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MCQ Questions for Class 11 Maths Chapter 9 Sequences and Series with Answers

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 9 Sequences and Series with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Sequences and Series Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 9 Sequences and Series Objective Questions.

Sequences and Series Class 11 MCQs Questions with Answers

Students are advised to solve the Sequences and Series Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Sequences and Series Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Sequences and Series Class 11 with answers provided with detailed solutions by looking below.

Question 1.
Let Tr be the rth term of an A.P. whose first term is a and the common difference is d. If for some positive integers m, n, m ≠ n, T m = 1/n and T n = 1/m then (a-d) equals to
(a) 0
(b) 1
(c) 1/mn
(d) 1/m + 1/n

Answer

Answer: (a) 0
Given the first term is a and the common difference is d of the AP
Now, Tm = 1/n
⇒ a + (m – 1)d = 1/n ………… 1
and Tn = 1/m
⇒ a + (n – 1)d = 1/m ………. 2
From equation 2 – 1, we get
(m – 1)d – (n – 1)d = 1/n – 1/m
⇒ (m – n)d = (m – n)/mn
⇒ d = 1/mn
From equation 1, we get
a + (m – 1)/mn = 1/n
⇒ a = 1/n – (m – 1)/mn
⇒ a = {m – (m – 1)}/mn
⇒ a = {m – m + 1)}/mn
⇒ a = 1/mn
Now, a – d = 1/mn – 1/mn
⇒ a – d = 0

Question 2.
The first term of a GP is 1. The sum of the third term and fifth term is 90. The common ratio of GP is
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (c) 3
Let first term of the GP is a and common ratio is r.
3rd term = ar²
5th term = ar4
Now
⇒ ar² + ar4 = 90
⇒ a(r² + r4) = 90
⇒ r² + r4 = 90
⇒ r² × (r² + 1) = 90
⇒ r² (r² + 1) = 3² × (3² + +1)
⇒ r = 3
So the common ratio is 3

Question 3.
If a is the first term and r is the common ratio then the nth term of GP is
(a) (ar)n-1
(b) a × rⁿ
(c) a × rn-1
(d) None of these

Answer

Answer: (c) a × rn-1
Given, a is the first term and r is the common ratio.
Now, nth term of GP = a × rn-1

Question 4.
The sum of odd integers from 1 to 2001 is
(a) 10201
(b) 102001
(c) 100201
(d) 1002001

Answer

Answer: (d) 1002001
The odd numbers from 1 to 2001 are:
1, 3, 5, ………, 2001
This froms an AP
where first term a = 1
Common difference d = 3 – 1 = 2
last term l = 2001
Let number of terms = n
Now, l = a + (n – 1)d
⇒ 2001 = 1 + (n – 1)2
⇒ 2001 – 1 = (n – 1)2
⇒ 2(n – 1) = 2000
⇒ n – 1 = 2000/2
⇒ n – 1 = 1000
⇒ n = 1001
Now, sum = (n/2) × (a + 1)
= (1001/2) × (1 + 2001)
= (1001/2) × 2002
= 1001 × 1001
= 1002001
So, the sum of odd integers from 1 to 2001 is 1002001

Question 5.
If a, b, c are in AP and x, y, z are in GP then the value of xb-c × yc-a × za-b is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (b) 1
Given, a, b, c are in AP
⇒ 2b = a + c ………. 1
and x, y, z are in GP
⇒ y² = xz ……….. 2
Now, xb-c × yc-a × za-b = xb-c × (√xz)c-a × za-b
= xb-c × x(c-a)/2 × z(c-a)/2 × za-b
= xb-c + x(c-a)/2 × z(c-a)/2+ a -b
= x2b+(c+a) × z(c+a)-2b
= x° × z°
= 1
So, the value of xb-c × yc-a × za-b is 1

Question 6.
An example of geometric series is
(a) 9, 20, 21, 28
(b) 1, 2, 4, 8
(c) 1, 2, 3, 4
(d) 3, 5, 7, 9

Answer

Answer: (b) 1, 2, 4, 8
1, 2, 4, 8 is the example of geometric series
Here common ratio = 2/1 = 4/2 = 8/4 = 2

Question 7.
Three numbers from an increasing GP of the middle number is doubled, then the new numbers are in AP. The common ratio of the GP is
(a) 2
(b) √3
(c) 2 + √3
(d) 2 – √3

Answer

Answer: (c) 2 + √3
Given that three numbers from an increasing GP
Let the 3 number are: a, ar, ar² (r > 1)
Now, according to question,
a, 2ar, ar² are in AP
So, 2ar – a = ar² – 2ar
⇒ a(2r – 1) = a(r² – 2r)
⇒ 2r – 1 = r² – 2r
⇒ r² – 2r – 2r + 1 = 0
⇒ r² – 4r + 1 = 0
⇒ r = [4 ± √{16 – 4 × 1 × 1}]/2
⇒ r = [4 ± √{16 – 4}]/2
⇒ r = {4 ± √12}/2
⇒ r = {4 ± 2√3}/2
⇒ r = {2 ± √3}
Since r > 1
So, the common ratio of the GP is (2 + √3)

Question 8.
An arithmetic sequence has its 5th term equal to 22 and its 15th term equal to 62. Then its 100th term is equal to
(a) 410
(b) 408
(c) 402
(d) 404

Answer

Answer: (c) 402
Let ais the first term and d is the common difference of the AP
Given,
a5 = a + (5 – 1)d = 22
⇒ a + 4d = 22 ………….1
and a15 = a + (15 – 1)d = 62
⇒ a + 14d = 62 ………2
From equation 2 – 1, we get
62 – 22 = 14d – 4d
⇒ 10d = 40
⇒ d = 4
From equation 1, we get
a + 4 × 4 = 22
⇒ a + 16 = 22
⇒ a = 6
Now,
a100 = 6 + 4(100 – 1 )
⇒ a100 = 6 + 4 × 99
⇒ a100 = 6 + 396
⇒ a100 = 402

Question 9.
Suppose a, b, c are in A.P. and a², b², c² are in G.P. If a < b < c and a + b + c = 3/2, then the value of a is
(a) 1/2√2
(b) 1/2√3
(c) 1/2 – 1/√3
(d) 1/2 – 1/√2

Answer

Answer: (d) 1/2 – 1/√2
Given, a, b, c are in AP
⇒ 2b = a + c
⇒ b = (a + c)/2 ………….. 1
Again given, a², b², c² are in GP then b4 = a² c²
⇒ b² = ± ac ………… 2
Using 1 in a + b + c = 3/2, we get
3b = 3/2
⇒ b = 1/2
hence a + c = 1
and ac = ± 1/4
So a & c are roots of either x2 −x + 1/4 = 0 or x² − x − 1/4 = 0
The first has equal roots of x = 1/2 and second gives x= (1 ± √2)/2 for a and c
Since a < c,
we must have a = (1−√2)/2
⇒ a – 1/2 – √2/2
⇒ a – 1/2 – √2/(√2×√2)
⇒ a – 1/2 – 1/√2

Question 10.
If the positive numbers a, b, c, d are in A.P., then abc, abd, acd, bcd are
(a) not in A.P. / G.P. / H. P.
(b) in A.P.
(c) in G.P.
(d) in H.P.

Answer

Answer: (d) in H.P.
Given, the positive numbers a, b, c, d are in A.P.
⇒ 1/a, 1/b, 1/c, 1/d are in H.P.
⇒ 1/d, 1/c, 1/b, 1/a are in H.P.
Now, Multiply by abcd, we get
abcd/d, abcd/c, abcd/b, abcd/a are in H.P.
⇒ abc, abd, acd, bcd are in H.P.

Question 11.
Let Tr be the rth term of an A.P. whose first term is a and the common difference is d. If for some positive integers m, n, m ≠ n, T m = 1/n and T n = 1/m then (a-d) equals to
(a) 0
(b) 1
(c) 1/mn
(d) 1/m + 1/n

Answer

Answer: (a) 0
Given the first term is a and the common difference is d of the AP
Now, Tm = 1/n
⇒ a + (m – 1)d = 1/n ………… 1
and Tn = 1/m
⇒ a + (n – 1)d = 1/m ………. 2
From equation 2 – 1, we get
(m – 1)d – (n – 1)d = 1/n – 1/m
⇒ (m – n)d = (m – n)/mn
⇒ d = 1/mn
From equation 1, we get
a + (m – 1)/mn = 1/n
⇒ a = 1/n – (m – 1)/mn
⇒ a = {m – (m – 1)}/mn
⇒ a = {m – m + 1)}/mn
⇒ a = 1/mn
Now, a – d = 1/mn – 1/mn
⇒ a – d = 0

Question 12.
In the sequence obtained by omitting the perfect squares from the sequence of natural numbers, then 2011th term is
(a) 2024
(b) 2036
(c) 2048
(d) 2055

Answer

Answer: (d) 2055
Before 2024, there are 44 squares,
So, 1980th term is 2024
Hence, 2011th term is 2055

Question 13.
If the first term minus third term of a G.P. = 768 and the third term minus seventh term of the same G.P. = 240, then the product of first 21 terms =
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (a) 1
Let first term = a
and common ratio = r
Given, a – ar² = 768
⇒ a(1 – r²) = 768
and ar² – ar6 = 240
⇒ ar² (1 – r4) = 240
Dividing the above 2 equations, we get
ar² (1 – r4)/a(1 – r²) = 240/768
⇒ {ar² (1 – r²) × (1 + r²)}/a(1 – r²) = 240/768
⇒ 1 + r² = 0.3125
⇒ r² = 0.25
⇒ r² = 25/100
⇒ r² = √(1/4)
⇒ r = ± 1/2
Now, a(1 – r²) = 768
⇒ a(1 – 1/4 ) = 768
⇒ 3a/4 = 768
⇒ 3a = 4 × 768
⇒ a = (4 × 768)/3
⇒ a = 4 × 256
⇒ a = 1024
⇒ a = 210
Now product of first 21 terms = (a² × r20)10 × a × r10
= a21 × r210
= (210)21 × (1/2)210
= 2210 /2210
= 1

Question 14.
If the sum of the first 2n terms of the A.P. 2, 5, 8, ….., is equal to the sum of the first n terms of the A.P. 57, 59, 61, ….., then n equals
(a) 10
(b) 12
(c) 11
(d) 13

Answer

Answer: (c) 11
Given, the sum of the first 2n terms of the A.P. 2, 5, 8, ….. = the sum of the first n terms of the A.P. 57, 59, 61, ….
⇒ (2n/2) × {2 × 2 + (2n – 1)3} = (n/2) × {2 × 57 + (n – 1)2}
⇒ n × {4 + 6n – 3} = (n/2) × {114 + 2n – 2}
⇒ 6n + 1 = {2n + 112}/2
⇒ 6n + 1 = n + 56
⇒ 6n – n = 56 – 1
⇒ 5n = 55
⇒ n = 55/5
⇒ n = 11

Question 15.
If a, b, c are in GP then log aⁿ, log bⁿ, log cⁿ are in
(a) AP
(b) GP
(c) Either in AP or in GP
(d) Neither in AP nor in GP

Answer

Answer: (a) AP
Given, a, b, c are in GP
⇒ b² = ac
⇒ (b²)ⁿ = (ac)ⁿ
⇒ (b2 )ⁿ= aⁿ × cⁿ
⇒ log (b²)ⁿ = log(an × cn )
⇒ log b²ⁿ = log aⁿ + log cⁿ
⇒ log (bⁿ)² = log aⁿ + log cⁿ
⇒ 2 × log bⁿ = log aⁿ + log cⁿ
⇒ log aⁿ, log bⁿ, log cⁿ are in AP

Question 16.
If the nth term of an AP is 3n – 4, the 10th term of AP is
(a) 12
(b) 22
(c) 28
(d) 30

Answer

Answer: (c) 28
Given, an = 3n – 2
Put n = 10, we get
a10 = 3 × 10 – 2
⇒ a10 = 30 – 2
⇒ a10 = 28
So, the 10th term of AP is 28

Question 17.
If the third term of an A.P. is 7 and its 7 th term is 2 more than three times of its third term, then the sum of its first 20 terms is
(a) 228
(b) 74
(c) 740
(d) 1090

Answer

Answer: (c) 740
Let a is the first term and d is the common difference of AP
Given the third term of an A.P. is 7 and its 7th term is 2 more than three times of its third term
⇒ a + 2d = 7 ………….. 1
and
3(a + 2d) + 2 = a + 6d
⇒ 3 × 7 + 2 = a + 6d
⇒ 21 + 2 = a + 6d
⇒ a + 6d = 23 ………….. 2
From equation 1 – 2, we get
4d = 16
⇒ d = 16/4
⇒ d = 4
From equation 1, we get
a + 2 × 4 = 7
⇒ a + 8 = 7
⇒ a = -1
Now, the sum of its first 20 terms
= (20/2) × {2 × (-1) + (20-1) × 4}
= 10 × {-2 + 19 × 4)}
= 10 × {-2 + 76)}
= 10 × 74
= 740

Question 18.
If a, b, c are in AP then
(a) b = a + c
(b) 2b = a + c
(c) b² = a + c
(d) 2b² = a + c

Answer

Answer: (b) 2b = a + c
Given, a, b, c are in AP
⇒ b – a = c – b
⇒ b + b = a + c
⇒ 2b = a + c

Question 19.
If 1/(b + c), 1/(c + a), 1/(a + b) are in AP then
(a) a, b, c are in AP
(b) a², b², c² are in AP
(c) 1/1, 1/b, 1/c are in AP
(d) None of these

Answer

Answer: (b) a², b², c² are in AP
Given, 1/(b + c), 1/(c + a), 1/(a + b)
⇒ 2/(c + a) = 1/(b + c) + 1/(a + b)
⇒ 2b2 = a² + c²
⇒ a², b², c² are in AP

Question 20.
3, 5, 7, 9, …….. is an example of
(a) Geometric Series
(b) Arithmetic Series
(c) Rational Exponent
(d) Logarithm

Answer

Answer: (b) Arithmetic Series
3, 5, 7, 9, …….. is an example of Arithmetic Series.
Here common difference = 5 – 3 = 7 – 5 = 9 – 7 = 2

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MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Conic Sections Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 11 Conic Sections Objective Questions.

Conic Sections Class 11 MCQs Questions with Answers

Students are advised to solve the Conic Sections Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Conic Sections Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Conic Sections Class 11 with answers provided with detailed solutions by looking below.

Question 1.
The straight line y = mx + c cuts the circle x² + y² = a² in real points if
(a) √{a² × (1 + m²)} < c
(b) √{a² × (1 – m²)} < c
(c) √{a² × (1 + m²)} > c
(d) √{a² × (1 – m²)} > c

Answer

Answer: (c) √{a² × (1 + m²)} > c
The straight line y = mx + c cuts the circle x² + y² = a² in real points if
√{a² × (1 + m²)} > c

Question 2.
Equation of the directrix of the parabola x² = 4ay is
(a) x = -a
(b) x = a
(c) y = -a
(d) y = a

Answer

Answer: (c) y = -a
Given, parabola x² = 4ay
Now, its equation of directrix = y = -a

Question 3.
The equation of parabola with vertex at origin and directrix x – 2 = 0 is
(a) y² = -4x
(b) y² = 4x
(c) y² = -8x
(d) y² = 8x

Answer

Answer: (c) y² = -8x
Since the line passing through the focus and perpendicular to the directrix is x-axis,
therefore axis of the required parabola is x-axis.
Let the coordinate of the focus is S(a, 0).
Since the vertex is the mid point of the line joining the focus and the point (2, 0) where the directix is x – 2 = 0 meets the axis.
So, 0 = {a – (-2)}/2
⇒ 0 = (a + 2)/2
⇒ a + 2 = 0
⇒ a = -2
Thus the coordinate of focus is (-2, 0)
Let P(x, y) be a point on the parabola.
Then by definition of parabola
(x + 2)² + (y – 0)² = (x – 2)²
⇒ x² + 4 + 4x + y² = x² + 4 – 4x
⇒ 4x + y² = – 4x
⇒ y² = -4x – 4x
⇒ y² = -8x
This is the required equation of the parabola.

Question 4.
The perpendicular distance from the point (3, -4) to the line 3x – 4y + 10 = 0
(a) 7
(b) 8
(c) 9
(d) 10

Answer

Answer: (a) 7
The perpendicular distance = {3 × 3 – 4 × (-4) + 10}/√(3² + 4²)
= {9 + 16 + 10}/√(9 + 16)
= 35/√25
= 35/5
= 7

Question 5.
The equation of a hyperbola with foci on the x-axis is
(a) x²/a² + y²/b² = 1
(b) x²/a² – y²/b² = 1
(c) x² + y² = (a² + b²)
(d) x² – y² = (a² + b²)

Answer

Answer: (b) x²/a² – y²/b² = 1
The equation of a hyperbola with foci on the x-axis is defined as
x²/a² – y²/b² = 1

Question 6.
If the line 2x – y + λ = 0 is a diameter of the circle x² + y² + 6x − 6y + 5 = 0 then λ =
(a) 5
(b) 7
(c) 9
(d) 11

Answer

Answer: (c) 9
Given equation of the circle is
x² + y² + 6x − 6y + 5 = 0
Center O = (-3, 3)
radius r = √{(-3)² + (3)² – 5} = √{9 + 9 – 5} = √13
Since diameter of the circle passes through the center of the circle.
So (-3, 3) satisfies the equation 2x – y + λ = 0
⇒ -3 × 2 – 3 + λ = 0
⇒ -6 – 3 + λ = 0
⇒ -9 + λ = 0
⇒ λ = 9

Question 7.
The number of tangents that can be drawn from (1, 2) to x² + y² = 5 is
(a) 0
(b) 1
(c) 2
(d) More than 2

Answer

Answer: (b) 1
Given point (1, 2) and equation of circle is x² + y² = 5
Now, x² + y² – 5 = 0
Put (1, 2) in this equation, we get
1² + 2² – 5 = 1 + 4 – 5 = 5 – 5 = 0
So, the point (1, 2) lies on the circle.
Hence, only one tangent can be drawn.

Question 8.
The equation of the circle x² + y² + 2gx + 2fy + c = 0 will represent a real circle if
(a) g² + f² – c < 0
(b) g² + f² – c ≥ 0
(c) always
(d) None of these

Answer

Answer: (b) g² + f² – c ≥ 0
Given, equation of the circle is: x² + y² + 2gx + 2fy + c = 0
This equation can be written as
{x – (-g)}² + {y – (-f)}² + = √{g² + f² – c}²
So, the circle is real is g² + f² – c ≥ 0

Question 9.
The equation of parabola whose focus is (3, 0) and directrix is 3x + 4y = 1 is
(a) 16x² – 9y² – 24xy – 144x + 8y + 224 = 0
(b) 16x² + 9y² – 24xy – 144x + 8y – 224 = 0
(c) 16x² + 9y² – 24xy – 144x – 8y + 224 = 0
(d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0

Answer

Answer: (d) 16x² + 9y² – 24xy – 144x + 8y + 224 = 0
Given focus S(3, 0)
and equation of directrix is: 3x + 4y = 1
⇒ 3x + 4y – 1 = 0
Let P (x, y) be any point on the required parabola and let PM be the length of the perpendicular from P on the directrix
Then, SP = PM
⇒ SP² = PM²
⇒ (x – 3)² + (y – 0)² = {(3x + 4y – 1) /{√(3² + 4²)}²
⇒ x² + 9 – 6x + y² = (9x² + 16y² + 1 + 24xy – 8y – 6x)/25
⇒ 25(x² + 9 – 6x + y²) = 9x² + 16y² + 1 + 24xy – 8y – 6x
⇒ 25x² + 225 – 150x + 25y² = 9x² + 16y² + 1 + 24xy – 8y – 6x
⇒ 25x² + 225 – 150x + 25y² – 9x² – 16y² – 1 – 24xy + 8y + 6x = 0
⇒ 16x² + 9y² – 24xy – 144x + 8y + 224 = 0
This is the required equation of parabola.

Question 10.
If the parabola y² = 4ax passes through the point (3, 2), then the length of its latusrectum is
(a) 2/3
(b) 4/3
(c) 1/3
(d) 4

Answer

Answer: (b) 4/3
Since, the parabola y² = 4ax passes through the point (3, 2)
⇒ 2² = 4a × 3
⇒ 4 = 12a
⇒ a = 4/12
⇒ a = 1/3
So, the length of latusrectum = 4a = 4 × (1/3) = 4/3

Question 11.
The eccentricity of an ellipse is?
(a) e = 1
(b) e < 1
(c) e > 1
(d) 0 < e < 1

Answer

Answer: (d) 0 < e < 1
The eccentricity of an ellipse e = √(1 – a²/b²) and 0 < e < 1

Question 12.
If the length of the tangent from the origin to the circle centered at (2, 3) is 2 then the equation of the circle is
(a) (x + 2)² + (y – 3)² = 3²
(b) (x – 2)² + (y + 3)² = 3²
(c) (x – 2)² + (y – 3)² = 3²
(d) (x + 2)² + (y + 3)² = 3²

Answer

Answer: (c) (x – 2)² + (y – 3)² = 3²
Radius of the circle = √{(2 – 0)² + (3 – 0)² – 2²}
= √(4 + 9 – 4)
= √9
= 3
So, the equation of the circle = (x – 2)² + (y – 3)² = 3²

Question 13.
If the length of the major axis of an ellipse is three times the length of the minor axis then its eccentricity is
(a) 1/3
(b) 1/√3
(c) 1/√2
(d) 2√2/√3

Answer

Answer: (d) 2√2/√3
Given, the length of the major axis of an ellipse is three times the length of the minor axis
⇒ 2a = 3(2b)
⇒ 2a = 6b
⇒ a = 3b
⇒ a² = 9b²
⇒ a² = 9a² (1 – e²) {since b² = a²(1 – e²)}
⇒ 1 = 9(1 – e²)
⇒ 1/9 = 1 – e²
⇒ e² = 1 – 1/9
⇒ e² = 8/9
⇒ e = √(8/9)
⇒ e = 2√2/√3
So, the eccentricity of the ellipse is 2√2/√3

Question 14.
The equation of parabola with vertex at origin and directrix x – 2 = 0 is
(a) y² = -4x
(b) y² = 4x
(c) y² = -8x
(d) y² = 8x

Answer

Answer: (c) y² = -8x
Since the line passing through the focus and perpendicular to the directrix is x-axis,
therefore axis of the required parabola is x-axis.
Let the coordinate of the focus is S(a, 0).
Since the vertex is the mid point of the line joining the focus and the point (2, 0) where the directix is x – 2 = 0 meets the axis.
So, 0 = {a – (-2)}/2
⇒ 0 = (a + 2)/2
⇒ a + 2 = 0
⇒ a = -2
Thus the coordinate of focus is (-2, 0)
Let P(x, y) be a point on the parabola.
Then by definition of parabola
(x + 2)² + (y – 0)² = (x – 2)²
⇒ x² + 4 + 4x + y² = x² + 4 – 4x
⇒ 4x + y² = – 4x
⇒ y² = -4x – 4x
⇒ y² = -8x
This is the required equation of the parabola.

Question 15.
In an ellipse, the distance between its foci is 6 and its minor axis is 8 then its eccentricity is
(a) 4/5
(b) 1/√52
(c) 3/5
(d) 1/2

Answer

Answer: (c) 3/5
Given, distance between foci = 6
⇒ 2ae = 6
⇒ ae = 3
Again minor axis = 8
⇒ 2b = 8
⇒ b = 4
⇒ b² = 16
⇒ a² (1 – e²) = 16
⇒ a² – a² e² = 16
⇒ a² – (ae)² = 16
⇒ a² – 3² = 16
⇒ a² – 9 = 16
⇒ a² = 9 + 16
⇒ a² = 25
⇒ a = 5
Now, ae = 3
⇒ 5e = 3
⇒ e = 3/5
So, the eccentricity is 3/5

Question 16.
One of the diameters of the circle x² + y² – 12x + 4y + 6 = 0 is given by
(a) x + y = 0
(b) x + 3y = 0
(c) x = y
(d) 3x + 2y = 0

Answer

Answer: (b) x + 3y = 0
The coordinate of the centre of the circle x² + y² – 12x + 4y + 6 = 0 are (6, -2)
Clearly, the line x + 3y passes through this point.
Hence, x + 3y = 0 is a diameter of the given circle.

Question 17.
The center of the circle 4x² + 4y² – 8x + 12y – 25 = 0 is?
(a) (2, -3)
(b) (-2, 3)
(c) (-4, 6)
(d) (4, -6)

Answer

Answer: (a) (2, -3)
Given, equation of the circle is 4x² + 4y² – 8x + 12y – 25 = 0
⇒ x² + y² – 8x/4 + 12y/4 – 25/4 = 0
⇒ x² + y² – 2x + 3y – 25/4 = 0
Now, center = {-(-2), -3} = (2, -3)

Question 18.
If the parabola y² = 4ax passes through the point (3, 2), then the length of its latusrectum is
(a) 2/3
(b) 4/3
(c) 1/3
(d) 4

Answer

Answer: (b) 4/3
Since, the parabola y² = 4ax passes through the point (3, 2)
⇒ 2² = 4a × 3
⇒ 4 = 12a
⇒ a = 4/12
⇒ a = 1/3
So, the length of latusrectum = 4a = 4 × (1/3) = 4/3

Question 19.
The equation of ellipse whose one focus is at (4, 0) and whose eccentricity is 4/5 is
(a) x²/5 + y²/9 = 1
(b) x2 /25 + y² /9 = 1
(c) x²/9 + y²/5 = 1
(d) x²/9 + y²/25 = 1

Answer

Answer: (b) x² /25 + y² /9 = 1
Given focus is (4, 0)
⇒ ae = 4
and e = 4/5
a × (4/5) = 4
⇒ a = 5
Now, b² = a² (1 – e²)
⇒ b² = 5² {1 – (4/5)²}
⇒ b² = 25{1 – 16/25}
⇒ b² = 25{(25 – 16)/25}
⇒ b² = 9
Hence, the equation of the ellipse is x²/a² + y²/b² = 1
⇒ x²/5² + y²/9 = 1
⇒ x²/25 + y²/9 = 1

Question 20.
The focus of parabola y² = 8x is
(a) (2, 0)
(b) (-2, 0)
(c) (0, 2)
(d) (0, -2)

Answer

Answer: (a) (2, 0)
Given, y² = 8x
General equation is y² = 4ax
Now, 4a = 8
⇒ a = 2
Now, focus = (a, 0) = (2, 0)

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MCQ Questions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry with Answers

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Introduction to Three Dimensional Geometry Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry Objective Questions.

Introduction to Three Dimensional Geometry Class 11 MCQs Questions with Answers

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Explore numerous MCQ Questions of Introduction to Three Dimensional Geometry Class 11 with answers provided with detailed solutions by looking below.

Question 1.
The projections of a directed line segment on the coordinate axes are 12, 4, 3. The DCS of the line are
(a) 12/13, -4/13, 3/13
(b) -12/13, -4/13, 3/13
(c) 12/13, 4/13, 3/13
(d) None of these

Answer

Answer: (c) 12/13, 4/13, 3/13
Let AB be the given line and the DCs of AB be l, m, n. Then
Projection on x-axis = AB . l = 12 (Given)
Projection on y-axis = AB . m = 4 (Given)
Projection on z-axis = AB . n = 3 (Given)
⇒ (AB²) (l² + m² + n²) = 144 + 16 + 9
⇒ (AB²) = 169 {since l² + m² + n² = 1}
⇒ AB = 13
Hence, DCs of AB are 12/13, 4/13, 3/13

Question 2.
The angle between the planes r . n1 = d1 and r . n1 = d2 is
(a) cos θ ={|n1| × |n2|}/ (n1. n2)
(b) cos θ = (n1 . n2)/{|n1| × |n2|}²
(c) cos θ = (n1 . n2)/{|n1| × |n2|}
(d) cos θ = (n1 . n2)² /{|n1| × |n2|}

Answer

Answer: (c) cos θ = (n1 . n2)/{|n1| × |n2|}
The angle between the planes r . n1 = d1 and r . n2 = d2 is defined as
cos θ = (n1 . n2)/{|n1| × |n2|}

Question 3.
For every point P(x, y, z) on the xy-plane
(a) x = 0
(b) y = 0
(c) z = 0
(d) None of these

Answer

Answer: (c) z = 0
The perpendicular distance of P(x, y, z) from xy-plane is zero.

Question 4.
The locus of a point P(x, y, z) which moves in such a way that x = a and y = b, is a
(a) Plane parallel to xy-plane
(b) Line parallel to x-axis
(c) Line parallel to y-axis
(d) Line parallel to z-axis

Answer

Answer: (b) Line parallel to x-axis
Since x = 0 and y = 0 together represent x-axis, therefore x = a and y = b represent a line parallel to x-axis.

Question 5.
The equation of the plane containing the line 2x – 5y + z = 3, x + y + 4z = 5 and parallel to the plane x + 3y + 6z = 1 is
(a) x + 3y + 6z + 7 = 0
(b) x + 3y – 6z – 7 = 0
(c) x – 3y + 6z – 7 = 0
(d) x + 3y + 6z – 7 = 0

Answer

Answer: (d) x + 3y + 6z – 7 = 0
Let the equation of the plane is
(2x – 5y + z – 3) + λ(x + y + 4z – 5) = 0
⇒ (2 + λ)x + (λ – 5)y + (4λ + 1)z – (3 + 5λ) = 0
Since the plane is parallel to x + 3y + 6z – 1 = 0
⇒ (2 + λ)/1 = (λ – 5)/3 = (1 + 4λ)/6
⇒ 6 + 3λ = λ – 5
⇒ 2λ = -11
⇒ λ = -11/2
Again,
6λ – 30 = 3 + 12λ
⇒ -6λ = -33
⇒ λ = -33/6
⇒ λ = -11/2
So, the required equation of plane is
(2x – 5y + z – 3) + (-11/2) × (x + y + 4z – 5) = 0
⇒ 2(2x – 5y + z – 3) + (-11) × (x + y + 4z – 5) = 0
⇒ 4x – 10y + 2z – 6 – 11x – 11y – 44z + 55 = 0
⇒ -7x – 21y – 42z + 49 = 0
⇒ x + 3y + 6z – 7 = 0

Question 6.
The coordinate of foot of perpendicular drawn from the point A(1, 0, 3) to the join of the point B(4, 7, 1) and C(3, 5, 3) are
(a) (5/3, 7/3, 17/3)
(b) (5, 7, 17)
(c) (5/3, -7/3, 17/3)
(d) (5/7, -7/3, -17/3)

Answer

Answer: (a) (5/3, 7/3, 17/3)
Let D be the foot of perpendicular and let it divide BC in the ration m : 1
Then the coordinates of D are {(3m + 4)/(m + 1), (5m + 7)/(m + 1), (3m + 1)/(m + 1)}
Now, AD ⊥ BC
⇒ AD . BC = 0
⇒ -(2m + 3) – 2(5m + 7) – 4 = 0
⇒ m = -7/4
So, the coordinate of D are (5/3, 7/3, 17/3)

Question 7.
The coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ plane is
(a) (0, 17/2, 13/2)
(b) (0, -17/2, -13/2)
(c) (0, 17/2, -13/2)
(d) None of these

Answer

Answer: (c) (0, 17/2, -13/2)
The line passing through the points (5, 1, 6) and (3, 4, 1) is given as
(x – 5)/(3 – 5) = (y – 1)/(4 – 1) = (z – 6)/(1 – 6)
⇒ (x – 5)/(-2) = (y – 1)/3 = (z – 6)/(-5) = k(say)
⇒ (x – 5)/(-2) = k
⇒ x – 5 = -2k
⇒ x = 5 – 2k
(y – 1)/3 = k
⇒ y – 1 = 3k
⇒ y = 3k + 1
and (z – 6)/(-5) = k
⇒ z – 6 = -5k
⇒ z = 6 – 5k
Now, any point on the line is of the form (5 – 2k, 3k + 1, 6 – 5k)
The equation of YZ-plane is x = 0
Since the line passes through YZ-plane
So, 5 – 2k = 0
⇒ k = 5/2
Now, 3k + 1 = 3 × 5/2 + 1 = 15/2 + 1 = 17/2
and 6 – 5k = 6 – 5 × 5/2 = 6 – 25/2 = -13/2
Hence, the required point is (0, 17/2, -13/2)

Question 8.
If P is a point in space such that OP = 12 and OP inclined at angles 45 and 60 degrees with OX and OY respectively, then the position vector of P is
(a) 6i + 6j ± 6√2k
(b) 6i + 6√2j ± 6k
(c) 6√2i + 6j ± 6k
(d) None of these

Answer

Answer: (c) 6√2i + 6j ± 6k
Let l, m, n be the DCs of OP.
Then it is given that l = cos 45 = 1/√2
m = cos 60 = 1/2
Now, l² + m² + n² = 1
⇒ 1/2 + 1/4 + n² = 1
⇒ n² = 1/4
⇒ n = ±1/2
Now, r = |r|(li + mj + nk)
⇒ r = 12(i/√2 + j/2 ± k/√2)
⇒ r = 6√2i + 6j ± 6k

Question 9.
The image of the point P(1,3,4) in the plane 2x – y + z = 0 is
(a) (-3, 5, 2)
(b) (3, 5, 2)
(c) (3, -5, 2)
(d) (3, 5, -2)

Answer

Answer: (a) (-3, 5, 2)
Let image of the point P(1, 3, 4) is Q in the given plane.
The equation of the line through P and normal to the given plane is
(x – 1)/2 = (y – 3)/-1 = (z – 4)/1
Since the line passes through Q, so let the coordinate of Q are (2r + 1, -r + 3, r + 4)
Now, the coordinate of the mid-point of PQ is
(r + 1, -r/2 + 3, r/2 + 4)
Now, this point lies in the given plane.
2(r + 1) – (-r/2 + 3) + (r/2 + 4) + 3 = 0
⇒ 2r + 2 + r/2 – 3 + r/2 + 4 + 3 = 0
⇒ 3r + 6 = 0
⇒ r = -2
Hence, the coordinate of Q is (2r + 1, -r + 3, r + 4) = (-4 + 1, 2 + 3, -2 + 4)
= (-3, 5, 2)

Question 10.
There is one and only one sphere through
(a) 4 points not in the same plane
(b) 4 points not lie in the same straight line
(c) none of these
(d) 3 points not lie in the same line

Answer

Answer: (a) 4 points not in the same plane
Sphere is referred to its center and it follows a quadratic equation with 2 roots. The mid-point of chords of a sphere and parallel to fixed direction lies in the normal diametrical plane.
Now, general equation of the plane depends on 4 constants. So, one sphere passes through 4 points and they need not be in the same plane.

Question 11.
The points on the y- axis which are at a distance of 3 units from the point ( 2, 3, -1) is
(a) either (0, -1, 0) or (0, -7, 0)
(b) either (0, 1, 0) or (0, 7, 0)
(c) either (0, 1, 0) or (0, -7, 0)
(d) either (0, -1, 0) or (0, 7, 0)

Answer

Answer: (d) either (0, -1, 0) or (0, 7, 0)
Let the point on y-axis is O(0, y, 0)
Given point is A(2, 3, -1)
Given OA = 3
⇒ OA² = 9
⇒ (2 – 0)² + (3 – y)² + (-1 – 0)² = 9
⇒ 4 + (3 – y)² + 1 = 9
⇒ 5 + (3 – y)² = 9
⇒ (3 – y)² = 9 – 5
⇒ (3 – y)² = 4
⇒ 3 – y = √4
⇒ 3 – y = ±4
⇒ 3 – y = 4 and 3 – y = -4
⇒ y = -1, 7
So, the point is either (0, -1, 0) or (0, 7, 0)

Question 12.
The coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ plane is
(a) (0, 17/2, 13/2)
(b) (0, -17/2, -13/2)
(c) (0, 17/2, -13/2)
(d) None of these

Answer

Answer: (c) (0, 17/2, -13/2)
The line passing through the points (5,1,6) and (3,4,1) is given as
(x – 5)/(3 – 5) = (y – 1)/(4 – 1) = (z – 6)/(1 – 6)
⇒ (x – 5)/(-2) = (y – 1)/3 = (z – 6)/(-5) = k(say)
⇒ (x – 5)/(-2) = k
⇒ x – 5 = -2k
⇒ x = 5 – 2k
(y – 1)/3 = k
⇒ y – 1 = 3k
⇒ y = 3k + 1
and (z-6)/(-5) = k
⇒ z – 6 = -5k
⇒ z = 6 – 5k
Now, any point on the line is of the form (5 – 2k, 3k + 1, 6 – 5k)
The equation of YZ-plane is x = 0
Since the line passes through YZ-plane
So, 5 – 2k = 0
⇒ k = 5/2
Now, 3k + 1 = 3 × 5/2 + 1 = 15/2 + 1 = 17/2
and 6 – 5k = 6 – 5 × 5/2 = 6 – 25/2 = -13/2
Hence, the required point is (0, 17/2, -13/2)

Question 13.
he equation of plane passing through the point i + j + k and parallel to the plane r . (2i – j + 2k) = 5 is
(a) r . (2i – j + 2k) = 2
(b) r . (2i – j + 2k) = 3
(c) r . (2i – j + 2k) = 4
(d) r . (2i – j + 2k) = 5

Answer

Answer: (b) r . (2i – j + 2k) = 3
The equation of plane parallel to the plane r . (2i – j + 2k) = 5 is
r . (2i – j + 2k) = d
Since it passes through the point i + j + k, therefore
(i + j + k) . (2i – j + 2k) = d
⇒ d = 2 – 1 + 2
⇒ d = 3
So, the required equation of the plane is
r . (2i – j + 2k) = 3

Question 14.
The cartesian equation of the line is 3x + 1 = 6y – 2 = 1 – z then its direction ratio are
(a) 1/3, 1/6, 1
(b) -1/3, 1/6, 1
(c) 1/3, -1/6, 1
(d) 1/3, 1/6, -1

Answer

Answer: (a) 1/3, 1/6, 1
Give 3x + 1 = 6y – 2 = 1 – z
= (3x + 1)/1 = (6y – 2)/1 = (1 – z)/1
= (x + 1/3)/(1/3) = (y – 2/6)/(1/6) = (1 – z)/1
= (x + 1/3)/(1/3) = (y – 1/3)/(1/6) = (1 – z)/1
Now, the direction ratios are: 1/3, 1/6, 1

Question 15.
Under what condition does the equation x² + y² + z² + 2ux + 2vy + 2wz + d represent a real sphere
(a) u² + v² + w² = d²
(b) u² + v² + w² > d
(c) u² + v² + w² < d
(d) u² + v² + w² < d²

Answer

Answer: (b) u² + v² + w² > d
Equation x² + y² + z² + 2ux + 2vy + 2wz + d represent a real sphere if
u² + v² + w² – d > 0
⇒ u² + v² + w² > d

Question 16.
The locus of a first-degree equation in x, y, z is a
(a) sphere
(b) straight line
(c) plane
(d) none of these

Answer

Answer: (c) plane
In an x-y-z cartesian coordinate system, the general form of the equation of a plane is
ax + by + cz + d = 0
It is an equation of the first degree in three variables.

Question 17.
The image of the point P(1,3,4) in the plane 2x – y + z = 0 is
(a) (-3, 5, 2)
(b) (3, 5, 2)
(c) (3, -5, 2)
(d) (3, 5, -2)

Answer

Answer: (a) (-3, 5, 2)
Let image of the point P(1, 3, 4) is Q in the given plane.
The equation of the line through P and normal to the given plane is
(x – 1)/2 = (y – 3)/-1 = (z – 4)/1
Since the line passes through Q, so let the coordinate of Q are (2r + 1, -r + 3, r + 4)
Now, the coordinate of the mid-point of PQ is
(r + 1, -r/2 + 3, r/2 + 4)
Now, this point lies in the given plane.
2(r + 1) – (-r/2 + 3) + (r/2 + 4) + 3 = 0
⇒ 2r + 2 + r/2 – 3 + r/2 + 4 + 3 = 0
⇒ 3r + 6 = 0
⇒ r = -2
Hence, the coordinate of Q is (2r + 1, -r + 3, r + 4) = (-4 + 1, 2 + 3, -2 + 4)
= (-3, 5, 2)

Question 18.
The distance of the point P(a, b, c) from the x-axis is
(a) √(a² + c²)
(b) √(a² + b²)
(c) √(b² + c²)
(d) None of these

Answer

Answer: (c) √(b² + c²)
The coordinate of the foot of the perpendicular from P on x-axis are (a, 0, 0).
So, the required distance = √{(a – a)² + (b – 0)² + (c – 0)²}
= √(b² + c²)

Question 19.
The vector equation of a sphere having centre at origin and radius 5 is
(a) |r| = 5
(b) |r| = 25
(c) |r| = √5
(d) none of these

Answer

Answer: (a) |r| = 5
We know that the vector equation of a sphere having center at the origin and radius R
= |r| = R
Here R = 5
Hence, the equation of the required sphere is |r| = 5

Question 20.
The ratio in which the line joining the points(1, 2, 3) and (-3, 4, -5) is divided by the xy-plane is
(a) 2 : 5
(b) 3 : 5
(c) 5 : 2
(d) 5 : 3

Answer

Answer: (b) 3 : 5
Let the points are P(1, 2, 3) and Q(-3, 4, -5)
Let the line joining the points P(1, 2, 3) and Q(-3, 4, -5) is divided by the xy-plane at point R in the ratio k : 1
Now, the coordinate of R is
{(-3k + 1)/(k + 1), (4k + 2)/(k + 1), (-5k + 3)/(k + 1)}
Since R lies on the xy-plane.
So, z-coordinate is zero
⇒ (-5k + 3)/(k + 1) = 0
⇒ k = 3/5
So, the ratio = 3/5 : 1 = 3 : 5

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MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives with Answers

MCQ Questions /

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Limits and Derivatives Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 13 Limits and Derivatives Objective Questions.

Limits and Derivatives Class 11 MCQs Questions with Answers

Students are advised to solve the Limits and Derivatives Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Limits and Derivatives Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Limits and Derivatives Class 11 with answers provided with detailed solutions by looking below.

Question 1.
The expansion of log(1 – x) is
(a) x – x² /2 + x³ /3 – ……..
(b) x + x² /2 + x³ /3 + ……..
(c) -x + x² /2 – x³ /3 + ……..
(d) -x – x² /2 – x³ /3 – ……..

Answer

Answer: (d) -x – x² /2 – x³ /3 – ……..
log(1 – x) = -x – x² /2 – x³ /3 – ……..

Question 2.
The value of Limx→a (a × sin x – x × sin a)/(ax² – xa²) is
(a) = (a × cos a + sin a)/a²
(b) = (a × cos a – sin a)/a²
(c) = (a × cos a + sin a)/a
(d) = (a × cos a – sin a)/a

Answer

Answer: (b) = (a × cos a – sin a)/a²
Given,
Limx→a (a × sin x – x × sin a)/(ax² – xa²)
When we put x = a in the expression, we get 0/0 form.
Now apply L. Hospital rule, we get
Limx→a (a × cos x – sin a)/(2ax – a²)
= (a × cos a – sin a)/(2a × a – a²)
= (a × cos a – sin a)/(2a² – a²)
= (a × cos a – sin a)/a²
So, Limx→a (a × sin x – x × sin a)/(ax² – xa²) = (a × cos a – sin a)/a²

Question 3.
Limx→-1 [1 + x + x² + ……….+ x10] is
(a) 0
(b) 1
(c) -1
(d) 2

Answer

Answer: (b) 1
Given, Limx→-1 [1 + x + x² + ……….+ x10]
= 1 + (-1) + (-1)² + ……….+ (-1)10
= 1 – 1 + 1 – ……. + 1
= 1

Question 4.
The value of the limit Limx→0 {log(1 + ax)}/x is
(a) 0
(b) 1
(c) a
(d) 1/a

Answer

Answer: (c) a
Given, Limx→0 {log(1 + ax)}/x
= Limx→0 {ax – (ax)² /2 + (ax)³ /3 – (ax)4 /4 + …….}/x
= Limx→0 {ax – a² x² /2 + a³ x³ /3 – a4 x4 /4 + …….}/x
= Limx→0 {a – a² x /2 + a³ x² /3 – a4 x³ /4 + …….}
= a – 0
= a

Question 5.
The value of the limit Limx→0 (cos x)cot² x is
(a) 1
(b) e
(c) e1/2
(d) e-1/2

Answer

Answer: (d) e-1/2
Given, Limx→0 (cos x)cot² x
= Limx→0 (1 + cos x – 1)cot² x
= eLimx→0 (cos x – 1) × cot² x
= eLimx→0 (cos x – 1)/tan² x
= e-1/2

Question 6.
Then value of Limx→1 (1 + log x – x)}/(1 – 2x + x²) is
(a) 0
(b) 1
(c) 1/2
(d) -1/2

Answer

Answer: (d) -1/2
Given, Limx→1 (1 + log x – x)}/(1 – 2x + x²)
= Limx→1 (1/x – 1)}/(-2 + 2x) {Using L. Hospital Rule}
= Limx→1 (1 – x)/{2x(x – 1)}
= Limx→1 (-1/2x)
= -1/2

Question 7.
The value of limy→0 {(x + y) × sec (x + y) – x × sec x}/y is
(a) x × tan x × sec x
(b) x × tan x × sec x + x × sec x
(c) tan x × sec x + sec x
(d) x × tan x × sec x + sec x

Answer

Answer: (d) x × tan x × sec x + sec x
Given, limy→0 {(x + y) × sec (x + y) – x × sec x}/y
= limy→0 {x sec (x + y) + y sec (x + y) – x × sec x}/y
= limy→0 [x{ sec (x + y) – sec x} + y sec (x + y)]/y
= limy→0 x{ sec (x + y) – sec x}/y + limy→0 {y sec (x + y)}/y
= limy→0 x{1/cos (x + y) – 1/cos x}/y + limy→0 {y sec (x + y)}/y
= limy→0 [{cos x – cos (x + y)} × x/{y × cos (x + y) × cos x}] + limy→0 {y sec (x + y)}/y
= limy→0 [{2sin (x + y/2) × sin(y/2)} × 2x/{2y × cos (x + y) × cos x}] + limy→0 {y sec (x + y)}/y
= limy→0 {sin (x + y/2) × limy→0 {sin(y/2)/(2y/2)} × limy→0 { x/{y × cos (x + y) × cos x}] + sec x
= sin x × 1 × x/cos² x + sec x
= x × tan x × sec x + sec x
So, limy→0 {(x + y) × sec (x + y) – x × sec x}/y = x × tan x × sec x + sec x

Question 8.
Limx→0 (ex² – cos x)/x² is equals to
(a) 0
(b) 1
(c) 2/3
(d) 3/2

Answer

Answer: (d) 3/2
Given, Limx→0 (ex² – cos x)/x²
= Limx→0 (ex² – cos x – 1 + 1)/x²
= Limx→0 {(ex² – 1)/x² + (1 – cos x)}/x²
= Limx→0 {(ex² – 1)/x² + Limx→0 (1 – cos x)}/x²
= 1 + 1/2
= (2 + 1)/2
= 3/2

Question 9.
The expansion of ax is
(a) ax = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..
(b) ax = 1 – x/1! × (log a) + x² /2! × (log a)² – x³ /3! × (log a)³ + ………..
(c) ax = 1 + x/1 × (log a) + x² /2 × (log a)² + x³ /3 × (log a)³ + ………..
(d) ax = 1 – x/1 × (log a) + x² /2 × (log a)² – x³ /3 × (log a)³ + ………..

Answer

Answer: (a) ax = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..
ax = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..

Question 10.
The value of the limit Limn→0 (1 + an)b/n is
(a) ea
(b) eb
(c) eab
(d) ea/b

Answer

Answer: (c) eab
Given, Limn→0 (1 + an)b/n
= eLimn→0(an × b/n)
= eLimn→0(ab)
= eab

Question 11.
The value of Limx→0 cos x/(1 + sin x) is
(a) 0
(b) -1
(c) 1
(d) None of these

Answer

Answer: (c) 1
Given, Limx→0 cos x/(1 + sin x)
= cos 0/(1 + sin 0)
= 1/(1 + 0)
= 1/1
= 1

Question 12.
Lim tanx→π/4 tan 2x × tan(π/4 – x) is
(a) 0
(b) 1
(c) 1/2
(d) 3/2

Answer

Answer: (c) 1/2
Given, Lim tanx→π/4 tan 2x × tan(π/4 – x)
= Lim tanh→0 tan 2(π/4 – x) × tan(-h)
= Lim tanh→0 -cot 2h/(-cot h)
= Lim tanh→0 tan h/tan 2h
= (1/2) × Lim tanh→0 (tan h/h)/(2h/tan 2h)
= (1/2) × {Lim tanh→0 (tan h/h)}/{Lim tanh→0 (2h/tan 2h)}
= (1/2) × 1
= 1/2

Question 13.
Limx→2 (x³ – 6x² + 11x – 6)/(x² – 6x + 8) =
(a) 0
(b) 1
(c) 1/2
(d) Limit does not exist

Answer

Answer: (c) 1/2
When x = 2, the expression
(x³ – 6x² + 11x – 6)/(x² – 6x + 8) assumes the form 0/0
Now,
Limx→2 (x³ – 6x² + 11x – 6)/(x² – 6x + 8) = Limx→2 {(x – 1) × (x – 2) × (x – 3)}/{(x – 2) × (x – 4)}
= Limx→2 {(x – 1) × (x – 3)}/(x – 4)
= {(2 – 1) × (2 – 3)}/(2 – 4)
= 1/2

Question 14.
The value of the limit Limx→2 (x – 2)/√(2 – x) is
(a) 0
(b) 1
(c) -1
(d) 2

Answer

Answer: (a) 0
Given, Limx→2 (x – 2)/√(2 – x)
= Limx→2 -(2 – x)/√(2 – x)
= Limx→2 -{√(2 – x) × √(2 – x)}/√(2 – x)
= Limx→2 -√(2 – x)
= -√(2 – 2)
= 0

Question 15.
The derivative of the function f(x) = 3x³ – 2x³ + 5x – 1 at x = -1 is
(a) 0
(b) 1
(c) -18
(d) 18

Answer

Answer: (d) 18
Given, function f(x) = 3x³ – 2x² + 5x – 1
Differentiate w.r.t. x, we get
df(x)/dx = 3 × 3 × x² – 2 × 2 × x + 5
⇒ df(x)/dx = 9x² – 4x + 5
⇒ {df(x)/dx}x =-1 = 9 × (-1)² – 4 × (-1) + 5
⇒ {df(x)/dx}x =-1 = 9 + 4 + 5
⇒ {df(x)/dx}x =-1 = 18

Question 16.
Limx→0 sin²(x/3)/x² is equals to
(a) 1/2
(b) 1/3
(c) 1/4
(d) 1/9

Answer

Answer: (d) 1/9
Given, Limx→0 sin² (x/3)/ x²
= Limx→0 [sin² (x/3)/ (x/3)² × {(x/3)² /x²}]
= Limx→0 [{sin (x/3)/ (x/3)}² × {(x² /9)/x²}]
= 1 × 1/9
= 1/9

Question 17.
The expansion of ax is
(a) ax = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..
(b) ax = 1 – x/1! × (log a) + x² /2! × (log a)² – x³ /3! × (log a)³ + ………..
(c) ax = 1 + x/1 × (log a) + x² /2 × (log a)² + x³ /3 × (log a)³ + ………..
(d) ax = 1 – x/1 × (log a) + x² /2 × (log a)² – x³ /3 × (log a)³ + ………..

Answer

Answer: (a) ax = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..
ax = 1 + x/1! × (log a) + x² /2! × (log a)² + x³ /3! × (log a)³ + ………..

Question 18.
Differentiation of cos √x with respect to x is
(a) sin x /2√x
(b) -sin x /2√x
(c) sin √x /2√x
(d) -sin √x /2√x

Answer

Answer: (d) -sin √x /2√x
Let y = cos √x
Put u = √x
du/dx = 1/2√x
Now, y = cos u
dy/du = -sin u
dy/dx = (dy/du) × (du/dx)
= -sin u × (1/2√x)
= -sin √x /2√x

Question 19.
Differentiation of log(sin x) is
(a) cosec x
(b) cot x
(c) sin x
(d) cos x

Answer

Answer: (b) cot x
Let y = log(sin x)
Again let u = sin x
du/dx = cos x
Now, y = log u
dy/du = 1/u = 1/sin x
Now, dy/dx = (dy/du) × (du/dx)
⇒ dy/dx = (1/sin x) × cos x
⇒ dy/dx = cos x/sin x
⇒ dy/dx = cot x

Question 20.
Limx→∞ {(x + 5)/(x + 1)}x equals
(a) e²
(b) e4
(c) e6
(d) e8

Answer

Answer: (c) e6
Given, Limx→∞ {(x + 5)/(x + 1)}x
= Limx→∞ {1 + 6/(x + 1)}x
= eLimx→∞6x/(x + 1)
= eLimx→∞ 6/(1 + 1/x)
= e6/(1 + 1/∞)
= e6/(1 + 0)
= e6

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MCQ Questions for Class 11 Maths Chapter 14 Mathematical Reasoning with Answers

MCQ Questions /

Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 14 Mathematical Reasoning with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Mathematical Reasoning Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 14 Mathematical Reasoning Objective Questions.

Mathematical Reasoning Class 11 MCQs Questions with Answers

Students are advised to solve the Mathematical Reasoning Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Mathematical Reasoning Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Mathematical Reasoning Class 11 with answers provided with detailed solutions by looking below.

Question 1.
Which of the following statement is a conjunction
(a) Ram and Shyam are friends
(b) Both Ram and Shyam are friends
(c) Both Ram and Shyam are enemies
(d) None of these

Answer

Answer: (d) None of these
All the statements are conjunction. So, option 4 is the correct answer.

Question 2.
Which of the following is true
(a) A prime number is either even or odd
(b) √3 is irrational number.
(c) 24 is a multiple of 2, 4 and 8
(d) Everyone in India speaks Hindi.

Answer

Answer: (d) Everyone in India speaks Hindi.
The statement Everyone in India speaks Hindi is not true.
This is because, there are some states like Tamilnadu, Kerala, etc. where the person does not speak Hindi.

Question 3.
The contrapositive of the statement If a number is not divisible by 3, it is not divisible by 9 is
(a) If a number is divisible by 3, it is not divisible by 9
(b) If a number is not divisible by 3, it is divisible by 9
(c) If a number is divisible by 3, it is divisible by 9
(d) If a number is not divisible by 3, it is not divisible by 9

Answer

Answer: (c) If a number is divisible by 3, it is divisible by 9
Given, statement is: If a number is not divisible by 3, it is not divisible by 9
Now, contrapositive is:
If a number is divisible by 3, it is divisible by 9

Question 4.
The negation of the statement The product of 3 and 4 is 9 is
(a) It is false that the product of 3 and 4 is 9
(b) The product of 3 and 4 is 12
(c) The product of 3 and 4 is not 12
(d) It is false that the product of 3 and 4 is not 9

Answer

Answer: (a) It is false that the product of 3 and 4 is 9
Given, statement is The product of 3 and 4 is 9
The negation of the statement is:
It is false that the product of 3 and 4 is 9

Question 5.
The connective in the statement Earth revolves round the Sun and Moon is a satellite of earth is
(a) or
(b) Earth
(c) Sun
(d) and

Answer

Answer: (d) and
Given, statement is Earth revolves round the Sun and Moon is a satellite of earth.
Here, and is the connective.
It connects two statements Earth revolves round the Sun, Moon is a satellite of earth.

Question 6.
If p is a statement then the negation of p is
(a) p
(b) -p
(c) ~p
(d) p~

Answer

Answer: (c) ~p
If p is a statement then the negation of p is ~p

Question 7.
If (p or q) is false when
(a) p is true and q is false
(b) p is true and q is true
(c) p is false and q is false
(d) p is false and q is true

Answer

Answer: (a) p is true and q is false
(p or q) is false when both p and q are false otherwise it is true.

Question 8.
If (p and q) is true then
(a) p is true and q is false
(b) p is false and q is false
(c) p is false and q is true
(d) p is true and q is true

Answer

Answer: (d) p is true and q is true
(p and q) is true when both p and q are true otherwise it is false.

Question 9.
If (p and q) is false then
(a) p is true and q is false
(b) p is false and q is false
(c) p is false and q is true
(d) all of the above

Answer

Answer: (d) all of the above
(p and q) is true when both p and q are true otherwise it is false.

Question 10.
The compound statement with AND is true if all its component statements are
(a) true
(b) false
(c) either true or false
(d) None of these

Answer

Answer: (a) true
The compound statement with AND is true if all its component statements are true.
Ex: Sheela is an honest and nice girl
If this statement is true when the statements
Sheela is an honest girl, Sheela is a nice girl are true separately.

Question 11.
Which of the following is not a negation of the statement A natural number is greater than zero
(a) A natural number is not greater than zero
(b) It is false that a natural number is greater than zero
(c) It is false that a natural number is not greater than zero
(d) None of these

Answer

Answer: (c) It is false that a natural number is not greater than zero
Given statement is:
A natural number is greater than zero
Negation of the statement:
A natural number is not greater than zero
It is false that a natural number is greater than zero
So, option 3 is not true.

Question 12.
If a recipe calls for 2/3 cups of sugar but I only want to make 1/4 of the recipe, how much sugar do I need?
(a) 2/3
(b) 1/4
(c) 1/6
(d) 1/3

Answer

Answer: (c) 1/6
The sugar I need = (1/4)*(2/3) = 2/12 = 1/6

Question 13.
Which of the following is the conditional p → q
(a) q is sufficient for p
(b) p is necessary for q
(c) p only if q
(d) if q then p

Answer

Answer: (c) p only if q
Given, p → q
Now, conditional of the statement is
p only if q

Question 14.
Which of the following is true
(a) A prime number is either even or odd
(b) √3 is irrational number.
(c) 24 is a multiple of 2, 4 and 8
(d) Everyone in India speaks Hindi.

Answer

Answer: (d) Everyone in India speaks Hindi.
The statement Everyone in India speaks Hindi is not true.
This is because, there are some states like Tamilnadu, Kerala, etc. where the person does not speak Hindi.

Question 15.
Which of the following statement is a conjunction
(a) Ram and Shyam are friends
(b) Both Ram and Shyam are friends
(c) Both Ram and Shyam are enemies
(d) None of these

Answer

Answer: (d) None of these
All the statements are conjunction. So, option 4 is the correct answer.

Question 16.
The converse of the statement If a number n is even, then n² is even is
(a) If a number n is even, then n² is even
(b) If a number n is even, then n² is not even
(c) If a number n² is even, then n is even
(d) If a number n² is not even, then n is even

Answer

Answer: (c) If a number n² is even, then n is even
Given, statement is: If a number n is even, then n² is even
Now, the converse of the number is:
If a number n² is even, then n is even

Question 17.
Which of the following is a statement
(a) 8 is less than 5
(b) Every set is finite set
(c) Mathematics is fun.
(d) 2 is the only even prime number

Answer

Answer: (d) 2 is the only even prime number
The statement 2 is the only even prime number is true.
So, it is a statement.

Question 18.
Which of the following is a statement
(a) I will go tomorrow
(b) She will come today
(c) 3 is a prime number
(d) Tomorrow is Friday

Answer

Answer: (c) 3 is a prime number
The statement 3 is a prime number is true.
So, it is a statement.

Question 19.
if p then q is written as
(a) p ⇒ q
(b) q ⇒ p
(c) p ⇔ q
(d) q ⇔ p

Answer

Answer: (a) p ⇒ q
if p then q is written as
p ⇒ q

Question 20.
Which of the following is not a statement
(a) The product of (-1) and 8 is 8
(b) All complex number are real number
(c) Today is windy day
(d) All of the above

Answer

Answer: (d) All of the above
A sentence is a statement if it is true.
None of the above sentence is true.
So, option 4 is the correct answer.

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MCQ Questions for Class 11 Maths Chapter 15 Statistics with Answers

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Students can access the NCERT MCQ Questions for Class 11 Maths Chapter 15 Statistics with Answers Pdf free download aids in your exam preparation and you can get a good hold of the chapter. Use MCQ Questions for Class 11 Maths with Answers during preparation and score maximum marks in the exam. Students can download the Statistics Class 11 MCQs Questions with Answers from here and test their problem-solving skills. Clear all the fundamentals and prepare thoroughly for the exam taking help from Class 11 Maths Chapter 15 Statistics Objective Questions.

Statistics Class 11 MCQs Questions with Answers

Students are advised to solve the Statistics Multiple Choice Questions of Class 11 Maths to know different concepts. Practicing the MCQ Questions on Statistics Class 11 with answers will boost your confidence thereby helping you score well in the exam.

Explore numerous MCQ Questions of Statistics Class 11 with answers provided with detailed solutions by looking below.

Question 1.
If the varience of the data is 121 then the standard deviation of the data is
(a) 121
(b) 11
(c) 12
(d) 21

Answer

Answer: (b) 11
Given, varience of the data = 121
Now, the standard deviation of the data = √(121)
= 11

Question 2.
The mean deviation from the mean for the following data: 4, 7, 8, 9, 10, 12, 13 and 17 is
(a) 2
(b) 3
(c) 4
(d) 5

Answer

Answer: (b) 3
Mean = (4 + 7 + 8 + 9 + 10 + 12 + 13 + 17)/10 = 80/10 = 8
|xi – mean|= |4 – 10| + |7 – 10| + |8 – 10| + |9 – 10| + |10 – 10| + |12 – 10| + |13 – 10| + |17 – 10|
= 6 + 3 + 2 + 1 + 0 + 2 + 3 + 7 = 24
Now, mean deviation form mean = 24/8 = 3

Question 3.
The mean of 1, 3, 4, 5, 7, 4 is m the numbers 3, 2, 2, 4, 3, 3, p have mean m – 1 and median q. Then, p + q =
(a) 4
(b) 5
(c) 6
(d) 7

Answer

Answer: (d) 7
The mean of 1, 3, 4, 5, 7, 4 is m
⇒ (1 + 3 + 4 + 5 + 7 + 4)/6 = m
⇒ m = 24/6
⇒ m = 4
The numbers 3, 2, 2, 4, 3, 3, p have mean m – 1
⇒ (3 + 2 + 2 + 4 + 3 + 3 + p)/7 = m – 1
⇒ (17 + p)/7 = 4 – 1
⇒ (17 + p)/7 = 3
⇒ 17 + p = 7 × 3
⇒ 17 + p = 21
⇒ p = 21 – 17
⇒ p = 4
The numbers 3, 2, 2, 4, 3, 3, p have median q.
⇒ The numbers 2, 2, 3, 3, 3, 4, 4 have median q
⇒ (7 + 1)/2 th term = q
⇒ 4th term = q
⇒ q = 3
Now p + q = 4 + 3 = 7

Question 4.
If the difference of mode and median of a data is 24, then the difference of median and mean is
(a) 12
(b) 24
(c) 8
(d) 36

Answer

Answer: (a) 12
Given the difference of mode and median of a data is 24
⇒ Mode – Median = 24
⇒ Mode = Median + 24
Now, Mode = 3 × Median – 2 × Mean
⇒ Median + 24 = 3 × Median – 2 × Mean
⇒ 24 = 3 × Median – 2 × Mean – Median
⇒ 24 = 2 × Median – 2 × Mean
⇒ Median – Mean = 24/2
⇒ Median – Mean = 12

Question 5.
The coefficient of variation is computed by
(a) S.D/.Mean × 100
(b) S.D./Mean
(c) Mean./S.D × 100
(d) Mean/S.D.

Answer

Answer: (b) S.D./Mean
The coefficient of variation = S.D./Mean

Question 6.
The geometric mean of series having mean = 25 and harmonic mean = 16 is
(a) 16
(b) 20
(c) 25
(d) 30

Answer

Answer: (b) 20
The relationship between Arithmetic mean (AM), Geometric mean (GM) And Harmonic mean (HM) is
GM² = AM × HM
Given AM = 25
HM = 16
So GM² = 25 × 16
⇒GM = √(25 × 16)
= 5 × 4
= 20
So, Geometric mean = 20

Question 7.
When tested the lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623. The mean of the lives of 5 bulbs is
(a) 1445
(b) 1446
(c) 1447
(d) 1448

Answer

Answer: (b) 1446
Given, lives (in hours) of 5 bulbs were noted as follows: 1357, 1090, 1666, 1494, 1623
Now, mean = (1357 + 1090 + 1666 + 1494 + 1623)/5
= 7230/5
= 1446

Question 8.
Mean of the first n terms of the A.P. a + (a + d) + (a + 2d) + ……… is
(a) a + nd/2
(b) a + (n – 1)d
(c) a + (n − 1)d/2
(d) a + nd

Answer

Answer: (c) a + (n − 1)d/2
Mean of the first n terms of the A.P. {a + (a + d) + (a + 2d) + ……… a + (n-1)d}/n
= (n/2){2a + (n – 1)d}/n
= (1/2){2a + (n – 1)d}
= a + (n – 1)d/2

Question 9.
The mean of a group of 100 observations was found to be 20. Later on, it was found that three observations were incorrect, which was recorded as 21, 21 and 18. Then the mean if the incorrect observations are omitted is
(a) 18
(b) 20
(c) 22
(d) 24

Answer

Answer: (b) 20
Given mean of 100 observations is 20
Now
∑ xi/100 = 20 (1 <= i <= 100)
⇒ ∑ xi = 100 × 20
⇒ ∑ xi = 2000
3 observations 21, 21 and 18 are recorded incorrectly.
So ∑ xi = 2000 – 21 – 21 – 18
⇒ ∑ xi = 2000 – 60
⇒ ∑ xi = 1940
Now new mean is
∑ xi/100 = 1940/97 = 20
So, the new mean is 20

Question 10.
If covariance between two variables is 0, then the correlation coefficient between them is
(a) nothing can be said
(b) 0
(c) positive
(d) negative

Answer

Answer: (b) 0
The relationship between the correlation coefficient and covariance for two variables as shown below:
r(x, y) = COV (x, y)/{sx × sy}
r(x, y) = correlation of the variables x and y
COV (x, y) = covariance of the variables x and y
sx = sample standard deviation of the random variable x
sy = sample standard deviation of the random variable y
Now given COV (x, y) = 0
Then r(x, y) = 0

Question 11.
The mean of 1, 3, 4, 5, 7, 4 is m the numbers 3, 2, 2, 4, 3, 3, p have mean m – 1 and median q. Then, p + q =
(a) 4
(b) 5
(c) 6
(d) 7

Answer

Answer: (d) 7
The mean of 1, 3, 4, 5, 7, 4 is m
⇒ (1 + 3 + 4 + 5 + 7 + 4)/6 = m
⇒ m = 24/6
⇒ m = 4
The numbers 3, 2, 2, 4, 3, 3, p have mean m – 1
⇒ (3 + 2 + 2 + 4 + 3 + 3 + p)/7 = m – 1
⇒ (17 + p)/7 = 4 – 1
⇒ (17 + p)/7 = 3
⇒ 17 + p = 7 × 3
⇒ 17 + p = 21
⇒ p = 21 – 17
⇒ p = 4
The numbers 3, 2, 2, 4, 3, 3, p have median q.
⇒ The numbers 2, 2, 3, 3, 3, 4, 4 have median q
⇒ (7 + 1)/2th term = q
⇒ 4th term = q
⇒ q = 3
Now p + q = 4 + 3 = 7

Question 12.
In a series, the coefficient of variation is 50 and standard deviation is 20 then the arithmetic mean is
(a) 20
(b) 40
(c) 50
(d) 60

Answer

Answer: (b) 40
Given, in a series, the coefficient of variation is 50 and standard deviation is 20
⇒ (standard deviation/AM) × 100 = 50
⇒ 20/AM = 50/100
⇒ 20/AM = 1/2
⇒ AM = 2 × 20
⇒ AM = 40
So, the arithmetic mean is 40

Question 13.
The coefficient of correlation between two variables is independent of
(a) both origin and the scale
(b) scale but not origin
(c) origin but not scale
(d) neither scale nor origin

Answer

Answer: (a) both origin and the scale
The coefficient of correlation between two variables is independent of both origin and the scale.

Question 14.
The geometric mean of series having mean = 25 and harmonic mean = 16 is
(a) 16
(b) 20
(c) 25
(d) 30

Answer

Answer: (b) 20
The relationship between Arithmetic mean (AM), Geometric mean (GM) And Harmonic mean (HM) is
GM² = AM × HM
Given AM = 25
HM = 16
So GM² = 25 × 16
⇒ GM = √(25 × 16)
= 5 × 4
= 20
So, Geometric mean = 20

Question 15.
One of the methods of determining mode is
(a) Mode = 2 Median – 3 Mean
(b) Mode = 2 Median + 3 Mean
(c) Mode = 3 Median – 2 Mean
(d) Mode = 3 Median + 2 Mean

Answer

Answer: (c) Mode = 3 Median – 2 Mean
We can calculate the mode as
Mode = 3 Median – 2 Mean

Question 16.
If the correlation coefficient between two variables is 1, then the two least square lines of regression are
(a) parallel
(b) none of these
(c) coincident
(d) at right angles

Answer

Answer: (c) coincident
If the correlation coefficient between two variables is 1, then the two least square lines of regression are coincident

Question 17.
The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. then the remaining two observations are
(a) 4, 6
(b) 6, 8
(c) 8, 10
(d) 10, 12

Answer

Answer: (b) 6, 8
Given mean and variance of 7 observations are 8 and 16.
Five observations are 2, 4, 10, 12, 14.
Let the other two observations are x and y.
So 7 observations are : 2, 4, 10, 12, 14 ,x ,y
Now
Mean = (2 + 4 + 10 + 12 + 14 + x + y)/7
⇒ 8 = (2 + 4 + 10 + 12 + 14 + x + y)/7
⇒ 8 × 7 = 2 + 4 + 10 + 12 + 14 + x + y
⇒ 56 = 42 + x + y
⇒ x + y = 56 – 42
⇒ x + y = 14 ………….. 1
Again Given varience = 16
⇒ (1/7) × ∑ (xi – mean)² = 16 (7 <= i <= 1)
⇒ ∑ (xi – mean)² = 16 × 7
⇒ ∑ (xi – mean)² = 112
⇒ {(2 – 8)² + (4 – 8)² + (10 – 8)² + (12 – 8)² + (14 – 8)² + (x – 8)² + (y – 8)²} = 112
⇒ {(-6)² +(-4)² + (2)² + (4)² + (6)² + x² + 64 – 16x + y² + 64 – 16y } = 112
⇒ {36 + 16 + 4 + 16 + 36 + x² + y² + 64 + 64 – 16(x + y) } = 112
⇒ {108 + x² + y² + 128 – (16 × 14)} = 112 (since x + y = 14)
⇒ {108 + x² + y² + 128 – 224} = 112
⇒ x² + y² + 236 – 224 = 112
⇒ x² + y² + 12 = 112
⇒ x² + y² = 12 – 12
⇒ x² + y² = 100…………….2
Squaring equation 1, we get
(x + y)² = 196
⇒ x² + y² + 2xy = 196
⇒ 100 + 2xy = 196
⇒ 2xy = 196 – 100
⇒2xy = 96
⇒ xy = 96/2
⇒ xy = 48 …………. 3
Now (x – y)² = x² + y² – 2xy
= 100 – 2 × 48
= 100 – 96
= 4
⇒ x – y = √2
⇒ x – y = 2, -2
case 1: when x – y = 2 and x + y = 14
After solving it, we get x = 8, y = 6
case 2: when x – y = -2 and x + y = 14
After solving it, we get x = 6, y = 8
So, the two numbers are 6 and 8

Question 18.
Range of a data is calculated as
(a) Range = Max Value – Min Value
(b) Range = Max Value + Min Value
(c) Range = (Max Value – Min Value)/2
(d) Range = (Max Value + Min Value)/2

Answer

Answer: (a) Range = Max Value – Min Value
Range of a data is calculated as
Range = Max Value – Min Value

Question 19.
Mean deviation for n observations x1, x2, ……….., xn from their mean x is given by
(a) ∑(xi – x) where (1 ≤ i ≤ n)
(b) {∑|xi – x|}/n where (1 ≤ i ≤ n)
(c) ∑(xi – x)² where (1 ≤ i ≤ n)
(d) {∑(xi – x)²}/n where (1 ≤ i ≤ n)

Answer

Answer: (b) {∑|xi – x|}/n where (1 ≤ i ≤ n)
Mean deviation for n observations x1, x2, ……….., xn from their mean x is calculated as
{∑|xi – x|}/n where (1 ≤ i ≤ n)

Question 20.
If the mean of the following data is 20.6, then the value of p is
x: 10 15 p 25 35
f: 3 10 25 7 5
(a) 30
(b) 20
(c) 25
(d) 10

Answer

Answer: (b) 20
Mean = ∑ fi × xi /∑ fi
⇒ 20.6 = (10 × 3 + 15 × 10 + p × 25 + 25 × 7 + 35 × 5)/(3 + 10 + 25 + 7 + 5)
⇒ 20.6 = (30 + 150 + 25p + 175 + 175)/50
⇒ 20.6 = (530 + 25p)/50
⇒ 530 + 25p = 20.6 × 50
⇒ 530 + 25p = 1030
⇒ 25p = 1030 – 530
⇒ 25p = 500
⇒ p = 500/25
⇒ p = 20
So, the value of p is 20

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