MCQ Questions

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 polynomial remains same.
Here, α = —2 and β = +5
α + β = —2 + 5 = 3 and αβ = — 2 x 5 =- 10
So, required polynomial is
x2 — (α + β)x + αβ = x2 — 3x —10
If we multiply this polynomial by any real number, let 5 and 2, we get 5x² — 15x — 50 and 2x² — 6x — 20 which are different polynomials but having same zeroes -2 and 5. So, we can obtain so many (infinite polynomials) from two given zeroes.

Question 2.

Given that one of the zeroes of the cubic polynomial ax³+ bx² + cx + d is zero, the product of the other two zeroes is:

(A) \(-\frac{c}{a}\)
(B) \(\frac{c}{a}\)
(C) 0
(D)\(-\frac{b}{a}\)

Answer:
(B) \(\frac{c}{a}\)

Explanation:
Let/(x) = ax³+ bx² + cx + d If a, p, y are the zeroes of f(x), then
αβ + βγ + γα =\(=\frac{c}{a}\)
One root is zero (given) so, α = 0. βγ =\(=\frac{c}{a}\)

Question 3.

If one of the zeroes of the cubic polynomial x³ + ax² + bx + c is — 1, then the product of the other two zeroes is:

(A) b – a +1
(B) b – a – 1
(C) a – b + 1
(D) a – b – 1
Answer:
(A) b – a +1

Explanation:
Let f(x) = x³ + ax² + bx + c
∴ One of the zeroes of f(x) is – I so
f(-1) = 0
(-1)³ + a(-1)² + b(-1) + c = 0
-1 + a – b + c = 0
a – b + c = 1
c = 1 + b – a
Now, αβγ = \(\frac{-d}{a}\) [∴a = 1, d = c]
\(-1 \beta y=\frac{-c}{1}\)
βγ = c
βγ = 1 + b – a

Question 4.

The zeroes of the quadratic polynomial x² + 99x 127 are:

(A) both positive
(B) both negative
(C) one positive and one negative
(D) both equal
Answer:
(B) both negative

Explanation:
Let given quadratic polynomial
p(x) = x²+ 99x + 127
On comparing p(x) with ax²+ bx + c. we get a = 1,b = 99 and c = 127
We know that,
x = \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\)
= \(\frac{-99 \pm \sqrt{9801-508}}{2}\)
= \(\frac{-99 \pm \sqrt{9293}}{2}=\frac{-99 \pm 96.4}{2}\)
= \(\frac{-99+96.4}{2}, \frac{-99-96.4}{2}\)
= \(\frac{-2.6}{2}, \frac{-195.4}{2}\)
= -1.3, – 97.7
Hence, both zeroes of the given quadratic polynomial p(x) are negative.

Question 5.

The zeroes of the quadratic polynomial x² + kx + k, k ≠ 0,

(A) cannot both be positive
(B) cannot both be negative
(C) are always unequal
(D) are always equal
Answer:
(A) cannot both be positive

Explanation:
Let f(x) = x ² + kx + k, k 0.
On comparing the given polynomial with ax² + bx + c, we get a = 1, b – k,c = k
If α and β be the zeroes of the polynomial (x).
We know thaL,
Sum of zeroes, α + β = \(-\frac{b}{a}\)
α + β = \(-\frac{k}{1}\) = -k ……(i)
And ofzeroes, αβ = \(\frac{c}{a}\)
αβ = \(\frac{k}{1}\)=k
Case I: If k is negative, αβ[from equation (ii)] is negative. It means α and β are of opposite sign.
Case II: If k is positive, then αβ [from equation (ii)] is positive but α + β 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.

Question 6.

If the zeroes of the quadratic polynomial ax² + bx + c, a ≠ 0 are equal, then :

(A) c and a have opposite signs
(B) c and b have opposite signs
(C) c and a have the same sign
(D) c and b have the same sign
Answer:
(C) c and a have the same sign

Explanation:
L Tor equal roots b² – 4ac = 0 b² = 4ac
b² 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.

Question 7.

If one of the zeroes of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it

(A) has no linear term and the constant term is negative.
(B) has no linear term and the constant term is positive.
(C) can have a linear term but the constant term is negative.
(D) can have a linear term but the constant term is positive. [U]
Answer:
(A) has no linear term and the constant term is negative.

Explanation:
Let f(x) = x² + ax + b and α, β are tha roots of it.
then, β = – α(given)
α + β = \(-\frac{b}{a}\) and αβ = \(\frac{c}{a}\)
α – α = \(-\frac{a}{1}\) and α(-α) = \(\frac{b}{1}\)
– a = 0 – α² = b
a = 0 b<0 or b is negative
so, f(x) = x² + b shows that it has no linear term

Question 8.

Which of the following is not the graph of a quadratic polynomial?

Answer:

Explanation:
Graph (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.

Assertion and Reason Based MCQs

Directions: In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is NOT the correct explanation of A
(C) A is true but R is false
(D) A is false and R is True

Question 1.

Assertion (A): If the zeroes of a quadratic polynomial ax² + bx + c are both positive, then a, b and c all have the same sign.
Reason (R): If two of the zeroes of a cubic polynomial are zero, then it does not have linear and constant terms.

Answer:
(D) A is false and R is True

Explanation:
In case of assertion: Let α and β be the roots of the quadratic polynomial. If α and β are positive then
α + β = \(\frac{-b}{a}\) it shows that \(\frac{-b}{a}\) is negetive a sum of two positive numbers (α , β) must be Tve i.c., either b or a must be negative. So, a, b and c will have different signs.
∴ Given statement is incorrect.
In case of reason:
Let β = 0, γ =0
f(x) = (x – α) (x – β) (x – γ)
= (x – α) x^.x
f(x) = x³– ax²
which has no linear (coefficient of x) and constant terms. Given statement is correct. Thus, assertion is incorrect but reason is correct.

Question 2.

Assertion (A): The value of k for which the quadratic polynomial kx² + x + k has equal zeroes are ±\(\frac{1}{2}\)
Reason (R): If all the three zeroes of a cubic polynomial x ³ + ax² – bx + c are positive, then at least one of a, b and c is non-negative.

Answer:
(C) A is true but R is false

Explanation:
In case of assertion:
f(x) = kx² + x + k (a = k, b = 1, c = k)
Lor equal roots b²– 4ac = 0
(1)² — 4(k) (k) = 0
4k²=1
k² = \(\frac{1}{4}\)
k = \(\pm \frac{1}{2}\)
So, there are \(+\frac{1}{2}\) and \(-\frac{1}{2}\) values of k so that the given equation has equal roots.
∴ Given statement is correct.
In case of reason:
All the zeroes’of cubic polynomial are positive only when all the constants a, b, and c are i negative.
∴Given statement is incorrect.
Thus, assertion is correct but reason is incorrect.
Question 3.

Assertion (A): The graph of y = p(x), where p(x) is a polynomial in variable x, is as follows:

The number of zeroes of p(x) is 5.
Reason (R): If the graph of a polynomial intersects the x-axis at exactly two points, it need not be a quadratic polynomial.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In case of assertion: Since the graph touches the r-axis 5 times. So, the number of zeroes of p{x) is 5.
∴Given statement is correct.
In case of reason:
If a polynomial of degree more than two hastwo . real zeroes aqd other zeroes are not real or are • imaginary, and then graph of the polynomial jj will intersect at two points on x-axis.
∴Given statement is correct:
Thus, both assertion and reason are correct but reason is not the correct explanation for assertion.

Question 4.

Assertion (A): If the zeroes of the quadratic polynomial (k – 1) x² + kx + 1 is – 3, then the value of k is \(\frac{4}{3}\)
Reason (R): If – 1 is a zero of the polynomial p(x) = kx² -4x + k, then the value of k is -2.

Answer:
(B) Both A and R are true but R is NOT the correct explanation of A

Explanation:
In Case of assertion:
Let p(x) = (k – 1)x² + kx + 1
As -3 is a zero of p(x) then p(-3) = 0
(k-1)(-3)2 + k(-3) + 1 = 0
9k- 9-3k + 1 = 0
9k – 3k = + 9 – 1
6k = 8
k = \(\frac{4}{3}\)

∴ Given statement is correct.
In case of reason:
Since,-1is a zero of the polynomial and
p (x) = kx²– 4x + k,
P (-1) = 0
Ck(-1) ² -4(-1) + k = 0
k + 4 + k = 0
2k + 4 = 0
2k = -4
k = -2
∴ Given statement is correct.
Thus, both assertion and reason are correct but reason is not the correct explanation for assertion.

Case -Based MCQs

Attempt any four sub-parts from each question. Each sub-part carries 1 mark.
I. Read the following text and answer the following questions on the basis of the same:
The 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.

Question 1.

In the standard form of quadratic polynomial, ax² + bx, c, a, b and c are

(A) All are real numbers.
(B) All are rational numbers.
(C) ‘a’ is a non zero real number and b and c are any real numbers.
(D) All are integers.
Answer:
(C) ‘a’ is a non zero real number and b and c are any real numbers.

Question 2.

If the roots of the quadratic polynomial are equal, where the discriminant D = b²– 4ac, then

(A) D > 0
(B) D < 0
(C) D
(D) D = 0
Answer:
(D) D = 0

Explanation:
If the roots of the quadratic polynomial are equal, then discriminant is equal to zero
D = b² – 4ac = 0

Question 3.

If a are \(\frac{1}{\alpha}\) the zeroes of the quadratic polynomial a 2x²– a + 8k, then k is

(A) 4
(B) \(\frac{1}{4}\)
(C)\(\frac{-1}{4}\)
(D) 2
Answer:
(B) \(\frac{1}{4}\)

Explanation:
Given equation, 2x² – x + 8k
Sum of zeroes = α +\(+\frac{1}{\alpha}\)
Product of zeroes = α. \(\frac{1}{\alpha}\)=1
Product of zeroes = \(\frac{c}{a}=\frac{8 k}{2}\)
\(\begin{aligned}
\frac{8 k}{2} &=1 \\
k &=\frac{2}{8} \\
k &=\frac{1}{4}
\end{aligned}\)

Question 4.

The graph of x² + 1 = 0

(A) Intersects a-axis at two distinct points.
(B) Touches a-axis at a point.
(C) Neither touches nor intersects a-axis.
(D) Either touches or intersects a-axis.
Answer:
(C) Neither touches nor intersects a-axis.

Question 5.

If the sum of the roots is -p and product of the roots is \(-\frac{1}{p}\) , then the quadratic polynomial is

\(\text { (A) } k\left(-p x^{2}+\frac{x}{p}+1\right)\)
\(\text { (B) } k\left(p x^{2}-\frac{x}{p}-1\right)\)
\(\text { (C) } k\left(x^{2}+p x-\frac{1}{p}\right)\)
\(\text { (D) } k\left(x^{2}+p x+\frac{1}{p}\right)\)

Answer:
\(\text { (C) } k\left(x^{2}+p x-\frac{1}{p}\right)\)

II. Read the following text and answer the following questions on the basis of the same:
An 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.

Question 1.

The shape of the poses shown is

(A) Spiral
(B) Ellipse
(C) Linear
(D) Parabola
Answer:
(D) Parabola

Question 2.

The graph of parabola opens downwards, if

(A) a ≥ 0
(B) a = 0
(C) a < 0
(D) a > 0
Answer:
(C) a < 0

Question 3.

In the graph, how many zeroes are there for the polynomial?

(A) 0
(B) 1
(C) 2
(D) 3
Answer:
(C) 2

Question 4.

The two zeroes in the above shown graph are

(A) 2,4
(B) -2,4
(C) -8,4
(D) 2,-8
Answer:
(B) -2,4

Question 5.

The zeroes of the quadratic polynomial \(4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}\)are

\(
[latex]\text { (B) }-\frac{2}{\sqrt{3}}, \frac{\sqrt{3}}{4}\)
\(\text { (C) } \frac{2}{\sqrt{3}},-\frac{\sqrt{3}}{4}\)
\(\text { (D) }-\frac{2}{\sqrt{3}},-\frac{\sqrt{3}}{4}\)
Answer:
\(\text { (B) }-\frac{2}{\sqrt{3}}, \frac{\sqrt{3}}{4}\)

Explanation:
\(4 \sqrt{3} x^{2}+5 x-2 \sqrt{3}\)
\(\begin{aligned}
&=4 \sqrt{3} x^{2}+(8-3) x-2 \sqrt{3} \\
&=4 \sqrt{3} x^{2}+8 x-3 x-2 \sqrt{3} \\
&=4 x(\sqrt{3} x+2)-\sqrt{3}(\sqrt{3} x+2) \\
&=(\sqrt{3} x+2)(4 x-\sqrt{3})
\end{aligned}\)
Hence, zeroes of polynomial\(\frac{-2}{\sqrt{3}}, \frac{\sqrt{3}}{4}\)

III. Read the following text and answer the following questions on the basis of the same:
Basketball 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.

Question 1.

The shape of the path traced shown is

(A) Spiral
(B) Ellipse
(C) Linear
(D) Parabola
Answer:
(D) Parabola

Question 2.

The graph of parabola opens upwards, if ………..

(A) a = 0
(B) a < 0 (C) a>0
(D) o ≥ 0
Answer:
(C) a>0

Question 3.

Observe the following graph and answer

In the above graph, how many zeroes are there for the polynomial?

(A) 0
(B) 1
(C) 2
(D) 3
Answer:
(D) 3

Explanation:
The 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.

Question 4.

The three zeroes in the above shown graph are

(A) 2,3, -1
(B) -2, 3,1
(C) -3, -1,2
(D) -2, -3, -1
Answer:
(C) -3, -1,2

Question 5.

What will be the expression of the polynomial?

(A) x³ + 2 x² – 5 x – 6
(B) x³ + 2 x² – 5 x – 6
(C) x³ + 2 x²+ 5 x-6
(D) x³ + 2 x² + 5 x + 6
Answer:
(A) x³ + 2 x² – 5 x – 6

Explanation:
Since, the three zeroes = – 3, -1,2 Hence, the expression is (x + 3)(x + l)(x – 2)
= [x² + x + 3 x + 3] (x – 2)
= x³+ 4x² + 3x-2x²-8x-6
= x³+ 2x² – 5x – 6

lV. Read the following text and answer the following questions on the basis of the same: Applications of Parabolas: Highway Overpasses/ Underpasses
A highway underpass is parabolic in shape.


Parabolic camber equation y = 2x²/nw
Parabola
A parabola is the graph that results from p(x) = ax² + bx + c.
Parabolas 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.

Question 1.

If the highway overpass is represented by x²– 2x – 8. Then its zeroes are

(A) (2,-4)
(B) (4,-2)
(C) (-2,-2)
(D) (- 4, – 4)
Answer:
(C) (-2,-2)

Explanation:
x²– 2 x – 8 = 0
or, x² -4x + 2x- 8 = 0
or, x(x – 4) + 2(x – 4) = 0
or, (x- 4) (x + 2) = 0
or, x = 4, x = – 2

Question 2.

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:

(A) Intersects X-axis
(B) Intersects Y-axis
(C) Intersects Y-axis or X-axis
(D) None of the above
Answer:
(A) Intersects X-axis

Explanation:
We know that the number of zeroes of polynomial is equal to number of points where the graph of polynomial intersects X-axis.

Question 3.

Graph of a quadratic polynomial is a:

(A) straight line
(B) circle
(C) parabola
(D) ellipse
Answer:
(C) parabola

Explanation:
Here, the given graph of a quadratic polynomial is a parabola.

Question 4.

The representation of Highway Underpass whose one zero is 6 and sum of the zeroes is 0, is:

(A) x² – 6x + 2
(B) x²– 36
(C) x² – 6
(D) x² – 3
Answer:
(B) x² – 36

Explanation:
x²-36 =0 => x² = 36
=>x 6, – 6. => x = \(\pm \sqrt{36}\)

Question 5.

The number of zeroes that polynomial f(x) = (x – 2)²+ 4 can have is:

(A) 1
(B) 2
(C) 0
(D) 3
Answer:
(C) 0

Explanation:
We have,
f(x) = (x – 2)² + 4
= x² + 4 – 4 x + 4
=x²-4x + 8. i.e., It has no factorisation:
Hence no real value of x is possible, i.e., no zero.

V. Read the following text and answer the following questions on the basis of the same:
For a linear polynomial kx + c, k ≠ 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 ≠ 0, has exactly one zero, namely, the X-coordinate of the point where the graph of y = kx + c intersects the x-axis.

Question 1.

If a linear polynomial is 2x + 3, then the zero of 2x + 3 is:

\(\text { (A) } \frac{3}{2}\)
\(\text { (B) }-\frac{3}{2}\)
\(\text { (C) } \frac{2}{3}\)
\(\text { (D) }-\frac{2}{3}\)
Answer:
\(\text { (B) }-\frac{3}{2}\)

Explanation:
Given, polynomial = 2+ 3
Let p(x) = 2x + 3
Fora zero of p(x),
2x + 3 =0
2x = -3
x = \(-\frac{3}{2}\)

Question 2.

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:

(A) 1
(B) 2
(C) 3
(D) 0
img-10
Answer:
(D) 0

Explanation:
Since the graph does not intersect the x-axis therefore it has no zero.

Question 3.

If a and p are the zeroes of the quadratic polynomial x² – 5x + k such that α – β = 1, then the value of k is:

(A) 4
(B) 5
(C) 6
(D) 3
Answer:
(C) 6

Explanation:
S ∴p (x) = x²-5x + k
α – β = \(\frac{-(-5)}{1}\) = 5
αβ = \(\frac{k}{1}\) = k
Also given, α – β = 1
\(\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}=1\)

Question 4.

If α and β are the zeroes of the quadratic polynomial p(x) = 4x² + 5x + 1, then the product of zeroes is:

(A) -1
(B) \(\frac{1}{4}\)
(C) -2
(D) \(-\frac{5}{4}\)
Answer:
(B) \(\frac{1}{4}\)

Explanation:
We have, p(x) = 4x² + 5x + 1
αβ = \(\frac{c}{a}=\frac{1}{4}\)

Question 5.

If the product of the zeroes of the quadratic polynomial p(x) = ax²– 6x – 6 is 4, then the value of a is: –

\(\text { (A) }-\frac{3}{2}\)
\(\text { (B) } \frac{3}{2}\)
\(\text { (C) } \frac{2}{3}\)
\(\text { (D) }-\frac{2}{3}\)
Answer:
\(\text { (A) }-\frac{3}{2}\)

Explanation:
WWe have,
p (x) = ax² – 6x – 6
Let a and p be the zeroes of the given polynomial, then
\(\begin{aligned}
\alpha \beta &=\frac{c}{a} \\
4 &=\frac{-6}{a} \\
4 a &=-6 \\
a &=-\frac{6}{4}=-\frac{3}{2}
\end{aligned}\)

MCQ Questions for Class 10 Maths with Answers