Question

Class 11 Maths MCQ Questions of Binomial Theorem with Answers?

Answer 1

Practice the Important MCQ Questions for Class 11, which are given here. In this part, students can figure out how to determine the answer for the Binomial Theorem. Clear every one of the essentials and prepare altogether for the test-taking assistance from Class 11 Maths Binomial Theorem Objective Questions.

Practicing the MCQ Questions for Class 11 Maths with answers will boost your certainty consequently assisting you with scoring admirably in the exam. Students are encouraged to solve the Class 11 Maths MCQ Questions of Binomial Theorem with Answers to know various ideas and concepts.

Practice MCQ Questions for class 11 Maths Chapter-Wise

1. The coefficient of y in the expansion of (y² + c/y)⁵ is

(a) 10c
(b) 10c²
(c) 10c³
(d) None of these

2. (1.1)¹⁰⁰⁰⁰ is _____ 1000

(a) greater than
(b) less than
(c) equal to
(d) None of these

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

4. 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

5. 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!)

6. 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

7. The coefficient of xn in the expansion (1 + x + x² + …..)^-n is

(a) 1
(b) (-1)ⁿ
(c) n
(d) n+1

8. 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

9. 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

10. if n is a positive ineger then 2³ⁿn – 7n – 1 is divisible by

(a) 7
(b) 9
(c) 49
(d) 81

11. The coefficient of the middle term in the expansion of (2+3x)⁴ is:

(a) 5!
(b) 6
(c) 216
(d) 8!

12. The value of (126)^1/3 up to three decimal places is

(a) 5.011
(b) 5.012
(c) 5.013
(d) 5.014

13. The coefficient of x³y⁴ in (2x+3y²)⁵ is

(a) 360
(b) 720
(c) 240
(d) 1080

14. The integral part of \((8+3\sqrt7)^n\) is

(a) an odd integer
(b) an even integer
(c) zero
(d) nothing can be said.

15. If the third term in the expansion of \([x+x^{log_{10}}\;x]^5\),is 10⁶ then x may be

(a) 1
(b) \(\sqrt{10}\)
(c) 10
(d) 10^-2/5

16. The number of terms in the expansion of (y^1/5+x^1/10)⁵⁵, in which powers of x and y are free from radical signs are

(a) six
(b) twelve
(c) seven
(d) five

17. If x = 99⁵⁰ +100⁵⁰ and y= (101)⁵⁰ then

(a) x = y
(b) x<y 
(c) x>y
(d) None of these

18. If A and B are coefficients of xⁿ in the expansions of (1+x)²ⁿ and (1+x)²ⁿ⁻¹ respectively, then A/B is equal to

(a) 4
(b) 2
(c) 9
(d) 6

19. In the expansion of (1 + x)⁵⁰, the sum of the coefficients of odd powers of x is 

(a) 2²⁶
(b) 2⁴⁹
(c) 2⁵⁰
(d) 2⁵¹

20. Find an approximate value of (0.99)5 using the first three terms of its binomial expansion.

(a) 0.591
(b) 0.951
(c) 0.195
(d) None

Answer 2

1. Answer: (c) 10c³

Explanation: Given, binomial expression is (y² + c/y)⁵

Now, T_r+1 = 5Cr × (y²)^5-r × (c/y)r

= 5Cr × y^10-3r × Cr

Now, 10 – 3r = 1

⇒ 3r = 9

⇒ r = 3

So, the coefficient of y = 5C³ × c³ = 10c³

2. Answer: (a) greater than

Explanation: 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

3. Answer: (c) -1760 x⁹ × y³

Explanation: 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

4. Answer: (b) 1/2

Explanation: (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

5. Answer: (b) 10!/(5!)²

Explanation: The coefficient of xr in the expansion of (1 + x)¹⁰ is ¹⁰C_r and ¹⁰C_r is maximum for

r = 10/ 2

= 5

Hence, the greatest coefficient = ¹⁰C₅

= 10!/(5!)²

6. Answer: (b) (2n)!/(n!)²

Explanation: (1 – 2x + 3x² – 4x³ + ……..)^-n = {(1 + x)^-2}^-n

= (1 + x)²ⁿ

So, the coefficient of xnC₃ = ²ⁿC_n = (2n)!/(n!)²

7. Answer: (b) (-1)ⁿ

Explanation: (1 + x + x² + …..)-n = (1 – x)^-n

Now, the coefficient of x = (-1)ⁿ × ⁿC_n

= (-1)ⁿ

8. Answer: (c) 2

Explanation: 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.

9. Answer: (b) 4851

Explanation: 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+3-1C_3-1

= ⁹⁹C₂

= (99 × 98)/2

= 4851

10. Answer: (c) 49

Explanation: 2³ⁿ – 7n – 1 

= 2^{3 × n} – 7n – 1

= 8ⁿ – 7n – 1

= (1 + 7)ⁿ – 7n – 1

= {ⁿC₀ + ⁿC₁ 7 + ⁿC₂ 7² + …….. + ⁿC_n 7ⁿ} – 7n – 1

= {1 + 7n + nC₂ 7² + …….. + ⁿC_n 7ⁿ} – 7n – 1

= ⁿC₂ 7² + …….. + ⁿC_n 7ⁿ

= 49(ⁿC₂ + …….. + ⁿC_n 7^n-2) 

which is divisible by 49

So, 2³ⁿ – 7n – 1 is divisible by 49

Answer 3

11. Answer: (c) 216

Explanation: If the exponent of the expression is n, then the total number of terms is n+1.

Hence, the total number of terms is 4+1 = 5.

Hence, the middle term is the 3rd term.

Therefore, T₃ = ⁴C₂.(2)².(3x)²

T₃ = (6).(4).(9x²)

T₃ = 216x².

Therefore, the coefficient of the middle term is 216.

12. Answer: (c) 5.013

Explanation: (126)^1/3 can also be written as the cube root of 126.

Hence, (126)^1/3 is approximately equal to 5.013.

Hence, option (c) 5.013 is the correct answer.

13. Answer: (b) 720

Explanation: Given: (2x+3y²)⁵

Therefore, the general form for the expression (2x+3y²)⁵ is T_r+1 = ⁵Cr. (2x)^r.(3y²)^5-r

Hence, T₃₊₁ = ⁵C₃ (2x)³.(3y²)^5-3

T₄ = ⁵C₃ (2x)³.(3y²)²

T₄ = ⁵C₃.8x³.9y⁴

On simplification, we get

T₄ = 720x³y⁴

Therefore, the coefficient of x³y⁴ in (2x+3y²)² is 720. 

14. Answer: (b) an even integer

Explanation: Let (8+\(3\sqrt{1}\))ⁿ = p+f, where p∈I and f is a proper fraction and let (8+\(3\sqrt 1\))ⁿ = f′, a proper  fraction [∵0<8−\(3\sqrt 7\)<1]

Since (8+\(3\sqrt 7\))ⁿ+(8−\(3\sqrt 7\))ⁿ = p+f+f′ is an even integer

∴p+1 is even

∴p is an odd integer

15. Answer: (c) 10

Explanation: Put log10x =y, the given expression becomes (x+xy)⁵.

T₃ = ⁵C₂.x³(xy)² = 10x^3+2y =10⁶

⇒(3+2y)log₁₀x = 5log₁₀ 10 = 5

⇒ (3+2y)y = 5

⇒ y=1,−5/2

⇒log₁₀ x = 1log₁₀ x =−5/2

16. Answer: (a) six

Explanation: Given expansion is (y^1/5+x^1/10)⁵⁵

The general term is

T_r+1= ⁵⁵Cr(y^1/5)^55−r. (x^1/10)r

T_r+1 would free from radical sign if powers of j and x are integers.

i.e \(\frac{55-r}{5}\) and r/10 are integer.

⇒ r is multiple of 10.

Hence, r = 0,10,20,30,40,50

It is an A.P.

Thus, 50 = 0+(k−1)10

50 =10k−10

k = 6

Thus, the six terms of the given expansion in which x andy are free from radical signs.

17. Answer: (b) x<y

Explanation: (101)⁵⁰ − (99)⁵⁰ =(100+1)⁵⁰ − (100−1)⁵⁰

=2[⁵⁰C₁(100)⁴⁹+⁵⁰C₃(100)⁴⁷+......+⁵⁰C₄₉(100)]

>2.⁵⁰C₁.(100)⁴⁹ =2×50(100)⁴⁹=(100)⁵⁰

⇒(101)⁵⁰ >(99)⁵⁰ +(100)⁵⁰

⇒ y>x

⇒ x<y.

18. Answer: (b) 2

Explanation: Given, A= Coefficient of xⁿ in (1+x)²ⁿ

and B= Coefficient of xⁿ in (1+x)²ⁿ⁻¹

∴ A = ²ⁿC_n and  B=²ⁿ⁻¹C_n

\(\therefore \frac{A}{B}=\frac{^{2n}C_n}{2n-1}C_n=\frac{2n}{n}\)

= 2. 

19. Answer: (b) 2⁴⁹

Explanation: C₀ + C₁ + C₂ + C₃ + C₄ + C₅ + …..C_n = 2n 

Therefore C₁ + C₃ + C₅ + ….. C_n = 1/2 × 2n 

= 1/2 × 2⁵⁰ 

= 2⁴⁹

20. Answer: (b) 0.951

Explanation: (0.99)5 =(1−0.01)⁵

= 1 - ⁵C₁ x (0.01) + ⁵C₂ x (0.01)²........

= 1 - 0.05 + 10 x 0.0001......

= 1.001 - 0.05

= 0.951

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