a zero of the polynomial p(x) = kx2<\/sup> -4x + k, then the value of k is -2.<\/h2>\nAnswer:
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
Explanation:
\nIn Case of assertion:
\nLet p(x) = (k – 1)x2<\/sup> + kx + 1
\nAs -3 is a zero of p(x) then p(-3) = 0
\n(k-1)(-3)2 + k(-3) + 1 = 0
\n9k- 9-3k + 1 = 0
\n9k – 3k = + 9 – 1
\n6k = 8
\nk = \\(\\frac{4}{3}\\)<\/p>\n\u2234 Given statement is correct.
\nIn case of reason:
\nSince,-1is a zero of the polynomial and
\np (x) = kx2<\/sup>– 4x + k,
\nP (-1) = 0
\nCk(-1) 2<\/sup> -4(-1) + k = 0
\nk + 4 + k = 0
\n2k + 4 = 0
\n2k = -4
\nk = -2
\n\u2234 Given statement is correct.
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\nCase -Based MCQs<\/span><\/p>\nAttempt any four sub-parts from each question. Each sub-part carries 1 mark.
\nI. Read the following text and answer the following questions on the basis of the same:
\nThe below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.
\n
\n
\n
\n<\/p>\n
Question 1.<\/p>\n
In the standard form of quadratic polynomial, ax2<\/sup> + bx, c, a, b and c are<\/h2>\n(A) All are real numbers.
\n(B) All are rational numbers.
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.
\n(D) All are integers.
\nAnswer:
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.<\/p>\n
Question 2.<\/p>\n
If the roots of the quadratic polynomial are equal, where the discriminant D = b2<\/sup>– 4ac, then<\/h2>\n(A) D > 0
\n(B) D < 0
\n(C) D
\n(D) D = 0
\nAnswer:
\n(D) D = 0<\/p>\n
Explanation:
\nIf the roots of the quadratic polynomial are equal, then discriminant is equal to zero
\nD = b2<\/sup> – 4ac = 0<\/p>\nQuestion 3.<\/p>\n
If a are \\(\\frac{1}{\\alpha}\\) the zeroes of the quadratic polynomial a 2x2<\/sup>– a + 8k, then k is<\/h2>\n(A) 4
\n(B) \\(\\frac{1}{4}\\)
\n(C)\\(\\frac{-1}{4}\\)
\n(D) 2
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
Explanation:
\nGiven equation, 2x2<\/sup> – x + 8k
\nSum of zeroes = \u03b1 +\\(+\\frac{1}{\\alpha}\\)
\nProduct of zeroes = \u03b1. \\(\\frac{1}{\\alpha}\\)=1
\nProduct of zeroes = \\(\\frac{c}{a}=\\frac{8 k}{2}\\)
\n\\(\\begin{aligned}
\n\\frac{8 k}{2} &=1 \\\\
\nk &=\\frac{2}{8} \\\\
\nk &=\\frac{1}{4}
\n\\end{aligned}\\)<\/p>\n<\/p>\n
Question 4.<\/p>\n
The graph of x2<\/sup> + 1 = 0<\/h2>\n(A) Intersects a-axis at two distinct points.
\n(B) Touches a-axis at a point.
\n(C) Neither touches nor intersects a-axis.
\n(D) Either touches or intersects a-axis.
\nAnswer:
\n(C) Neither touches nor intersects a-axis.<\/p>\n
Question 5.<\/p>\n
If the sum of the roots is -p and product of the roots is \\(-\\frac{1}{p}\\) , then the quadratic polynomial is<\/h2>\n
\\(\\text { (A) } k\\left(-p x^{2}+\\frac{x}{p}+1\\right)\\)
\n\\(\\text { (B) } k\\left(p x^{2}-\\frac{x}{p}-1\\right)\\)
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)
\n\\(\\text { (D) } k\\left(x^{2}+p x+\\frac{1}{p}\\right)\\)<\/p>\n
Answer:
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)<\/p>\n
II. Read the following text and answer the following questions on the basis of the same:
\nAn asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.
\n
\n
\n<\/p>\n
Question 1.<\/p>\n
The shape of the poses shown is<\/h2>\n
(A) Spiral
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\n
Question 2.<\/p>\n
The graph of parabola opens downwards, if<\/h2>\n
(A) a \u2265 0
\n(B) a = 0
\n(C) a < 0
\n(D) a > 0
\nAnswer:
\n(C) a < 0<\/p>\n
Question 3.<\/p>\n
In the graph, how many zeroes are there for the polynomial?<\/h2>\n
\n(A) 0
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(C) 2<\/p>\n
Question 4.<\/p>\n
The two zeroes in the above shown graph are<\/h2>\n
(A) 2,4
\n(B) -2,4
\n(C) -8,4
\n(D) 2,-8
\nAnswer:
\n(B) -2,4<\/p>\n
Question 5.<\/p>\n
The zeroes of the quadratic polynomial \\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)are<\/h2>\n
\\(
\n[latex]\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (C) } \\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (D) }-\\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
Explanation:
\n\\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)
\n\\(\\begin{aligned}
\n&=4 \\sqrt{3} x^{2}+(8-3) x-2 \\sqrt{3} \\\\
\n&=4 \\sqrt{3} x^{2}+8 x-3 x-2 \\sqrt{3} \\\\
\n&=4 x(\\sqrt{3} x+2)-\\sqrt{3}(\\sqrt{3} x+2) \\\\
\n&=(\\sqrt{3} x+2)(4 x-\\sqrt{3})
\n\\end{aligned}\\)
\nHence, zeroes of polynomial\\(\\frac{-2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
III. Read the following text and answer the following questions on the basis of the same:
\nBasketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoor on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing quadratic polynomial.
\n
\n<\/p>\n
Question 1.<\/p>\n
The shape of the path traced shown is<\/h2>\n
(A) Spiral
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\n
<\/p>\n
Question 2.<\/p>\n
The graph of parabola opens upwards, if ………..<\/h2>\n
(A) a = 0
\n(B) a < 0 (C) a>0
\n(D) o \u2265 0
\nAnswer:
\n(C) a>0<\/p>\n
Question 3.<\/p>\n
Observe the following graph and answer
\n
\nIn the above graph, how many zeroes are there for the polynomial?<\/h2>\n
(A) 0
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(D) 3<\/p>\n
Explanation:
\nThe number of zeroes of polynomial is the number of times the curve intersects the x-axis, i.e. attains the value 0. Here, the polynomial meets the x-axis at 3 points. So, number of zeroes = 3.<\/p>\n
Question 4.<\/p>\n
The three zeroes in the above shown graph are<\/h2>\n
(A) 2,3, -1
\n(B) -2, 3,1
\n(C) -3, -1,2
\n(D) -2, -3, -1
\nAnswer:
\n(C) -3, -1,2<\/p>\n
Question 5.<\/p>\n
What will be the expression of the polynomial?<\/h2>\n
(A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6
\n(B) x3<\/sup> + 2 x2<\/sup> – 5 x – 6
\n(C) x3<\/sup> + 2 x2<\/sup>+ 5 x-6
\n(D) x3<\/sup> + 2 x2<\/sup> + 5 x + 6
\nAnswer:
\n(A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6<\/p>\nExplanation:
\nSince, the three zeroes = – 3, -1,2 Hence, the expression is (x + 3)(x + l)(x – 2)
\n= [x2<\/sup> + x + 3 x + 3] (x – 2)
\n= x3<\/sup>+ 4x2<\/sup> + 3x-2x2<\/sup>-8x-6
\n= x3<\/sup>+ 2x2<\/sup> – 5x – 6<\/p>\nlV. Read the following text and answer the following questions on the basis of the same: Applications of Parabolas: Highway Overpasses\/ Underpasses
\nA highway underpass is parabolic in shape.
\n
\n
\nParabolic camber equation y = 2x2<\/sup>\/nw
\nParabola
\nA parabola is the graph that results from p(x) = ax2<\/sup> + bx + c.
\nParabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the vertex.
\n<\/p>\nQuestion 1.<\/p>\n
If the highway overpass is represented by x2<\/sup>– 2x – 8. Then its zeroes are<\/h2>\n(A) (2,-4)
\n(B) (4,-2)
\n(C) (-2,-2)
\n(D) (- 4, – 4)
\nAnswer:
\n(C) (-2,-2)<\/p>\n
Explanation:
\nx2<\/sup>– 2 x – 8 = 0
\nor, x2 <\/sup>-4x + 2x- 8 = 0
\nor, x(x – 4) + 2(x – 4) = 0
\nor, (x- 4) (x + 2) = 0
\nor, x = 4, x = – 2<\/p>\nQuestion 2.<\/p>\n
The highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial:<\/h2>\n
(A) Intersects X-axis
\n(B) Intersects Y-axis
\n(C) Intersects Y-axis or X-axis
\n(D) None of the above
\nAnswer:
\n(A) Intersects X-axis<\/p>\n
Explanation:
\nWe know that the number of zeroes of polynomial is equal to number of points where the graph of polynomial intersects X-axis.<\/p>\n
Question 3.<\/p>\n
Graph of a quadratic polynomial is a:<\/h2>\n
(A) straight line
\n(B) circle
\n(C) parabola
\n(D) ellipse
\nAnswer:
\n(C) parabola<\/p>\n
Explanation:
\nHere, the given graph of a quadratic polynomial is a parabola.<\/p>\n
<\/p>\n
Question 4.<\/p>\n
The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is:<\/h2>\n
(A) x2<\/sup> – 6x + 2
\n(B) x2<\/sup>– 36
\n(C) x2 <\/sup>– 6
\n(D) x2 <\/sup>– 3
\nAnswer:
\n(B) x2 <\/sup>– 36<\/p>\nExplanation:
\nx2<\/sup>-36 =0 => x2<\/sup> = 36
\n=>x 6, – 6. => x = \\(\\pm \\sqrt{36}\\)<\/p>\nQuestion 5.<\/p>\n
The number of zeroes that polynomial f(x) = (x – 2)2<\/sup>+ 4 can have is:<\/h2>\n(A) 1
\n(B) 2
\n(C) 0
\n(D) 3
\nAnswer:
\n(C) 0<\/p>\n
Explanation:
\nWe have,
\nf(x) = (x – 2)2<\/sup> + 4
\n= x2<\/sup> + 4 – 4 x + 4
\n=x2<\/sup>-4x + 8. i.e., It has no factorisation:
\nHence no real value of x is possible, i.e., no zero.<\/p>\nV. Read the following text and answer the following questions on the basis of the same:
\nFor a linear polynomial kx + c, k \u2260 0, the graph of y = kx + c is a straight line which intersects the x-axis at exactly one point, namely, \\(\\left(\\frac{-c}{k}, 0\\right),\\) the linear polynomial kx + c, k \u2260 0, has exactly one zero, namely, the X-coordinate of the point where the graph of y = kx + c intersects the x-axis.
\n<\/p>\n
Question 1.<\/p>\n
If a linear polynomial is 2x + 3, then the zero of 2x + 3 is:<\/h2>\n
\\(\\text { (A) } \\frac{3}{2}\\)
\n\\(\\text { (B) }-\\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{3}{2}\\)<\/p>\n
Explanation:
\nGiven, polynomial = 2+ 3
\nLet p(x) = 2x + 3
\nFora zero of p(x),
\n2x + 3 =0
\n2x = -3
\nx = \\(-\\frac{3}{2}\\)<\/p>\n
Question 2.<\/p>\n
The graph of y = p(x) is given in figure below for some polynomial p(x). The number of zero\/zeroes of p(x) is\/are:
\n<\/h2>\n
(A) 1
\n(B) 2
\n(C) 3
\n(D) 0
\nimg-10
\nAnswer:
\n(D) 0<\/p>\n
Explanation:
\nSince the graph does not intersect the x-axis therefore it has no zero.<\/p>\n
Question 3.<\/p>\n
If a and p are the zeroes of the quadratic polynomial x2<\/sup> – 5x + k such that \u03b1 – \u03b2 = 1, then the value of k is:<\/h2>\n(A) 4
\n(B) 5
\n(C) 6
\n(D) 3
\nAnswer:
\n(C) 6<\/p>\n
Explanation:
\nS \u2234p (x) = x2<\/sup>-5x + k
\n\u03b1 – \u03b2 = \\(\\frac{-(-5)}{1}\\) = 5
\n\u03b1\u03b2 = \\(\\frac{k}{1}\\) = k
\nAlso given, \u03b1 – \u03b2 = 1
\n\\(\\sqrt{(\\alpha+\\beta)^{2}-4 \\alpha \\beta}=1\\)<\/p>\nQuestion 4.<\/p>\n
If \u03b1 and \u03b2 are the zeroes of the quadratic polynomial p(x) = 4x2<\/sup> + 5x + 1, then the product of zeroes is:<\/h2>\n(A) -1
\n(B) \\(\\frac{1}{4}\\)
\n(C) -2
\n(D) \\(-\\frac{5}{4}\\)
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
Explanation:
\nWe have, p(x) = 4x2<\/sup> + 5x + 1
\n\u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{1}{4}\\)<\/p>\n<\/p>\n
Question 5.<\/p>\n
If the product of the zeroes of the quadratic polynomial p(x) = ax2<\/sup>– 6x – 6 is 4, then the value of a is: –<\/h2>\n\\(\\text { (A) }-\\frac{3}{2}\\)
\n\\(\\text { (B) } \\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (A) }-\\frac{3}{2}\\)<\/p>\n
Explanation:
\nWWe have,
\np (x) = ax2<\/sup> – 6x – 6
\nLet a and p be the zeroes of the given polynomial, then
\n\\(\\begin{aligned}
\n\\alpha \\beta &=\\frac{c}{a} \\\\
\n4 &=\\frac{-6}{a} \\\\
\n4 a &=-6 \\\\
\na &=-\\frac{6}{4}=-\\frac{3}{2}
\n\\end{aligned}\\)<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"Polynomials Class 10 MCQ Questions with Answers Question 1. The number of polynomials having zeroes as -2 and 5 is: (A) 1 (B) 2 (C) 3 (D) more than 3 Answer: (D) more than 3 Explanation: We know that if we divide or multiply a polynomial by any constant (real number), then the zeroes of …<\/p>\n
\n(B) cannot both be negative
\n(C) are always unequal
\n(D) are always equal
\nAnswer:
\n(A) cannot both be positive<\/p>\n
\nLet f(x) = x 2<\/sup> + kx + k, k 0.
\nOn comparing the given polynomial with ax2<\/sup> + bx + c, we get a = 1, b – k,c = k
\nIf \u03b1 and \u03b2 be the zeroes of the polynomial (x).
\nWe know thaL,
\nSum of zeroes, \u03b1 + \u03b2 = \\(-\\frac{b}{a}\\)
\n\u03b1 + \u03b2 = \\(-\\frac{k}{1}\\) = -k ……(i)
\nAnd ofzeroes, \u03b1\u03b2 = \\(\\frac{c}{a}\\)
\n\u03b1\u03b2 = \\(\\frac{k}{1}\\)=k
\nCase I: If k is negative, \u03b1\u03b2[from equation (ii)] is negative. It means \u03b1 and \u03b2 are of opposite sign.
\nCase II: If k is positive, then \u03b1\u03b2 [from equation (ii)] is positive but \u03b1 + \u03b2 is negative. If, the product of two numbers is positive, then either both are negative or both are positive. But the sum of these numbers is negative, so numbers must be negative. Hence, in any case zeroes of the given quadratic polynomial cannoL both be positive.<\/p>\n
Question 6.<\/p>\n
If the zeroes of the quadratic polynomial ax2<\/sup> + bx + c, a \u2260 0 are equal, then :<\/h2>\n(A) c and a have opposite signs
\n(B) c and b have opposite signs
\n(C) c and a have the same sign
\n(D) c and b have the same sign
\nAnswer:
\n(C) c and a have the same sign<\/p>\n
Explanation:
\nL Tor equal roots b2<\/sup> – 4ac = 0 b2<\/sup> = 4ac
\nb2<\/sup> is always positive so 4ac must be positive, i.e., product of a and r must be positive, i.e., a and r must have same sign either positive or negative.<\/p>\nQuestion 7.<\/p>\n
If one of the zeroes of a quadratic polynomial of the form x2<\/sup> + ax + b is the negative of the other, then it<\/h2>\n(A) has no linear term and the constant term is negative.
\n(B) has no linear term and the constant term is positive.
\n(C) can have a linear term but the constant term is negative.
\n(D) can have a linear term but the constant term is positive. [U]
\nAnswer:
\n(A) has no linear term and the constant term is negative.<\/p>\n
Explanation:
\nLet f(x) = x2<\/sup> + ax + b and \u03b1, \u03b2 are tha roots of it.
\nthen, \u03b2 = – \u03b1(given)
\n\u03b1 + \u03b2 = \\(-\\frac{b}{a}\\) and \u03b1\u03b2 = \\(\\frac{c}{a}\\)
\n\u03b1 – \u03b1 = \\(-\\frac{a}{1}\\) and \u03b1(-\u03b1) = \\(\\frac{b}{1}\\)
\n– a = 0 – \u03b12<\/sup> = b
\na = 0 b<0 or b is negative
\nso, f(x) = x2<\/sup> + b shows that it has no linear term<\/p>\n<\/p>\n
Question 8.<\/p>\n
Which of the following is not the graph of a quadratic polynomial?<\/h2>\n
<\/p>\n
Answer:
\n<\/p>\n
Explanation:
\nGraph (D) intersect at three points cm .v-axis so the roots ot polynomial of graph is three, so it is cubic polynomial. Other graphs are of quadratic polynomial. Graph a have no real zeroes and Graph b has coincident zeroes.<\/p>\n
Assertion and Reason Based MCQs<\/span><\/p>\nDirections: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\n
Question 1.<\/p>\n
Assertion (A): If the zeroes of a quadratic polynomial ax2<\/sup> + bx + c are both positive, then a, b and c all have the same sign.
\nReason (R): If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.<\/h2>\nAnswer:
\n(D) A is false and R is True<\/p>\n
Explanation:
\nIn case of assertion: Let \u03b1 and \u03b2 be the roots of the quadratic polynomial. If \u03b1 and \u03b2 are positive then
\n\u03b1 + \u03b2 = \\(\\frac{-b}{a}\\) it shows that \\(\\frac{-b}{a}\\) is negetive a sum of two positive numbers (\u03b1 , \u03b2) must be Tve i.c., either b or a must be negative. So, a, b and c will have different signs.
\n\u2234 Given statement is incorrect.
\nIn case of reason:
\nLet \u03b2 = 0, \u03b3 =0
\nf(x) = (x – \u03b1) (x – \u03b2) (x – \u03b3)
\n= (x – \u03b1) x.<\/sup>x
\nf(x) = x3<\/sup>– ax2<\/sup>
\nwhich has no linear (coefficient of x) and constant terms. Given statement is correct. Thus, assertion is incorrect but reason is correct.<\/p>\nQuestion 2.<\/p>\n
Assertion (A): The value of k for which the quadratic polynomial kx2<\/sup> + x + k has equal zeroes are \u00b1\\(\\frac{1}{2}\\)
\nReason (R): If all the three zeroes of a cubic polynomial x 3<\/sup> + ax2<\/sup> – bx + c are positive, then at least one of a, b and c is non-negative.<\/h2>\nAnswer:
\n(C) A is true but R is false<\/p>\n
Explanation:
\nIn case of assertion:
\nf(x) = kx2<\/sup> + x + k (a = k, b = 1, c = k)
\nLor equal roots b2<\/sup>– 4ac = 0
\n(1)2<\/sup> \u2014 4(k) (k) = 0
\n4k2<\/sup>=1
\nk2<\/sup> = \\(\\frac{1}{4}\\)
\nk = \\(\\pm \\frac{1}{2}\\)
\nSo, there are \\(+\\frac{1}{2}\\) and \\(-\\frac{1}{2}\\) values of k so that the given equation has equal roots.
\n\u2234 Given statement is correct.
\nIn case of reason:
\nAll the zeroes’of cubic polynomial are positive only when all the constants a, b, and c are i negative.
\n\u2234Given statement is incorrect.
\nThus, assertion is correct but reason is incorrect.
\nQuestion 3.<\/p>\nAssertion (A): The graph of y = p(x), where p(x) is a polynomial in variable x, is as follows:
\n
\nThe number of zeroes of p(x) is 5.
\nReason (R): If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.<\/h2>\n
Answer:
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
Explanation:
\nIn case of assertion: Since the graph touches the r-axis 5 times. So, the number of zeroes of p{x) is 5.
\n\u2234Given statement is correct.
\nIn case of reason:
\nIf a polynomial of degree more than two hastwo . real zeroes aqd other zeroes are not real or are \u2022 imaginary, and then graph of the polynomial jj will intersect at two points on x-axis.
\n\u2234Given statement is correct:
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\n
<\/p>\n
Question 4.<\/p>\n
Assertion (A): If the zeroes of the quadratic polynomial (k – 1) x2<\/sup> + kx + 1 is – 3, then the value of k is \\(\\frac{4}{3}\\)
\nReason (R): If – 1 is a zero of the polynomial p(x) = kx2<\/sup> -4x + k, then the value of k is -2.<\/h2>\nAnswer:
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
Explanation:
\nIn Case of assertion:
\nLet p(x) = (k – 1)x2<\/sup> + kx + 1
\nAs -3 is a zero of p(x) then p(-3) = 0
\n(k-1)(-3)2 + k(-3) + 1 = 0
\n9k- 9-3k + 1 = 0
\n9k – 3k = + 9 – 1
\n6k = 8
\nk = \\(\\frac{4}{3}\\)<\/p>\n\u2234 Given statement is correct.
\nIn case of reason:
\nSince,-1is a zero of the polynomial and
\np (x) = kx2<\/sup>– 4x + k,
\nP (-1) = 0
\nCk(-1) 2<\/sup> -4(-1) + k = 0
\nk + 4 + k = 0
\n2k + 4 = 0
\n2k = -4
\nk = -2
\n\u2234 Given statement is correct.
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\nCase -Based MCQs<\/span><\/p>\nAttempt any four sub-parts from each question. Each sub-part carries 1 mark.
\nI. Read the following text and answer the following questions on the basis of the same:
\nThe below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.
\n
\n
\n
\n<\/p>\n
Question 1.<\/p>\n
In the standard form of quadratic polynomial, ax2<\/sup> + bx, c, a, b and c are<\/h2>\n(A) All are real numbers.
\n(B) All are rational numbers.
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.
\n(D) All are integers.
\nAnswer:
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.<\/p>\n
Question 2.<\/p>\n
If the roots of the quadratic polynomial are equal, where the discriminant D = b2<\/sup>– 4ac, then<\/h2>\n(A) D > 0
\n(B) D < 0
\n(C) D
\n(D) D = 0
\nAnswer:
\n(D) D = 0<\/p>\n
Explanation:
\nIf the roots of the quadratic polynomial are equal, then discriminant is equal to zero
\nD = b2<\/sup> – 4ac = 0<\/p>\nQuestion 3.<\/p>\n
If a are \\(\\frac{1}{\\alpha}\\) the zeroes of the quadratic polynomial a 2x2<\/sup>– a + 8k, then k is<\/h2>\n(A) 4
\n(B) \\(\\frac{1}{4}\\)
\n(C)\\(\\frac{-1}{4}\\)
\n(D) 2
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
Explanation:
\nGiven equation, 2x2<\/sup> – x + 8k
\nSum of zeroes = \u03b1 +\\(+\\frac{1}{\\alpha}\\)
\nProduct of zeroes = \u03b1. \\(\\frac{1}{\\alpha}\\)=1
\nProduct of zeroes = \\(\\frac{c}{a}=\\frac{8 k}{2}\\)
\n\\(\\begin{aligned}
\n\\frac{8 k}{2} &=1 \\\\
\nk &=\\frac{2}{8} \\\\
\nk &=\\frac{1}{4}
\n\\end{aligned}\\)<\/p>\n<\/p>\n
Question 4.<\/p>\n
The graph of x2<\/sup> + 1 = 0<\/h2>\n(A) Intersects a-axis at two distinct points.
\n(B) Touches a-axis at a point.
\n(C) Neither touches nor intersects a-axis.
\n(D) Either touches or intersects a-axis.
\nAnswer:
\n(C) Neither touches nor intersects a-axis.<\/p>\n
Question 5.<\/p>\n
If the sum of the roots is -p and product of the roots is \\(-\\frac{1}{p}\\) , then the quadratic polynomial is<\/h2>\n
\\(\\text { (A) } k\\left(-p x^{2}+\\frac{x}{p}+1\\right)\\)
\n\\(\\text { (B) } k\\left(p x^{2}-\\frac{x}{p}-1\\right)\\)
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)
\n\\(\\text { (D) } k\\left(x^{2}+p x+\\frac{1}{p}\\right)\\)<\/p>\n
Answer:
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)<\/p>\n
II. Read the following text and answer the following questions on the basis of the same:
\nAn asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.
\n
\n
\n<\/p>\n
Question 1.<\/p>\n
The shape of the poses shown is<\/h2>\n
(A) Spiral
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\n
Question 2.<\/p>\n
The graph of parabola opens downwards, if<\/h2>\n
(A) a \u2265 0
\n(B) a = 0
\n(C) a < 0
\n(D) a > 0
\nAnswer:
\n(C) a < 0<\/p>\n
Question 3.<\/p>\n
In the graph, how many zeroes are there for the polynomial?<\/h2>\n
\n(A) 0
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(C) 2<\/p>\n
Question 4.<\/p>\n
The two zeroes in the above shown graph are<\/h2>\n
(A) 2,4
\n(B) -2,4
\n(C) -8,4
\n(D) 2,-8
\nAnswer:
\n(B) -2,4<\/p>\n
Question 5.<\/p>\n
The zeroes of the quadratic polynomial \\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)are<\/h2>\n
\\(
\n[latex]\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (C) } \\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (D) }-\\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
Explanation:
\n\\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)
\n\\(\\begin{aligned}
\n&=4 \\sqrt{3} x^{2}+(8-3) x-2 \\sqrt{3} \\\\
\n&=4 \\sqrt{3} x^{2}+8 x-3 x-2 \\sqrt{3} \\\\
\n&=4 x(\\sqrt{3} x+2)-\\sqrt{3}(\\sqrt{3} x+2) \\\\
\n&=(\\sqrt{3} x+2)(4 x-\\sqrt{3})
\n\\end{aligned}\\)
\nHence, zeroes of polynomial\\(\\frac{-2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
III. Read the following text and answer the following questions on the basis of the same:
\nBasketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoor on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing quadratic polynomial.
\n
\n<\/p>\n
Question 1.<\/p>\n
The shape of the path traced shown is<\/h2>\n
(A) Spiral
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\n
<\/p>\n
Question 2.<\/p>\n
The graph of parabola opens upwards, if ………..<\/h2>\n
(A) a = 0
\n(B) a < 0 (C) a>0
\n(D) o \u2265 0
\nAnswer:
\n(C) a>0<\/p>\n
Question 3.<\/p>\n
Observe the following graph and answer
\n
\nIn the above graph, how many zeroes are there for the polynomial?<\/h2>\n
(A) 0
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(D) 3<\/p>\n
Explanation:
\nThe number of zeroes of polynomial is the number of times the curve intersects the x-axis, i.e. attains the value 0. Here, the polynomial meets the x-axis at 3 points. So, number of zeroes = 3.<\/p>\n
Question 4.<\/p>\n
The three zeroes in the above shown graph are<\/h2>\n
(A) 2,3, -1
\n(B) -2, 3,1
\n(C) -3, -1,2
\n(D) -2, -3, -1
\nAnswer:
\n(C) -3, -1,2<\/p>\n
Question 5.<\/p>\n
What will be the expression of the polynomial?<\/h2>\n
(A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6
\n(B) x3<\/sup> + 2 x2<\/sup> – 5 x – 6
\n(C) x3<\/sup> + 2 x2<\/sup>+ 5 x-6
\n(D) x3<\/sup> + 2 x2<\/sup> + 5 x + 6
\nAnswer:
\n(A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6<\/p>\nExplanation:
\nSince, the three zeroes = – 3, -1,2 Hence, the expression is (x + 3)(x + l)(x – 2)
\n= [x2<\/sup> + x + 3 x + 3] (x – 2)
\n= x3<\/sup>+ 4x2<\/sup> + 3x-2x2<\/sup>-8x-6
\n= x3<\/sup>+ 2x2<\/sup> – 5x – 6<\/p>\nlV. Read the following text and answer the following questions on the basis of the same: Applications of Parabolas: Highway Overpasses\/ Underpasses
\nA highway underpass is parabolic in shape.
\n
\n
\nParabolic camber equation y = 2x2<\/sup>\/nw
\nParabola
\nA parabola is the graph that results from p(x) = ax2<\/sup> + bx + c.
\nParabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the vertex.
\n<\/p>\nQuestion 1.<\/p>\n
If the highway overpass is represented by x2<\/sup>– 2x – 8. Then its zeroes are<\/h2>\n(A) (2,-4)
\n(B) (4,-2)
\n(C) (-2,-2)
\n(D) (- 4, – 4)
\nAnswer:
\n(C) (-2,-2)<\/p>\n
Explanation:
\nx2<\/sup>– 2 x – 8 = 0
\nor, x2 <\/sup>-4x + 2x- 8 = 0
\nor, x(x – 4) + 2(x – 4) = 0
\nor, (x- 4) (x + 2) = 0
\nor, x = 4, x = – 2<\/p>\nQuestion 2.<\/p>\n
The highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial:<\/h2>\n
(A) Intersects X-axis
\n(B) Intersects Y-axis
\n(C) Intersects Y-axis or X-axis
\n(D) None of the above
\nAnswer:
\n(A) Intersects X-axis<\/p>\n
Explanation:
\nWe know that the number of zeroes of polynomial is equal to number of points where the graph of polynomial intersects X-axis.<\/p>\n
Question 3.<\/p>\n
Graph of a quadratic polynomial is a:<\/h2>\n
(A) straight line
\n(B) circle
\n(C) parabola
\n(D) ellipse
\nAnswer:
\n(C) parabola<\/p>\n
Explanation:
\nHere, the given graph of a quadratic polynomial is a parabola.<\/p>\n
<\/p>\n
Question 4.<\/p>\n
The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is:<\/h2>\n
(A) x2<\/sup> – 6x + 2
\n(B) x2<\/sup>– 36
\n(C) x2 <\/sup>– 6
\n(D) x2 <\/sup>– 3
\nAnswer:
\n(B) x2 <\/sup>– 36<\/p>\nExplanation:
\nx2<\/sup>-36 =0 => x2<\/sup> = 36
\n=>x 6, – 6. => x = \\(\\pm \\sqrt{36}\\)<\/p>\nQuestion 5.<\/p>\n
The number of zeroes that polynomial f(x) = (x – 2)2<\/sup>+ 4 can have is:<\/h2>\n(A) 1
\n(B) 2
\n(C) 0
\n(D) 3
\nAnswer:
\n(C) 0<\/p>\n
Explanation:
\nWe have,
\nf(x) = (x – 2)2<\/sup> + 4
\n= x2<\/sup> + 4 – 4 x + 4
\n=x2<\/sup>-4x + 8. i.e., It has no factorisation:
\nHence no real value of x is possible, i.e., no zero.<\/p>\nV. Read the following text and answer the following questions on the basis of the same:
\nFor a linear polynomial kx + c, k \u2260 0, the graph of y = kx + c is a straight line which intersects the x-axis at exactly one point, namely, \\(\\left(\\frac{-c}{k}, 0\\right),\\) the linear polynomial kx + c, k \u2260 0, has exactly one zero, namely, the X-coordinate of the point where the graph of y = kx + c intersects the x-axis.
\n<\/p>\n
Question 1.<\/p>\n
If a linear polynomial is 2x + 3, then the zero of 2x + 3 is:<\/h2>\n
\\(\\text { (A) } \\frac{3}{2}\\)
\n\\(\\text { (B) }-\\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{3}{2}\\)<\/p>\n
Explanation:
\nGiven, polynomial = 2+ 3
\nLet p(x) = 2x + 3
\nFora zero of p(x),
\n2x + 3 =0
\n2x = -3
\nx = \\(-\\frac{3}{2}\\)<\/p>\n
Question 2.<\/p>\n
The graph of y = p(x) is given in figure below for some polynomial p(x). The number of zero\/zeroes of p(x) is\/are:
\n<\/h2>\n
(A) 1
\n(B) 2
\n(C) 3
\n(D) 0
\nimg-10
\nAnswer:
\n(D) 0<\/p>\n
Explanation:
\nSince the graph does not intersect the x-axis therefore it has no zero.<\/p>\n
Question 3.<\/p>\n
If a and p are the zeroes of the quadratic polynomial x2<\/sup> – 5x + k such that \u03b1 – \u03b2 = 1, then the value of k is:<\/h2>\n(A) 4
\n(B) 5
\n(C) 6
\n(D) 3
\nAnswer:
\n(C) 6<\/p>\n
Explanation:
\nS \u2234p (x) = x2<\/sup>-5x + k
\n\u03b1 – \u03b2 = \\(\\frac{-(-5)}{1}\\) = 5
\n\u03b1\u03b2 = \\(\\frac{k}{1}\\) = k
\nAlso given, \u03b1 – \u03b2 = 1
\n\\(\\sqrt{(\\alpha+\\beta)^{2}-4 \\alpha \\beta}=1\\)<\/p>\nQuestion 4.<\/p>\n
If \u03b1 and \u03b2 are the zeroes of the quadratic polynomial p(x) = 4x2<\/sup> + 5x + 1, then the product of zeroes is:<\/h2>\n(A) -1
\n(B) \\(\\frac{1}{4}\\)
\n(C) -2
\n(D) \\(-\\frac{5}{4}\\)
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
Explanation:
\nWe have, p(x) = 4x2<\/sup> + 5x + 1
\n\u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{1}{4}\\)<\/p>\n<\/p>\n
Question 5.<\/p>\n
If the product of the zeroes of the quadratic polynomial p(x) = ax2<\/sup>– 6x – 6 is 4, then the value of a is: –<\/h2>\n\\(\\text { (A) }-\\frac{3}{2}\\)
\n\\(\\text { (B) } \\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (A) }-\\frac{3}{2}\\)<\/p>\n
Explanation:
\nWWe have,
\np (x) = ax2<\/sup> – 6x – 6
\nLet a and p be the zeroes of the given polynomial, then
\n\\(\\begin{aligned}
\n\\alpha \\beta &=\\frac{c}{a} \\\\
\n4 &=\\frac{-6}{a} \\\\
\n4 a &=-6 \\\\
\na &=-\\frac{6}{4}=-\\frac{3}{2}
\n\\end{aligned}\\)<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"Polynomials Class 10 MCQ Questions with Answers Question 1. The number of polynomials having zeroes as -2 and 5 is: (A) 1 (B) 2 (C) 3 (D) more than 3 Answer: (D) more than 3 Explanation: We know that if we divide or multiply a polynomial by any constant (real number), then the zeroes of …<\/p>\n
\n(B) c and b have opposite signs
\n(C) c and a have the same sign
\n(D) c and b have the same sign
\nAnswer:
\n(C) c and a have the same sign<\/p>\n
\nL Tor equal roots b2<\/sup> – 4ac = 0 b2<\/sup> = 4ac
\nb2<\/sup> is always positive so 4ac must be positive, i.e., product of a and r must be positive, i.e., a and r must have same sign either positive or negative.<\/p>\n
Question 7.<\/p>\n
If one of the zeroes of a quadratic polynomial of the form x2<\/sup> + ax + b is the negative of the other, then it<\/h2>\n(A) has no linear term and the constant term is negative.
\n(B) has no linear term and the constant term is positive.
\n(C) can have a linear term but the constant term is negative.
\n(D) can have a linear term but the constant term is positive. [U]
\nAnswer:
\n(A) has no linear term and the constant term is negative.<\/p>\n
Explanation:
\nLet f(x) = x2<\/sup> + ax + b and \u03b1, \u03b2 are tha roots of it.
\nthen, \u03b2 = – \u03b1(given)
\n\u03b1 + \u03b2 = \\(-\\frac{b}{a}\\) and \u03b1\u03b2 = \\(\\frac{c}{a}\\)
\n\u03b1 – \u03b1 = \\(-\\frac{a}{1}\\) and \u03b1(-\u03b1) = \\(\\frac{b}{1}\\)
\n– a = 0 – \u03b12<\/sup> = b
\na = 0 b<0 or b is negative
\nso, f(x) = x2<\/sup> + b shows that it has no linear term<\/p>\n<\/p>\n
Question 8.<\/p>\n
Which of the following is not the graph of a quadratic polynomial?<\/h2>\n
<\/p>\n
Answer:
\n<\/p>\n
Explanation:
\nGraph (D) intersect at three points cm .v-axis so the roots ot polynomial of graph is three, so it is cubic polynomial. Other graphs are of quadratic polynomial. Graph a have no real zeroes and Graph b has coincident zeroes.<\/p>\n
Assertion and Reason Based MCQs<\/span><\/p>\nDirections: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\n
Question 1.<\/p>\n
Assertion (A): If the zeroes of a quadratic polynomial ax2<\/sup> + bx + c are both positive, then a, b and c all have the same sign.
\nReason (R): If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.<\/h2>\nAnswer:
\n(D) A is false and R is True<\/p>\n
Explanation:
\nIn case of assertion: Let \u03b1 and \u03b2 be the roots of the quadratic polynomial. If \u03b1 and \u03b2 are positive then
\n\u03b1 + \u03b2 = \\(\\frac{-b}{a}\\) it shows that \\(\\frac{-b}{a}\\) is negetive a sum of two positive numbers (\u03b1 , \u03b2) must be Tve i.c., either b or a must be negative. So, a, b and c will have different signs.
\n\u2234 Given statement is incorrect.
\nIn case of reason:
\nLet \u03b2 = 0, \u03b3 =0
\nf(x) = (x – \u03b1) (x – \u03b2) (x – \u03b3)
\n= (x – \u03b1) x.<\/sup>x
\nf(x) = x3<\/sup>– ax2<\/sup>
\nwhich has no linear (coefficient of x) and constant terms. Given statement is correct. Thus, assertion is incorrect but reason is correct.<\/p>\nQuestion 2.<\/p>\n
Assertion (A): The value of k for which the quadratic polynomial kx2<\/sup> + x + k has equal zeroes are \u00b1\\(\\frac{1}{2}\\)
\nReason (R): If all the three zeroes of a cubic polynomial x 3<\/sup> + ax2<\/sup> – bx + c are positive, then at least one of a, b and c is non-negative.<\/h2>\nAnswer:
\n(C) A is true but R is false<\/p>\n
Explanation:
\nIn case of assertion:
\nf(x) = kx2<\/sup> + x + k (a = k, b = 1, c = k)
\nLor equal roots b2<\/sup>– 4ac = 0
\n(1)2<\/sup> \u2014 4(k) (k) = 0
\n4k2<\/sup>=1
\nk2<\/sup> = \\(\\frac{1}{4}\\)
\nk = \\(\\pm \\frac{1}{2}\\)
\nSo, there are \\(+\\frac{1}{2}\\) and \\(-\\frac{1}{2}\\) values of k so that the given equation has equal roots.
\n\u2234 Given statement is correct.
\nIn case of reason:
\nAll the zeroes’of cubic polynomial are positive only when all the constants a, b, and c are i negative.
\n\u2234Given statement is incorrect.
\nThus, assertion is correct but reason is incorrect.
\nQuestion 3.<\/p>\nAssertion (A): The graph of y = p(x), where p(x) is a polynomial in variable x, is as follows:
\n
\nThe number of zeroes of p(x) is 5.
\nReason (R): If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.<\/h2>\n
Answer:
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
Explanation:
\nIn case of assertion: Since the graph touches the r-axis 5 times. So, the number of zeroes of p{x) is 5.
\n\u2234Given statement is correct.
\nIn case of reason:
\nIf a polynomial of degree more than two hastwo . real zeroes aqd other zeroes are not real or are \u2022 imaginary, and then graph of the polynomial jj will intersect at two points on x-axis.
\n\u2234Given statement is correct:
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\n
<\/p>\n
Question 4.<\/p>\n
Assertion (A): If the zeroes of the quadratic polynomial (k – 1) x2<\/sup> + kx + 1 is – 3, then the value of k is \\(\\frac{4}{3}\\)
\nReason (R): If – 1 is a zero of the polynomial p(x) = kx2<\/sup> -4x + k, then the value of k is -2.<\/h2>\nAnswer:
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
Explanation:
\nIn Case of assertion:
\nLet p(x) = (k – 1)x2<\/sup> + kx + 1
\nAs -3 is a zero of p(x) then p(-3) = 0
\n(k-1)(-3)2 + k(-3) + 1 = 0
\n9k- 9-3k + 1 = 0
\n9k – 3k = + 9 – 1
\n6k = 8
\nk = \\(\\frac{4}{3}\\)<\/p>\n\u2234 Given statement is correct.
\nIn case of reason:
\nSince,-1is a zero of the polynomial and
\np (x) = kx2<\/sup>– 4x + k,
\nP (-1) = 0
\nCk(-1) 2<\/sup> -4(-1) + k = 0
\nk + 4 + k = 0
\n2k + 4 = 0
\n2k = -4
\nk = -2
\n\u2234 Given statement is correct.
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\nCase -Based MCQs<\/span><\/p>\nAttempt any four sub-parts from each question. Each sub-part carries 1 mark.
\nI. Read the following text and answer the following questions on the basis of the same:
\nThe below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.
\n
\n
\n
\n<\/p>\n
Question 1.<\/p>\n
In the standard form of quadratic polynomial, ax2<\/sup> + bx, c, a, b and c are<\/h2>\n(A) All are real numbers.
\n(B) All are rational numbers.
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.
\n(D) All are integers.
\nAnswer:
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.<\/p>\n
Question 2.<\/p>\n
If the roots of the quadratic polynomial are equal, where the discriminant D = b2<\/sup>– 4ac, then<\/h2>\n(A) D > 0
\n(B) D < 0
\n(C) D
\n(D) D = 0
\nAnswer:
\n(D) D = 0<\/p>\n
Explanation:
\nIf the roots of the quadratic polynomial are equal, then discriminant is equal to zero
\nD = b2<\/sup> – 4ac = 0<\/p>\nQuestion 3.<\/p>\n
If a are \\(\\frac{1}{\\alpha}\\) the zeroes of the quadratic polynomial a 2x2<\/sup>– a + 8k, then k is<\/h2>\n(A) 4
\n(B) \\(\\frac{1}{4}\\)
\n(C)\\(\\frac{-1}{4}\\)
\n(D) 2
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
Explanation:
\nGiven equation, 2x2<\/sup> – x + 8k
\nSum of zeroes = \u03b1 +\\(+\\frac{1}{\\alpha}\\)
\nProduct of zeroes = \u03b1. \\(\\frac{1}{\\alpha}\\)=1
\nProduct of zeroes = \\(\\frac{c}{a}=\\frac{8 k}{2}\\)
\n\\(\\begin{aligned}
\n\\frac{8 k}{2} &=1 \\\\
\nk &=\\frac{2}{8} \\\\
\nk &=\\frac{1}{4}
\n\\end{aligned}\\)<\/p>\n<\/p>\n
Question 4.<\/p>\n
The graph of x2<\/sup> + 1 = 0<\/h2>\n(A) Intersects a-axis at two distinct points.
\n(B) Touches a-axis at a point.
\n(C) Neither touches nor intersects a-axis.
\n(D) Either touches or intersects a-axis.
\nAnswer:
\n(C) Neither touches nor intersects a-axis.<\/p>\n
Question 5.<\/p>\n
If the sum of the roots is -p and product of the roots is \\(-\\frac{1}{p}\\) , then the quadratic polynomial is<\/h2>\n
\\(\\text { (A) } k\\left(-p x^{2}+\\frac{x}{p}+1\\right)\\)
\n\\(\\text { (B) } k\\left(p x^{2}-\\frac{x}{p}-1\\right)\\)
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)
\n\\(\\text { (D) } k\\left(x^{2}+p x+\\frac{1}{p}\\right)\\)<\/p>\n
Answer:
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)<\/p>\n
II. Read the following text and answer the following questions on the basis of the same:
\nAn asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.
\n
\n
\n<\/p>\n
Question 1.<\/p>\n
The shape of the poses shown is<\/h2>\n
(A) Spiral
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\n
Question 2.<\/p>\n
The graph of parabola opens downwards, if<\/h2>\n
(A) a \u2265 0
\n(B) a = 0
\n(C) a < 0
\n(D) a > 0
\nAnswer:
\n(C) a < 0<\/p>\n
Question 3.<\/p>\n
In the graph, how many zeroes are there for the polynomial?<\/h2>\n
\n(A) 0
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(C) 2<\/p>\n
Question 4.<\/p>\n
The two zeroes in the above shown graph are<\/h2>\n
(A) 2,4
\n(B) -2,4
\n(C) -8,4
\n(D) 2,-8
\nAnswer:
\n(B) -2,4<\/p>\n
Question 5.<\/p>\n
The zeroes of the quadratic polynomial \\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)are<\/h2>\n
\\(
\n[latex]\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (C) } \\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (D) }-\\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
Explanation:
\n\\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)
\n\\(\\begin{aligned}
\n&=4 \\sqrt{3} x^{2}+(8-3) x-2 \\sqrt{3} \\\\
\n&=4 \\sqrt{3} x^{2}+8 x-3 x-2 \\sqrt{3} \\\\
\n&=4 x(\\sqrt{3} x+2)-\\sqrt{3}(\\sqrt{3} x+2) \\\\
\n&=(\\sqrt{3} x+2)(4 x-\\sqrt{3})
\n\\end{aligned}\\)
\nHence, zeroes of polynomial\\(\\frac{-2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
III. Read the following text and answer the following questions on the basis of the same:
\nBasketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoor on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing quadratic polynomial.
\n
\n<\/p>\n
Question 1.<\/p>\n
The shape of the path traced shown is<\/h2>\n
(A) Spiral
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\n
<\/p>\n
Question 2.<\/p>\n
The graph of parabola opens upwards, if ………..<\/h2>\n
(A) a = 0
\n(B) a < 0 (C) a>0
\n(D) o \u2265 0
\nAnswer:
\n(C) a>0<\/p>\n
Question 3.<\/p>\n
Observe the following graph and answer
\n
\nIn the above graph, how many zeroes are there for the polynomial?<\/h2>\n
(A) 0
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(D) 3<\/p>\n
Explanation:
\nThe number of zeroes of polynomial is the number of times the curve intersects the x-axis, i.e. attains the value 0. Here, the polynomial meets the x-axis at 3 points. So, number of zeroes = 3.<\/p>\n
Question 4.<\/p>\n
The three zeroes in the above shown graph are<\/h2>\n
(A) 2,3, -1
\n(B) -2, 3,1
\n(C) -3, -1,2
\n(D) -2, -3, -1
\nAnswer:
\n(C) -3, -1,2<\/p>\n
Question 5.<\/p>\n
What will be the expression of the polynomial?<\/h2>\n
(A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6
\n(B) x3<\/sup> + 2 x2<\/sup> – 5 x – 6
\n(C) x3<\/sup> + 2 x2<\/sup>+ 5 x-6
\n(D) x3<\/sup> + 2 x2<\/sup> + 5 x + 6
\nAnswer:
\n(A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6<\/p>\nExplanation:
\nSince, the three zeroes = – 3, -1,2 Hence, the expression is (x + 3)(x + l)(x – 2)
\n= [x2<\/sup> + x + 3 x + 3] (x – 2)
\n= x3<\/sup>+ 4x2<\/sup> + 3x-2x2<\/sup>-8x-6
\n= x3<\/sup>+ 2x2<\/sup> – 5x – 6<\/p>\nlV. Read the following text and answer the following questions on the basis of the same: Applications of Parabolas: Highway Overpasses\/ Underpasses
\nA highway underpass is parabolic in shape.
\n
\n
\nParabolic camber equation y = 2x2<\/sup>\/nw
\nParabola
\nA parabola is the graph that results from p(x) = ax2<\/sup> + bx + c.
\nParabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the vertex.
\n<\/p>\nQuestion 1.<\/p>\n
If the highway overpass is represented by x2<\/sup>– 2x – 8. Then its zeroes are<\/h2>\n(A) (2,-4)
\n(B) (4,-2)
\n(C) (-2,-2)
\n(D) (- 4, – 4)
\nAnswer:
\n(C) (-2,-2)<\/p>\n
Explanation:
\nx2<\/sup>– 2 x – 8 = 0
\nor, x2 <\/sup>-4x + 2x- 8 = 0
\nor, x(x – 4) + 2(x – 4) = 0
\nor, (x- 4) (x + 2) = 0
\nor, x = 4, x = – 2<\/p>\nQuestion 2.<\/p>\n
The highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial:<\/h2>\n
(A) Intersects X-axis
\n(B) Intersects Y-axis
\n(C) Intersects Y-axis or X-axis
\n(D) None of the above
\nAnswer:
\n(A) Intersects X-axis<\/p>\n
Explanation:
\nWe know that the number of zeroes of polynomial is equal to number of points where the graph of polynomial intersects X-axis.<\/p>\n
Question 3.<\/p>\n
Graph of a quadratic polynomial is a:<\/h2>\n
(A) straight line
\n(B) circle
\n(C) parabola
\n(D) ellipse
\nAnswer:
\n(C) parabola<\/p>\n
Explanation:
\nHere, the given graph of a quadratic polynomial is a parabola.<\/p>\n
<\/p>\n
Question 4.<\/p>\n
The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is:<\/h2>\n
(A) x2<\/sup> – 6x + 2
\n(B) x2<\/sup>– 36
\n(C) x2 <\/sup>– 6
\n(D) x2 <\/sup>– 3
\nAnswer:
\n(B) x2 <\/sup>– 36<\/p>\nExplanation:
\nx2<\/sup>-36 =0 => x2<\/sup> = 36
\n=>x 6, – 6. => x = \\(\\pm \\sqrt{36}\\)<\/p>\nQuestion 5.<\/p>\n
The number of zeroes that polynomial f(x) = (x – 2)2<\/sup>+ 4 can have is:<\/h2>\n(A) 1
\n(B) 2
\n(C) 0
\n(D) 3
\nAnswer:
\n(C) 0<\/p>\n
Explanation:
\nWe have,
\nf(x) = (x – 2)2<\/sup> + 4
\n= x2<\/sup> + 4 – 4 x + 4
\n=x2<\/sup>-4x + 8. i.e., It has no factorisation:
\nHence no real value of x is possible, i.e., no zero.<\/p>\nV. Read the following text and answer the following questions on the basis of the same:
\nFor a linear polynomial kx + c, k \u2260 0, the graph of y = kx + c is a straight line which intersects the x-axis at exactly one point, namely, \\(\\left(\\frac{-c}{k}, 0\\right),\\) the linear polynomial kx + c, k \u2260 0, has exactly one zero, namely, the X-coordinate of the point where the graph of y = kx + c intersects the x-axis.
\n<\/p>\n
Question 1.<\/p>\n
If a linear polynomial is 2x + 3, then the zero of 2x + 3 is:<\/h2>\n
\\(\\text { (A) } \\frac{3}{2}\\)
\n\\(\\text { (B) }-\\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{3}{2}\\)<\/p>\n
Explanation:
\nGiven, polynomial = 2+ 3
\nLet p(x) = 2x + 3
\nFora zero of p(x),
\n2x + 3 =0
\n2x = -3
\nx = \\(-\\frac{3}{2}\\)<\/p>\n
Question 2.<\/p>\n
The graph of y = p(x) is given in figure below for some polynomial p(x). The number of zero\/zeroes of p(x) is\/are:
\n<\/h2>\n
(A) 1
\n(B) 2
\n(C) 3
\n(D) 0
\nimg-10
\nAnswer:
\n(D) 0<\/p>\n
Explanation:
\nSince the graph does not intersect the x-axis therefore it has no zero.<\/p>\n
Question 3.<\/p>\n
If a and p are the zeroes of the quadratic polynomial x2<\/sup> – 5x + k such that \u03b1 – \u03b2 = 1, then the value of k is:<\/h2>\n(A) 4
\n(B) 5
\n(C) 6
\n(D) 3
\nAnswer:
\n(C) 6<\/p>\n
Explanation:
\nS \u2234p (x) = x2<\/sup>-5x + k
\n\u03b1 – \u03b2 = \\(\\frac{-(-5)}{1}\\) = 5
\n\u03b1\u03b2 = \\(\\frac{k}{1}\\) = k
\nAlso given, \u03b1 – \u03b2 = 1
\n\\(\\sqrt{(\\alpha+\\beta)^{2}-4 \\alpha \\beta}=1\\)<\/p>\nQuestion 4.<\/p>\n
If \u03b1 and \u03b2 are the zeroes of the quadratic polynomial p(x) = 4x2<\/sup> + 5x + 1, then the product of zeroes is:<\/h2>\n(A) -1
\n(B) \\(\\frac{1}{4}\\)
\n(C) -2
\n(D) \\(-\\frac{5}{4}\\)
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
Explanation:
\nWe have, p(x) = 4x2<\/sup> + 5x + 1
\n\u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{1}{4}\\)<\/p>\n<\/p>\n
Question 5.<\/p>\n
If the product of the zeroes of the quadratic polynomial p(x) = ax2<\/sup>– 6x – 6 is 4, then the value of a is: –<\/h2>\n\\(\\text { (A) }-\\frac{3}{2}\\)
\n\\(\\text { (B) } \\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (A) }-\\frac{3}{2}\\)<\/p>\n
Explanation:
\nWWe have,
\np (x) = ax2<\/sup> – 6x – 6
\nLet a and p be the zeroes of the given polynomial, then
\n\\(\\begin{aligned}
\n\\alpha \\beta &=\\frac{c}{a} \\\\
\n4 &=\\frac{-6}{a} \\\\
\n4 a &=-6 \\\\
\na &=-\\frac{6}{4}=-\\frac{3}{2}
\n\\end{aligned}\\)<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"Polynomials Class 10 MCQ Questions with Answers Question 1. The number of polynomials having zeroes as -2 and 5 is: (A) 1 (B) 2 (C) 3 (D) more than 3 Answer: (D) more than 3 Explanation: We know that if we divide or multiply a polynomial by any constant (real number), then the zeroes of …<\/p>\n
\n(B) has no linear term and the constant term is positive.
\n(C) can have a linear term but the constant term is negative.
\n(D) can have a linear term but the constant term is positive. [U]
\nAnswer:
\n(A) has no linear term and the constant term is negative.<\/p>\n
\nLet f(x) = x2<\/sup> + ax + b and \u03b1, \u03b2 are tha roots of it.
\nthen, \u03b2 = – \u03b1(given)
\n\u03b1 + \u03b2 = \\(-\\frac{b}{a}\\) and \u03b1\u03b2 = \\(\\frac{c}{a}\\)
\n\u03b1 – \u03b1 = \\(-\\frac{a}{1}\\) and \u03b1(-\u03b1) = \\(\\frac{b}{1}\\)
\n– a = 0 – \u03b12<\/sup> = b
\na = 0 b<0 or b is negative
\nso, f(x) = x2<\/sup> + b shows that it has no linear term<\/p>\n
<\/p>\n
Question 8.<\/p>\n
Which of the following is not the graph of a quadratic polynomial?<\/h2>\n
<\/p>\n
Answer:
\n<\/p>\n
Explanation:
\nGraph (D) intersect at three points cm .v-axis so the roots ot polynomial of graph is three, so it is cubic polynomial. Other graphs are of quadratic polynomial. Graph a have no real zeroes and Graph b has coincident zeroes.<\/p>\n
Assertion and Reason Based MCQs<\/span><\/p>\n Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. Question 1.<\/p>\n Answer: Explanation: Question 2.<\/p>\n Answer: Explanation: Answer: Explanation: Question 4.<\/p>\n Answer: Explanation: \u2234 Given statement is correct. Case -Based MCQs<\/span><\/p>\n Attempt any four sub-parts from each question. Each sub-part carries 1 mark. Question 1.<\/p>\n (A) All are real numbers. Question 2.<\/p>\n (A) D > 0 Explanation: Question 3.<\/p>\n (A) 4 Explanation: Question 4.<\/p>\n (A) Intersects a-axis at two distinct points. Question 5.<\/p>\n \\(\\text { (A) } k\\left(-p x^{2}+\\frac{x}{p}+1\\right)\\) Answer: II. Read the following text and answer the following questions on the basis of the same: Question 1.<\/p>\n (A) Spiral Question 2.<\/p>\n (A) a \u2265 0 Question 3.<\/p>\n Question 4.<\/p>\n (A) 2,4 Question 5.<\/p>\n \\( Explanation: III. Read the following text and answer the following questions on the basis of the same: Question 1.<\/p>\n (A) Spiral Question 2.<\/p>\n (A) a = 0 Question 3.<\/p>\n (A) 0 Explanation: Question 4.<\/p>\n (A) 2,3, -1 Question 5.<\/p>\n (A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6 Explanation: lV. Read the following text and answer the following questions on the basis of the same: Applications of Parabolas: Highway Overpasses\/ Underpasses Question 1.<\/p>\n (A) (2,-4) Explanation: Question 2.<\/p>\n (A) Intersects X-axis Explanation: Question 3.<\/p>\n (A) straight line Explanation: Question 4.<\/p>\n (A) x2<\/sup> – 6x + 2 Explanation: Question 5.<\/p>\n (A) 1 Explanation: V. Read the following text and answer the following questions on the basis of the same: Question 1.<\/p>\n \\(\\text { (A) } \\frac{3}{2}\\) Explanation: Question 2.<\/p>\n (A) 1 Explanation: Question 3.<\/p>\n (A) 4 Explanation: Question 4.<\/p>\n (A) -1 Explanation: Question 5.<\/p>\n \\(\\text { (A) }-\\frac{3}{2}\\) Explanation: Polynomials Class 10 MCQ Questions with Answers Question 1. The number of polynomials having zeroes as -2 and 5 is: (A) 1 (B) 2 (C) 3 (D) more than 3 Answer: (D) more than 3 Explanation: We know that if we divide or multiply a polynomial by any constant (real number), then the zeroes of …<\/p>\n
\n(A) Both A and R are true and R is the correct explanation of A
\n(B) Both A and R are true but R is NOT the correct explanation of A
\n(C) A is true but R is false
\n(D) A is false and R is True<\/p>\nAssertion (A): If the zeroes of a quadratic polynomial ax2<\/sup> + bx + c are both positive, then a, b and c all have the same sign.
\nReason (R): If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIn case of assertion: Let \u03b1 and \u03b2 be the roots of the quadratic polynomial. If \u03b1 and \u03b2 are positive then
\n\u03b1 + \u03b2 = \\(\\frac{-b}{a}\\) it shows that \\(\\frac{-b}{a}\\) is negetive a sum of two positive numbers (\u03b1 , \u03b2) must be Tve i.c., either b or a must be negative. So, a, b and c will have different signs.
\n\u2234 Given statement is incorrect.
\nIn case of reason:
\nLet \u03b2 = 0, \u03b3 =0
\nf(x) = (x – \u03b1) (x – \u03b2) (x – \u03b3)
\n= (x – \u03b1) x.<\/sup>x
\nf(x) = x3<\/sup>– ax2<\/sup>
\nwhich has no linear (coefficient of x) and constant terms. Given statement is correct. Thus, assertion is incorrect but reason is correct.<\/p>\nAssertion (A): The value of k for which the quadratic polynomial kx2<\/sup> + x + k has equal zeroes are \u00b1\\(\\frac{1}{2}\\)
\nReason (R): If all the three zeroes of a cubic polynomial x 3<\/sup> + ax2<\/sup> – bx + c are positive, then at least one of a, b and c is non-negative.<\/h2>\n
\n(C) A is true but R is false<\/p>\n
\nIn case of assertion:
\nf(x) = kx2<\/sup> + x + k (a = k, b = 1, c = k)
\nLor equal roots b2<\/sup>– 4ac = 0
\n(1)2<\/sup> \u2014 4(k) (k) = 0
\n4k2<\/sup>=1
\nk2<\/sup> = \\(\\frac{1}{4}\\)
\nk = \\(\\pm \\frac{1}{2}\\)
\nSo, there are \\(+\\frac{1}{2}\\) and \\(-\\frac{1}{2}\\) values of k so that the given equation has equal roots.
\n\u2234 Given statement is correct.
\nIn case of reason:
\nAll the zeroes’of cubic polynomial are positive only when all the constants a, b, and c are i negative.
\n\u2234Given statement is incorrect.
\nThus, assertion is correct but reason is incorrect.
\nQuestion 3.<\/p>\nAssertion (A): The graph of y = p(x), where p(x) is a polynomial in variable x, is as follows:
\n
\nThe number of zeroes of p(x) is 5.
\nReason (R): If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn case of assertion: Since the graph touches the r-axis 5 times. So, the number of zeroes of p{x) is 5.
\n\u2234Given statement is correct.
\nIn case of reason:
\nIf a polynomial of degree more than two hastwo . real zeroes aqd other zeroes are not real or are \u2022 imaginary, and then graph of the polynomial jj will intersect at two points on x-axis.
\n\u2234Given statement is correct:
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\n<\/p>\nAssertion (A): If the zeroes of the quadratic polynomial (k – 1) x2<\/sup> + kx + 1 is – 3, then the value of k is \\(\\frac{4}{3}\\)
\nReason (R): If – 1 is a zero of the polynomial p(x) = kx2<\/sup> -4x + k, then the value of k is -2.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A<\/p>\n
\nIn Case of assertion:
\nLet p(x) = (k – 1)x2<\/sup> + kx + 1
\nAs -3 is a zero of p(x) then p(-3) = 0
\n(k-1)(-3)2 + k(-3) + 1 = 0
\n9k- 9-3k + 1 = 0
\n9k – 3k = + 9 – 1
\n6k = 8
\nk = \\(\\frac{4}{3}\\)<\/p>\n
\nIn case of reason:
\nSince,-1is a zero of the polynomial and
\np (x) = kx2<\/sup>– 4x + k,
\nP (-1) = 0
\nCk(-1) 2<\/sup> -4(-1) + k = 0
\nk + 4 + k = 0
\n2k + 4 = 0
\n2k = -4
\nk = -2
\n\u2234 Given statement is correct.
\nThus, both assertion and reason are correct but reason is not the correct explanation for assertion.<\/p>\n
\nI. Read the following text and answer the following questions on the basis of the same:
\nThe below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.
\n
\n
\n
\n<\/p>\nIn the standard form of quadratic polynomial, ax2<\/sup> + bx, c, a, b and c are<\/h2>\n
\n(B) All are rational numbers.
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.
\n(D) All are integers.
\nAnswer:
\n(C) ‘a’ is a non zero real number and b and c are any real numbers.<\/p>\nIf the roots of the quadratic polynomial are equal, where the discriminant D = b2<\/sup>– 4ac, then<\/h2>\n
\n(B) D < 0
\n(C) D
\n(D) D = 0
\nAnswer:
\n(D) D = 0<\/p>\n
\nIf the roots of the quadratic polynomial are equal, then discriminant is equal to zero
\nD = b2<\/sup> – 4ac = 0<\/p>\nIf a are \\(\\frac{1}{\\alpha}\\) the zeroes of the quadratic polynomial a 2x2<\/sup>– a + 8k, then k is<\/h2>\n
\n(B) \\(\\frac{1}{4}\\)
\n(C)\\(\\frac{-1}{4}\\)
\n(D) 2
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
\nGiven equation, 2x2<\/sup> – x + 8k
\nSum of zeroes = \u03b1 +\\(+\\frac{1}{\\alpha}\\)
\nProduct of zeroes = \u03b1. \\(\\frac{1}{\\alpha}\\)=1
\nProduct of zeroes = \\(\\frac{c}{a}=\\frac{8 k}{2}\\)
\n\\(\\begin{aligned}
\n\\frac{8 k}{2} &=1 \\\\
\nk &=\\frac{2}{8} \\\\
\nk &=\\frac{1}{4}
\n\\end{aligned}\\)<\/p>\n<\/p>\nThe graph of x2<\/sup> + 1 = 0<\/h2>\n
\n(B) Touches a-axis at a point.
\n(C) Neither touches nor intersects a-axis.
\n(D) Either touches or intersects a-axis.
\nAnswer:
\n(C) Neither touches nor intersects a-axis.<\/p>\nIf the sum of the roots is -p and product of the roots is \\(-\\frac{1}{p}\\) , then the quadratic polynomial is<\/h2>\n
\n\\(\\text { (B) } k\\left(p x^{2}-\\frac{x}{p}-1\\right)\\)
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)
\n\\(\\text { (D) } k\\left(x^{2}+p x+\\frac{1}{p}\\right)\\)<\/p>\n
\n\\(\\text { (C) } k\\left(x^{2}+p x-\\frac{1}{p}\\right)\\)<\/p>\n
\nAn asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.
\n
\n
\n<\/p>\nThe shape of the poses shown is<\/h2>\n
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\nThe graph of parabola opens downwards, if<\/h2>\n
\n(B) a = 0
\n(C) a < 0
\n(D) a > 0
\nAnswer:
\n(C) a < 0<\/p>\nIn the graph, how many zeroes are there for the polynomial?<\/h2>\n
\n(A) 0
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(C) 2<\/p>\nThe two zeroes in the above shown graph are<\/h2>\n
\n(B) -2,4
\n(C) -8,4
\n(D) 2,-8
\nAnswer:
\n(B) -2,4<\/p>\nThe zeroes of the quadratic polynomial \\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)are<\/h2>\n
\n[latex]\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (C) } \\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\n\\(\\text { (D) }-\\frac{2}{\\sqrt{3}},-\\frac{\\sqrt{3}}{4}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
\n\\(4 \\sqrt{3} x^{2}+5 x-2 \\sqrt{3}\\)
\n\\(\\begin{aligned}
\n&=4 \\sqrt{3} x^{2}+(8-3) x-2 \\sqrt{3} \\\\
\n&=4 \\sqrt{3} x^{2}+8 x-3 x-2 \\sqrt{3} \\\\
\n&=4 x(\\sqrt{3} x+2)-\\sqrt{3}(\\sqrt{3} x+2) \\\\
\n&=(\\sqrt{3} x+2)(4 x-\\sqrt{3})
\n\\end{aligned}\\)
\nHence, zeroes of polynomial\\(\\frac{-2}{\\sqrt{3}}, \\frac{\\sqrt{3}}{4}\\)<\/p>\n
\nBasketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoor on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing quadratic polynomial.
\n
\n<\/p>\nThe shape of the path traced shown is<\/h2>\n
\n(B) Ellipse
\n(C) Linear
\n(D) Parabola
\nAnswer:
\n(D) Parabola<\/p>\n<\/p>\nThe graph of parabola opens upwards, if ………..<\/h2>\n
\n(B) a < 0 (C) a>0
\n(D) o \u2265 0
\nAnswer:
\n(C) a>0<\/p>\nObserve the following graph and answer
\n
\nIn the above graph, how many zeroes are there for the polynomial?<\/h2>\n
\n(B) 1
\n(C) 2
\n(D) 3
\nAnswer:
\n(D) 3<\/p>\n
\nThe number of zeroes of polynomial is the number of times the curve intersects the x-axis, i.e. attains the value 0. Here, the polynomial meets the x-axis at 3 points. So, number of zeroes = 3.<\/p>\nThe three zeroes in the above shown graph are<\/h2>\n
\n(B) -2, 3,1
\n(C) -3, -1,2
\n(D) -2, -3, -1
\nAnswer:
\n(C) -3, -1,2<\/p>\nWhat will be the expression of the polynomial?<\/h2>\n
\n(B) x3<\/sup> + 2 x2<\/sup> – 5 x – 6
\n(C) x3<\/sup> + 2 x2<\/sup>+ 5 x-6
\n(D) x3<\/sup> + 2 x2<\/sup> + 5 x + 6
\nAnswer:
\n(A) x3<\/sup> + 2 x2<\/sup> – 5 x – 6<\/p>\n
\nSince, the three zeroes = – 3, -1,2 Hence, the expression is (x + 3)(x + l)(x – 2)
\n= [x2<\/sup> + x + 3 x + 3] (x – 2)
\n= x3<\/sup>+ 4x2<\/sup> + 3x-2x2<\/sup>-8x-6
\n= x3<\/sup>+ 2x2<\/sup> – 5x – 6<\/p>\n
\nA highway underpass is parabolic in shape.
\n
\n
\nParabolic camber equation y = 2x2<\/sup>\/nw
\nParabola
\nA parabola is the graph that results from p(x) = ax2<\/sup> + bx + c.
\nParabolas are symmetric about a vertical line known as the Axis of Symmetry. The Axis of Symmetry runs through the maximum or minimum point of the parabola which is called the vertex.
\n<\/p>\nIf the highway overpass is represented by x2<\/sup>– 2x – 8. Then its zeroes are<\/h2>\n
\n(B) (4,-2)
\n(C) (-2,-2)
\n(D) (- 4, – 4)
\nAnswer:
\n(C) (-2,-2)<\/p>\n
\nx2<\/sup>– 2 x – 8 = 0
\nor, x2 <\/sup>-4x + 2x- 8 = 0
\nor, x(x – 4) + 2(x – 4) = 0
\nor, (x- 4) (x + 2) = 0
\nor, x = 4, x = – 2<\/p>\nThe highway overpass is represented graphically. Zeroes of a polynomial can be expressed graphically. Number of zeroes of polynomial is equal to number of points where the graph of polynomial:<\/h2>\n
\n(B) Intersects Y-axis
\n(C) Intersects Y-axis or X-axis
\n(D) None of the above
\nAnswer:
\n(A) Intersects X-axis<\/p>\n
\nWe know that the number of zeroes of polynomial is equal to number of points where the graph of polynomial intersects X-axis.<\/p>\nGraph of a quadratic polynomial is a:<\/h2>\n
\n(B) circle
\n(C) parabola
\n(D) ellipse
\nAnswer:
\n(C) parabola<\/p>\n
\nHere, the given graph of a quadratic polynomial is a parabola.<\/p>\n<\/p>\nThe representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is:<\/h2>\n
\n(B) x2<\/sup>– 36
\n(C) x2 <\/sup>– 6
\n(D) x2 <\/sup>– 3
\nAnswer:
\n(B) x2 <\/sup>– 36<\/p>\n
\nx2<\/sup>-36 =0 => x2<\/sup> = 36
\n=>x 6, – 6. => x = \\(\\pm \\sqrt{36}\\)<\/p>\nThe number of zeroes that polynomial f(x) = (x – 2)2<\/sup>+ 4 can have is:<\/h2>\n
\n(B) 2
\n(C) 0
\n(D) 3
\nAnswer:
\n(C) 0<\/p>\n
\nWe have,
\nf(x) = (x – 2)2<\/sup> + 4
\n= x2<\/sup> + 4 – 4 x + 4
\n=x2<\/sup>-4x + 8. i.e., It has no factorisation:
\nHence no real value of x is possible, i.e., no zero.<\/p>\n
\nFor a linear polynomial kx + c, k \u2260 0, the graph of y = kx + c is a straight line which intersects the x-axis at exactly one point, namely, \\(\\left(\\frac{-c}{k}, 0\\right),\\) the linear polynomial kx + c, k \u2260 0, has exactly one zero, namely, the X-coordinate of the point where the graph of y = kx + c intersects the x-axis.
\n<\/p>\nIf a linear polynomial is 2x + 3, then the zero of 2x + 3 is:<\/h2>\n
\n\\(\\text { (B) }-\\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (B) }-\\frac{3}{2}\\)<\/p>\n
\nGiven, polynomial = 2+ 3
\nLet p(x) = 2x + 3
\nFora zero of p(x),
\n2x + 3 =0
\n2x = -3
\nx = \\(-\\frac{3}{2}\\)<\/p>\nThe graph of y = p(x) is given in figure below for some polynomial p(x). The number of zero\/zeroes of p(x) is\/are:
\n<\/h2>\n
\n(B) 2
\n(C) 3
\n(D) 0
\nimg-10
\nAnswer:
\n(D) 0<\/p>\n
\nSince the graph does not intersect the x-axis therefore it has no zero.<\/p>\nIf a and p are the zeroes of the quadratic polynomial x2<\/sup> – 5x + k such that \u03b1 – \u03b2 = 1, then the value of k is:<\/h2>\n
\n(B) 5
\n(C) 6
\n(D) 3
\nAnswer:
\n(C) 6<\/p>\n
\nS \u2234p (x) = x2<\/sup>-5x + k
\n\u03b1 – \u03b2 = \\(\\frac{-(-5)}{1}\\) = 5
\n\u03b1\u03b2 = \\(\\frac{k}{1}\\) = k
\nAlso given, \u03b1 – \u03b2 = 1
\n\\(\\sqrt{(\\alpha+\\beta)^{2}-4 \\alpha \\beta}=1\\)<\/p>\nIf \u03b1 and \u03b2 are the zeroes of the quadratic polynomial p(x) = 4x2<\/sup> + 5x + 1, then the product of zeroes is:<\/h2>\n
\n(B) \\(\\frac{1}{4}\\)
\n(C) -2
\n(D) \\(-\\frac{5}{4}\\)
\nAnswer:
\n(B) \\(\\frac{1}{4}\\)<\/p>\n
\nWe have, p(x) = 4x2<\/sup> + 5x + 1
\n\u03b1\u03b2 = \\(\\frac{c}{a}=\\frac{1}{4}\\)<\/p>\n<\/p>\nIf the product of the zeroes of the quadratic polynomial p(x) = ax2<\/sup>– 6x – 6 is 4, then the value of a is: –<\/h2>\n
\n\\(\\text { (B) } \\frac{3}{2}\\)
\n\\(\\text { (C) } \\frac{2}{3}\\)
\n\\(\\text { (D) }-\\frac{2}{3}\\)
\nAnswer:
\n\\(\\text { (A) }-\\frac{3}{2}\\)<\/p>\n
\nWWe have,
\np (x) = ax2<\/sup> – 6x – 6
\nLet a and p be the zeroes of the given polynomial, then
\n\\(\\begin{aligned}
\n\\alpha \\beta &=\\frac{c}{a} \\\\
\n4 &=\\frac{-6}{a} \\\\
\n4 a &=-6 \\\\
\na &=-\\frac{6}{4}=-\\frac{3}{2}
\n\\end{aligned}\\)<\/p>\nMCQ Questions for Class 10 Maths with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"