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

Sequences and Series Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 9 Sequences and Series with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Sequences and Series Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 9 MCQ With Answers

Maths Class 11 Chapter 9 MCQs On Sequences and Series

Sequence And Series Class 11 MCQ Question 1.
If a, b, c are in G.P., then the equations ax² + 2bx + c = 0 and dx² + 2ex + f = 0 have a common root if d/a, e/b, f/c are in
(a) AP
(b) GP
(c) HP
(d) none of these

Answer

Answer: (a) AP
Hint:
Given a, b, c are in GP
⇒ b² = ac
⇒ b² – ac = 0
So, ax² + 2bx + c = 0 have equal roots.
Now D = 4b² – 4ac
and the root is -2b/2a = -b/a
So -b/a is the common root.
Now,
dx² + 2ex + f = 0
⇒ d(-b/a)² + 2e×(-b/a) + f = 0
⇒ db2 /a² – 2be/a + f = 0
⇒ d×ac /a² – 2be/a + f = 0
⇒ dc/a – 2be/a + f = 0
⇒ d/a – 2be/ac + f/c = 0
⇒ d/a + f/c = 2be/ac
⇒ d/a + f/c = 2be/b²
⇒ d/a + f/c = 2e/b
⇒ d/a, e/b, f/c are in AP

MCQ On Sequence And Series Class 11 Question 2.
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
Hint:
Given, a, b, c are in AP
⇒ b – a = c – b
⇒ b + b = a + c
⇒ 2b = a + c

Calculate the common ratio calculator of geometric sequence using our geometric sequence calculator tool in split seconds provided with steps.

MCQ Of Sequence And Series Class 11 Question 3:
Three numbers form an increasing GP. If the middle term is doubled, then the new numbers are in Ap. The common ratio of GP is
(a) 2 + √3
(b) 2 – √3
(c) 2 ± √3
(d) None of these

Answer

Answer: (a) 2 + √3
Hint:
Let the three numbers be a/r, a, ar
Since the numbers form an increasing GP, So r > 1
Now, it is given that a/r, 2a, ar are in AP
⇒ 4a = a/r + ar
⇒ r² – 4r + 1 = 0
⇒ r = 2 ± √3
⇒ r = 2 + √3 {Since r > 1}

Class 11 Sequence And Series MCQ Question 4:
The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is
(a) n/(n+1)
(b) 1/(n+1)
(c) 1/n
(d) None of these

Answer

Answer: (a) n/(n+1)
Hint:
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) – ………………. 1/n.(n+1)
⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -……… (1/n – 1/(n+1))
⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)
⇒ S = 1 – 1/(n+1)
⇒ S = (n + 1 – 1)/(n+1)
⇒ S = n/(n+1)

Class 11 Maths Chapter 9 MCQ Question 5:
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
Hint:
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

Sequence And Series Class 11 MCQ Questions Question 6:
The sum of series 1/2! + 1/4! + 1/6! + ….. is
(a) e² – 1 / 2
(b) (e – 1)² /2 e
(c) e² – 1 / 2 e
(d) e² – 2 / e

Answer

Answer: (b) (e – 1)² /2 e
Hint:
We know that,
e^x = 1 + x/1! + x² /2! + x³ /3! + x⁴ /4! + ………..
Now,
e¹ = 1 + 1/1! + 1/2! + 1/3! + 1/4! + ………..
e^-1 = 1 – 1/1! + 1/2! – 1/3! + 1/4! + ………..
e¹ + e^-1 = 2(1 + 1/2! + 1/4! + ………..)
⇒ e + 1/e = 2(1 + 1/2! + 1/4! + ………..)
⇒ (e² + 1)/e = 2(1 + 1/2! + 1/4! + ………..)
⇒ (e² + 1)/2e = 1 + 1/2! + 1/4! + ………..
⇒ (e² + 1)/2e – 1 = 1/2! + 1/4! + ………..
⇒ (e² + 1 – 2e)/2e = 1/2! + 1/4! + ………..
⇒ (e – 1)² /2e = 1/2! + 1/4! + ………..

MCQ Questions On Sequence And Series Class 11 Question 7:
The third term of a geometric progression is 4. The product of the first five terms is
(a) 4³
(b) 4⁵
(c) 4⁴
(d) none of these

Answer

Answer: (b) 4⁵
Hint:
here it is given that T₃ = 4.
⇒ ar² = 4
Now product of first five terms = a.ar.ar².ar³.ar⁴
= a⁵r¹⁰
= (ar²)⁵
= 4⁵

Class 11 Maths Ch 9 MCQ Question 8:
Let Tr be the r th term of an A.P., for r = 1, 2, 3, … If for some positive integers m, n, we have Tm = 1/n and Tn = 1/m, then Tm n equals
(a) 1/m n
(b) 1/m + 1/n
(c) 1
(d) 0

Answer

Answer: (c) 1
Hint:
Let first term is a and the common difference is d of the AP
Now, T_m = 1/n
⇒ a + (m-1)d = 1/n ………… 1
and T_n = 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, T_mn = 1/mn + (mn-1)/mn
⇒ T_mn = 1/mn + 1 – 1/mn
⇒ T_mn = 1

MCQ Of Chapter 9 Maths Class 11 Question 9.
The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is
(a) 2
(b) 4
(c) 6
(d) 8

Answer

Answer: (c) 6
Hint:
Let a and b are two numbers such that
a + b = 13/6
Let A₁, A₂, A₃, ………A_2n be 2n arithmetic means between a and b
Then, A₁ + A₂ + A₃ + ………+ A_2n = 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A₁ + A₂ + A₃ + ………+ A_2n = 2n + 1
⇒ 13n/6 = 2n + 1
⇒ n = 6

Class 11 Maths Chapter 9 MCQ With Answers Question 10.
If the sum of the roots of the quadratic equation ax² + bx + c = 0 is equal to the sum of the squares of their reciprocals, then a/c, b/a, c/b are in
(a) A.P.
(b) G.P.
(c) H.P.
(d) A.G.P.

Answer

Answer: (c) H.P.
Hint:
Given, equation is
ax² + bx + c = 0
Let p and q are the roots of this equation.
Now p+q = -b/a
and pq = c/a
Given that
p + q = 1/p² + 1/q²
⇒ p + q = (p² + q²)/(p² ×q²)
⇒ p + q = {(p + q)² – 2pq}/(pq)²
⇒ -b/a = {(-b/a)² – 2c/a}/(c/a)²
⇒ (-b/a)×(c/a)² = {b²/a² – 2c/a}
⇒ -bc²/a³ = {b² – 2ca}/a²
⇒ -bc²/a = b² – 2ca
Divide by bc on both side, we get
⇒ -c /a = b/c – 2a/b
⇒ 2a/b = b/c + c/a
⇒ b/c, a/b, c/a are in AP
⇒ c/a, a/b, b/c are in AP
⇒ 1/(c/a), 1/(a/b), 1/(b/c) are in HP
⇒ a/c, b/a, c/b are in HP

Ch 9 Maths Class 11 MCQ Question 11.
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
Hint:
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

Sequence And Series MCQ Questions Class 11 Question 12.
The 35th partial sum of the arithmetic sequence with terms a_n = n/2 + 1
(a) 240
(b) 280
(c) 330
(d) 350

Answer

Answer: (d) 350
Hint:
The 35th partial sum of this sequence is the sum of the first thirty-five terms.
The first few terms of the sequence are:
a₁ = 1/2 + 1 = 3/2
a₂ = 2/2 + 1 = 2
a₃ = 3/2 + 1 = 5/2
Here common difference d = 2 – 3/2 = 1/2
Now, a₃₅ = a₁ + (35 – 1)d = 3/2 + 34 ×(1/2) = 17/2
Now, the sum = (35/2) × (3/2 + 37/2)
= (35/2) × (40/2)
= (35/2) × 20
= 35 × 10
= 350

Chapter 9 Maths Class 11 MCQs Question 13.
The sum of two numbers is 13/6 An even number of arithmetic means are being inserted between them and their sum exceeds their number by 1. Then the number of means inserted is
(a) 2
(b) 4
(c) 6
(d) 8

Answer

MCQs On Sequence And Series Class 11 Question 14.
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
Hint:
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

Class 11 Maths Sequence And Series MCQ Question 15.
The sum of AP 2, 5, 8, …..up to 50 terms is
(a) 3557
(b) 3775
(c) 3757
(d) 3575

Answer

Answer: (b) 3775
Hint:
Given, AP is 2, 5, 8, …..up to 50
Now, first term a = 2
common difference d = 5 – 2 = 3
Number of terms = 50
Now, Sum = (n/2)×{2a + (n – 1)d}
= (50/2)×{2×2 + (50 – 1)3}
= 25×{4 + 49×3}
= 25×(4 + 147)
= 25 × 151
= 3775

Sequence And Series MCQ Questions Question 16.
If 2/3, k, 5/8 are in AP then the value of k is
(a) 31/24
(b) 31/48
(c) 24/31
(d) 48/31

Answer

Answer: (b) 31/48
Hint:
Given, 2/3, k, 5/8 are in AP
⇒ 2k = 2/3 + 5/8
⇒ 2k = 31/24
⇒ k = 31/48
So, the value of k is 31/48

Sequence And Series Class 11 MCQ Pdf Question 17.
The sum of n terms of the series (1/1.2) + (1/2.3) + (1/3.4) + …… is
(a) n/(n+1)
(b) 1/(n+1)
(c) 1/n
(d) None of these

Answer

Answer: (a) n/(n+1)
Hint:
Given series is:
S = (1/1·2) + (1/2·3) + (1/3·4) – ……………….1/n.(n+1)
⇒ S = (1 – 1/2) + (1/2 – 1/3) + (1/3 – 1.4) -………(1/n – 1/(n+1))
⇒ S = 1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 – ……….. 1/n – 1/(n+1)
⇒ S = 1 – 1/(n+1)
⇒ S = (n + 1 – 1)/(n+1)
⇒ S = n/(n+1)

Sequence And Series Class 11 MCQs Question 18.
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
Hint:
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

MCQ Of Ch 9 Maths Class 11 Question 19.
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
Hint:
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

Sequences And Series Class 11 MCQ Question 20.
If a is the A.M. of b and c and G₁ and G₂ are two GM between them then the sum of their cubes is
(a) abc
(b) 2abc
(c) 3abc
(d) 4abc

Answer

Answer: (b) 2abc
Hint:
Given, a is the A.M. of b and c
⇒ a = (b + c)
⇒ 2a = b + c ………… 1
Again, given G₁ and G₁ are two GM between b and c,
⇒ b, G₁, G₂, c are in the GP having common ration r, then
⇒ r = (c/b)^1/(2+1) = (c/b)^1/3
Now,
G₁ = br = b×(c/b)^1/3
and G₁ = br = b×(c/b)^2/3
Now,
(G₁)³ + (G₂)3 = b³ ×(c/b) + b³ ×(c/b)²
⇒ (G₁)³ + (G₂)³ = b³ ×(c/b)×( 1 + c/b)
⇒ (G₁)³ + (G₂)³ = b³ ×(c/b)×( b + c)/b
⇒ (G₁)³ + (G₂)³ = b² ×c×( b + c)/b
⇒ (G₁)³ + (G₂)³ = b² ×c×( b + c)/b ………….. 2
From equation 1
2a = b + c
⇒ 2a/b = (b + c)/b
Put value of(b + c)/b in eqaution 2, we get
(G₁)³ + (G₂)³ = b² × c × (2a/b)
⇒ (G₁)³ + (G₂)³ = b × c × 2a
⇒ (G₁)³ + (G₂)³ = 2abc

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Class 11 Maths MCQ:

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

Introduction to Three Dimensional Geometry Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Introduction to Three Dimensional Geometry Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 12 MCQ With Answers

Maths Class 11 Chapter 12 MCQs On Introduction to Three Dimensional Geometry

MCQ On Three Dimensional Geometry Class 11 Question 1.
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
Hint:
Given 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

MCQ On Three Dimensional Geometry Question 2.
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)
Hint:
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)

Introduction To 3d Geometry Class 11 Extra Questions Question 3.
Three planes x + y = 0, y + z = 0, and x + z = 0
(a) none of these
(b) meet in a line
(c) meet in a unique point
(d) meet taken two at a time in parallel lines

Answer

Answer: (c) meet in a unique point
Hint:
Given, three planes are
x + y = 0 …….. 1
y + z = 0 …….. 2
and x + z = 0 ……… 3
add these planes, we get
2(x + y + z) = 0
⇒ x + y + z = 0 ……… 4
From equation 1
0 + z = 0
⇒ z = 0
From equation 2
x + 0 = 0
⇒ x = 0
From equation 3
y + 0 = 0
⇒ y = 0
So, (x, y, z) = (0, 0, 0)
Hence, the three planes meet in a unique point.

Important Questions Of 3d Geometry Class 11 Question 4.
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)
Hint:
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)

MCQ On Introduction To Three Dimensional Geometry Class 11 Question 5.
The locus of a point which moves so that the difference of the squares of its distances from two given points is constant, is a
(a) Straight line
(b) Plane
(c) Sphere
(d) None of these

Answer

Answer: (b) Plane
Hint:
Let the position vectors of the given points A and B be a and b respectively and that of the variable point be r.
Now, given that
PA² – PB² = k (constant)
⇒ |AP|² – |BP|² = k
⇒ |r – a|² – |r – b|² = k
⇒ (|r|² + |a|² – 2r.a) – (|r|² + |b|² – 2r.b) = k
⇒ 2r.(b – a) = k + |b|² – |a|²
⇒ r.(b – a) = (k + |b|² – |a|²)/2
⇒ r.(b – a) = C where C = (k + |b|² – |a|²)/2 = constant
So, it represents the equation of a plane.

3d Geometry Class 11 Questions Question 6.
The equation of the set of point P, the sum of whose distance from A(4, 0, 0) and B(-4, 0, 0) is equal to 10 is
(a) 9x² + 25y² + 25z² + 225 = 0
(b) 9x² + 25y² + 25z² – 225 = 0
(c) 9x² + 25y² – 25z² – 225 = 0
(d) 9x² – 25y² – 25z² – 225 = 0

Answer

Answer: (b) 9x² + 25y² + 25z² – 225 = 0
Hint:
Let the point P is (x, y, z)
Now given that
PA + PB = 10
⇒ √{(x-4)² + y² + z²} + √{(x+4)² + y² + z²} = 10
⇒ √{(x-4)² + y² + z²} = 10 – √{(x+4)² + y² + z²}
Now square both side
[√{(x-4)² + y² + z²}]² = (10)² + [{(x+4)² + y² + z²}]² – 2 ×10×√{(x+4)² + y² + z²}
⇒ {(x-4)² + y² + z²} = 100 + {(x+4)² + y² + z²} – 20×√{(x+4)² + y² + z²}
⇒ x² + 16 – 8x + y² + z² = 100 + x² + 16 + 8x + y² + z² – 20×√{(x+4)² + y² + z²}
⇒ – 8x = 100 + 8x – 20×√{(x+4)² + y² + z²}
⇒ -8x -8x – 100 = – 20×√{(x+4)² + y² + z²}
⇒ -16x -100 = – 20×√{(x+4)² + y² + z²}
⇒ 4x + 25 = 5×√{(x+4)² + y² + z²}
Again square both side,
(4x + 25)² = 25 ×[√{(x+4)² + y² + z²}]²
⇒ 16x² + 625 + 200x = 25×{(x+4)² + y² + z²}
⇒ 16x² + 625 + 200x = 25×(x² + 16 + 8x + y² + z²)
⇒ 16x² + 625 + 200x = 25x² + 400 + 200x + 25y² + 25z²
⇒ 25x² + 400 + 200x + 25y² + 25z² – 16x² – 625 – 200x = 0
⇒ 9x² + 25y² + 25z² – 225 = 0

Important Questions Of Three Dimensional Geometry Class 11 Question 7.
The maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is
(a) 3
(b) 4
(c) 5
(d) Can not be find

Answer

Answer: (c) 5
Hint:
Given two points are (3sin θ, 0, 0) and (4cos θ, 0, 0)
Now distance = √{(4cos θ – 3sin θ)² + (0 – 0)² + (0 – 0)²}
⇒ distance = √{(4cos θ – 3sin θ)²}
⇒ distance = 4cos θ – 3sin θ ……………. 1
Now, maximum value of 4cos θ – 3sin θ = √{(4² + (-3)²}
= √(16 + 9)
= √25
= 5
From equation 1, we get
distance = 5
So, the maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is 5

MCQ On 3d Geometry Class 11 Question 8.
A vector r is equally inclined with the coordinate axes. If the tip of r is in the positive octant and |r| = 6, then r is
(a) 2√3(i – j + k)
(b) 2√3(-i + j + k)
(c) 2√3(i + j – k)
(d) 2√3(i + j + k)

Answer

Answer: (d) 2√3(i + j + k)
Hint:
Let l, m, n are DCs of r.
Given, l = m = n
⇒ l² + m² + n² = 1
⇒ 3l² = 1
⇒ l² = 1/3
⇒ l = m = n = 1/√3
So, r = |r|(li + mj + nk)
⇒ r = 6(i/√3 + j/√3 + k/√3)
⇒ r = 2√3(i + j + k)

Class 11 Maths Chapter 12 Important Questions Question 9.
The plane 2x – (1 + a)y + 3az = 0 passes through the intersection of the planes
2x – y = 0 and y + 3z = 0
2x – y = 0 and y – 3z = 0
2x + 3z = 0 and y = 0
2x – 3z = 0 and y = 0

Answer

Answer: (d) A
Hint:
Given, equation of plane is:
2x – (1 + a)y + 3az = 0
=> (2x – y) + a(-y + 3z) = 0
which is passing through the intersection of the planes
2x – y = 0 and -y + 3z = 0
2x – y = 0 and y – 3z = 0

Introduction To 3d Geometry Class 11 Questions Question 10.
If the end points of a diagonal of a square are (1, -2, 3) and (2, -3, 5) then the length of the side of square is
(a) √3 unit
(b) 2√3 unit
(c) 3√3 unit
(d) 4√3 unit

Answer

Answer: (a) √3 unit
Hint:
Let a is the length of the side of a square.
Given, the diagonal of a square are (1,–2,3) and (2, -3, 5)
Now, length of the diagonal of square = √{(1 – 2)² + (-2 + 3)² + (3 – 5)²}
= √{1 + 1 + 4}
= √6
Again length of the diagonal of square is √2 times the length of side of the square.
⇒ a√2 = √6
⇒ a√2 = √3×√2
⇒ a = √3
So, the length of side of square is √3 unit

MCQ Questions For Class 11 Maths With Answers Pdf Download Question 11.
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)
Hint:
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)

Class 11 Maths MCQ Questions Question 12.
The angle between the vectors with direction ratios are 4, -3, 5 and 3, 4, 5 is
(a) π/2
(b) π/3
(c) π/4
(d) π/6

Answer

Answer: (b) π/3
Hint:
Let a is a vector parallel to the vector having direction ratio is 4, -3, 5
⇒ a = 4i – 3j + 5k
Let b is a vector parallel to the vector having direction ratio is 3 ,4, 5
⇒ b = 3i + 4j + 5k
Let θ be the angle between the given vectors.
Now, cos θ = (a . b)/(|a|×|b|)
⇒ cos θ = (12 – 12 + 25)/{√(16 + 9 + 25)×√(9 + 16 + 25)}
⇒ cos θ = 25/{√(50)×√(50)}
⇒ cos θ = 25/50
⇒ cos θ = 1/2
⇒ cos θ = π/3
⇒ θ = π/3
So, the angle between the vectors with direction ratios are 4, -3, 5 and 3, 4, 5 is π/3

Chapter 12 Class 11 Maths Question 13.
The 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
Hint:
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

Questions On 3d Geometry Question 14.
A vector r is equally inclined with the coordinate axes. If the tip of r is in the positive octant and |r| = 6, then r is
(a) 2√3(i – j + k)
(b) 2√3(-i + j + k)
(c) 2√3(i + j – k)
(d) 2√3(i + j + k)

Answer

Chapter 12 Maths Class 11 Question 15.
The maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is
(a) 3
(b) 4
(c) 5
(d) Can not be find

Answer

Answer: (c) 5
Hint:
Given two points are (3sin θ, 0, 0) and (4cos θ, 0, 0)
Now distance = √{(4cos θ – 3sin θ)² + (0 – 0)² + (0 – 0)²}
⇒ distance = √{(4cos θ – 3sin θ)²}
⇒ distance = 4cos θ – 3sin θ …………….1
Now, maximum value of 4cos θ – 3sin θ = √{(4² + (-3)²}
= √(16 + 9)
= √25
= 5
From equation 1, we get
distance = 5
So, the maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is 5

Introduction To 3d Geometry Class 11 Formulas Question 16.
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

Introduction To Three Dimensional Geometry Ncert Solutions Question 17.
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)
Hint:
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 18.
If α, β, γ are the angles made by a half ray of a line respectively with positive directions of X-axis Y-axis and Z-axis, then sin² α + sin² β + sin² γ =
(a) 1
(b) 0
(c) -1
(d) None of these

Answer

Answer: (d) None of these
Hint:
Let l, m, n be the direction cosines of the given vector.
Then, α, β, γ
l = cos α
m = cos β
n = cos γ
Now, l² + m² + n² = 1
⇒ cos² α + cos² β + cos² γ = 1
⇒ 1 – sin² α + 1 – sin² β + 1 – sin² γ = 1
⇒ 3 – sin² α – sin² β – sin² γ = 1
⇒ 3 – 1 = sin² α + sin² β + sin² γ
⇒ sin² α + sin² β + sin² γ = 2

Question 19.
If P(x, y, z) is a point on the line segment joining Q(2, 2, 4) and R(3, 5, 6) such that the projections of OP on the axes are 13/5, 19/5, 26/5 respectively, then P divides QR in the ration
(a) 1 : 2
(b) 3 : 2
(c) 2 : 3
(d) 1 : 3

Answer

Answer: (b) 3 : 2
Hint:
Since OP has projections 13/5, 19/5 and 26/5 on the coordinate axes, therefore
OP = 13i/5 + 19j/5 + 26/5k
Let P divides the join of Q(2, 2, 4) and R(3, 5, 6) in the ratio m : 1
Then the position vector of P is
{(3m + 2)/(m + 1), (5m + 2)/(m + 1), (6m + 4)/(m + 1)}
So, 13i/5 + 19j/5 + 26/5k = (3m + 2)/(m + 1)+ (5m + 2)/(m + 1)+ (6m + 4)/(m + 1)
⇒ (3m + 2)/(m + 1) = 13/5
⇒ 2m = 3
⇒ m = 3/2
⇒ m : 1 = 3 : 2
Hence, P divides QR in the ration 3 : 2

Question 20.
In a three dimensional space, the equation 3x – 4y = 0 represents
(a) a plane containing Y axis
(b) none of these
(c) a plane containing Z axis
(d) a plane containing X axis

Answer

Answer: (c) a plane containing Z axis
Hint:
Given, equation is 3x – 4y = 0
Here z = 0
So, the given equation 3x – 4y = 0 represents a plane containing Z axis.

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MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction with Answers

Principle of Mathematical Induction Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Principle of Mathematical Induction Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 4 MCQ With Answers

Maths Class 11 Chapter 4 MCQs On Principle of Mathematical Induction

Mathematical Induction MCQ Question 1.
The sum of the series 1³ + 2³ + 3³ + ………..n³ is
(a) {(n + 1)/2}²
(b) {n/2}²
(c) n(n + 1)/2
(d) {n(n + 1)/2}²

Answer

Answer: (d) {n(n + 1)/2}²
Hint:
Given, series is 1³ + 2³ + 3³ + ……….. n³
Sum = {n(n + 1)/2}²

MCQ On Mathematical Induction Question 2.
If n is an odd positive integer, then an+ bnis divisible by :
(a) a² + b²
(b) a + b
(c) a – b
(d) none of these

Answer

Answer: (b) a + b
Hint:
Given number = an + bn
Let n = 1, 3, 5, ……..
an + bn = a + b
an + bn = a³ + b³ = (a + b) × (a² + b² + ab) and so on.
Since, all these numbers are divisible by (a + b) for n = 1, 3, 5,…..
So, the given number is divisible by (a + b)

MCQ Questions On Mathematical Induction Question 3.
1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{n(n + 1)}
(a) n(n + 1)
(b) n/(n + 1)
(c) 2n/(n + 1)
(d) 3n/(n + 1)

Answer

Answer: (b) n/(n + 1)
Hint:
Let the given statement be P(n). Then,
P(n): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{n(n + 1)} = n/(n + 1).
Putting n = 1 in the given statement, we get
LHS = 1/(1 ∙ 2) = and RHS = 1/(1 + 1) = 1/2.
LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)} = k/(k + 1) ..…(i)
Now 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)} + 1/{(k + 1)(k + 2)}
[1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ….. + 1/{k(k + 1)}] + 1/{(k + 1)(k + 2)}
= k/(k + 1)+1/{ (k + 1)(k + 2)}.
{k(k + 2) + 1}/{(k + 1)²/[(k + 1)k + 2)] using …(ii)
= {k(k + 2) + 1}/{(k + 1)(k + 2}
= {(k + 1)² }/{(k + 1)(k + 2)}
= (k + 1)/(k + 2) = (k + 1)/(k + 1 + 1)
⇒ P(k + 1): 1/(1 ∙ 2) + 1/(2 ∙ 3) + 1/(3 ∙ 4) + ……… + 1/{ k(k + 1)} + 1/{(k + 1)(k + 2)}
= (k + 1)/(k + 1 + 1)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1)is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Mathematical Induction MCQs Pdf Question 4.
The sum of the series 1² + 2² + 3² + ………..n² is
(a) n(n + 1)(2n + 1)
(b) n(n + 1)(2n + 1)/2
(c) n(n + 1)(2n + 1)/3
(d) n(n + 1)(2n + 1)/6

Answer

Answer: (d) n(n + 1)(2n + 1)/6
Hint:
Given, series is 1² + 2² + 3² + ………..n²
Sum = n(n + 1)(2n + 1)/6

Class 11 Maths Chapter 4 MCQ With Answers Question 5.
{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} =
(a) 1/(n + 1) for all n ∈ N.
(b) 1/(n + 1) for all n ∈ R
(c) n/(n + 1) for all n ∈ N.
(d) n/(n + 1) for all n ∈ R

Answer

Answer: (a) 1/(n + 1) for all n ∈ N.
Hint:
Let the given statement be P(n). Then,
P(n): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} = 1/(n + 1).
When n = 1, LHS = {1 – (1/2)} = ½ and RHS = 1/(1 + 1) = ½.
Therefore LHS = RHS.
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 1)
Now, [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] ∙ [1 – {1/(k + 2)}]
= [1/(k + 1)] ∙ [{(k + 2 ) – 1}/(k + 2)}]
= [1/(k + 1)] ∙ [(k + 1)/(k + 2)]
= 1/(k + 2)
Therefore p(k + 1): [{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. [1 – {1/(k + 1)}] = 1/(k + 2)
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Mathematical Induction MCQs Question 6.
For any natural number n, 7ⁿ – 2ⁿ is divisible by
(a) 3
(b) 4
(c) 5
(d) 7

Answer

Answer: (c) 5
Hint:
Given, 7ⁿ – 2ⁿ
Let n = 1
7ⁿ – 2ⁿ = 7¹ – 2¹ = 7 – 2 = 5
which is divisible by 5
Let n = 2
7ⁿ – 2ⁿ = 7² – 2² = 49 – 4 = 45
which is divisible by 5
Let n = 3
7ⁿ – 2ⁿ = 7³ – 2³ = 343 – 8 = 335
which is divisible by 5
Hence, for any natural number n, 7ⁿ – 2ⁿ is divisible by 5

Principle Of Mathematical Induction Class 11 MCQs Question 7.
1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + …….. + 1/{n(n + 1)(n + 2)} =
(a) {n(n + 3)}/{4(n + 1)(n + 2)}
(b) (n + 3)/{4(n + 1)(n + 2)}
(c) n/{4(n + 1)(n + 2)}
(d) None of these

Answer

Answer: (a) {n(n + 3)}/{4(n + 1)(n + 2)}
Hint:
Let P (n): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……. + 1/{n(n + 1)(n + 2)} = {n(n + 3)}/{4(n + 1)(n + 2)} .
Putting n = 1 in the given statement, we get
LHS = 1/(1 ∙ 2 ∙ 3) = 1/6 and RHS = {1 × (1 + 3)}/[4 × (1 + 1)(1 + 2)] = ( 1 × 4)/(4 × 2 × 3) = 1/6.
Therefore LHS = RHS.
Thus, the given statement is true for n = 1, i.e., P(1) is true.
Let P(k) be true. Then,
P(k): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……… + 1/{k(k + 1)(k + 2)} = {k(k + 3)}/{4(k + 1)(k + 2)}. ……. (i)
Now, 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………….. + 1/{k(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}
= [1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ………..…. + 1/{ k(k + 1)(k + 2}] + 1/{(k + 1)(k + 2)(k + 3)}
= [{k(k + 3)}/{4(k + 1)(k + 2)} + 1/{(k + 1)(k + 2)(k + 3)}] [using(i)]
= {k(k + 3)² + 4}/{4(k + 1)(k + 2)(k + 3)}
= (k³ + 6k² + 9k + 4)/{4(k + 1)(k + 2)(k + 3)}
= {(k + 1)(k + 1)(k + 4)}/{4 (k + 1)(k + 2)(k + 3)}
= {(k + 1)(k + 4)}/{4(k + 2)(k + 3)
⇒ P(k + 1): 1/(1 ∙ 2 ∙ 3) + 1/(2 ∙ 3 ∙ 4) + ……….….. + 1/{(k + 1)(k + 2)(k + 3)}
= {(k + 1)(k + 2)}/{4(k + 2)(k + 3)}
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

MCQs On Mathematical Induction Question 8.
The nth terms of the series 3 + 7 + 13 + 21 +………. is
(a) 4n – 1
(b) n² + n + 1
(c) none of these
(d) n + 2

Answer

Answer: (b) n² + n + 1
Hint:
Let S = 3 + 7 + 13 + 21 +……….a_n-1 + a_n …………1
and S = 3 + 7 + 13 + 21 +……….a_n-1 + a_n …………2
Subtract equation 1 and 2, we get
S – S = 3 + (7 + 13 + 21 +……….a_n-1 + a_n) – (3 + 7 + 13 + 21 +……….a_n-1 + a_n)
⇒ 0 = 3 + (7 – 3) + (13 – 7) + (21 – 13) + ……….+ (a_n – a_n-1) – a_n
⇒ 0 = 3 + {4 + 6 + 8 + ……(n-1)terms} – a_n
⇒ a_n = 3 + {4 + 6 + 8 + ……(n-1)terms}
⇒ a_n = 3 + (n – 1)/2 × {2 ×4 + (n – 1 – 1)2}
⇒ a_n = 3 + (n – 1)/2 × {8 + (n – 2)2}
⇒ a_n = 3 + (n – 1) × {4 + n – 2}
⇒ a_n = 3 + (n – 1) × (n + 2)
⇒ a_n = 3 + n² + n – 2
⇒ a_n = n² + n + 1
So, the nth term is n² + n + 1

MCQ On Principle Of Mathematical Induction Question 9.
n(n + 1)(n + 5) is a multiple of ____ for all n ∈ N
(a) 2
(b) 3
(c) 5
(d) 7

Answer

Answer: (b) 3
Hint:
Let P(n) : n(n + 1)(n + 5) is a multiple of 3.
For n = 1, the given expression becomes (1 × 2 × 6) = 12, which is a multiple of 3.
So, the given statement is true for n = 1, i.e. P(1) is true.
Let P(k) be true. Then,
P(k) : k(k + 1)(k + 5) is a multiple of 3
⇒ K(k + 1)(k + 5) = 3m for some natural number m, … (i)
Now, (k + 1)(k + 2)(k + 6) = (k + 1)(k + 2)k + 6(k + 1)(k + 2)
= k(k + 1)(k + 2) + 6(k + 1)(k + 2)
= k(k + 1)(k + 5 – 3) + 6(k + 1)(k + 2)
= k(k + 1)(k + 5) – 3k(k + 1) + 6(k + 1)(k + 2)
= k(k + 1)(k + 5) + 3(k + 1)(k +4) [on simplification]
= 3m + 3(k + 1 )(k + 4) [using (i)]
= 3[m + (k + 1)(k + 4)], which is a multiple of 3
⇒ P(k + 1) : (k + 1 )(k + 2)(k + 6) is a multiple of 3
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Mathematical Induction MCQ Questions Question 10.
Find the number of shots arranged in a complete pyramid the base of which is an equilateral triangle, each side containing n shots.
(a) n(n+1)(n+2)/3
(b) n(n+1)(n+2)/6
(c) n(n+2)/6
(d) (n+1)(n+2)/6

Answer

Answer: (b) n(n+1)(n+2)/6
Hint:
Let each side of the base contains n shots,
then the number of shots in the lowest layer = n + (n – 1) + (n – 2) + ………..+ 1
= n(n + 1)/2
= (n² + n)/2
Now, write (n – 1), (n – 2), ….. for n, then we obtain the number of shots in 2nd, 3rd…layers
So, Total shots = ∑(n² + n)/2
= (1/2)×{∑n² + ∑n}
= (1/2)×{n(n+1)(2n+1)/6 + n(n+1)/2}
= n(n+1)(n+2)/6

Principle Of Mathematical Induction MCQs Question 11.
For any natural number n, 7ⁿ – 2ⁿ is divisible by
(a) 3
(b) 4
(c) 5
(d) 7

Answer

Principle Of Mathematical Induction MCQ Question 12.
(n² + n) is ____ for all n ∈ N.
(a) Even
(b) odd
(c) Either even or odd
(d) None of these

Answer

Answer: (a) Even
Hint:
Let P(n): (n² + n) is even.
For n = 1, the given expression becomes (1² + 1) = 2, which is even.
So, the given statement is true for n = 1, i.e., P(1)is true.
Let P(k) be true. Then,
P(k): (k² + k) is even
⇒ (k² + k) = 2m for some natural number m. ….. (i)
Now, (k + 1)² + (k + 1) = k² + 3k + 2
= (k² + k) + 2(k + 1)
= 2m + 2(k + 1) [using (i)]
= 2[m + (k + 1)], which is clearly even.
Therefore, P(k + 1): (k + 1)² + (k + 1) is even
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n)is true for all n ∈ N.

Principle Of Mathematical Induction Class 11 Extra Questions Question 13.
For all n ∈ N, 3×5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by
(a) 19
(b) 17
(c) 23
(d) 25

Answer

Answer: (b) 17
Hint:
Given, 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹
Let n = 1,
3 × 5^2×1+1 + 2^3×1+1 = 3 × 5²⁺¹ + 2³⁺¹ = 3 × 5³ + 24 = 3 × 125 + 16 = 375 + 16 = 391
Which is divisible by 17
Let n = 2,
3 × 5^2×2+1 + 2^3×2+1 = 3 × 5⁴⁺¹ + 2⁶⁺¹ = 3 × 5⁵ + 2⁷ = 3 × 3125 + 128 = 9375 + 128
= 9503
Which is divisible by 17
Hence, For all n ∈ N, 3 × 5²ⁿ⁺¹ + 2³ⁿ⁺¹ is divisible by 17

Maths MCQs For Class 11 With Answers Pdf Question 14.
Find the number of shots arranged in a complete pyramid the base of which is an equilateral triangle, each side containing n shots.
(a) n(n+1)(n+2)/3
(b) n(n+1)(n+2)/6
(c) n(n+2)/6
(d) (n+1)(n+2)/6

Answer

Answer: (b) n(n+1)(n+2)/6
Hint:
Let each side of the base contains n shots,
then the number of shots in the lowest layer = n + (n – 1) + (n – 2) + ………..+ 1
= n(n + 1)/2
= (n² + n)/2
Now, write (n – 1), (n – 2), ….. for n, then we obtain the number of shots in 2nd, 3rd…layers
So, Total shots = ∑(n² + n)/2
= (1/2) × {∑n² + ∑n}
= (1/2) × {n(n+1)(2n+1)/6 + n(n+1)/2}
= n(n+1)(n+2)/6

Question 15.
{1 – (1/2)}{1 – (1/3)}{1 – (1/4)} ……. {1 – 1/(n + 1)} =
(a) 1/(n + 1) for all n ∈ N.
(b) 1/(n + 1) for all n ∈ R
(c) n/(n + 1) for all n ∈ N.
(d) n/(n + 1) for all n ∈ R

Answer

Question 16.
(1 + x)n ≥ ____ for all n ∈ N,where x > -1
(a) 1 + nx
(b) 1 – nx
(c) 1 + nx/2
(d) 1 – nx/2

Answer

Answer: (a) 1 + nx
Hint:
Let P(n): (1 + x) )ⁿ ≥ (1 + nx).
For n = 1, we have LHS = (1 + x))¹ = (1 + x), and
RHS = (1 + 1 ∙ x) = (1 + x).
Therefore LHS ≥ RHS is true.
Thus, P(1) is true.
Let P(k) is true. Then,
P(k): (1 + x)¹ ≥ (1 + kx). …….. (i)
Now,(1 + x)^k+1 = (1 + x)^k (1 + x)
≥ (1 + kx)(1 + x) [using (i)]
=1 + (k + 1)x + kx²
≥ 1 + (k + 1)x + x [Since kx² ≥ 0]
Therefore P(k + 1) : (1 + x)k + 1 ≥ 1 + (k + 1)x
⇒ P(k +1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 17.
10^2n-1 + 1 is divisible by ____ for all N ∈ N
(a) 9
(b) 10
(c) 11
(d) 13

Answer

Answer: (c) 11
Hint:
Let P (n): (10^2n-1 + 1) is divisible by 11.
For n=1, the given expression becomes {10^(2×1-1) + 1} = 11, which is divisible by 11.
So, the given statement is true for n = 1, i.e., P (1) is true.
Let P(k) be true. Then,
P(k): (10^2k-1 + 1) is divisible by 11
⇒ (10^2k-1 + 1) = 11 m for some natural number m.
Now, {10^2(k-1)-1 – 1 + 1} = (10^2k+1 + 1) = {10² ∙ 10^(2k+1)+ 1}
= 100 × {10^2k-1 + 1 } – 99
= (100 × 11 m) – 99
= 11 × (100 m – 9), which is divisible by 11
⇒ P (k + 1) : {10^2(k-1) – 1 + 1} is divisible by 11
⇒ P (k + 1) is true, whenever P(k) is true.
Thus, P (1) is true and P(k + 1) is true , whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for all n ∈ N.

Question 18.
For all n∈N, 7²ⁿ − 48n−1 is divisible by :
(a) 25
(b) 2304
(c) 1234
(d) 26

Answer

Answer: (b) 2304
Hint:
Given number = 7²ⁿ − 48n − 1
Let n = 1, 2 ,3, 4, ……..
7²ⁿ − 48n − 1 = 7² − 48 − 1 = 49 – 48 – 1 = 49 – 49 = 0
7²ⁿ − 48n − 1 = 7⁴ − 48 × 2 − 1 = 2401 – 96 – 1 = 2401 – 97 = 2304
7²ⁿ − 48n − 1 = 7⁶ − 48 × 3 − 1 = 117649 – 144 – 1 = 117649 – 145 = 117504 = 2304 × 51
Since, all these numbers are divisible by 2304 for n = 1, 2, 3,…..
So, the given number is divisible by 2304

Question 19.
The sum of the series 1² + 2² + 3² + ………..n² is
(a) n(n + 1)(2n + 1)
(b) n(n + 1)(2n + 1)/2
(c) n(n + 1)(2n + 1)/3
(d) n(n + 1)(2n + 1)/6

Answer

Answer: (d) n(n + 1)(2n + 1)/6
Hint:
Given, series is 1² + 2² + 3² + ………..n²
Sum = n(n + 1)(2n + 1)/6

Question 20.
{1/(3 ∙ 5)} + {1/(5 ∙ 7)} + {1/(7 ∙ 9)} + ……. + 1/{(2n + 1)(2n + 3)} =
(a) n/(2n + 3)
(b) n/{2(2n + 3)}
(c) n/{3(2n + 3)}
(d) n/{4(2n + 3)}

Answer

Answer: (c) n/{3(2n + 3)}
Hint:
Let the given statement be P(n). Then,
P(n): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + ……. + 1/{(2n + 1)(2n + 3)} = n/{3(2n + 3).
Putting n = 1 in the given statement, we get
and LHS = 1/(3 ∙ 5) = 1/15 and RHS = 1/{3(2 × 1 + 3)} = 1/15.
LHS = RHS
Thus, P(1) is true.
Let P(k) be true. Then,
P(k): {1/(3 ∙ 5) + 1/(5 ∙ 7) + 1/(7 ∙ 9) + …….. + 1/{(2k + 1)(2k + 3)} = k/{3(2k + 3)} ….. (i)
Now, 1/(3 ∙ 5) + 1/(5 ∙ 7) + ..…… + 1/[(2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}2(k + 1) + 3
= {1/(3 ∙ 5) + 1/(5 ∙ 7) + ……. + [1/(2k + 1)(2k + 3)]} + 1/{(2k + 3)(2k + 5)}
= k/[3(2k + 3)] + 1/[2k + 3)(2k + 5)] [using (i)]
= {k(2k + 5) + 3}/{3(2k + 3)(2k + 5)}
= (2k² + 5k + 3)/[3(2k + 3)(2k + 5)]
= {(k + 1)(2k + 3)}/{3(2k + 3)(2k + 5)}
= (k + 1)/{3(2k + 5)}
= (k + 1)/[3{2(k + 1) + 3}]
= P(k + 1) : 1/(3 ∙ 5) + 1/(5 ∙ 7) + …….. + 1/[2k + 1)(2k + 3)] + 1/[{2(k + 1) + 1}{2(k + 1) + 3}]
= (k + 1)/{3{2(k + 1) + 3}]
⇒ P(k + 1) is true, whenever P(k) is true.
Thus, P(1) is true and P(k + 1) is true, whenever P(k) is true.
Hence, by the principle of mathematical induction, P(n) is true for n ∈ N.

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Principle of Mathematical Induction MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Maths MCQ:

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

Conic Sections Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Conic Sections Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 11 MCQ With Answers

Maths Class 11 Chapter 11 MCQs On Conic Sections

MCQ On Conic Sections Class 11 Chapter 11 Question 1.
The locus of the point from which the tangent to the circles x² + y² – 4 = 0 and x² + y² – 8x + 15 = 0 are equal is given by the equation
(a) 8x + 19 = 0
(b) 8x – 19 = 0
(c) 4x – 19 = 0
(d) 4x + 19 = 0

Answer

Answer: (b) 8x – 19 = 0
Hint:
Given equation of circles are x² + y² – 4 = 0 and x² + y² – 8x + 15 = 0
Now, the required line is the radical axis of the two circles are
(x² + y² – 4) – (x² + y² – 8x + 15) = 0
⇒ x² + y² – 4 – x² – y² + 8x – 15 = 0
⇒ 8x – 19 = 0

MCQ On Circle Class 11 Chapter 11 Question 2.
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
Hint:
The perpendicular distance = {3 × 3 – 4 × (-4) + 10}/√(3² + 4²)
= {9 + 16 + 10}/√(9 + 16)
= 35/√25
= 35/5
= 7

Conic Sections Class 11 MCQ Chapter 11 Question 3.
A man running a race course notes that the sum of the distances from the two flag posts from him is always 10 meter and the distance between the flag posts is 8 meter. The equation of posts traced by the man is
(a) x²/9 + y²/5 = 1
(b) x²/9 + y2 /25 = 1
(c) x²/5 + y²/9 = 1
(d) x²/25 + y²/9 = 1

Answer

Answer: (d) x²/25 + y²/9 = 1
Hint:
MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers 1
From the question, it is clear that the path traced by the man is an ellipse having its foci at two posts.
Let the equation of the ellipse be
x²/a² + y²/b² = 1
It is given that the sum of the distances of the man from the two flag posts is 10 m
This means that the sum of focal distances of a point on the ellipse is 10 m
⇒ PS + PS₁ = 10
⇒ 2a = 10
⇒ a = 5
Again, given that the distance between the flag posts is 8 meters
⇒ 2ae = 8
⇒ ae = 4
Now, b² = a² (1 – e²)
⇒ b² = a² – a² e²
⇒ b² = a² – (ae)²
⇒ b² = 5² – 4²
⇒ b² = 25 – 16
⇒ b² = 9
⇒ b = 3
Hence, the equation of the path is x²/5² + y²/3² = 1
⇒ x²/25 + y²/9 = 1

Conic Section MCQ Chapter 11 Question 4.
The center of the ellipse (x + y – 2)² /9 + (x – y)² /16 = 1 is
(a) (0, 0)
(b) (0, 1)
(c) (1, 0)
(d) (1, 1)

Answer

Answer: (d) (1, 1)
Hint:
The center of the given ellipse is the point of intersection of the lines
x + y – 2 = 0 and x – y = 0
After solving, we get
x = 1, y = 1
So, the center of the ellipse is (1, 1)

Question 5.
The parametric coordinate of any point of the parabola y² = 4ax is
(a) (-at², -2at)
(b) (-at², 2at)
(c) (a sin²t, -2a sin t)
(d) (a sin t, -2a sin t)

Answer

Answer: (c) (a sin²t, -2a sin t)
Hint:
The point (a sin²t, -2a sin t) satisfies the equation of the parabola y² = 4ax for all
values of t. So, the parametric coordinate of any point of the parabola y² = 4ax is
(a sin²t, -2a sin t)

Question 6.
The equation of parabola with vertex at origin the axis is along x-axis and passing through the point (2, 3) is
(a) y² = 9x
(b) y² = 9x/2
(c) y² = 2x
(b) y² = 2x/9

Answer

Answer: (b) y² = 9x/2
Hint:
A parabola with its axis along the x-axis and vertex(0, 0) and direction x = -a has the equation:
y² = 4ax ………….. 1
Given, point (2,3) lies on the parabola,
⇒ 3² = 4a × 2
⇒ 9 = 4a × 2
⇒ 9/2 = 4a
From equation 1, we get
y² = (9/2)x
⇒ y² = 9x/2
This is the required equation of the parabola.

Question 7.
At what point of the parabola x² = 9y is the abscissa three times that of ordinate
(a) (1, 1)
(b) (3, 1)
(c) (-3, 1)
(d) (-3, -3)

Answer

Answer: (b) (3, 1)
Hint:
Given, parabola is x² = 9y
Let P(h, k) is the point on the parabola such that abscissa is 3 times the ordinate.
So, h = 3k ……… 1
Since P(h, k) lies on the parabola
So, h² = 9k ……… 2
From equation 1 and 2, we get
(3k)² = 9k
⇒ 9k² = 9k
⇒ 9k² – 9k = 0
⇒ 9k(k – 1) = 0
⇒ k = 0, 1
When k = 0, h = 0
So k = 1
Now, from equation 1,
h = 3 × 1 = 3
So, the point is (3, 1)

Question 8.
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
Hint:
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 9.
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
Hint:
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 10.
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
Hint:
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 11.
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
Hint:
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 12.
The parametric representation (2 + t², 2t + 1) represents
(a) a parabola
(b) a hyperbola
(c) an ellipse
(d) a circle

Answer

Answer: (a) a parabola
Hint:
Let x = 2 + t²
⇒ x – 2 = t² ……….. 1
and y = 2t + 1
⇒ y – 1 = 2t
⇒ (y – 1)/2 = t
From equation 1, we get
x – 2 = {(y – 1)/2}²
⇒ x – 2 = (y – 1)²/4
⇒ (y – 1)² = 4(x – 2)
This represents the equation of a parabola.

Question 13.
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
Hint:
The equation of a hyperbola with foci on the x-axis is defined as
x²/a² – y²/b² = 1

Question 14.
The equation of parabola with vertex (-2, 1) and focus (-2, 4) is
(a) 10y = x² + 4x + 16
(b) 12y = x² + 4x + 16
(c) 12y = x² + 4x
(d) 12y = x² + 4x + 8

Answer

Answer: (b) 12y = x² + 4x + 16
Hint:
Given, parabola having vertex is (-2, 1) and focus is (-2, 4)
As the vertex and focus share the same abscissa i.e. -2,
parabola axis of symmetry as x = -2
⇒ x + 2 = 0
Hence, the equation of a parabola is of the type
(y – k) = a(x – h)² where (h, k) is vertex
Now, focus = (h, k + 1/4a)
Since, vertex is (-2, 1) and parabola passes through vertex
So, focus = (-2, 1 + 1/4a)
Now, 1 + 1/4a = 4
⇒ 1/4a = 4 -1
⇒ 1/4a = 3
⇒ 4a = 1/3
⇒ a = /1(3 × 4)
⇒ a = 1/12
Now, equation of parabola is
(y – 1) = (1/12) × (x + 2)²
⇒ 12(y – 1) = (x + 2)²
⇒ 12y – 12 = x² + 4x + 4
⇒ 12y = x² + 4x + 4 + 12
⇒ 12y = x² + 4x + 16
This is the required equation of parabola.

Question 15.
If a parabolic reflector is 20 cm in diameter and 5 cm deep then the focus of parabolic reflector is
(a) (0 0)
(b) (0, 5)
(c) (5, 0)
(d) (5, 5)

Answer

Answer: (c) (5, 0)
Hint:
MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers 2
given diameter of the parabola is 20 m.
The equation of parabola is y² = 4ax.
Since this parabola passes through the point A(5,10) then
10² = 4a×5
⇒ 20a = 100
⇒ a = 100/20
⇒ a = 5
So focus of parabola is (a, 0) = (5, 0)

Question 16.
The radius of the circle 4x² + 4y² – 8x + 12y – 25 = 0 is?
(a) √57/4
(b) √77/4
(c) √77/2
(d) √87/4

Answer

Answer: (c) √77/2
Hint:
Given, equation fo the 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, radius = √{(-2)² + (3)² – (-25/4)}
= √{4 + 9 + 25/4}
= √{13 + 25/4}
= √{(13×4 + 25)/4}
= √{(52 + 25)/4}
= √{77/4}
= √77/2

Question 17.
If (a, b) is the mid point of a chord passing through the vertex of the parabola y² = 4x, then
(a) a = 2b
(b) 2a = b
(c) a² = 2b
(d) 2a = b²

Answer

Answer: (d) 2a = b²
Hint:
Let P(x, y) be the coordinate of the other end of the chord OP where O(0, 0)
Now, (x + 0)/2 = a
⇒ x = 2a
and (y + 0)/2 = b
⇒ y = 2b
Now, y² = 4x
⇒ (2b)² = 4 × 2a
⇒ 4b² = 8a
⇒ b² = 2a

Question 18.
A rod of length 12 CM moves with its and always touching the co-ordinate Axes. Then the equation of the locus of a point P on the road which is 3 cm from the end in contact with the x-axis is
(a) x²/81 + y²/9 = 1
(b) x²/9 + y²/81 = 1
(c) x²/169 + y²/9 = 1
(d) x²/9 + y²/169 = 1

Answer

Answer: (a) x²/81 + y²/9 = 1
Hint:
Given a rod of length 12 cm moves with its ends always touching the coordinate axes.
Again given a point P on the rod, which is 3 cm from the end in contact with the x-axis.
It is shown in the figure.
MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers 3
Here AP = 3 cm, AB = 12
Now BP = AB – AP
⇒ BP = 12 – 3
⇒ BP = 9 cm
Again from figure,
∠PAO = ∠BPO = θ (since PQ || OA and are corresponding angles)
Now in ΔBPO,
cosθ = QP/BP
⇒ cosθ = x/9 …………. 1
Again in ΔPAr,
sinθ = PR/PA
⇒ sinθ = y/3 …….. 2
Now square equation 1 and 2 and then add them, we get
cos² θ + sin² θ = x²/81 + y²/9
⇒ x²/81 + y²/9 = 1 (since cos² θ + sin² θ = 1 )
So, the equation of the locus of a point P is x²/81 + y²/9 = 1

Question 19.
The line lx + my + n = 0 will touches the parabola y² = 4ax if
(a) ln = am²
(b) ln = am
(c) ln = a² m²
(d) ln = a² m

Answer

Answer: (a) ln = am²
Hint:
Given, lx + my + n = 0
⇒ my = -lx – n
⇒ y = (-l/m)x + (-n/m)
This will touches the parabola y² = 4ax if
(-n/m) = a/(-l/m)
⇒ (-n/m) = (-am/l)
⇒ n/m = am/l
⇒ ln = am²

Question 20.
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)
Hint:
Given, equation fo the 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)

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 11 Conic Sections with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Conic Sections MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Maths MCQ:

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

Mathematical Reasoning Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 14 Mathematical Reasoning with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Mathematical Reasoning Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 14 MCQ With Answers

Maths Class 11 Chapter 14 MCQs On Mathematical Reasoning

Mathematical Reasoning MCQ Question 1.
The connective in the statement 2 + 7 > 9 or 2 + 7 < 9 is
(a) and
(b) or
(c) >
(d) <

Answer

Answer: (b) or
Hint:
Given, statement is 2 + 7 > 9 or 2 + 7 < 9 Here, connective is or. It connects two statement 2 + 7 > 9, 2 + 7 < 9

MCQ On Mathematical Reasoning Class 11 Question 2.
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
Hint:
Gievn 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.

Mathematical Reasoning Class 11 MCQ Question 3.
Which of the following is a statement
(a) x is a real number
(b) Switch of the fan
(c) 6 is a natural number
(d) Let me go

Answer

Answer: (c) 6 is a natural number
Hint:
The statement 6 is a natural number is true.
So, it is a statement.

Mathematical Reasoning Questions With Answers For Class 11 Question 4.
The contra-positive of the statement If a triangle is not equilateral, it is not isosceles is
(a) If a triangle is not equilateral, it is not isosceles
(b) If a triangle is equilateral, it is not isosceles
(c) If a triangle is not equilateral, it is isosceles
(d) If a triangle is equilateral, it is isosceles

Answer

Answer: (d) If a triangle is equilateral, it is isosceles
Hint:
Given, statement is:
If a triangle is not equilateral, it is not isosceles.
Now, contra-positive is:
If a triangle is equilateral, it is isosceles.

MCQ On Mathematical Reasoning Question 5.
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
Hint:
The statement 3 is a prime number is true.
So, it is a statement.

Mathematical Reasoning Questions With Answers Pdf Question 6.
The contra-positive of the statement if p then q is
(a) if ~p then q
(b) if p then ~q
(c) if q then p
(d) if ~q then ~p

Answer

Answer: (d) if ~q then ~p
Hint:
Given statement is if p then q
Now, contra-positive of the statement is:
if ~q then ~p

Mathematical Reasoning MCQs Question 7.
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
Hint:
A sentence is a statement if it is true.
None of the above sentence is true.
So, option 4 is the correct answer.

Mathematical Reasoning Questions Class 11 Question 8.
If (p or q) is true, then
(a) p is true and q is false
(b) p is true and q is true
(c) p is false and q is true
(d) All of the above

Answer

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

Mathematical Reasoning Class 11 Questions Question 9.
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
Hint:
All the statements are conjunction. So, option 4 is the correct answer.

MCQ Questions For Class 11th Maths Question 10.
Which of the following is a compound statement
(a) Sun is a star
(b) I am a very strong boy
(c) There is something wrong in the room
(d) 7 is both odd and prime number.

Answer

Answer: (d) 7 is both odd and prime number.
Hint:
A compound statement is connected with And , or , etc.
So, the statement 7 is both odd and prime number is a compound statement.

Reasoning Questions For Class 11 Question 11.
Which of the following is a statement
(a) x is a real number
(b) Switch of the fan
(c) 6 is a natural number
(d) Let me go

Answer

Answer: (c) 6 is a natural number
Hint:
The statement 6 is a natural number is true.
So, it is a statement.

Mathematical Reasoning Questions Question 12.
Which of the following is not a statement
(a) 8 is less than 6.
(b) Every set is finite set.
(c) The sun is a star.
(d) Mathematics is fun.

Answer

Answer: (d) Mathematics is fun.
Hint:
8 is less than 6 if false. So it is a statement.
Every set is finite set is false. So it is a statement.
The sun is a star is true. So it is a statement.
Mathematics is fun. This sentence is not always true. Hence, it is not a statement.

Maths MCQs For Class 11 With Answers Pdf Question 13.
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.
Hint:
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.

MCQ Questions For Class 11 Maths With Answers Pdf Download Question 14.
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
Hint:
(p and q) is true when both p and q are true otherwise it is false.

Mathematical Reasoning Class 11 Question 15.
The converse of the statement p ⇒ q is
(a) p ⇒ q
(b) q ⇒ p
(c) ~p ⇒ q
(d) ~q ⇒ p

Answer

Answer: (b) q ⇒ p
Hint:
The converse of the statement p ⇒ q is
q ⇒ p

Mathematical Reasoning Questions Pdf Question 16.
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
Hint:
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

MCQ Questions For Class 11 Maths With Answers Pdf Question 17.
Sentence involving variable time such as today, tomorrow, or yesterday are
(a) Statements
(b) Not statements
(c) may or may not be statements
(d) None of these

Answer

Answer: (b) Not statements
Hint:
Sentence involving variable time such as today, tomorrow, or yesterday are not statements. This is because it is not known what time is referred here.

Question 18.
The converse of the statement if a number is divisible by 10, then it is divisible by 5 is
(a) if a number is not divisible by 5, then it is not divisible by 10
(b) if a number is divisible by 5, then it is not divisible by 10
(c) if a number is not divisible by 5, then it is divisible by 10
(d) if a number is divisible by 5, then it is divisible by 10

Answer

Answer: (d) if a number is divisible by 5, then it is divisible by 10
Hint:
Given, statement is if a number is divisible by 10, then it is divisible by 5
Now, converse of the statement is:
if a number is divisible by 5, then it is divisible by 10

Question 19.
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) 6 is a natural number
Hint:
Given, p → q
Now, conditional of the statement is
p only if q

Question 20.
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
Hint:
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.

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 14 Mathematical Reasoning with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Mathematical Reasoning MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Maths MCQ:

MCQ Questions for Class 11 Maths Chapter 2 Relations and Functions with Answers

Relations and Functions Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 2 Relations and Functions with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Relations and Functions Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 2 MCQ With Answers

Maths Class 11 Chapter 2 MCQs On Relations and Functions

Relation And Function Class 11 MCQ Question 1.
If f(x) = (a – x)^1/n, a > 0 and n ∈ N, then the value of f(f(x)) is
(a) 1/x
(b) x
(c) x²
(d) x^1/2

Answer

Answer: (b) x
Hint:
Given, f(x) = (a – x)^1/n
Now, f(f(x)) = [(a – f(x))ⁿ]^1/n
⇒ f(f(x)) = [(a – {(a – xⁿ)^1/n }ⁿ ]^1/n
⇒ f(f(x)) = [a – (a – xⁿ)]^1/n
⇒ f(f(x)) = [a – a + xⁿ)]^1/n
⇒ f(f(x)) = (xⁿ)^1/n
⇒ f(f(x)) = x

MCQ On Relation And Function Class 11 Question 2.
The domain of the definition of the real function f(x) = √(log₁₂ x² ) of the real variable x is
(a) x > 0
(b) |x| ≥ 1
(c) |x| > 4
(d) x ≥ 4

Answer

Answer: (b) |x| ≥ 1
Hint:
We have f(x) = √(log₁₂ x²)
Since, log_a k ≥ 0 if a > 1, k ≥ 1
or 0 < a < 1 and 0 < k ≤ 1
So, the function f(x) exists if
log₁₂ x² ≥ 0
⇒ x² ≥ 1
⇒ |x| ≥ 1

Relations And Functions Class 11 MCQ Question 3.
If f(x) = e^x and g(x) = log_e x then the value of fog(1) is
(a) 0
(b) 1
(c) -1
(d) None of these

Answer

Answer: (b) 1
Hint:
Given, f(x) = e^x
and g(x) = log x
fog(x) = f(g(x))
= f (log x)
= e^{log x}
= x
So, fog(1) = 1

Class 11 Maths Chapter 2 MCQ With Answers Question 4.
Two functions f and g are said to be equal if f
(a) the domain of f = the domain of g
(b) the co-domain of f = the co-domain of g
(c) f(x) = g(x) for all x
(d) all of above

Answer

Answer: (d) all of above
Hint:
Two functions f and g are said to be equal if f
1. the domain of f = the domain of g
2. the co-domain of f = the co-domain of g
3. f(x) = g(x) for all x

Relations And Functions Class 11 MCQ Questions Question 5.
A function f(x) is said to be an odd function if
(a) f(-x) = f(x)
(b) f(-x) = -f(x)
(c) f(-x) = k * f(x) where k is a constant
(d) None of these

Answer

Answer: (b) f(-x) = -f(x)
Hint:
A function f(x) is said to be an odd function if
f(-x) = -f(x) for all x

Class 11 Maths Chapter 2 MCQ Question 6.
If f(x) is an odd differentiable function on R, then df(x)/dx is a/an
(a) Even function
(b) Odd function
(c) Either even or odd function
(d) Neither even nor odd function

Answer

Answer: (a) Even function
Hint:
Given, f(x) is an odd differentiable function on R
⇒ f(-x) = -f(x) for all x ∈ R
differentiate on both side, we get
⇒ -df(-x)/dx = -df(x)/dx for all x ∈ R
⇒ df(-x)/dx = df(x)/dx for all x ∈ R
⇒ df(x)/dx is an even function on R.

MCQ Of Relation And Function Class 11 Question 7.
The function f(x) = sin (‎πx/2) + cos (πx/2) is periodic with period
(a) 4
(b) 6
(c) 12
(d) 24

Answer

Answer: (a) 4
Hint:
Period of sin (‎πx/2) = 2π/(π/2) = 4
Period of cos (πx/2) = 2π/(π/2) = 4
So, period of f(x) = LCM (4, 4) = 4

Relations And Functions MCQ Class 11 Question 8.
If f(x) = log₃ x and A = (3, 27) then f(A) =
(a) (1, 1)
(b) (3, 3)
(c) (1, 3)
(d) (2, 3)

Answer

Answer: (c) (1, 3)
Hint:
Since f(x) = log₃ x is an increasing function
So, f(A) = (log₃ 3, log₃ 27) = (1, 3)

Relation And Function MCQ Class 11 Question 9.
The domain of tan^-1 (2x + 1) is
(a) R
(b) R -{1/2}
(c) R -{-1/2}
(d) None of these

Answer

Answer: (a) R
Hint:
Since tan^-1 x exists if x ∈ (-∞, ∞)
So, tan^-1 (2x + 1) is defined if
-∞ < 2x + 1 < ∞
⇒ -∞ < x < ∞
⇒ x ∈ (-∞, ∞)
⇒ x ∈ R
So, domain of tan^-1 (2x + 1) is R.

Class 11 Maths Ch 2 MCQ Question 10.
the function f(x) = x – [x] has period of
(a) 0
(b) 1
(c) 2
(d) 3

Answer

Answer: (b) 1
Hint:
Let T is a positive real number.
Let f(x) is periodic with period T.
Now, f(x + T) = f(x), for all x ∈ R
⇒ x + T – [x + T] = x – [x], for all x ∈ R
⇒ [x + T] – [x] = T, for all x ∈ R
Thus, there exist T > 0 such that f(x + T) = f(x) for all x ∈ R
Now, the smallest value of T satisfying f(x + T) = f(x) for all x ∈ R is 1
So, f(x) = x – [x] has period 1

MCQ Questions On Relations And Functions Class 11 Question 11.
If f(x) =(3x – 2)/(2x – 3) then the value of f(f(x)) is
(a) x
(b) x²
(c) x³
(d) None of these

Answer

Answer: (a) x
Hint:
Given, f(x) = (3x – 2)/(2x – 3)
Now, f(f(x)) = f{(3x – 2)/(2x – 3)}
= {(3×(3x – 2)/(2x – 3) – 2)}/{(2(3x – 2)/(2x – 3) – 3)}
= {(9x – 6)/(2x – 3) – 2)}/{((6x – 4)/(2x – 3) – 3)}
= [{(9x – 6) – 2(2x – 3)}/(2x – 3)]/[{(6x – 4) – 3(2x – 3)}/(2x – 3)]
= {(9x – 6) – 2(2x – 3)}/{(6x – 4) – 3(2x – 3)}
= (9x – 6 – 4x + 6)/(6x – 4 – 6x + 9)
= 5x/5
= x
So, f(f(x)) = x

Ch 2 Maths Class 11 MCQ Question 12.
Let R be the set of real numbers. If f(x) = x² and g(x) = 2x + 1, then fog(x) is equal to
(a) 2x + 1
(b) 2x² + 1
(c) (2x + 1)²
(d) None of these

Answer

Answer: (b) 2x² + 1
Hint:
Given, f(x) = x² and g(x) = 2x + 1
Now gof(x) = g(f(x)) = f(x²) = 2x² + 1

MCQ Questions For Class 11 Maths Chapter 2 Question 13.
A relation R is defined from the set of integers to the set of real numbers as (x, y) = R if x² + y² = 16 then the domain of R is
(a) (0, 4, 4)
(b) (0, -4, 4)
(c) (0, -4, -4)
(d) None of these

Answer

Answer: (b) (0, -4, 4)
Hint:
Given that:
(x, y) ∈ R ⇔ x² + y² = 16
⇔ y = ±√(16 – x² )
when x = 0 ⇒ y = ±4
(0, 4) ∈ R and (0, -4) ∈ R
when x = ±4 ⇒ y = 0
(4, 0) ∈ R and (-4, 0) ∈ R
Now for other integral values of x, y is not an integer.
Hence R = {(0, 4), (0, -4), (4, 0), (-4, 0)}
So, Domain(R) = {0, -4, 4}

MCQ Questions For Class 11 Maths With Answers Chapter 2 Question 14.
The number of binary operations on the set {a, b} are
(a) 2
(b) 4
(c) 8
(d) 16

Answer

Answer: (d) 16
Hint:
Let S is a finite set containing n elements.
Since binary operation on S is a function from S×S to S, therefore total number of
binary operations on S is the
total number of functions from S×S to S = (nⁿ)²
Given Set = {a, b}
Total number of elements = 2
Total number of binary operations = (2²)² = 2⁴ = 16

Class 11 Relations And Functions MCQ Questions Question 15.
If f is an even function and g is an odd function the fog is a/an
(a) Even function
(b) Odd function
(c) Either even or odd function
(d) Neither even nor odd function

Answer

Answer: (a) Even function
Hint:
Given, f is an even function and g is an odd function.
Now, fog(-x) = f{g(-x)}
= f{-g(x)} {since g is an odd function}
= f{g(x)} for all x {since f is an even function}
So, fog is an even function.

Question 16.
The domain of the function f(x) = 1/(x² – 3x + 2) is
(a) {1, 2}
(b) R
(c) R – {1, 2}
(d) R – {1, -2}

Answer

Answer: (c) R – {1, 2}
Hint:
Given, function is f(x) = 1/(x² – 3x + 2)
Clearly, f(x) is not defined when x² – 3x + 2 = 0
⇒ (x – 1)×(x – 1) = 0
⇒ x = 1, 2
So, f(x) is not defined when x = 1, 2
So, domain of function is R – {1, 2}

Question 17.
The domain of the function f(x) = sin^-1 (tan x) is
(a) -π/4 ≤ x ≤ π/4
(b) nπ – π/4 ≤ x ≤ nπ + π/4
(c) nπ – π/3 ≤ x ≤ nπ + π/3
(d) -π/3 ≤ x ≤ π/3

Answer

Answer: (b) nπ – π/4 ≤ x ≤ nπ + π/4
Hint:
sin^-1 (tan x) is defined for -1 ≤ tan x ≤ 1
= -π/4 ≤ x ≤ π/4
The general solution of the above inequality is
nπ -π/4 ≤ x ≤ nπ + π/4

Question 18.
Let A = {-2, -1, 0} and f(x) = 2x – 3 then the range of f is
(a) {7, -5, -3}
(b) {-7, 5, -3}
(c) {-7, -5, 3}
(d) {-7, -5, -3}

Answer

Answer: (d) {-7, -5, -3}
Hint:
Given, A = {-2, -1, 0}
and f(x) = 2x – 3
Now, f(-2) = 2 × (-2) – 3 = -4 – 3 = -7
f(-1) = 2 × (-1) – 3 = -2 – 3 = -5
f(0) = 2 × 0 – 3 = -3
So, range of f = {-7, -5, -3}

Question 19.
The range of the function ^7-xP_x-3 is
(a) {1, 2, 3, 4, 5}
(b) {3, 4, 5}
(c) None of these
(d) {1, 2, 3}

Answer

Answer: (d) {1, 2, 3}
Hint:
The function f(x) = ^7-xP_x-3 is defined only if x is an integer satisfying the following inequalities:
1. 7 – x ≥ 0
2. x – 3 ≥ 0
3. 7 – x ≥ x – 3
Now, from 1, we get x ≤ 7 ……… 4
from 2, we get x ≥ 3 ……………. 5
and from 2, we get x ≤ 5 ………. 6
From 4, 5 and 6, we get
3 ≤ x ≤ 5
So, the domain is {3, 4, 5}
Now, f(3) = ^7-3P_3-3 = ⁴P₀ = 1
⇒ f(4) = ^7-4P_4-3 = ³P₁ = 3
⇒ f(5) = ^7-5P_5-3 = ²P₂ = 2
So, the range of the function is {1, 2, 3}

Question 20.
The period of the function f(x) = sin⁴ 3x + cos⁴ 3x is
(a) π/2
(b) π/3
(c) π/4
(d) π/6

Answer

Answer: (d) π/6
Hint:
Since g(x) = sin⁴ x + cos⁴ x is periodic with period π/2
So, f(x) = sin⁴ 3x + cos⁴ 3x is periodic with period (π/2)/3 = π/6

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 2 Relations and Functions with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Relations and Functions MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Maths MCQ:

MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers

Linear Inequalities Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Linear Inequalities Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 11 Maths Chapter 6 MCQ With Answers

Maths Class 11 Chapter 6 MCQs On Linear Inequalities

Linear Inequalities Class 11 MCQ Question 1.
Sum of two rational numbers is ______ number.
(a) rational
(b) irrational
(c) Integer
(d) Both 1, 2 and 3

Answer

Answer: (a) rational
Hint:
The sum of two rational numbers is a rational number.
Ex: Let two rational numbers are 1/2 and 1/3
Now, 1/2 + 1/3 = 5/6 which is a rational number.

MCQ On Linear Inequalities Class 11 Question 2.
If x² = -4 then the value of x is
(a) (-2, 2)
(b) (-2, ∞)
(c) (2, ∞)
(d) No solution

Answer

Answer: (d) No solution
Hint:
Given, x² = -4
Since LHS ≥ 0 and RHS < 0
So, No solution is possible.

Class 11 Maths Chapter 6 MCQ With Answers Question 3.
Solve: (x + 1)² + (x² + 3x + 2)² = 0
(a) x = -1, -2
(b) x = -1
(c) x = -2
(d) None of these

Answer

Answer: (b) x = -1
Hint:
Given, (x + 1)² + (x² + 3x + 2)² = 0
This is true when each term is equal to zero simultaneously,
So, (x + 1)² = 0 and (x² + 3x + 2)² = 0
⇒ x + 1 = 0 and x² + 3x + 2 = 0
⇒ x = -1, and x = -1, -2
Now, the common solution is x = -1
So, solution of the equation is x = -1

Linear Inequalities Objective Questions Class 11 Question 4.
If (x + 3)/(x – 2) > 1/2 then x lies in the interval
(a) (-8, ∞)
(b) (8, ∞)
(c) (∞, -8)
(d) (∞, 8)

Answer

Answer: (a) (-8, ∞)
Hint:
Given,
(x + 3)/(x – 2) > 1/2
⇒ 2(x + 3) > x – 2
⇒ 2x + 6 > x – 2
⇒ 2x – x > -2 – 6
⇒ x > -8
⇒ x ∈ (-8, ∞)

Linear Inequalities MCQ Questions Question 5.
The region of the XOY-plane represented by the inequalities x ≥ 6, y ≥ 2, 2x + y ≤ 10 is
(a) unbounded
(b) a polygon
(c) none of these
(d) exterior of a triangle

Answer

Answer: (c) none of these
Hint:
Given inequalities x ≥ 6, y ≥ 2, 2x + y ≤ 10
Now take x = 6, y = 2 and 2x + y = 10
when x = 0, y = 10
when y = 0, x = 5
So, the points are A(6, 2), B(0, 10) and C(5, 0)
MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers 1
So, the region of the XOY-plane represented by the inequalities x ≥ 6, y ≥ 2, 2x + y ≤ 10 is not defined.

MCQ Of Chapter 6 Maths Class 11 Question 6.
The interval in which f(x) = (x – 1) × (x – 2) × (x – 3) is negative is
(a) x > 2
(b) 2 < x and x < 1
(c) 2 < x < 1 and x < 3
(d) 2 < x < 3 and x < 1

Answer

Answer: (d) 2 < x < 3 and x < 1
Hint:
Given, f(x) = (x – 1) × (x – 2) × (x – 3) has all factors with odd powers.
So, put them zero
i.e. x – 1 = 0, x – 2 = 0, x – 3 = 0
⇒ x = 1, 2, 3
Now, f(x) < 0 when 2 < x < 3 and x < 1

MCQ Of Linear Inequalities Class 11 Question 7.
If -2 < 2x – 1 < 2 then the value of x lies in the interval
(a) (1/2, 3/2)
(b) (-1/2, 3/2)
(c) (3/2, 1/2)
(d) (3/2, -1/2)

Answer

Answer: (b) (-1/2, 3/2)
Hint:
Given, -2 < 2x – 1 < 2
⇒ -2 + 1 < 2x < 2 + 1
⇒ -1 < 2x < 3
⇒ -1/2 < x < 3/2
⇒ x ∈(-1/2, 3/2)

Linear Inequalities Class 11 MCQ Questions Question 8.
The solution of the inequality |x – 1| < 2 is
(a) (1, ∞)
(b) (-1, 3)
(c) (1, -3)
(d) (∞, 1)

Answer

Answer: (b) (-1, 3)
Hint:
Given, |x – 1| < 2
⇒ -2 < x – 1 < 2
⇒ -2 + 1 < x < 2 + 1
⇒ -1 < x < 3
⇒ x ∈ (-1, 3)

Linear Inequalities Class 11 MCQ Pdf Question 9.
If | x − 1| > 5, then
(a) x∈(−∞, −4)∪(6, ∞]
(b) x∈[6, ∞)
(c) x∈(6, ∞)
(d) x∈(−∞, −4)∪(6, ∞)

Answer

Answer: (d) x∈(−∞, −4)∪(6, ∞)
Hint:
Given |x−1| >5
Case 1:
(x – 1) > 5
⇒ x > 6
⇒ x ∈ (6,∞)
Case 2:
-(x – 1) > 5
⇒ -x + 1 > 5
⇒ -x > 4
⇒ x < -4
⇒ x ∈ (−∞, −4)
So the range of x is (−∞, −4)∪(6, ∞)

MCQ Questions On Linear Inequalities Class 11 Question 10.
The solution of |2/(x – 4)| > 1 where x ≠ 4 is
(a) (2, 6)
(b) (2, 4) ∪ (4, 6)
(c) (2, 4) ∪ (4, ∞)
(d) (-∞, 4) ∪ (4, 6)

Answer

Answer: (b) (2, 4) ∪ (4, 6)
Hint:
Given, |2/(x – 4)| > 1
⇒ 2/|x – 4| > 1
⇒ 2 > |x – 4|
⇒ |x – 4| < 2
⇒ -2 < x – 4 < 2
⇒ -2 + 4 < x < 2 + 4
⇒ 2 < x < 6
⇒ x ∈ (2, 6) , where x ≠ 4
⇒ x ∈ (2, 4) ∪ (4, 6)

Linear Inequalities MCQs Pdf Question 11.
If (|x| – 1)/(|x| – 2) ‎≥ 0, x ∈ R, x ‎± 2 then the interval of x is
(a) (-∞, -2) ∪ [-1, 1]
(b) [-1, 1] ∪ (2, ∞)
(c) (-∞, -2) ∪ (2, ∞)
(d) (-∞, -2) ∪ [-1, 1] ∪ (2, ∞)

Answer

Answer: (d) (-∞, -2) ∪ [-1, 1] ∪ (2, ∞)
Hint:
Given, (|x| – 1)/(|x| – 2) ‎≥ 0
Let y = |x|
So, (y – 1)/(y – 2) ‎≥ 0
⇒ y ≤ 1 or y > 2
⇒ |x| ≤ 1 or |x| > 2
⇒ (-1 ≤ x ≤ 1) or (x < -2 or x > 2)
⇒ x ∈ [-1, 1] ∪ (-∞, -2) ∪ (2, ∞)
Hence the solution set is:
x ∈ (-∞, -2) ∪ [-1, 1] ∪ (2, ∞)

Class 11 Linear Inequalities MCQ Question 12.
The solution of the -12 < (4 -3x)/(-5) < 2 is
(a) 56/3 < x < 14/3
(b) -56/3 < x < -14/3
(c) 56/3 < x < -14/3
(d) -56/3 < x < 14/3

Answer

Answer: (d) -56/3 < x < 14/3
Hint:
Given inequality is :
-12 < (4 -3x)/(-5) < 2
⇒ -2 < (4-3x)/5 < 12
⇒ -2 × 5 < 4 – 3x < 12 × 5
⇒ -10 < 4 – 3x < 60
⇒ -10 – 4 < -3x < 60-4
⇒ -14 < -3x < 56
⇒ -56 < 3x < 14
⇒ -56/3 < x < 14/3

Inequalities MCQ Questions Question 13.
If x² = -4 then the value of x is
(a) (-2, 2)
(b) (-2, ∞)
(c) (2, ∞)
(d) No solution

Answer

Answer: (d) No solution
Hint:
Given, x² = -4
Since LHS ≥ 0 and RHS < 0
So, No solution is possible.

Question 14.
Solve: |x – 3| < 5
(a) (2, 8)
(b) (-2, 8)
(c) (8, 2)
(d) (8, -2)

Answer

Answer: (b) (-2, 8)
Hint:
Given, |x – 3| < 5
⇒ -5 < (x – 3) < 5
⇒ -5 + 3 < x < 5 + 3
⇒ -2 < x < 8
⇒ x ∈ (-2, 8)

Question 15.
The graph of the inequations x ≥ 0, y ≥ 0, 3x + 4y ≤ 12 is
(a) none of these
(b) interior of a triangle including the points on the sides
(c) in the 2nd quadrant
(d) exterior of a triangle

Answer

Answer: (b) interior of a triangle including the points on the sides
Hint:
Given inequalities x ≥ 0, y ≥ 0, 3x + 4y ≤ 12
Now take x = 0, y = 0 and 3x + 4y = 12
when x = 0, y = 3
when y = 0, x = 4
So, the points are A(0, 0), B(0, 3) and C(4, 0)
MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers 2
So, the graph of the inequations x ≥ 0, y ≥ 0, 3x + 4y ≤ 12 is interior of a triangle including the points on the sides.

Question 16.
If |x| < 5 then the value of x lies in the interval
(a) (-∞, -5)
(b) (∞, 5)
(c) (-5, ∞)
(d) (-5, 5)

Answer

Answer: (d) (-5, 5)
Hint:
Given, |x| < 5
It means that x is the number which is at distance less than 5 from 0
Hence, -5 < x < 5
⇒ x ∈ (-5, 5)

Question 17.
Solve: f(x) = {(x – 1)×(2 – x)}/(x – 3) ≥ 0
(a) (-∞, 1] ∪ (2, ∞)
(b) (-∞, 1] ∪ (2, 3)
(c) (-∞, 1] ∪ (3, ∞)
(d) None of these

Answer

Answer: (b) (-∞, 1] ∪ (2, 3)
Hint:
Given, f(x) = {(x – 1)×(2 – x)}/(x – 3) ≥ 0
or f(x) = -{(x – 1)×(2 – x)}/(x – 3)
which gives x – 3 ≠ 0
⇒ x ≠ 3
MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers 3
Using number line rule as shown in the figure,
which gives f(x) ≥ 0 when x ≤ 1 or 2 ≤ x < 3
i.e. x ∈ (-∞, 1] ∪ (2, 3)

Question 18.
If x² = 4 then the value of x is
(a) -2
(b) 2
(c) -2, 2
(d) None of these

Answer

Answer: (c) -2, 2
Hint:
Given, x² = 4
⇒ x² – 4 = 0
⇒ (x – 2)×(x + 2) = 0
⇒ x = -2, 2

Question 19.
The solution of the 15 < 3(x – 2)/5 < 0 is
(a) 27 < x < 2
(b) 27 < x < -2
(c) -27 < x < 2
(d) -27 < x < -2

Answer

Answer: (a) 27 < x < 2
Hint:
Given inequality is:
15 < 3(x-2)/5 < 0
⇒ 15 × 5 < 3(x-2) < 0 × 5
⇒ 75 < 3(x-2) < 0
⇒ 75/3 < x-2 < 0
⇒ 25 < x-2 < 0
⇒ 25 +2 < x <0+2
⇒ 27 < x < 2

Question 20.
Solve: 1 ≤ |x – 1| ≤ 3
(a) [-2, 0]
(b) [2, 4]
(c) [-2, 0] ∪ [2, 4]
(d) None of these

Answer

Answer: (c) [-2, 0] ∪ [2, 4]
Hint:
Given, 1 ≤ |x – 1| ≤ 3
⇒ -3 ≤ (x – 1) ≤ -1 or 1 ≤ (x – 1) ≤ 3
i.e. the distance covered is between 1 unit to 3 units
⇒ -2 ≤ x ≤ 0 or 2 ≤ x ≤ 4
Hence, the solution set of the given inequality is
x ∈ [-2, 0] ∪ [2, 4]

We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Linear Inequalities MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Maths MCQ:

MCQ Questions for Class 11 Business Studies Chapter 10 Internal Trade with Answers

Internal Trade Class 11 MCQ Online Test With Answers Questions

October 24, 2024

Check the below Online Education NCERT MCQ Questions for Class 11 Business Studies Chapter 10 Internal Trade with Answers Pdf free download. MCQ Questions for Class 11 Business Studies with Answers were prepared based on the latest exam pattern. We have provided Internal Trade Class 11 Business Studies MCQs Questions with Answers to help students understand the concept very well.

Class 11 Business Studies Chapter 10 Internal Trade MCQ With Answers

Business Studies Class 11 Chapter 10 MCQs On Internal Trade

Internal Trade Class 11 MCQ Question 1.
It helps disabled and elderly people.
(a) Tele-shopping
(b) E-commerce
(c) multiple shop
(d) installment system

Answer

Answer: (a) Tele-shopping

MCQ On Internal Trade Class 11 Question 2.
The aim is to economies by buying in common and to retain their profits by selling in common.
(a) multiple shop
(b) web marketing
(c) teleshopping
(d) consumer?s co-operative store

Answer

Answer: (d) consumer?s co-operative store

MCQ Questions For Class 11 Business Studies Chapter 10 Question 3.
This retail business acts as a universal supplier of a wide variety of products.
(a) Departmental store
(b) Multiple shop
(c) Mail order Business.
(d) Tele-shopping

Answer

Answer: (a) Departmental store

MCQ Of Internal Trade Class 11 Question 4.
____ are mobile traders who deal in low priced articles with no fixed place of business.
(a) Street stalls
(b) Retailers
(c) Itinerant traders
(d) Agents

Answer

Answer: (c) Itinerant traders

Question 5.
____________ can check the price fluctuations in the market by holding back the goods when prices fall and releasing the goods when prices raise.
(a) Wholesaler
(b) Agent
(c) Retailer
(d) Mercantile agent

Answer

Answer: (a) Wholesaler

Question 6.
A warehouse keeper accepts goods for the purpose of ____________
(a) Selling
(b) Packaging
(c) Export
(d) Storage

Answer

Answer: (d) Storage

Question 7.
When goods are imported for the purpose of export is called as _________
(a) Foreign trade
(b) Entrepot
(c) Home trade
(d) Trade

Answer

Answer: (b) Entrepot

Question 8.
The purchase of goods from a foreign country is called ____________
(a) Entreport
(b) Import
(c) Re-Export
(d) Export

Answer

Answer: (b) Import

Question 9.
The risk of bad debts in this business is eliminated particularly when payment is received through V.P.P.
(a) Tele-shopping
(b) mail order trading house
(c) Departmental store
(d) co-operative store

Answer

Answer: (b) mail order trading house

Question 10.
It is a network of a number of branches situated at different localities in the city or in the different parts of the country.
(a) Multiple shop
(b) Consumers? co-operative store
(c) Hire purchase system
(d) Internet marketing

Answer

Answer: (a) Multiple shop

Question 11.
Small scale Fixed retailers includes ____________
(a) Hawkers
(b) Pedlars
(c) Cheap Jacks
(d) General stores

Answer

Answer: (d) General stores

Question 12.
____________ are agents who merely bring the buyer and the seller into contact.
(a) Selling agent
(b) Commission agent
(c) Stockist
(d) Broker

Answer

Answer: (d) Broker

Question 13.
Wholesaler?s deals in ____________ quantity of goods
(a) Small
(b) Limited
(c) Large
(d) Medium

Answer

Answer: (c) Large

Question 14.
An agent is appointed by the ____________
(a) Principal
(b) Manufacturer
(c) Wholesaler
(d) Retailer

Answer

Answer: (a) Principal

Question 15.
The persons who come in between the primary producer and the final consumer to promote trade is called as ____________
(a) Trader
(b) Middleman
(c) Auctioneer
(d) Agent

Answer

Answer: (b) Middleman

One Mark Questions

1. What is internal trade?

Answer

Answer: Buying and selling of goods and services within the boundaries of a nation is referred as internal trade.

2. What is wholesale trade?

Answer

Answer: Wholesale trade refers to buying and selling of goods and services in large quantities for the purpose of resale or intermediate use.

3. Who is a wholesaler?

Answer

Answer: Wholesaler is a person who buys goods from the producers in bulk quantities and sells them in small quantities to a retailer.

4. What is retail trade?

Answer

Answer: Retail trade refers to sale of goods in small quantities for the final consumptions.

5. Who is a retailer?

Answer

Answer: Retailer is a person who buys the goods in large quantities from the wholesalers and sells ‘them in small quantities to the ultimate consumers.

6. Expand AVM.

Answer

Answer: Automatic Vending Machine

7. Expand FOB

Answer

Answer: Free on Board

8. Expand CIF

Answer

Answer: Cost, Insurance and Freight Price

9. Expand E&OE

Answer

Answer: Errors and Omissions excepted

10. Expand COD

Answer

Answer: Cash on Delivery

11. Name any one type of internal trade.

Answer

Answer: Wholesale Trade

12. State one type of Itinerant retailers.

Answer

Answer: Street Traders

13. Give an example for small scale fixed retail shop.

Answer

Answer: Soaps, Hair oil and Tooth Paste

14. Give an example for large scale fixed retail shop.

Answer

Answer: Spencer, Food world and reliance fresh

15. Give an example for departmental stores.

Answer

Answer: Spencer

16. Give an example for multiple shops.

Answer

Answer: Pizza Hut and Me. Donald

17. Give an example for Super markets.

Answer

Answer: Reliance Fresh and Food world

18. Give an example for malls.

Answer

Answer: Garuda Mall and Orion Mall

19. Who are peddlers?

Answer

Answer: Peddlers are those who carry their goods on their heads and backs from one door to another in the streets for selling the goods.

20. Who are Hawkers?

Answer

Answer: Hawkers are those who carry their goods on back of animals or by using wheeled carts from one door to another for selling the articles.

21. Who are cheap jacks?

Answer

Answer: Cheap jacks are petty retailers who have independent shops of a temporary nature in the business locality.

We hope the given NCERT MCQ Questions for Class 11 Business Studies Chapter 10 Internal Trade with Answers Pdf free download will help you. If you have any queries regarding Internal Trade CBSE Class 11 Business Studies MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

Class 11 Business Studies MCQ:

MCQ Questions for Class 12 Maths Chapter 10 Vector Algebra with Answers

Vector Algebra Class 12 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 12 Maths Chapter 10 Vector Algebra with Answers Pdf free download. MCQ Questions for Class 12 Maths with Answers were prepared based on the latest exam pattern. We have provided Vector Algebra Class 12 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 12 Maths Chapter 10 MCQ With Answers

Maths Class 12 Chapter 10 MCQs On Vector Algebra

Vector MCQ Chapter 10 Class 12 Question 1.
The position vector of the point (1, 0, 2) is
(a) \(\vec{i}\) +\(\vec{j}\) + 2\(\vec{k}\)
(b) \(\vec{i}\) + 2\(\vec{j}\)
(c) \(\vec{2}\) + 3\(\vec{k}\)
(d) \(\vec{i}\) + 2\(\vec{K}\)

Answer

Answer: (d) \(\vec{i}\) + 2\(\vec{K}\)

Vector MCQ Questions Chapter 10 Class 12 Question 2.
The modulus of 7\(\vec{i}\) – 2\(\vec{J}\) + \(\vec{K}\)
(a) \(\sqrt{10}\)
(b) \(\sqrt{55}\)
(c) 3\(\sqrt{6}\)
(d) 6

Answer

Answer: (c) 3\(\sqrt{6}\)

MCQ On Vectors Chapter 10 Class 12 Question 3.
If O be the origin and \(\vec{OP}\) = 2\(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\) and \(\vec{OQ}\) = 5\(\hat{i}\) + 4\(\hat{j}\) -3\(\hat{k}\), then \(\vec{PQ}\) is equal to
(a) 7\(\hat{i}\) + 7\(\hat{j}\) – 7\(\hat{k}\)
(b) -3\(\hat{i}\) + \(\hat{j}\) – \(\hat{k}\)
(c) -7\(\hat{i}\) – 7\(\hat{j}\) + 7\(\hat{k}\)
(d) 3\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)

Answer

Answer: (d) 3\(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)

MCQs On Vectors Chapter 10 Class 12 Question 4.
The scalar product of 5\(\hat{i}\) + \(\hat{j}\) – 3\(\hat{k}\) and 3\(\hat{i}\) – 4\(\hat{j}\) + 7\(\hat{k}\) is
(a) 10
(b) -10
(c) 15
(d) -15

Answer

Answer: (b) -10

MCQ On Vector Chapter 10 Class 12 Question 5.
If \(\vec{a}\).\(\vec{b}\) = 0, then
(a) a ⊥ b
(b) \(\vec{a}\) || \(\vec{b}\)
(c) \(\vec{a}\) + \(\vec{b}\) = 0
(d) \(\vec{a}\) – \(\vec{b}\) = 0

Answer

Answer: (a) a ⊥ b

Vectors MCQs Chapter 10 Class 12 Question 6.
\(\vec{i}\) – \(\vec{j}\) =
(a) 0
(b) 1
(c) \(\vec{k}\)
(d) –\(\vec{k}\)

Answer

Answer: (a) 0

MCQ On Vectors Class 12 Chapter 10 Question 7.
\(\vec{k}\) × \(\vec{j}\) =
(a) 0
(b) 1
(c) \(\vec{i}\)
(d) –\(\vec{i}\)

Answer

Answer: (d) –\(\vec{i}\)

Vectors MCQ Chapter 10 Class 12 Question 8.
\(\vec{a}\). \(\vec{a}\) =
(a) 0
(b) 1
(c) |\(\vec{a}\)|²
(d) |\(\vec{a}\)|

Answer

Answer: (c) |\(\vec{a}\)|²

Vector MCQ Questions Class 12 Chapter 10 Question 9.
The projection of the vector 2\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\) on the vector \(\hat{i}\) – 2\(\hat{j}\) + \(\hat{k}\) is
(a) \(\frac{4}{√6}\)
(b) \(\frac{5}{√6}\)
(c) \(\frac{4}{√3}\)
(d) \(\frac{7}{√6}\)

Answer

Answer: (b) \(\frac{5}{√6}\)

Vectors Are MCQ Chapter 10 Class 12 Question 10.
If \(\vec{a}\) = \(\vec{i}\) – \(\vec{j}\) + 2\(\vec{k}\) and b = 3\(\vec{i}\) + 2\(\vec{j}\) – \(\vec{k}\) then the value of (\(\vec{a}\) + 3\(\vec{b}\))(2\(\vec{a}\) – \(\vec{b}\))=.
(a) 15
(b) -15
(c) 18
(d) -18

Answer

Answer: (b) -15

MCQ On Vector Algebra Chapter 10 Class 12 Question 11.
If |\(\vec{a}\)|= \(\sqrt{26}\), |b| = 7 and |\(\vec{a}\) × \(\vec{b}\)| = 35, then \(\vec{a}\).\(\vec{b}\) =
(a) 8
(b) 7
(c) 9
(d) 12

Answer

Answer: (b) 7

Vector Algebra MCQ Chapter 10 Class 12 Question 12.
If \(\vec{a}\) = 2\(\vec{i}\) – 3\(\vec{j}\) + 4\(\vec{k}\) and \(\vec{b}\) = \(\vec{i}\) + 2\(\vec{j}\) + \(\vec{k}\) then \(\vec{a}\) + \(\vec{b}\) =
(a) \(\vec{i}\) + \(\vec{j}\) + 3\(\vec{k}\)
(b) 3\(\vec{i}\) – \(\vec{j}\) + 5\(\vec{k}\)
(c) \(\vec{i}\) – \(\vec{j}\) – 3\(\vec{k}\)
(d) 2\(\vec{i}\) + \(\vec{j}\) + \(\vec{k}\)

Answer

Answer: (b) 3\(\vec{i}\) – \(\vec{j}\) + 5\(\vec{k}\)

Vector MCQs Chapter 10 Class 12 Question 13.
If \(\vec{a}\) = \(\vec{i}\) + 2\(\vec{j}\) + 3\(\vec{k}\) and \(\vec{b}\) = 3\(\vec{i}\) + 2\(\vec{j}\) + \(\vec{k}\), then cos θ =
(a) \(\frac{6}{7}\)
(b) \(\frac{5}{7}\)
(c) \(\frac{4}{7}\)
(d) \(\frac{1}{2}\)

Answer

Answer: (b) \(\frac{5}{7}\)

Vectors MCQs With Solutions Chapter 10 Class 12 Question 14.
If |\(\vec{a}\) + \(\vec{b}\)| = |\(\vec{a}\) – \(\vec{b}\)|, then
(a) \(\vec{a}\) || \(\vec{a}\)
(b) \(\vec{a}\) ⊥ \(\vec{b}\)
(c) |\(\vec{a}\)| = |\(\vec{b}\)|
(d) None of these

Answer

Answer: (b) \(\vec{a}\) ⊥ \(\vec{b}\)

MCQ Questions On Vectors Chapter 10 Class 12 Question 15.
The projection of the vector 2\(\hat{i}\) + 3\(\hat{j}\) – 6\(\hat{k}\) on the line joining the points (3, 4, 2) and (5, 6,3) is
(a) \(\frac{2}{3}\)
(b) \(\frac{4}{3}\)
(c) –\(\frac{4}{3}\)
(d) \(\frac{5}{3}\)

Answer

Answer: (b) \(\frac{4}{3}\)

Question 16.
If |\(\vec{a}\) × \(\vec{b}\)| – |\(\vec{a}\).\(\vec{b}\)|, then the angle between \(\vec{a}\) and \(\vec{b}\), is
(a) 0
(b) \(\frac{π}{2}\)
(c) \(\frac{π}{4}\)
(d) π

Answer

Answer: (c) \(\frac{π}{4}\)

Question 17.
The angle between two vector \(\vec{a}\) and \(\vec{b}\) with magnitude √3 and 4, respectively and \(\vec{a}\).\(\vec{b}\) = 2√3 is
(a) \(\frac{π}{6}\)
(b) \(\frac{π}{3}\)
(c) \(\frac{π}{2}\)
(d) \(\frac{5π}{2}\)

Answer

Answer: (b) \(\frac{π}{3}\)

Question 18.
Unit vector perpendicular to each of the vector 3\(\hat{i}\) + \(\hat{j}\) + 2\(\hat{k}\) and 2\(\hat{i}\) – 2\(\hat{j}\) + 4\(\hat{k}\) is
(a) \(\frac{\hat{i}+\hat{j}+\hat{k}}{√3}\)
(b) \(\frac{\hat{i}-\hat{j}+\hat{k}}{√3}\)
(c) \(\frac{\hat{i}-\hat{j}-\hat{k}}{√3}\)
(d) \(\frac{\hat{i}+\hat{j}-\hat{k}}{√3}\)

Answer

Answer: (c) \(\frac{\hat{i}-\hat{j}-\hat{k}}{√3}\)

Question 19.
If \(\vec{a}\) = 2\(\vec{i}\) – 5\(\vec{j}\) + k and \(\vec{b}\) = 4\(\vec{i}\) + 2\(\vec{j}\) + \(\vec{k}\) then \(\vec{a}\).\(\vec{b}\) =
(a) 0
(b) -1
(c) 1
(d) 2

Answer

Answer: (b) -1

Question 20.
If 2\(\vec{i}\) + \(\vec{j}\) + \(\vec{k}\), 6\(\vec{i}\) – \(\vec{j}\) + 2\(\vec{k}\) and 14\(\vec{i}\) – 5\(\vec{j}\) + 4\(\vec{k}\) be the position vector of the points A, B and C respectively, then
(a) The A, B and C are collinear
(b) A, B and C are not colinear
(c) \(\vec{AB}\) ⊥ \(\vec{BC}\)
(d) None of these

Answer

Answer: (a) The A, B and C are collinear

Question 21.
According to the associative lass of addition of addition of s ector
(\(\vec{a}\) + …….) + \(\vec{c}\) = …… + (\(\vec{b}\) + \(\vec{c}\))
(a) \(\vec{b}\), \(\vec{a}\)
(b) \(\vec{a}\), \(\vec{b}\)
(c) \(\vec{a}\), 0
(d) \(\vec{b}\), 0

Answer

Answer: (a) \(\vec{b}\), \(\vec{a}\)

Question 22.
Which one of the following can be written for (\(\vec{a}\) – \(\vec{b}\)) × (\(\vec{a}\) + \(\vec{b}\))
(a) \(\vec{a}\) × \(\vec{b}\)
(b) 2\(\vec{a}\) × \(\vec{b}\)
(c) \(\vec{a}\)² – \(\vec{b}\)
(d) 2\(\vec{b}\) × \(\vec{b}\)

Answer

Answer: (b) 2\(\vec{a}\) × \(\vec{b}\)

Question 23.
The points with position vectors (2. 6), (1, 2) and (a, 10) are collinear if the of a is
(a) -8
(b) 4
(c) 3
(d) 12

Answer

Answer: (c) 3

Question 24.
|\(\vec{a}\) + \(\vec{b}\)| = |\(\vec{a}\) – \(\vec{b}\)| then the angle between \(\vec{a}\) and \(\vec{b}\)
(a) \(\frac{π}{2}\)
(b) 0
(c) \(\frac{π}{4}\)
(d) \(\frac{π}{6}\)

Answer

Answer: (a) \(\frac{π}{2}\)

Question 25.
|\(\vec{a}\) × \(\vec{b}\)| = |\(\vec{a}\).\(\vec{b}\)| then the angle between \(\vec{a}\) and \(\vec{b}\)
(a) 0
(b) \(\frac{π}{2}\)
(c) \(\frac{π}{4}\)
(d) π

Answer

Answer: (a) 0

Question 26.
If ABCDEF is a regular hexagon then \(\vec{AB}\) + \(\vec{EB}\) + \(\vec{FC}\) equals
(a) zero
(b) 2\(\vec{AB}\)
(c) 4\(\vec{AB}\)
(d) 3\(\vec{AB}\)

Answer

Answer: (d) 3\(\vec{AB}\)

Question 27.
Which one of the following is the modulus of x\(\hat{i}\) + y\(\hat{j}\) + z\(\hat{k}\)?
(a) \(\sqrt{x^2+y^2+z^2}\)
(b) \(\frac{1}{\sqrt{x^2+y^2+z^2}}\)
(c) x² + y² + z²
(d) none of these

Answer

Answer: (a) \(\sqrt{x^2+y^2+z^2}\)

Question 28.
If C is the mid point of AB and P is any point outside AB then,
(a) \(\vec{PA}\) + \(\vec{PB}\) = 2\(\vec{PC}\)
(b) \(\vec{PA}\) + \(\vec{PB}\) = \(\vec{PC}\)
(c) \(\vec{PA}\) + \(\vec{PB}\) = 2\(\vec{PC}\) = 0
(d) None of these

Answer

Answer: (a) \(\vec{PA}\) + \(\vec{PB}\) = 2\(\vec{PC}\)

Question 29.
If \(\vec{OA}\) = 2\(\vec{i}\) – \(\vec{j}\) + \(\vec{k}\), \(\vec{OB}\) = \(\vec{i}\) – 3\(\vec{j}\) – 5\(\vec{k}\) then |\(\vec{OA}\) × \(\vec{OB}\)| =
(a) 8\(\vec{i}\) + 11\(\vec{j}\) – 5\(\vec{k}\)
(b) \(\sqrt{210}\)
(c) sin θ
(d) \(\sqrt{40}\)

Answer

Answer: (b) \(\sqrt{210}\)

Question 30.
If |a| = |b| = |\(\vec{a}\) + \(\vec{b}\)| = 1 then |\(\vec{a}\) – \(\vec{b}\)| is equal to
(a) 1
(b) √3
(c) 0
(d) None of these

Answer

Answer: (b) √3

Question 31.
If \(\vec{a}\) and \(\vec{b}\) are any two vector then (\(\vec{a}\) × \(\vec{b}\))² is equal to
(a) (\(\vec{a}\))²(\(\vec{b}\))² – (\(\vec{a}\).\(\vec{b}\))²
(b) (\(\vec{a}\))² (\(\vec{b}\))² + (\(\vec{a}\).\(\vec{b}\))²
(c) (\(\vec{a}\).\(\vec{b}\))²
(d) (\(\vec{a}\))²(\(\vec{b}\))²

Answer

Answer: (a) (\(\vec{a}\))²(\(\vec{b}\))² – (\(\vec{a}\).\(\vec{b}\))²

Question 32.
If \(\hat{a}\) and \(\hat{b}\) be two unit vectors and 0 is the angle between them, then |\(\hat{a}\) – \(\hat{b}\)| is equal to
(a) sin \(\frac{θ}{2}\)
(b) 2 sin \(\frac{θ}{2}\)
(c) cos \(\frac{θ}{2}\)
(d) 2 cos \(\frac{θ}{2}\)

Answer

Answer: (b) 2 sin \(\frac{θ}{2}\)

Question 33.
The angle between the vector 2\(\hat{i}\) + 3\(\hat{j}\) + \(\hat{k}\) and 2\(\hat{i}\) – \(\hat{j}\) – \(\hat{k}\) is
(a) \(\frac{π}{2}\)
(b) \(\frac{π}{4}\)
(c) \(\frac{π}{3}\)
(d) 0

Answer

Answer: (a) \(\frac{π}{2}\)

Question 34.
If \(\vec{a}\) = \(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\), \(\vec{b}\) = \(\hat{i}\) + 2\(\hat{j}\) – \(\hat{k}\), \(\vec{c}\) = 3\(\hat{i}\) – p\(\hat{j}\) – 5\(\hat{k}\) are coplanar then P =
(a) 6
(b) -6
(c) 2
(d) -2

Answer

Answer: (a) 6

Question 35.
The distance of the point (- 3, 4, 5) from the origin
(a) 50
(b) 5√2
(c) 6
(d) None of these

Answer

Answer: (b) 5√2

Question 36.
If \(\vec{AB}\) = 2\(\hat{i}\) + \(\hat{j}\) – 3\(\hat{k}\) and the co-ordinates of A are (1, 2, -1) then coordinate of B are
(a) (2, 2, -3)
(b) (3, 2, -4)
(c) (4, 2, -1)
(d) (3, 3, -4)

Answer

Answer: (d) (3, 3, -4)

Question 37.
If \(\vec{b}\) is a unit vector in xy-plane making an angle of \(\frac{π}{4}\) with x-axis. then \(\vec{b}\) is equal to
(a) \(\hat{i}\) + \(\hat{j}\)
(b) \(\vec{i}\) – \(\vec{j}\)
(c) \(\frac{\vec{i}+\vec{j}}{√2}\)
(d) \(\frac{\vec{i}-\vec{j}}{√2}\)

Answer

Answer: (c) \(\frac{\vec{i}+\vec{j}}{√2}\)

Question 38.
\(\vec{a}\) = 2\(\hat{i}\) + \(\hat{j}\) – 8\(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) + 3\(\hat{j}\) – 4\(\hat{k}\) then the magnitude of \(\vec{a}\) + \(\vec{b}\) is equal to
(a) 13
(b) \(\frac{13}{4}\)
(c) \(\frac{3}{13}\)
(d) \(\frac{4}{13}\)

Answer

Answer: (a) 13

Question 39.
The vector in the direction of the vector \(\hat{i}\) – 2\(\hat{j}\) + 2\(\hat{k}\) that has magnitude 9 is
(a) \(\hat{i}\) – 2\(\hat{j}\) + 2\(\hat{k}\)
(b) \(\frac{\hat{i}-2\hat{j}+2\hat{k}}{3}\)
(c) 3(\(\hat{i}\) – 2\(\hat{j}\) + 2\(\hat{k}\))
(d) 9(\(\hat{i}\) – 2\(\hat{j}\) + 2\(\hat{k}\))

Answer

Answer: (c) 3(\(\hat{i}\) – 2\(\hat{j}\) + 2\(\hat{k}\))

Question 40.
The position vector of the point which divides the join of points 2\(\vec{a}\) – 3\(\vec{b}\) and \(\vec{a}\) + \(\vec{b}\) in the ratio 3 : 1 is
(a) \(\frac{3\vec{a}-2\vec{b}}{2}\)
(b) \(\frac{7\vec{a}-8\vec{b}}{2}\)
(c) \(\frac{3\vec{a}}{2}\)
(d) \(\frac{5\vec{a}}{4}\)

Answer

Answer: (d) \(\frac{5\vec{a}}{4}\)

Question 41.
The vector having, initial and terminal points as (2, 5, 0) and (- 3, 7, 4) respectively is
(a) –\(\hat{i}\) + 12\(\hat{j}\) + 4\(\hat{k}\)
(b) 5\(\hat{i}\) + 2\(\hat{j}\) – 4\(\hat{k}\)
(c) -5\(\hat{i}\) + 2\(\hat{j}\) + 4\(\hat{k}\)
(d) \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\)

Answer

Answer: (c) -5\(\hat{i}\) + 2\(\hat{j}\) + 4\(\hat{k}\)

Question 42.
Find the value of λ such that the vectors \(\vec{a}\) = 2\(\hat{i}\) + λ\(\hat{j}\) + \(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) + 2\(\hat{j}\) + 3\(\hat{k}\) are orthogonal
(a) 0
(b) 1
(c) \(\frac{3}{2}\)
(d) –\(\frac{5}{2}\)

Answer

Answer: (d) –\(\frac{5}{2}\)

Question 43.
The value of λ for which the vectors 3\(\hat{i}\) – 6\(\hat{j}\) + \(\hat{k}\) and 2\(\hat{i}\) – 4\(\hat{j}\) + λ\(\hat{k}\) are parallel is
(a) \(\frac{2}{3}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{5}{2}\)
(d) –\(\frac{2}{5}\)

Answer

Answer: (a) \(\frac{2}{3}\)

Question 44.
The vectors from origin to the points A and B are \(\vec{a}\) = 2\(\hat{i}\) – 3\(\hat{j}\) +2\(\hat{k}\) and \(\vec{b}\) = 2\(\hat{i}\) + 3\(\hat{j}\) + \(\hat{k}\) respectively, then the area of triangle OAB is
(a) 340
(b) \(\sqrt{25}\)
(c) \(\sqrt{229}\)
(d) \(\frac{1}{2}\) \(\sqrt{229}\)

Answer

Answer: (d) \(\frac{1}{2}\) \(\sqrt{229}\)

Question 45.
For any vector \(\vec{a}\) the value of (\(\vec{a}\) × \(\vec{i}\))² + (\(\vec{a}\) × \(\hat{j}\))² + (\(\vec{a}\) × \(\hat{k}\))² is equal to
(a) \(\vec{a}\)²
(b) 3\(\vec{a}\)²
(c) 4\(\vec{a}\)²
(d) 2\(\vec{a}\)²

Answer

Answer: (d) 2\(\vec{a}\)²

Question 46.
If |\(\vec{a}\)| = 10, |\(\vec{b}\)| = 2 and \(\vec{a}\).\(\vec{b}\) = 12, then the value of |\(\vec{a}\) × \(\vec{b}\)| is
(a) 5
(b) 10
(c) 14
(d) 16

Answer

Answer: (d) 16

Question 47.
The vectors λ\(\hat{i}\) + \(\hat{j}\) + 2\(\hat{k}\), \(\hat{i}\) + λ\(\hat{j}\) – \(\hat{k}\) and 2\(\hat{i}\) – \(\hat{j}\) + λ\(\hat{k}\) are coplanar if
(a) λ = -2
(b) λ = 0
(c) λ = 1
(d) λ = -1

Answer

Answer: (a) λ = -2

Question 48.
If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are unit vectors such that \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) = \(\vec{0}\), then the value of \(\vec{a}\).\(\vec{b}\) + \(\vec{b}\).\(\vec{c}\) + \(\vec{c}\).\(\vec{a}\)
(a) 1
(b) 3
(c) –\(\frac{3}{2}\)
(d) None of these

Answer

Answer: (c) –\(\frac{3}{2}\)

Question 49.
Projection vector of \(\vec{a}\) on \(\vec{b}\) is
(a) (\(\frac{\vec{a}.\vec{b}}{|\vec{b}|^2}\))\(\vec{b}\)
(b) \(\frac{\vec{a}.\vec{b}}{|\vec{b}|}\)
(c) \(\frac{\vec{a}.\vec{b}}{|\vec{a}|}\)
(d) (\(\frac{\vec{a}.\vec{b}}{|\vec{a}|^2}\))\(\hat{b}\)

Answer

Answer: (b) \(\frac{\vec{a}.\vec{b}}{|\vec{b}|}\)

Question 50.
If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are three vectors such that \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) = 5 and |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 3, |\(\vec{c}\)| = 5, then the value of \(\vec{a}\).\(\vec{b}\) +\(\vec{b}\).\(\vec{c}\) + \(\vec{c}\).\(\vec{a}\) is
(a) 0
(b) 1
(c) -19
(d) 38

Answer

Answer: (c) -19

Question 51.
If |\(\vec{a}\)| 4 and – 3 ≤ λ ≤ 2, then the range of |λ\(\vec{a}\)| is
(a) [0, 8]
(b) [-12, 8]
(c) [0, 12]
(d) [8, 12]

Answer

Answer: (b) [-12, 8]

Question 52.
The number of vectors of unit length perpendicular to the vectors \(\vec{a}\) = 2\(\hat{i}\) + \(\hat{j}\) + 2\(\hat{k}\) and \(\vec{b}\) = \(\hat{j}\) + \(\hat{k}\) is
(a) one
(b) two
(c) three
(d) infinite

Answer

Answer: (b) two

Question 53.
If (\(\frac{1}{2}\), \(\frac{1}{3}\), n) are the direction cosines of a line, then the value of n is
(a) \(\frac{\sqrt{23}}{6}\)
(b) \(\frac{23}{6}\)
(c) \(\frac{2}{3}\)
(d) –\(\frac{3}{2}\)

Answer

Answer: (a) \(\frac{\sqrt{23}}{6}\)

Question 54.
Find the magnitude of vector 3\(\hat{i}\) + 2\(\hat{j}\) + 12\(\hat{k}\)
(a) \(\sqrt{157}\)
(b) 4\(\sqrt{11}\)
(c) \(\sqrt{213}\)
(d) 9√3

Answer

Answer: (a) \(\sqrt{157}\)

Question 55.
Three points (2, -1, 3), (3, – 5, 1) and (-1, 11, 9) are
(a) Non-collinear
(b) Non-coplanar
(c) Collinear
(d) None of these

Answer

Answer: (c) Collinear

Question 56.
The vectors 3\(\hat{i}\) + 5\(\hat{j}\) + 2\(\hat{k}\), 2\(\hat{i}\) – 3\(\hat{j}\) – 5\(\hat{k}\) and 5\(\hat{i}\) + 2\(\hat{j}\) – 3\(\hat{k}\) form the sides of
(a) Isosceles triangle
(b) Right triangle
(c) Scalene triangle
(d) Equilateral triangle

Answer

Answer: (a) Isosceles triangle

Question 57.
The points with position vectors 60\(\hat{i}\) + 3\(\hat{j}\), 40\(\hat{i}\) – 8\(\hat{j}\) and a\(\hat{i}\) – 52\(\hat{j}\) are collinear if
(a) a = -40
(b) a = 40
(c) a = 20
(d) None of these

Answer

Answer: (a) a = -40

Question 58.
The ratio in which 2x + 3y + 5z = 1 divides the line joining the points (1, 0, -3) and (1, -5, 7) is
(a) 5 : 3
(b) 3 : 2
(c) 2 : 1
(d) 1 : 3

Answer

Answer: (a) 5 : 3

Question 59.
If O is origin and C is the mid point of A (2, -1) and B (-4, 3) then the value of \(\bar{OC}\) is
(a) \(\hat{i}\) + \(\hat{j}\)
(b) \(\hat{i}\) – \(\hat{j}\)
(c) –\(\hat{i}\) + \(\hat{j}\)
(d) –\(\hat{i}\) – \(\hat{j}\)

Answer

Answer: (c) –\(\hat{i}\) + \(\hat{j}\)

Question 60.
If ABCDEF is regular hexagon, then \(\vec{AD}\) + \(\vec{EB}\) + \(\vec{FC}\) is equal
(a) 0
(b) 2\(\vec{AB}\)
(c) 3\(\vec{AB}\)
(d) 4\(\vec{AB}\)

Answer

Answer: (d) 4\(\vec{AB}\)

Question 61.
If \(\vec{a}\) = \(\hat{i}\) + \(\hat{j}\) + \(\hat{k}\), \(\vec{b}\) = 2\(\hat{i}\) – 4\(\hat{k}\), \(\vec{c}\) = \(\hat{i}\) + λ\(\hat{j}\) + 3\(\hat{j}\) are coplanar, then the value of λ is
(a) \(\frac{5}{2}\)
(b) \(\frac{3}{5}\)
(c) \(\frac{7}{3}\)
(d) –\(\frac{5}{3}\)

Answer

Answer: (d) –\(\frac{5}{3}\)

Question 62.
The vectors \(\vec{a}\) = x\(\hat{i}\) – 2\(\hat{j}\) + 5\(\hat{k}\) and \(\vec{b}\) = \(\hat{i}\) + y\(\hat{j}\) – z\(\hat{k}\) are collinear, if
(a) x = 1, y = -2, z = -5
(b) x = \(\frac{3}{2}\), y = -4, z = -10
(c) x = \(\frac{3}{2}\), y = 4, z = 10
(d) All of these

Answer

Answer: (d) All of these

Question 63.
The vectors (x, x + 1, x + 2), (x + 3, x + 4, x + 5) and (x + 6, x + 7, x + 8) are coplanar for
(a) all values of x
(b) x < 0
(c) x ≤ 0
(d) None of these

Answer

Answer: (a) all values of x

Question 64.
The vectors \(\vec{AB}\) = 3\(\hat{i}\) +4\(\hat{k}\) and \(\vec{AC}\) = 5\(\hat{i}\) – 2\(\hat{j}\) + 4\(\hat{k}\) are the sides of ΔABC. The length of the median through A is
(a) \(\sqrt{18}\)
(b) \(\sqrt{72}\)
(c) \(\sqrt{33}\)
(d) \(\sqrt{288}\)

Answer

Answer: (c) \(\sqrt{33}\)

Question 65.
The summation of two unit vectors is a third unit vector, then the modulus of the difference of the unit vector is
(a) √3
(b) 1 – √3
(c) 1 + √3
(d) -√3

Answer

Answer: (a) √3

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

Three Dimensional Geometry Class 12 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 12 Maths Chapter 11 Three Dimensional Geometry with Answers Pdf free download. MCQ Questions for Class 12 Maths with Answers were prepared based on the latest exam pattern. We have provided Three Dimensional Geometry Class 12 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 12 Maths Chapter 11 MCQ With Answers

Maths Class 12 Chapter 11 MCQs On Three Dimensional Geometry

Three Dimensional Geometry Class 12 MCQ Questions Chapter 11 Question 1.
The direction cosines of the y-axis are
(a) (6, 0, 0)
(b) (1, 0, 0)
(c) (0, 1, 0)
(d) (0, 0, 1)

Answer

Answer: (c) (0, 1, 0)

MCQ On Three Dimensional Geometry Class 12 Chapter 11 Question 2.
The direction ratios of the line joining the points (x, y, z) and (x₂, y₂, z₁) are
(a) x₁ + x₂, y₁ + y₂, z₁ + z₂
(b) \(\sqrt{(x_1 – x_2)^2 + (y_1 – y_2)^2 + (z_1 + z_2)^2}\)
(c) \(\frac{x_1+x_2}{2}\), \(\frac{y_1+y_2}{2}\), \(\frac{z_1+z_2}{2}\)
(d) x₂ – x₁, y₂ – y₁, z₂ – z₁

Answer

Answer: (d) x₂ – x₁, y₂ – y₁, z₂ – z₁

3d Geometry Deals With MCQ Class 12 Chapter 11 Question 3.
The coordinates of the midpoints of the line segment joining the points (2, 3, 4) and (8, -3, 8) are
(a) (10, 0, 12)
(b) (5, 6, 0)
(c) (6, 5, 0)
(d) (5, 0, 6)

Answer

Answer: (d) (5, 0, 6)

MCQ On 3d Geometry Class 12 Chapter 11 Question 4.
If the planes a₁x + b, y + c, z + d₁ = 0 and a₂x + b, y + c₂z + d₂ = 0 are perpendicular to each other then
(a) \(\frac{a_1}{a_2}\) = \(\frac{b_1}{b_2}\) = \(\frac{c_1}{c_2}\)
(b) \(\frac{a_1}{a_2}\) + \(\frac{b_1}{b_2}\), \(\frac{c_1}{c_2}\)
(c) a₁a₂ + b₁b₂ + c₁c₂ = 0
(d) a\(_{1}^{2}\)a\(_{2}^{2}\) + b\(_{1}^{2}\)b\(_{2}^{2}\) + c\(_{1}^{2}\)c\(_{2}^{2}\) = 0

Answer

Answer: (c) a₁a₂ + b₁b₂ + c₁c₂ = 0

3d Geometry Class 12 MCQ Chapter 11 Question 5.
The distance of the plane 2x – 3y + 6z + 7 = 0 from the point (2, -3, -1) is
(a) 4
(b) 3
(c) 2
(d) \(\frac{1}{5}\)

Answer

Answer: (c) 2

MCQ Questions For Class 12 Maths With Answers Chapter 11 Question 6.
The direction cosines of the normal to the plane 2x – 3y – 6z – 3 = 0 are
(a) \(\frac{2}{7}\), \(\frac{-3}{7}\), \(\frac{-6}{7}\)
(b) \(\frac{2}{7}\), \(\frac{3}{7}\), \(\frac{6}{7}\)
(c) \(\frac{-2}{7}\), \(\frac{-3}{7}\), \(\frac{-6}{7}\)
(d) None of these

Answer

Answer: (a) \(\frac{2}{7}\), \(\frac{-3}{7}\), \(\frac{-6}{7}\)

MCQ Questions For Class 12 Maths With Answers Pdf Download Question 7.
If 2x + 5y – 6z + 3 = 0 be the equation of the plane, then the equation of any plane parallel to the given plane is
(a) 3x + 5y – 6z + 3 = 0
(b) 2x – 5y – 6z + 3 = 0
(c) 2x + 5y – 6z + k = 0
(d) None of these

Answer

Answer: (c) 2x + 5y – 6z + k = 0

Class 12 Maths MCQ Pdf Chapter 11 Question 8.
(2, – 3, – 1) 2x – 3y + 6z + 7 = 0
(a) 4
(b) 3
(c) 2
(d) \(\frac{1}{5}\)

Answer

Answer: (c) 2

Class 12 Maths Chapter 11 Important Questions Question 9.
The length of the ⊥^er from the point (0, – 1, 3) to the plane 2x + y – 2z + 1 = 0 is
(a) 0
(b) 2√3
(c) \(\frac{2}{3}\)
(d) 2

Answer

Answer: (d) 2

3d Geometry Class 12 Important Questions Chapter 11 Question 10.
The shortest distance between the lines \(\vec{r}\) = \(\vec{a}\) + k\(\vec{b}\) and r = \(\vec{a}\) + l\(\vec{c}\) is (\(\vec{b}\) and \(\vec{c}\) are non-collinear)
(a) 0
(b) |\(\vec{b}\).\(\vec{c}\)|
(c) \(\frac{|\vec{b}×\vec{c}|}{|\vec {a}|}\)
(d) \(\frac{|\vec{b}.\vec{c}|}{|\vec {a}|}\)

Answer

Answer: (a) 0

Class 12 Maths MCQ Pdf Free Download Chapter 11 Question 11.
The equation xy = 0 in three dimensional space is represented by
(a) a plane
(b) two plane are right angles
(c) a pair of parallel planes
(d) a pair of st. line

Answer

Answer: (b) two plane are right angles

Maths MCQ Questions Class 12 Chapter 11 Question 12.
The direction cosines of any normal to the xy plane are
(a) 1, 0 ,0
(b) 0, 1, 0
(c) 1, 1, 0
(d) 1, 1, 0

Answer

Answer: (d) 1, 1, 0

Class 12 Maths MCQs Chapter Wise Chapter 11 Question 13.
How many lines through the origin in make equal angles with the coordinate axis?
(a) 1
(b) 4
(c) 8
(d) 2

Answer

Answer: (c) 8

Class 12 Maths MCQ Pdf Download Chapter 11 Question 14.
The direction cosines of the line joining (1, -1, 1) and (-1, 1, 1) are
(a) 2, -2, 0
(b) 1, -1, 0
(c) \(\frac{1}{√2}\), – \(\frac{1}{√2}\)
(d) None of these

Answer

Answer: (c) \(\frac{1}{√2}\), – \(\frac{1}{√2}\)

MCQ Questions For Class 12 Maths With Solutions Chapter 11 Question 15.
The equation x² – x – 2 = 0 in three dimensional space is represented by
(a) A pair of parallel planes
(b) A pair of straight lines
(c) A pair of perpendicular plane
(d) None of these

Answer

Answer: (a) A pair of parallel planes

Maths MCQs For Class 12 With Answers Pdf Chapter 11 Question 16.
The distance of the point (-3, 4, 5) from the origin
(a) 50
(b) 5√2
(c) 6
(d) None of these

Answer

Answer: (b) 5√2

Three Dimensional Element Is MCQ Class 12 Chapter 11 Question 17.
If a line makes angles Q₁, Q₂₁ and Q₃ respectively with the coordinate axis then the value of cos² Q₁ + cos² Q₂ + cos² Q₃
(a) 2
(b) 1
(c) 4
(d) \(\frac{3}{2}\)

Answer

Answer: (b) 1

Ncert Solutions For Class 12 Maths Chapter 11 Question 18.
The direction ratios of a line are 1,3,5 then its direction cosines are
(a) \(\frac{1}{\sqrt{35}}\), \(\frac{3}{\sqrt{35}}\), \(\frac{5}{\sqrt{35}}\)
(b) \(\frac{1}{9}\), \(\frac{1}{3}\), \(\frac{5}{9}\)
(c) \(\frac{5}{\sqrt{35}}\), \(\frac{3}{\sqrt{35}}\), \(\frac{1}{\sqrt{35}}\)
(d) None of these

Answer

Answer: (a) \(\frac{1}{\sqrt{35}}\), \(\frac{3}{\sqrt{35}}\), \(\frac{5}{\sqrt{35}}\)

Class 12 Maths MCQs Chapter 11 Question 19.
The direction ratios of the normal to the plane 7x + 4y – 2z + 5 = 0 are
(a) 7, 4,-2
(b)7, 4, 5
(c) 7, 4, 2
(d) 4, -2, 5

Answer

Answer: (a) 7, 4,-2

MCQ Questions For Class 12th Maths Chapter 11 Question 20.
The direction ratios of the line of intersection of the planes 3x + 2y – z = 5 and x – y + 2z = 3 are
(a) 3, 2, -1
(b) -3, 7, 5
(c) 1, -1, 2
(d) – 11, 4, -5

Answer

Answer: (b) -3, 7, 5

Question 21.
The lines of intersection of the planes \(\vec{r}\)(3\(\hat{i}\) – \(\hat{j}\) + \(\hat{k}\)) = 1 and \(\vec{r}\)(\(\hat{i}\) +4\(\hat{j}\) – 2\(\hat{k}\)) = 2 is parallel to the vector
(a) 2\(\hat{i}\) + 7\(\hat{j}\) + 13\(\hat{k}\)
(b) -2\(\hat{i}\) + 7\(\hat{j}\) + 13\(\hat{k}\)
(c) 2\(\hat{i}\) – 7\(\hat{j}\) + 13\(\hat{i}\)
(b) -2\(\hat{i}\) – 7\(\hat{j}\) – 13\(\hat{k}\)

Answer

Answer: (b) -2\(\hat{i}\) + 7\(\hat{j}\) + 13\(\hat{k}\)

Question 22.
The equation of the plane through the origin and parallel to the plane 3x – 4y + 5z + 6 = 0
(a) 3x – 4y – 5z – 6 = 0
(b) 3x – 4y + 5z + 6 = 0
(c) 3x – 4y + 5z = 0
(d) 3x + 4y – 5z + 6 = 0

Answer

Answer: (c) 3x – 4y + 5z = 0

Question 23.
The locus of xy + yz = 0 is
(a) A pair of st. lines
(b) A pair of parallel lines
(c) A pair of parallel planes
(d) A pair of perpendicular planes

Answer

Answer: (d) A pair of perpendicular planes

Question 24.
The plane x + y = 0
(a) is parallel to z-axis
(b) is perpendicular to z-axis
(c) passes through z-axis
(d) None of these

Answer

Answer: (c) passes through z-axis

Question 25.
If α, β, γ are the angle which a half ray makes with the positive directions of the axis then sin²α + sin²β + sin²γ =
(a) 1
(b) 2
(c) 0
(d) -1

Answer

Answer: (b) 2

Question 26.
If a line makes angles α, β, γ with the axis then cos 2α + cos 2β + cos 2γ =
(a) -2
(b) -1
(c) 1
(d) 2

Answer

Answer: (b) -1

Question 27.
The line x = 1, y = 2 is
(a) parallel to x-axis
(b) parallel to y-axis
(c) parallel to z-axis
(d) None of these

Answer

Answer: (c) parallel to z-axis

Question 28.
The points A (1, 1, 0), B(0, 1, 1), C(1, 0, 1) and D(\(\frac{2}{3}\), \(\frac{2}{3}\), \(\frac{2}{3}\))
(a) Coplanar
(b) Non-coplanar
(c) Vertices of a parallelogram
(d) None of these

Answer

Answer: (a) Coplanar

Question 29.
The angle between the planes 2x – y + z = 6 and x + y + 2z = 7 is
(a) \(\frac{π}{4}\)
(b) \(\frac{π}{6}\)
(c) \(\frac{π}{3}\)
(d) \(\frac{π}{2}\)

Answer

Answer: (c) \(\frac{π}{3}\)

Question 30.
The distance of the points (2, 1, -1) from the plane x- 2y + 4z – 9 is
(a) \(\frac{\sqrt{31}}{21}\)
(b) \(\frac{13}{21}\)
(c) \(\frac{13}{\sqrt{21}}\)
(d) \(\sqrt{\frac{π}{2}}\)

Answer

Answer: (c) \(\frac{13}{\sqrt{21}}\)

Question 31.
The planes \(\vec{r}\)(2\(\hat{i}\) + 3\(\hat{j}\) – 6\(\hat{k}\)) = 7 and
\(\vec{r}\)(\(\frac{-2}{7}\)\(\vec{i}\) – \(\frac{3}{j}\)\(\vec{j}\) + \(\frac{6}{7}\)\(\vec{k}\)) = 0 are
(a) parallel
(b) at right angles
(c) equidistant front origin
(d) None of these

Answer

Answer: (a) parallel

Question 32.
The equation of the plane through point (1, 2, -3) which is parallel to the plane 3x- 5y + 2z = 11 is given by
(a) 3x – 5y + 2z – 13 = 0
(b) 5x – 3y + 2z + 13 = 0
(c) 3x – 2y + 5z + 13 = 0
(d) 3x – 5y + 2z + 13 = 0

Answer

Answer: (d) 3x – 5y + 2z + 13 = 0

Question 33.
Distance of the point (a, β, γ) from y-axis is
(a) β
(b) |β|
(c) |β + γ|
(d) \(\sqrt{α^2+γ^2}\)

Answer

Answer: (d) \(\sqrt{α^2+γ^2}\)

Question 34.
If the directions cosines of a line are A, k, k, then
(a) k > 0
(b) 0 < k < 1
(c) k = 1
(d) k = \(\frac{1}{√3}\) or –\(\frac{1}{√3}\)

Answer

Answer: (d) k = \(\frac{1}{√3}\) or –\(\frac{1}{√3}\)

Question 35.
The distance of the plane \(\vec{r}\)(\(\frac{-2}{7}\)\(\hat{i}\) – \(\frac{3}{7}\)\(\hat{j}\) + \(\frac{6}{7}\)\(\hat{k}\)) = 0 from the orgin is
(a) 1
(b) 7
(c) \(\frac{1}{7}\)
(d) None of these

Answer

Answer: (a) 1

Question 36.
The sine of the angle between the straight line \(\frac{x-2}{3}\) = \(\frac{y-3}{4}\) = \(\frac{z-4}{5}\) and the plane 2x – 2y + z = 5 is
(a) \(\frac{10}{6√5}\)
(b) \(\frac{4}{5√2}\)
(c) \(\frac{2√3}{5}\)
(d) \(\sqrt{\frac{√2}{10}}\)

Answer

Answer: (c) \(\frac{2√3}{5}\)

Question 37.
The reflection of the point (a, β, γ) in the xy-plane is
(a) (α, β, 0)
(b) (0, 0, γ)
(c) (- α, – β, γ)
(d) (α, β, γ)

Answer

Answer: (d) (α, β, γ)

Question 38.
The area of the quadrilateral ABCD, where A(0, 4, 1), B(2, 3, -1), C(4, 5, 0) and D(2, 6, 2) is equal to
(a) 9 sq. units
(b) 18 sq. units
(c) 27 sq. units
(d) 81 sq. units

Answer

Answer: (a) 9 sq. units

Question 39.
The plane 2x – 3y + 6z – 11 = 0 makes an angle sin^-1 (α) with .e-axis. The value of a is equal to
(a) \(\frac{√3}{2}\)
(b) \(\frac{√2}{3}\)
(c) \(\frac{2}{7}\)
(d) \(\frac{3}{7}\)

Answer

Answer: (c) \(\frac{2}{7}\)

Question 40.
The cosines of the angle between any two diagonals of a cube is
(a) \(\frac{1}{3}\)
(b) \(\frac{1}{2}\)
(c) \(\frac{2}{3}\)
(d) \(\frac{1}{√3}\)

Answer

Answer: (a) \(\frac{1}{3}\)

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MCQ Questions for Class 12 Maths Chapter 7 Integrals with Answers

Integrals Class 12 MCQ Online Test With Answers Questions

October 24, 2024

Check the below NCERT MCQ Questions for Class 12 Maths Chapter 7 Integrals with Answers Pdf free download. MCQ Questions for Class 12 Maths with Answers were prepared based on the latest exam pattern. We have provided Integrals Class 12 Maths MCQs Questions with Answers to help students understand the concept very well.

Class 12 Maths Chapter 7 MCQ With Answers

Maths Class 12 Chapter 7 MCQs On Integrals

Integration MCQ Class 12 Chapter 7 Question 1.
\(\int \frac{x+\sin x}{1+\cos x}\) dx is equal to
(a) log |1 + cos x | + c
(b) log | x + sin x | + c
(c) x – tan + c
(d) x. tan \(\frac{x}{2}\) + c

Answer

Answer: (d) x. tan \(\frac{x}{2}\) + c

MCQ On Integration Class 12 Chapter 7 Question 2.
∫1.dx =
(a) x + k
(b) 1 + k
(c) \(\frac{x^2}{2}\) + k
(d) log x + k

Answer

Answer: (a) x + k

Integration MCQ Questions Class 12 Chapter 7 Question 3.
∫\(\frac{dx}{√x}\) =
(a) √x + k
(b) 2√x + k
(c) x + k
(d) \(\frac{2}{3}\)x^3/2 + k

Answer

Answer: (b) 2√x + k

MCQ On Integration Class 12 Chapter 7 Question 4.
∫\(\frac{dx}{1+cos x}\) =
(a) tan \(\frac{x}{2}\) + k
(b) \(\frac{1}{2}\) tan \(\frac{x}{2}\) + k
(c) 2 tan \(\frac{x}{2}\) + k
(d) tan² \(\frac{x}{2}\) + k

Answer

Answer: (a) tan \(\frac{x}{2}\) + k

Integration MCQs With Answers Pdf Class 12 Chapter 7 Question 5.
\(\int_{a}^{b}\) x⁵ dx =
(a) tan \(\frac{x}{2}\) + k
(b) \(\frac{1}{2}\) tan \(\frac{x}{2}\) + k
(c) 2 tan \(\frac{x}{2}\) + k
(d) tan² \(\frac{x}{2}\) + k

Answer

Answer: (a) tan \(\frac{x}{2}\) + k

Integration MCQ With Answers Class 12 Chapter 7 Question 6.
If x > a, ∫\(\frac{dx}{x^2-a^2}\) =
(a) \(\frac{2}{2a}\) log \(\frac{x-a}{x+a}\) + k
(b) \(\frac{2}{2a}\) log \(\frac{x+a}{x-a}\) + k
(c) \(\frac{1}{a}\) log(x² – a²) + k
(d) log(x + \(\sqrt{x^2-a^2}\) + k)

Answer

Answer: (a) \(\frac{2}{2a}\) log \(\frac{x-a}{x+a}\) + k

Integration Objective Questions Class 12 Chapter 7 Question 7.
∫\(\frac{cos 2x dx}{(sinx+cosx)^2}\) =
(a) –\(\frac{1}{sinx+cosx}\) + c
(b) log | sin x + cos x | + c
(c) log | sin x – cos x | + c
(d) \(\frac{1}{(sinx+cosx)^2}\)

Answer

Answer: (b) log | sin x + cos x | + c

MCQ On Integration With Answers Class 12 Chapter 7 Question 8.
∫\(\frac{(1+logx)^2}{1+x^2}\) dx =
(a) \(\frac{1}{3}\)(1+log)³ + c
(b) \(\frac{1}{2}\)(1+log)² + c
(c) log (log 1 + x) + 2
(d) None of these

Answer

Answer: (a) \(\frac{1}{3}\)(1+log)³ + c

Integration MCQs Class 12 Chapter 7 Question 9.
\(\int_{0}^{1}\frac{(tan^{-1}x)^2}{1+x^2}\) dx =
(a) 1
(b) \(\frac{π^2}{64}\)
(c) \(\frac{π^2}{192}\)
(d) None of these

Answer

Answer: (c) \(\frac{π^2}{192}\)

MCQ Of Integration Class 12 Chapter 7 Question 10.
\(\int_{-2}^{2}\) |x|dx =
(a) 0
(b) 2
(c) 1
(d) 4

Answer

Answer: (d) 4

MCQs On Integration Class 12 Chapter 7 Question 11.
∫\(\frac{x^4+1}{x^2+1}\) dx is equal to
(a) \(\frac{x^3}{3}\) + x + tan^-1 x + c
(b) \(\frac{x^3}{3}\) – x + tan x + c
(c) \(\frac{x^3}{3}\) + x + 2tan^-1 x + c
(d) \(\frac{x^3}{3}\) – x + 2tan^-1 x + c

Answer

Answer: (d) \(\frac{x^3}{3}\) – x + 2tan^-1 x + c

Integration MCQ With Answers Pdf Class 12 Chapter 7 Question 12.
∫(√x + \(\frac{1}{√x}\)) dx =
(a) \(\frac{1}{x}\) x^{\(\frac{1}{3}\)} + 2x^{\(\frac{1}{2}\)} + c
(b) \(\frac{2}{3}\) x^{\(\frac{2}{3}\)} + \(\frac{1}{2}\)x² + c
(c) \(\frac{2}{3}\) x^{\(\frac{3}{2}\)} + 2x^{\(\frac{1}{2}\)} + c
(d) \(\frac{3}{2}\) x^{\(\frac{3}{2}\)} + \(\frac{1}{2}\)x^{\(\frac{1}{2}\)} + c

Answer

Answer: (c) \(\frac{2}{3}\) x^{\(\frac{3}{2}\)} + 2x^{\(\frac{1}{2}\)} + c

Integration MCQ Questions And Answers Class 12 Chapter 7 Question 13.
∫\(\frac{sin^2x-cos^2x}{sin^2xcos^2x}\) dx is equal to
(a) tan x + cos x + c
(b) tan x + cosec x + c
(c) tan x + cot x + c
(d) tan x+ sec x + c

Answer

Answer: (c) tan x + cot x + c

Integral MCQ Questions Class 12 Chapter 7 Question 14.
\(\frac{d}{dx}\)∫f(x)dx is equal to
(a) f'(x)
(b) f(x)
(c) f'(x’)
(d) f(x) + c

Answer

Answer: (b) f(x)

MCQ Integration Class 12 Chapter 7 Question 15.
∫\(\frac{xdx}{(x-1)(x-2)}\) equals

(d) log |(x – 1)(x – 2) + c

Answer

Answer: (b) log|\(\frac{(x-2)^2}{x-2}\)| + c

Integration MCQs With Answers Chapter 7 Question 16.
What is the value of \(\int_{0}^{\pi / 2}\) \(\frac{\sqrt{tan x}}{\sqrt{tan x}+\sqrt{cot x}}\) dx?
(a) \(\frac{π}{2}\)
(b) \(\frac{π}{4}\)
(c) \(\frac{π}{8}\)
(d) None of these

Answer

Answer: (b) \(\frac{π}{4}\)

Integrals MCQs Class 12 Chapter 7 Question 17.
What is the value of \(\int_{0}^{\pi / 2}\) \(\frac{sinx – cos x}{1+sin xcos x}\) dx?
(a) 1
(b) \(\frac{π}{2}\)
(c) 0
(d) –\(\frac{π}{2}\)

Answer

Answer: (c) 0

Integration MCQs Class 12 Chapter 7 Question 18.
What is the value of \(\int_{\pi / 6}^{\pi / 3}\) \(\frac{dx}{sin2x}\)?
(a) \(\frac{1}{2}\) log(-l)
(b) log(- 1)
(c) log 3
(d) log √3

Answer

Answer: (c) log 3

Integration MCQ Class 12 Question 19.
What is the value of \(\int_{-1}^{1}\) sin³ x cos² xdx?
(a) 0
(b) 1
(c) \(\frac{1}{2}\)
(d) 2

Answer

Answer: (a) 0

MCQ Questions On Integration Question 20.
What is the value of \(\int_{1}^{e} \frac{1+\log x}{x}\) dx?
(a) \(\frac{3}{2}\)
(b) \(\frac{1}{2}\)
(c) e
(d) \(\frac{1}{e}\)

Answer

Answer: (a) \(\frac{3}{2}\)

Question 21.
\(\int_{-\pi / 2}^{\pi / 2}\) sin⁹ xdx =
(a) -1
(b) 0
(c) 1
(d) None of these

Answer

Answer: (b) 0

Question 22.
\(\int_{0}^{\pi^{2} / 4} \frac{\sin \sqrt{y}}{\sqrt{y}}\)
(a) 1
(b) 2
(c) \(\frac{π}{4}\)
(d) \(\frac{π^2}{8}\)

Answer

Answer: (b) 2

Question 23.
\(\int_{0}^{\infty} \frac{1}{1+e^{x}}\) dx =
(a) log 2
(b) -log 2
(c) log 2 – 1
(d) log 4 – 1

Answer

Answer: (a) log 2

Question 24.
\(\int_{0}^{1}\) x(1 – x)⁹⁹ is equal to
(a) \(\frac{1}{10010}\)
(b) \(\frac{1}{10100}\)
(c) \(\frac{1}{1010}\)
(d) \(\frac{11}{10100}\)

Answer

Answer: (b) \(\frac{1}{10100}\)

Question 25.
What is the value of \(\int_{0}^{1}\) \(\frac{d}{dx}\){sin^-1(\(\frac{2x}{1+x^2}\))}dx?
(a) 0
(b) π
(c) -π
(d) \(\frac{π}{2}\)

Answer

Answer: (d) \(\frac{π}{2}\)

Question 26.
\(\int_{0}^{1}\) \(\frac{x}{1+x}\) dx =
(a) 1 – log 2
(b) 2
(c) 1 + log 2
(d) log 2

Answer

Answer: (a) 1 – log 2

Question 27.
∫\(\frac{sin x + cos x}{\sqrt{1+2sin x}}\) dx =
(a) log(sin x – cos x)
(b) x
(c) log x
(d) log sin (cos x)

Answer

Answer: (b) x

Question 28.
∫log₁₀ xdx =
(a) log_e 10.x log_e (\(\frac{x}{e}\)) + c
(b) log₁₀ e.x log_e (\(\frac{x}{e}\)) + c
(c) (x – 1) log_e x + c
(d) \(\frac{1}{x}\) + c

Answer

Answer: (b) log₁₀ e.x log_e (\(\frac{x}{e}\)) + c

Question 29.
∫(\(\frac{cos 2θ – 1}{cos 2θ + 1}\)) dθ =
(a) tan θ – θ + c
(b) θ + tan θ + c
(c) θ – tan θ + c
(d) -θ – cot θ + c

Answer

Answer: (c) θ – tan θ + c

Question 30.
∫\(\frac{2dx}{\sqrt{1-4x^2}}\) =
(a) tan^-1 (2x) + c
(b) cot^-1 (2x) + c
(c) cos^-1 (2x) + c
(d) sin^-1 (2x) + c

Answer

Answer: (d) sin^-1 (2x) + c

Question 31.
Value of ∫\(\frac{dx}{\sqrt{2x – x^2}}\)
(a) sin^-1 (x – 1) + c
(b) sin^-1 (1 + x) + c
(c) sin^-1 (1 + x²) + c
(d) –\(\sqrt{2x-x^2}\) + c

Answer

Answer: (a) sin^-1 (x – 1) + c

Question 32.
∫x² sin x³ dx =
(a) \(\frac{1}{3}\) cos x³ + c
(b) –\(\frac{1}{3}\) cos x + c
(c) \(\frac{-1}{3}\) cos x³ + c
(d) \(\frac{1}{2}\) sin² x³ + c

Answer

Answer: (c) \(\frac{-1}{3}\) cos x³ + c

Question 33.
∫\(\frac{cos 2x- cos 2θ}{cos x – cos θ}\)dx is equal to
(a) 2 (sin x + x cos θ) + C
(b) 2 (sin x – x cos θ) + C
(c) 2 (sin x + 2x cos θ) + C
(d) 2 (sin x – 2x cos θ) + C

Answer

Answer: (a) 2 (sin x + x cos θ) + C

Question 34.
∫\(\frac{dx}{sin(x-a)sin(x-b)}\) is equal to
(a) sin(b – a) log |\(\frac{sin (x-b)}{sin(x-a)}\)| + C
(b) cosec (b – a) log |\(\frac{sin (x-b)}{sin(x-b)}\)| + C
(c) cosec (b – a) log |\(\frac{sin (x-b)}{sin(x-a)}\)| + C
(d) sin (b – a) log |\(\frac{sin (x-a)}{sin(x-b)}\)| + C

Answer

Answer: (c) cosec (b – a) log |\(\frac{sin (x-b)}{sin(x-a)}\)| + C

Question 35.
∫tan^-1 √xdx is equal to
(a) (x + 1)tan^-1 √x – √x + C
(b) x tan^-1 √x – √x + C
(c) √x – x tan^-1 √x + C
(d) √x – (x + 1)tan^-1 √x + C

Answer

Answer: (a) (x + 1)tan^-1 √x – √x + C

Question 36.
∫e^x(\(\frac{1-x}{1+x^2}\))² dx is equal to
(a) \(\frac{e^x}{1+x^2}\) + C
(b) –\(\frac{-e^x}{1+x^2}\) + C
(c) \(\frac{e^x}{(1+x^2)^2}\) + C
(d) \(\frac{-e^x}{(1+x^2)^2}\) + C

Answer

Answer: (a) \(\frac{e^x}{1+x^2}\) + C

Question 37.
∫\(\frac{x^9}{(4x^2+1)^6}\) dx is equal to

Answer

Answer: (d) \(\frac{1}{10}\) (\(\frac{1}{x^2}\) + 4)^-5 + C

Question 38.
If ∫\(\frac{dx}{(x+2)(x^2+1)}\) = a log |1 + x²| + b tan^-1 x + \(\frac{1}{5}\) log |x + 2| + C, then
(a) a = \(\frac{-1}{10}\), b = \(\frac{-2}{5}\)
(b) a = \(\frac{1}{10}\), b = \(\frac{-2}{5}\)
(c) a = \(\frac{-1}{10}\), b = \(\frac{2}{5}\)
(d) a = \(\frac{1}{10}\), b = \(\frac{2}{5}\)

Answer

Answer: (c) a = \(\frac{-1}{10}\), b = \(\frac{2}{5}\)

Question 39.
∫ \(\frac{x^3}{x+1}\) is equal to
(a) x + \(\frac{x^2}{2}\) + \(\frac{x^3}{3}\) – log |1 – x| + C
(b) x + \(\frac{x^2}{2}\) – \(\frac{x^3}{3}\) – log |1 – x| + C
(c) x + \(\frac{x^2}{2}\) – \(\frac{x^3}{3}\) – log |1 + x| + C
(d) x + \(\frac{x^2}{2}\) + \(\frac{x^3}{3}\) – log |1 + x| + C

Answer

Answer: (d) x + \(\frac{x^2}{2}\) + \(\frac{x^3}{3}\) – log |1 + x| + C

Question 40.
If ∫\(\frac{x^3dx}{\sqrt{1+x^2}}\) = a(1 + x²)^3/2 + b\(\sqrt{1 + x^2}\) + C, then
(a) a = \(\frac{1}{3}\), b = 1
(b) a = \(\frac{-1}{3}\), b = 1
(c) a = \(\frac{-1}{3}\), b = -1
(d) a = \(\frac{1}{3}\), b = -1

Answer

Answer: (d) a = \(\frac{1}{3}\), b = -1

Question 41.
\(\int_{-\pi / 4}^{\pi / 4}\) \(\frac{dx}{1+cos 2x}\) dx is equal to
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (a) 1

Question 42.
\(\int_{0}^{\pi / 2}\) \(\sqrt{1+sin 2x}\) dx is equal to
(a) 2√2
(b) 2(√2 + 1)
(c) 0
(d) 2(√2 – 1)

Answer

Answer: (c) 0

Question 43.
Evaluate: ∫(2 tan x – 3 cot x)² dx
(a) -4tan x – cot x – 25x + C
(b) 4 tan x – 9 cot x – 25x + C
(c) – 4 tan x + 9 cot x + 25x + C
(d) 4 tan x + 9 cot x + 25x + C

Answer

Answer: (b) 4 tan x – 9 cot x – 25x + C

Question 44.
Evaluate: ∫ sec²(7 – 4x)dx
(a) –\(\frac{1}{4}\) tan(7 – 4x) + C
(b) \(\frac{1}{4}\) tan(7 – 4x) + C
(c) \(\frac{1}{4}\) tan(7 + 4x) + C
(d) –\(\frac{1}{4}\) tan(7x – 4) + C

Answer

Answer: (a) –\(\frac{1}{4}\) tan(7 – 4x) + C

Question 45.
∫ \(\frac{10x^9+10^xlog_e 10}{10^x+x^{10}}\) dx is equal to
(a) 10^x – x¹⁰ + C
(b) 10^x + x¹⁰ + C
(c) (10^x – x¹⁰)^-1 + C
(d) log_e(10^x + x¹⁰) + C

Answer

Answer: (d) log_e(10^x + x¹⁰) + C

Question 46.
Evaluate: ∫ sec^4/3 x cosec^8/3 xdx
(a) \(\frac{3}{5}\) tan^-5/3 x – 3 tan^1/3 x + C
(b) –\(\frac{3}{5}\) tan^-5/3 x + 3 tan^1/3 + C
(c) –\(\frac{3}{5}\) tan^-05/3 x – 3 tan^1/3 + C
(d) None of these

Answer

Answer: (b) –\(\frac{3}{5}\) tan^-5/3 x + 3 tan^1/3 + C

Question 47.
∫ \(\frac{a}{(1+x^2)tan^{-1}x}\) dx =
(a) a log |tan^-1 x| + C
(b) \(\frac{1}{2}\)(tan^-1 x)² + C
(c) a log (1 + x²) + C
(d) None of these

Answer

Answer: (a) a log |tan^-1 x| + C

Question 48.
∫ \(\frac{cot x}{\sqrt[3]{sin x}}\) dx =
(a) \(\frac{-3}{\sqrt[3]{sin x}}\) + C
(b) \(\frac{-2}{sin^3 x}\) + C
(c) \(\frac{3}{sin^{1/3}x}\) + C
(d) None of these

Answer

Answer: (a) \(\frac{-3}{\sqrt[3]{sin x}}\) + C

Question 49.
Evaluate: ∫ \(\frac{1}{1+3sin^2x+8cos^2x}\) dx
(a) \(\frac{1}{6}\) tan^-1 (2 tan x) + C
(b) tan^-1 (2 tan x) + C
(c) \(\frac{1}{6}\) tan^-1(\(\frac{2 tan x}{3}\)) + C
(d) None of these

Answer

Answer: (c) \(\frac{1}{6}\) tan^-1(\(\frac{2 tan x}{3}\)) + C

Question 50.
Evaluate: ∫ \(\frac{1}{\sqrt{9+8x-x^2}}\) dx
(a) -sin^-1 (\(\frac{x-4}{5}\)) + C
(b) sin^-1 (\(\frac{x+4}{5}\)) + C
(c) sin^-1 (\(\frac{x-4}{5}\)) + C
(d) None of these

Answer

Answer: (c) sin^-1 (\(\frac{x-4}{5}\)) + C

Question 51.
∫ \(\frac{dx}{1-cosx-sinx}\) is equal to
(a) log |1 + cot\(\frac{x}{2}\)| + C
(b) log |1 – tan\(\frac{x}{2}\)| + C
(c) log |1 – cot\(\frac{x}{2}\)| + C
(d) log |1 + tan\(\frac{x}{2}\)| + C

Answer

Answer: (c) log |1 – cot\(\frac{x}{2}\)| + C

Question 52.
Evaluate: ∫ \(\frac{1}{\sqrt{1-e^{2x}}}\) dx
(a) log |e^-x + \(\sqrt{e^{-2x} – 1}\)| + C
(b) -log |e^-x + \(\sqrt{e^{-2x} – 1}\)| + C
(c) -log |e^-x – \(\sqrt{e^{-2x} – 1}\)| + C
(d) None of these

Answer

Answer: (b) -log |e^-x + \(\sqrt{e^{-2x} – 1\)| + C

Question 53.
If ∫ \(\frac{3x+4}{x^3-2x-4}\) dx = log |x – 2| + k log f(x) + c, then
(a) f(x) = |x² + 2x + 2|
(b) f(x) = x² + 2x + 2
(c) k = –\(\frac{1}{2}\)
(d) All of these

Answer

Answer: (d) All of these

Question 54.
Evaluate: ∫ \(\frac{1-cosx}{cosx(1+cosx)}\) dx
(a) log|sec x + tan x| – 2 tan(x/2) + C
(b) log|sec x – tan x| – 2 tan(x/2) + C
(c) log|sec x + tan x| + 2 tan(x/2) + C
(d) None of these

Answer

Answer: (a) log|sec x + tan x| – 2 tan(x/2) + C

Question 55.
∫ cos(log_e.x)dx is equal to
(a) \(\frac{1}{2}\) x[cos (log_ex) + sin(log_ex)]
(b) x[cos (log_ex) + sin(log_ex)]
(c) \(\frac{1}{2}\) x[cos (log_ex) – sin(log_ex)]
(d) x[cos (log_ex) – sin(log_ex)]

Answer

Answer: (b) –\(\frac{3}{5}\) tan^-5/3 x + 3 tan^1/3 + C

Question 56.
∫ |x| dx is equal to
(a) \(\frac{1}{2}\) x² + C
(b) –\(\frac{x^2}{2}\) + C
(c) x|x| + C
(d) \(\frac{1}{2}\) x|x| + C

Answer

Answer: (d) \(\frac{1}{2}\) x|x| + C

Question 57.
∫ sin^-1 xdx is equal to
(a) cos^-1 x + C
(b) x sin^-1x + \(\sqrt{1-x^2}\) + C
(c) \(\frac{1}{\sqrt{1-x^2}}\) + C
(d) x sin^-1x – \(\sqrt{1-x^2}\) + C

Answer

Answer: (b) x sin^-1x + \(\sqrt{1-x^2}\) + C

Question 58.
∫ cos^-1(\(\frac{1}{x}\))dx equals
(a) x sec^-1 x + log |x + \(\sqrt{x^2-1}\)| + C
(b) x sec^-1 x – log |x + \(\sqrt{x^2-1}\)| + C
(c) -x sec^-1 x – log |x + \(\sqrt{x^2-1}\)| + C
(d) None of these

Answer

Answer: (b) x sec^-1 x – log |x + \(\sqrt{x^2-1}\)| + C

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