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

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. https://mcqquestions.guru/mcq-questions-for-class-11-maths-chapter-9/

Sequences and Series Class 11 MCQs Questions with Answers

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


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,
ex = 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, Tm = 1/n
⇒ a + (m-1)d = 1/n ………… 1
and Tn = 1/m
⇒ a + (n-1)d = 1/m ………. 2
From equation 2 – 1, we get
(m-1)d – (n-1)d = 1/n – 1/m
⇒ (m-n)d = (m-n)/mn
⇒ d = 1/mn
From equation 1, we get
a + (m-1)/mn = 1/n
⇒ a = 1/n – (m-1)/mn
⇒ a = {m – (m-1)}/mn
⇒ a = {m – m + 1)}/mn
⇒ a = 1/mn
Now, Tmn = 1/mn + (mn-1)/mn
⇒ Tmn = 1/mn + 1 – 1/mn
⇒ Tmn = 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, ………A2n be 2n arithmetic means between a and b
Then, A1 + A2 + A3 + ………+ A2n = 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A1 + A2 + A3 + ………+ A2n = 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 an = 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

Answer: (c) 6
Hint:
Let a and b are two numbers such that
a + b = 13/6
Let A1, A2, A3, ………A2n be 2n arithmetic means between a and b
Then, A1 + A2 + A3 + ………+ A2n = 2n{(n + 1)/2}
⇒ n(a + b) = 13n/6
Given that A1 + A2 + A3 + ………+ A2n = 2n + 1
⇒ 13n/6 = 2n + 1
⇒ n = 6


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


We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 9 Sequences and Series with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Sequences and Series MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

MCQ Questions for Class 11 Maths Chapter 8 Binomial Theorem with Answers

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 8 Binomial Theorem with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Binomial Theorem Class 11 Maths MCQs Questions with Answers to help students understand the concept very well. https://mcqquestions.guru/mcq-questions-for-class-11-maths-chapter-8/

Online Binomial Expansion Calculator and find the expansion of the given binomial term.

Binomial Theorem Class 11 MCQs Questions with Answers

MCQ On Binomial Theorem Class 11 Question 1.
The coefficient of y in the expansion of (y² + c/y)5 is
(a) 10c
(b) 10c²
(c) 10c³
(d) None of these

Answer

Answer: (c) 10c³
Hint:
Given, binomial expression is (y² + c/y)5
Now, Tr+1 = 5Cr × (y²)5-r × (c/y)r
= 5Cr × y10-3r × Cr
Now, 10 – 3r = 1
⇒ 3r = 9
⇒ r = 3
So, the coefficient of y = 5C3 × c³ = 10c³


Binomial Theorem MCQ Question 2.
(1.1)10000 is _____ 1000
(a) greater than
(b) less than
(c) equal to
(d) None of these

Answer

Answer: (a) greater than
Hint:
Given, (1.1)10000 = (1 + 0.1)10000
10000C0 + 10000C1 × (0.1) + 10000C2 ×(0.1)² + other +ve terms
= 1 + 10000×(0.1) + other +ve terms
= 1 + 1000 + other +ve terms
> 1000
So, (1.1)10000 is greater than 1000


MCQ On Binomial Theorem Question 3.
The fourth term in the expansion (x – 2y)12 is
(a) -1670 x9 × y³
(b) -7160 x9 × y³
(c) -1760 x9 × y³
(d) -1607 x9 × y³

Answer

Answer: (c) -1760 x9 × y³
Hint:
4th term in (x – 2y)12 = T4
= T3+1
= 12C3 (x)12-3 ×(-2y)³
= 12C3 x9 ×(-8y³)
= {(12×11×10)/(3×2×1)} × x9 ×(-8y³)
= -(2×11×10×8) × x9 × y³
= -1760 x9 × y³


MCQs On Binomial Theorem Question 4.
If n is a positive integer, then (√3+1)2n+1 + (√3−1)2n+1 is
(a) an even positive integer
(b) a rational number
(c) an odd positive integer
(d) an irrational number

Answer

Answer: (d) an irrational number
Hint:
Since n is a positive integer, assume n = 1
(√3+1)³ + (√3−1)³
= {3√3 + 1 + 3√3(√3 + 1)} + {3√3 – 1 – 3√3(√3 – 1)}
= 3√3 + 1 + 9 + 3√3 + 3√3 – 1 – 9 + 3√3
= 12√3, which is an irrational number.


Binomial Theorem Class 11 MCQ Question 5.
If the third term in the binomial expansion of (1 + x)m is (-1/8)x² then the rational value of m is
(a) 2
(b) 1/2
(c) 3
(d) 4

Answer

Answer: (b) 1/2
Hint:
(1 + x)m = 1 + mx + {m(m – 1)/2}x² + ……..
Now, {m(m – 1)/2}x² = (-1/8)x²
⇒ m(m – 1)/2 = -1/8
⇒ 4m² – 4m = -1
⇒ 4m² – 4m + 1 = 0
⇒ (2m – 1)² = 0
⇒ 2m – 1 = 0
⇒ m = 1/2


Binomial Theorem MCQ Pdf Question 6.
The greatest coefficient in the expansion of (1 + x)10 is
(a) 10!/(5!)
(b) 10!/(5!)²
(c) 10!/(5! × 4!)²
(d) 10!/(5! × 4!)

Answer

Answer: (b) 10!/(5!)²
Hint:
The coefficient of xr in the expansion of (1 + x)10 is 10Cr and 10Cr is maximum for
r = 10/ = 5
Hence, the greatest coefficient = 10C5
= 10!/(5!)²


Binomial Theorem MCQs With Answers Pdf Question 7.
The coefficient of xn in the expansion of (1 – 2x + 3x² – 4x³ + ……..)-n is
(a) (2n)!/n!
(b) (2n)!/(n!)²
(c) (2n)!/{2×(n!)²}
(d) None of these

Answer

Answer: (b) (2n)!/(n!)²
Hint:
We have,
(1 – 2x + 3x² – 4x³ + ……..)-n = {(1 + x)-2}-n
= (1 + x)2n
So, the coefficient of xnC3 = 2nCn = (2n)!/(n!)²


Binomial Theorem MCQ With Solution Question 8.
The value of n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively is
(a) 2
(b) 4
(c) 6
(d) 8

Answer

Answer: (c) 6
Hint:
Given that the first three terms of the expansion are 729, 7290 and 30375 respectively.
Now T1 = nC0 × an-0 × b0 = 729
⇒ an = 729 ……………. 1
T2 = nC1 × an-1 × b1 = 7290
⇒ n
an-1 × b = 7290 ……. 2
T3 = nC2 × an-2 × b² = 30375
⇒ {n(n-1)/2}
an-2 × b² = 30375 ……. 3
Now equation 2/equation 1
n
an-1 × b/an = 7290/729
⇒ n×b/n = 10 ……. 4
Now equation 3/equation 2
{n(n-1)/2}
an-2 × b² /n
an-1 × b = 30375/7290
⇒ b(n-1)/2a = 30375/7290
⇒ b(n-1)/a = (30375×2)/7290
⇒ bn/a – b/a = 60750/7290
⇒ 10 – b/a = 6075/729 (60750 and 7290 is divided by 10)
⇒ 10 – b/a = 25/3 (6075 and 729 is divided by 243)
⇒ 10 – 25/3 = b/a
⇒ (30-25)/3 = b/a
⇒ 5/3 = b/a
⇒ b/a = 5/3 …………….. 5
Put this value in equation 4, we get
n × 5/3 = 10
⇒ 5n = 30
⇒ n = 30/5
⇒ n = 6
So, the value of n is 6


Maths MCQs For Class 11 With Answers Pdf Question 9.
If α and β are the roots of the equation x² – x + 1 = 0 then the value of α2009 + β2009 is
(a) 0
(b) 1
(c) -1
(d) 10

Answer

Answer: (b) 1
Hint:
Given, x² – x + 1 = 0
Now, by Shridharacharya formula, we get
x = {1 ± √(1 – 4×1×1) }/2
⇒ x = {1 ± √(1 – 4) }/2
⇒ x = {1 ± √(-3)}/2
⇒ x = {1 ± √(3 × -1)}/2
⇒ x = {1 ± √3 × √-1}/2
⇒ x = {1 ± i√3}/2 {since i = √-1}
⇒ x = {1 + i√3}/2, {–1 – i√3}/2
⇒ x = -{-1 – i√3}/2, -{-1 + i√3}/2
⇒ x = w, w² {since w = {-1 + i√3}/2 and w² = {-1 – i√3}/2 }
Hence, α = -w, β = w²
Again we know that w³ = 1 and 1 + w + w² = 0
Now, α2009 + β2009 = α2007 × α² + β2007 × β²
= (-w)2007 × (-w)² + (-w²)2007 × (-w²)² {since 2007 is multiple of 3}
= -(w)2007 × (w)² – (w²)2007 × (w4)
= -1 × w² – 1 × w³ × w
= -1 × w² – 1 × 1 × w
= -w² – w
= 1 {since 1 + w + w² = 0}
So, α2009 + β2009 = 1


MCQ Questions For Class 11 Maths With Answers Pdf Question 10.
The general term of the expansion (a + b)n is
(a) Tr+1 = nCr × ar × br
(b) Tr+1 = nCr × ar × bn-r
(c) Tr+1 = nCr × an-r × bn-r
(d) Tr+1 = nCr × an-r × br

Answer

Answer: (d) Tr+1 = nCr × an-r × br
Hint:
The general term of the expansion (a + b)n is
Tr+1 = nCr × an-r × br


MCQ Questions For Class 11 Maths With Answers Question 11.
The coefficient of xn in the expansion (1 + x + x² + …..)-n is
(a) 1
(b) (-1)n
(c) n
(d) n+1

Answer

Answer: (b) (-1)n
Hint:
We know that
(1 + x + x² + …..)-n = (1 – x)-n
Now, the coefficient of x = (-1)n × nCn
= (-1)n


Question 12.
If n is a positive integer, then (√5+1)2n + 1 − (√5−1)2n + 1 is
(a) an odd positive integer
(b) not an integer
(c) none of these
(d) an even positive integer

Answer

Answer: (b) not an integer
Hint:
Since n is a positive integer, assume n = 1
(√5+1)² + 1 − (√5−1)² + 1
= (5 + 2√5 + 1) + 1 – (5 – 2√5 + 1) + 1 {since (x+y)² = x² + 2xy + y²}
= 4√5 + 2, which is not an integer


Question 13.
In the expansion of (a + b)n, if n is even then the middle term is
(a) (n/2 + 1)th term
(b) (n/2)th term
(c) nth term
(d) (n/2 – 1)th term

Answer

Answer: (a) (n/2 + 1)th term
Hint:
In the expansion of (a + b)n,
if n is even then the middle term is (n/2 + 1)th term


Question 14.
In the expansion of (a + b)n, if n is odd then the number of middle term is/are
(a) 0
(b) 1
(c) 2
(d) More than 2

Answer

Answer: (c) 2
Hint:
In the expansion of (a + b)n,
if n is odd then there are two middle terms which are
{(n + 1)/2}th term and {(n+1)/2 + 1}th term


Question 15.
if n is a positive ineger then 23nn – 7n – 1 is divisible by
(a) 7
(b) 9
(c) 49
(d) 81

Answer

Answer: (c) 49
Hint:
Given, 23n – 7n – 1 = 23×n – 7n – 1
= 8n – 7n – 1
= (1 + 7)n – 7n – 1
= {nC0 + nC1 7 + nC2 7² + …….. + nCn 7n} – 7n – 1
= {1 + 7n + nC2 7² + …….. + nCn 7n} – 7n – 1
= nC2 7² + …….. + nCn 7n
= 49(nC2 + …….. + nCn 7n-2)
which is divisible by 49
So, 23n – 7n – 1 is divisible by 49


Question 16.
In the binomial expansion of (71/2 + 51/3)37, the number of integers are
(a) 2
(b) 4
(c) 6
(d) 8

Answer

Answer: (c) 6
Hint:
Given, (71/2 + 51/3)37
Now, general term of this binomial Tr+1 = 37Cr × (71/2)37-r × (51/3)r
⇒ Tr+1 = 37Cr × 7(37-r)/2 × (5)r/3
This General term will be an integer if 37Cr is an integer, 7(37-r)/2 is an integer and (5)r/3 is an integer.
Now, 37Cr will always be a positive integer.
Since 37Cr denotes the number of ways of selecting r things out of 37 things, it can not be a fraction.
So, 37Cr is an integer.
Again, 7(37-r)/2Cr will be an integer if (37 – r)/2 is an integer.
So, r = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37 …………. 1
And if (5)r/3 is an integer, then r/3 should be an integer.
So, r = 0, 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36 ………….2
Now, take intersection of 1 and 2, we get
r = 3, 9, 15, 21, 27, 33
So, total possible value of r is 6
Hence, there are 6 integers are in the binomial expansion of (71/2 + 51/3)37


Question 17.
The number of ordered triplets of positive integers which are solution of the equation x + y + z = 100 is
(a) 4815
(b) 4851
(c) 8451
(d) 8415

Answer

Answer: (b) 4851
Hint:
Given, x + y + z = 100;
where x ≥ 1, y ≥ 1, z ≥ 1
Let u = x – 1, v = y – 1, w = z – 1
where u ≥ 0, v ≥ 0, w ≥ 0
Now, equation becomes
u + v + w = 97
So, the total number of solution = 97+3-1C3-1
= 99C2
= (99 × 98)/2
= 4851


Question 18.
The greatest coefficient in the expansion of (1 + x)10 is
(a) 10!/(5!)
(b) 10!/(5!)²
(c) 10!/(5! × 4!)²
(d) 10!/(5! × 4!)

Answer

Answer: (b) 10!/(5!)²
Hint:
The coefficient of xr in the expansion of (1 + x)10 is 10Cr and 10Cr is maximum for
r = 10/2 = 5
Hence, the greatest coefficient = 10C5
= 10!/(5!)²


Question 19.
If the third term in the binomial expansion of (1 + x)m is (-1/8)x² then the rational value of m is
(a) 2
(b) 1/2
(c) 3
(d) 4

Answer

Answer: (b) 1/2
Hint:
(1 + x)m = 1 + mx + {m(m – 1)/2}x² + ……..
Now, {m(m – 1)/2}x² = (-1/8)x²
⇒ m(m – 1)/2 = -1/8
⇒ 4m² – 4m = -1
⇒ 4m² – 4m + 1 = 0
⇒ (2m – 1)² = 0
⇒ 2m – 1 = 0
⇒ m = 1/2


Question 20.
In the binomial expansion of (a + b)n, the coefficient of fourth and thirteenth terms are equal to each other, then the value of n is
(a) 10
(b) 15
(c) 20
(d) 25

Answer

Answer: (b) 15
Hint:
Given, in the binomial expansion of (a + b)n, the coefficient of fourth and thirteenth terms are equal to each other
nC3 = nC12
This is possible when n = 15
Because 15C13 = 15C12


We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 8 Binomial Theorem with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Binomial Theorem MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations with Answers

Check the below NCERT MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have provided Permutations and Combinations Class 11 Maths MCQs Questions with Answers to help students understand the concept very well. https://mcqquestions.guru/mcq-questions-for-class-11-maths-chapter-7/

Permutations and Combinations Class 11 MCQs Questions with Answers

MCQ Questions On Permutation And Combination Class 11 Question 1.
There are 12 points in a plane out of which 5 are collinear. The number of triangles formed by the points as vertices is
(a) 185
(b) 210
(c) 220
(d) 175

Answer

Answer: (b) 210
Hint:
Total number of triangles that can be formed with 12 points (if none of them are collinear)
= 12C3
(this is because we can select any three points and form the triangle if they are not collinear)
With collinear points, we cannot make any triangle (as they are in straight line).
Here 5 points are collinear. Therefore we need to subtract 5C3 triangles from the above count.
Hence, required number of triangles = 12C35C3 = 220 – 10 = 210


Permutation And Combination Class 11 MCQ Question 2.
The number of combination of n distinct objects taken r at a time be x is given by
(a) n/2Cr
(b) n/2Cr/2
(c) nCr/2
(d) nCr

Answer

Answer: (d) nCr
Hint:
The number of combination of n distinct objects taken r at a time be x is given by
nCr = n!/{(n – r)! × r!}
Let the number of combination of n distinct objects taken r at a time be x.
Now consider one of these n ways. There are e objects in this selection which can be arranged in r! ways.
So, each of the x combinations gives rise to r! permutations. So, x combinations will give rise to x×(r!).
Consequently, the number of permutations of n things, taken r at a time is x×(r!) and it is equal to nPr
So, x×(r!) = nPr
⇒ x×(r!) = n!/(n – r)!
⇒ x = n!/{(n – r)! × r!}
nCr = n!/{(n – r)! × r!}


MCQ On Permutation And Combination Class 11 Question 3.
Four dice are rolled. The number of possible outcomes in which at least one dice show 2 is
(a) 1296
(b) 671
(c) 625
(d) 585

Answer

Answer: (b) 671
Hint:
No. of ways in which any number appearing in one dice = 6
No. of ways in which 2 appear in one dice = 1
No. of ways in which 2 does not appear in one dice = 5
There are 4 dice.
Getting 2 in at least one dice = Getting any number in all the 4 dice – Getting not 2 in any of the 4 dice.
= (6×6×6×6) – (5×5×5×5)
= 1296 – 625
= 671


Permutation And Combination MCQ Question 4.
If repetition of the digits is allowed, then the number of even natural numbers having three digits is
(a) 250
(b) 350
(c) 450
(d) 550

Answer

Answer: (c) 450
Hint:
In a 3 digit number, 1st place can be filled in 5 different ways with (0, 2, 4, 6, 8)
10th place can be filled in 10 different ways.
100th place can be filled in 9 different ways.
So, the total number of ways = 5 × 10 × 9 = 450


MCQ On Permutation And Combination Question 5.
The number of ways in which 8 distinct toys can be distributed among 5 children is
(a) 58
(b) 85
(c) 8P5
(d) 5P5

Answer

Answer: (a) 58
Hint:
Total number of toys = 8
Total number of children = 5
Now, each toy can be distributed in 5 ways.
So, total number of ways = 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5
= 58


Permutation And Combination MCQ Class 11 Question 6.
The value of P(n, n – 1) is
(a) n
(b) 2n
(c) n!
(d) 2n!

Answer

Answer: (c) n!
Hint:
Given,
Given, P(n, n – 1)
= n!/{(n – (n – 1)}
= n!/(n – n + 1)}
= n!
So, P(n, n – 1) = n!


Class 11 Maths Chapter 7 MCQ With Answers Question 7.
In how many ways can 4 different balls be distributed among 5 different boxes when any box can have any number of balls?
(a) 54 – 1
(b) 54
(c) 45 – 1
(d) 45

Answer

Answer: (b) 54
Hint:
Here, both balls and boxes are different.
Now, 1st ball can be placed into any of the 5 boxes.
2nd ball can be placed into any of the 5 boxes.
3rd ball can be placed into any of the 5 boxes.
4th ball can be placed into any of the 5 boxes.
So, the required number of ways = 5 × 5 × 5 × 5 = 54


Permutations And Combinations MCQ Question 8.
The number of ways of painting the faces of a cube with six different colors is
(a) 1
(b) 6
(c) 6!
(d) None of these

Answer

Answer: (a) 1
Hint:
Since the number of faces is same as the number of colors,
therefore the number of ways of painting them is 1


MCQ Of Chapter 7 Maths Class 11 Question 9.
Out of 5 apples, 10 mangoes and 13 oranges, any 15 fruits are to be distributed among 2 persons. Then the total number of ways of distribution is
(a) 1800
(b) 1080
(c) 1008
(d) 8001

Answer

Answer: (c) 1008
Hint:
Given there are 5 apples, 10 mangoes and 13 oranges.
Let x1 is for apple, x2 is for mango and x3 is for orange.
Now, first we have to select total 15 fruits out of them.
x1 + x2 + x3 = 15 (where 0 ⇐ x1 ⇐ 5, 0 ⇐ x2 ⇐ 10, 0 ⇐ x3 ⇐ 13)
= (x0 + x1 + x2 +………+ x5)×(x0 + x1 + x2 +………+ x110)×(x0 + x1 + x2 +………+ x13)
= {(1- x6)/(1 – x)}×{(1- x11)/(1 – x)}×{(1- x14)/(1 – x)}
= {(1- x6)×(1- x11)×{(1- x14)}/(1 – x)³
= {(1- x6)×(1- x11)×{(1- x14)} × ∑3+r+1Cr × xr
= {(1- x11 – x6 + x17)×{(1- x14)} × ∑3+r+1Cr × xr
= {(1- x11 – x6 + x17 – x14 + x25 + x20 – x31)} × ∑2+rCr × xr
= 1 × ∑2+rCr × xr – x11 × ∑2+rCr × xr – x6 × ∑2+rCr × xr + x17 × ∑2+rCr × xr – x14 × ∑2+rCr × xr + x25 × ∑2+rCr × xr + x20 × ∑2+rCr × xr – x31 × ∑2+rCr × xr
= ∑2+rCr × xr – ∑2+rCr × xr+11 – ∑2+rCr × xr+6 + ∑2+rCr × xr+17 – ∑2+rCr × xr+14 + ∑2+rCr × xr+25 + ∑2+rCr × xr+20 – ∑2+rCr × xr+25
Now we have to find co-efficeient of x15
= 2+15C152+4C42+9C92+1C1 (rest all terms have greater than x15, so its coefficients are 0)
= 17C156C411C93C1
= 17C26C211C23C1
= {(17×16)/2} – {(6×5)/2} – {(11×10)/2} – 3
= (17×8) – (3×5) – (11×5) – 3
= 136 – 15 – 55 – 3
= 136 – 73
= 63
Again we have to distribute 15 fruits between 2 persons.
So x1 + x2 = 15
= 2-1+15C15
= 16C15
= 16C1
= 16
Now total number of ways of distribution = 16 × 63 = 1008


Permutation And Combination MCQs With Answers Question 10.
6 men and 4 women are to be seated in a row so that no two women sit together. The number of ways they can be seated is
(a) 604800
(b) 17280
(c) 120960
(d) 518400

Answer

Answer: (a) 604800
Hint:
6 men can be sit as
× M × M × M × M × M × M ×
Now, there are 7 spaces and 4 women can be sit as 7P4 = 7P3 = 7!/3! = (7 × 6 × 5 × 4 × 3!)/3!
= 7 × 6 × 5 × 4 = 840
Now, total number of arrangement = 6! × 840
= 720 × 840
= 604800


MCQ Of Permutation And Combination Class 11 Question 11.
The number of ways can the letters of the word ASSASSINATION be arranged so that all the S are together is
(a) 152100
(b) 1512
(c) 15120
(d) 151200

Answer

Answer: (d) 151200
Hint:
Given word is : ASSASSINATION
Total number of words = 13
Number of A : 3
Number of S : 4
Number of I : 2
Number of N : 2
Number of T : 1
Number of O : 1
Now all S are taken together. So it forms a single letter.
Now total number of words = 10
Now number of ways so that all S are together = 10!/(3!×2!×2!)
= (10×9×8×7×6×5×4×3!)/(3! × 2×2)
= (10×9×8×7×6×5×4)/(2×2)
= 10×9×8×7×6×5
= 151200
So total number of ways = 151200


MCQs On Permutation And Combination Question 12.
If repetition of the digits is allowed, then the number of even natural numbers having three digits is
(a) 250
(b) 350
(c) 450
(d) 550

Answer

Answer: (c) 450
Hint:
In a 3 digit number, 1st place can be filled in 5 different ways with (0, 2, 4, 6, 8)
10th place can be filled in 10 different ways.
100th place can be filled in 9 different ways.
So, the total number of ways = 5 × 10 × 9 = 450


MCQs On Permutations And Combinations Class 11 Question 13.
Let Tn denote the number of triangles which can be formed using the vertices of a regular polygon on n sides. If Tn+1 – Tn = 21, then n equals
(a) 5
(b) 7
(c) 6
(d) 4

Answer

Answer: (b) 7
Hint:
The number of triangles that can be formed using the vertices of a regular polygon = nC3
Given, Tn+1 – Tn = 21
n+1C3nC3 = 21
nC2 + nC3nC3 = 21 {since n+1Cr = nCr-1 + nCr}
nC2 = 21
⇒ n(n – 1)/2 = 21
⇒ n(n – 1) = 21×2
⇒ n² – n = 42
⇒ n² – n – 42 = 0
⇒ (n – 7)×(n + 6) = 0
⇒ n = 7, -6
Since n can not be negative,
So, n = 7


Permutations And Combinations MCQ Class 11 Question 14.
How many ways are here to arrange the letters in the word GARDEN with the vowels in alphabetical order?
(a) 120
(b) 240
(c) 360
(d) 480

Answer

Answer: (c) 360
Hint:
Given word is GARDEN.
Total number of ways in which all letters can be arranged in alphabetical order = 6!
There are 2 vowels in the word GARDEN A and E.
So, the total number of ways in which these two vowels can be arranged = 2!
Hence, required number of ways = 6!/2! = 720/2 = 360


Permutations And Combinations Class 11 MCQ Questions Question 15.
How many factors are 25 × 36 × 52 are perfect squares
(a) 24
(b) 12
(c) 16
(d) 22

Answer

Answer: (a) 24
Hint:
Any factors of 25 × 36 × 52 which is a perfect square will be of the form 2a × 3b × 5c
where a can be 0 or 2 or 4, So there are 3 ways
b can be 0 or 2 or 4 or 6, So there are 4 ways
a can be 0 or 2, So there are 2 ways
So, the required number of factors = 3 × 4 × 2 = 24


Permutation And Combination Class 11 Extra Questions With Answers Question 16.
A student is to answer 10 out of 13 questions in an examination such that he must choose at least 4 from the first five questions. The number of choices available to him is
(a) 40
(b) 196
(c) 280
(d) 346

Answer

Answer: (b) 196
Hint:
There are two cases
1. When 4 is selected from the first 5 and rest 6 from remaining 8
Total arrangement = 5C4 × 8C6
= 5C1 × 8C2
= 5 × (8×7)/(2×1)
= 5 × 4 × 7
= 140
2. When all 5 is selected from the first 5 and rest 5 from remaining 8
Total arrangement = 5C5 × 8C5
= 1 × 8C3
= (8×7×6)/(3×2×1)
= 8×7
= 56
Now, total number of choices available = 140 + 56 = 196


Permutation And Combination MCQs With Answers Pdf Question 17.
Four dice are rolled. The number of possible outcomes in which at least one dice show 2 is
(a) 1296
(b) 671
(c) 625
(d) 585

Answer

Answer: (b) 671
Hint:
No. of ways in which any number appearing in one dice = 6
No. of ways in which 2 appear in one dice = 1
No. of ways in which 2 does not appear in one dice = 5
There are 4 dice.
Getting 2 in at least one dice = Getting any number in all the 4 dice – Getting not 2 in any of the 4 dice.
= (6×6×6×6) – (5×5×5×5)
= 1296 – 625
= 671


Question 18.
In how many ways in which 8 students can be sated in a line is
(a) 40230
(b) 40320
(c) 5040
(d) 50400

Answer

Answer: (b) 40320
Hint:
The number of ways in which 8 students can be sated in a line = 8P8
= 8!
= 40320


Question 19.
The number of squares that can be formed on a chess board is
(a) 64
(b) 160
(c) 224
(d) 204

Answer

Answer: (d) 204
Hint:
A chess board contains 9 lines horizontal and 9 lines perpendicular to them.
To obtain a square, we select 2 lines from each set lying at equal distance and this equal
distance may be 1, 2, 3, …… 8 units, which will be the length of the corresponding square.
Now, two lines from either set lying at 1 unit distance can be selected in 8C1 = 8 ways.
Hence, the number of squares with 1 unit side = 8²
Similarly, the number of squares with 2, 3, ….. 8 unit side will be 7², 6², …… 1²
Hence, total number of square = 8² + 7² + ……+ 1² = 204


Question 20.
How many 3-letter words with or without meaning, can be formed out of the letters of the word, LOGARITHMS, if repetition of letters is not allowed
(a) 720
(b) 420
(c) none of these
(d) 5040

Answer

Answer: (a) 720
Hint:
The word LOGARITHMS has 10 different letters.
Hence, the number of 3-letter words(with or without meaning) formed by using these letters
= 10P3
= 10 ×9 ×8
= 720


We hope the given NCERT MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations with Answers Pdf free download will help you. If you have any queries regarding CBSE Class 11 Maths Permutations and Combinations MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.

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

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. https://mcqquestions.guru/mcq-questions-for-class-11-maths-chapter-6/

Linear Inequalities Class 11 MCQs Questions with Answers

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.

MCQ Questions for Class 10 Maths Chapter 2 Polynomials with Answers

Check the below NCERT MCQ Questions for Class 10 Maths Chapter 2 Polynomials with Answers Pdf free download. MCQ Questions for Class 10 Maths with Answers were prepared based on the latest exam pattern. We have provided Polynomials Class 10 Maths MCQs Questions with Answers to help students understand the concept very well. https://mcqquestions.guru/mcq-questions-for-class-10-maths-chapter-2/

Students can also refer to NCERT Solutions for Class 10 Maths Chapter 2 Polynomials for better exam preparation and score more marks.

Polynomials Class 10 MCQs Questions with Answers

Question 1.
The maximum number of zeroes that a polynomial of degree 4 can have is
(a) One
(b) Two
(c) Three
(d) Four

Answer

Answer: (d) Four


Question 2.
The graph of the polynomial p(x) = 3x – 2 is a straight line which intersects the x-axis at exactly one point namely
(a) (\(\frac{-2}{3}\), 0)
(b) (0, \(\frac{-2}{3}\))
(c) (\(\frac{2}{3}\), 0)
(d) \(\frac{2}{3}\), \(\frac{-2}{3}\)

Answer

Answer: (c) (\(\frac{2}{3}\), 0)


Question 3.
In fig. given below, the number of zeroes of the polynomial f(x) is
MCQ Questions for Class 10 Maths Chapter 2 Polynomials with Answers
(a) 1
(b) 2
(c) 3
(d) None

Answer

Answer: (c) 3


Question 4.
The graph of the polynomial ax² + bx + c is an upward parabola if
(a) a > 0
(b) a < 0
(b) a = 0
(d) None

Answer

Answer: (a) a > 0


Question 5.
The graph of the polynomial ax² + bx + c is a downward parabola if
(a) a > 0
(b) a < 0
(c) a = 0
(d) a = 1

Answer

Answer: (b) a < 0


Question 6.
A polynomial of degree 3 is called
(a) a linear polynomial
(b) a quadratic polynomial
(c) a cubic polynomial
(d) a biquadratic polynomial

Answer

Answer: (c) a cubic polynomial


Question 7.
If α, β are the zeroes of the polynomial x² – 16, then αβ(α + β) is
(a) 0
(b) 4
(c) -4
(d) 16

Answer

Answer: (a) 0


Question 8.
If α and \(\frac{1}{α}\) are the zeroes of the polynomial ax² + bx + c, then value of c is
(a) 0
(b) a
(c) -a
(d) 1

Answer

Answer: (b) a


Question 9.
Zeroes of the polynomial x² – 11 are
(a) ±\(\sqrt{17}\)
(b) ±\(\sqrt{3}\)
(c) 0
(d) None

Answer

Answer: (a) ±\(\sqrt{17}\)


Question 10.
If α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d then α + β + γ is equal
(a) \(\frac{-b}{a}\)
(b) \(\frac{b}{a}\)
(c) \(\frac{c}{a}\)
(d) \(\frac{d}{a}\)

Answer

Answer: (a) \(\frac{-b}{a}\)


Question 11.
If α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d then αβ + βγ + αγ is equal to
(a) \(\frac{-b}{a}\)
(b) \(\frac{b}{a}\)
(c) \(\frac{c}{a}\)
(d) \(\frac{d}{a}\)

Answer

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


Question 12.
If the zeroes of the polynomial x³ – 3x² + x – 1 are \(\frac{s}{t}\), s and st then value of s is
(a) 1
(b) -1
(c) 2
(d) -3

Answer

Answer: (a) 1


Question 13.
If the sum of the zeroes of the polynomial f(x) = 2x³ – 3kx² + 4x – 5 is 6, then the value of k is
(a) 2
(b) 4
(c) -2
(d) -4

Answer

Answer: (b) 4


Question 14.
If a polynomial of degree 4 is divided by quadratic polynomial, the degree of the remainder is
(a) ≤ 1
(b) ≥ 1
(c) 2
(d) 4

Answer

Answer: (a) ≤ 1


Question 15.
If a – b, a and a + b are zeroes of the polynomial fix) = 2x³ – 6x² + 5x – 7, then value of a is
(a) 1
(b) 2
(c) -5
(d) 7

Answer

Answer: (a) 1


Question 16.
Dividend is equal to
(a) divisor × quotient + remainder
(b) divisior × quotient
(c) divisior × quotient – remainder
(d) divisor × quotient × remainder

Answer

Answer: (a) divisor × quotient + remainder


Question 17.
A quadratic polynomial whose sum of the zeroes is 2 and product is 1 is given by
(a) x² – 2x + 1
(b) x² + 2x + 1
(c) x² + 2x – 1
(d) x² – 2x – 1

Answer

Answer: (a) x² – 2x + 1


Question 18.
If one of the zeroes of a quadratic polynomial ax² + bx + c is 0, then the other zero is
(a) \(\frac{-b}{a}\)
(b) 0
(c) \(\frac{b}{a}\)
(d) \(\frac{-c}{a}\)

Answer

Answer: (a) \(\frac{-b}{a}\)


Question 19.
The sum and the product of the zeroes of polynomial 6x² – 5 respectively are
(a) 0, \(\frac{-6}{5}\)
(b) 0, \(\frac{6}{5}\)
(c) 0, \(\frac{5}{6}\)
(d) 0, \(\frac{-5}{6}\)

Answer

Answer: (d) 0, \(\frac{-5}{6}\)


Question 20.
What should be subtracted from x³ – 2x² + 4x + 1 to get 1?
(a) x³ – 2x² + 4x
(b) x³ – 2x² + 4 + 1
(c) -1
(d) 1

Answer

Answer: (a) x³ – 2x² + 4x


We hope the given NCERT MCQ Questions for Class 10 Maths Chapter 2 Polynomials with Answers Pdf free download will help you. If you have any queries regarding Polynomials CBSE Class 10 Maths MCQs Multiple Choice Questions with Answers, drop a comment below and we will get back to you soon.