MP Board Class 11th Maths Important Questions Chapter 8 Binomial Theorem

MP Board Class 11th Maths Important Questions Chapter 8 Binomial Theorem

MP Board Class 11th Maths Important Questions Chapter 8 Binomial Theorem

Binomial Theorem Objective Type Questions

(A) Choose the correct option :

Question 1.
The total number of terms in the expansion of
(a) 7
(b) 12
(c) 13
(d) 6.
Answer:
(c) 13

Question 2.
If y = 3x + 6x² + 10x³ + …………… ∞, then the correct relation will be :

Question 3.
The total number of terms in the expansion of (1+ x)^-1 will be :
(a) 0
(b) ∞
(c) 2
(d) It can not be expand
Answer:
(b) ∞

Question 4.
The mid – term in the expansion of (x – 1/x)¹⁰ will be :
(a) – ¹⁰C₅
(b) ¹⁰C₅
(c) 251
(d) 252
Answer:
(a) – ¹⁰C₅

An online remainder theorem calculator allows you to determine the remainder of given polynomial expressions by remainder theorem.

Question 5.
For all positive integer of n, n(n – 1) is :
(a) Integer
(b) Natural number
(c) Even positive integer
(d) Odd positive integer.
Answer:
(c) Even positive integer

Question 6.
Expansion of (a + x)ⁿ is :
(a) aⁿ + ⁿC₁a^n-1x + ⁿC₂a^n-2x² + ……….. + ⁿC_ra^n-rx^r + ………… + aⁿ
(b) xⁿ + ⁿC₁x^n-1a +ⁿC₂x^n-2a² + ……….. + ⁿC_rx^n-ra^r + ………… + aⁿ
(c) aⁿ – ⁿC₁a^n-1x + ⁿC₂a^n-2x² + ……….. + (-1)^rnC_ra^n-rx^r + ………… + (-1)ⁿaⁿ
(d) xⁿ – ⁿC₁a^n-1a + ⁿC₂a^n-2a² + ……….. + (-1)^rnC_ra^n-rx^r + ………… + (-1)ⁿxⁿ
Answer:
(a) aⁿ + ⁿC₁a^n-1x +ⁿC₂a^n-2x² + ……….. + ⁿC_ra^n-rx^r + ………… + aⁿ

Question 7.

[Hint: The total number of terms be 11 in the expansion of it.
∵ The 5^th term from end = (11 – 5)^th = 7^th term from the beginning.]

Question 8.

Question 9.
The value of ⁿC₀ + ⁿC₁ + ⁿC₂ + …………….. + ⁿC_n :
(a) 2ⁿ + 1
(b) 2^{n – 1}
(c) 2ⁿ – 1
(d) 2ⁿ
Answer:
(d) 2ⁿ

Question 10.
The value of ⁿC₀ + ⁿC₂ + ⁿC₄ + …………….. = ⁿC₁ + ⁿC₃ + ……………. will be :
(a) 2ⁿ + 1
(b) 2^{n – 1}
(c) 2ⁿ – 1
(d) 2ⁿ
Answer:
(b) 2^{n – 1}

The Chebyshev’s Theorem calculator, above, will allow you to enter any value of k greater than 1.

Question 11.
The total number of terms in the expansion of (a + b + c + d)ⁿ will be :

Question 12.
The necessary condition for expansion of (1 + x)^-1 is :
(a) | x | < 1
(b) | x | > 1
(c) | x | = 1
(d) | x | = – 1.
Answer:
(a) | x | < 1

Question 13.
The general term in the expansion of (x + a)ⁿ will be :
(a) r^th
(b) (r+1)^thterm
(c) (r-1)^th
(d) (r+2)^thterm
Answer:
(b) (r+1)^thterm

Question 14.

Question 15.
The value of ⁿC₀ + ⁿC₁ + ⁿC₂ + …………….. = ⁿC_n the expansion of (l + x)ⁿ is :
(a) 2ⁿ – 1
(b) 2ⁿ – 2
(c) 2ⁿ
(d) 2^n-1
Answer:
(c) 2ⁿ

Question 16.
The value of ¹⁵C₀ + ¹⁵C₂ + ¹⁵C₄ + ¹⁵C₆ + …………….. = ¹⁵C₁₄ is :
(a) 2¹⁴
(b) 2¹⁵
(c) 2¹⁵ – 1
(d) None of these
Answer:
(a) 2¹⁴

(B) Match the following :

Answer:

1. (d)
2. (a)
3. (e)
4. (c)
5. (b)
6. (g)
7. (f)
8. (i)
9. (b)

(C) Fill in the blanks :

Answer:

(D) Write true / false :

Answer:

1. True
2. True
3. True
4. True
5. False
6. True
7. False
8. False
9. True
10. False
11. True
12. True.

(E) Write answer in one word / sentence :

Answer:

1. 997002999
2. – 20 x⁵
3. ⁿC_r x^{n – r} a^r
4. 28
5. (n + 1)
6. 3
7. 16x²

Binomial Theorem Long Answer Type Questions

Question 1.

Question 2.
Expand : (2x – 3)⁶ (NCERT)
Solution:

Question 3.

Question 4.

Question 5.

Question 6.

Solution:

Question 7.

Question 8.
If coefficient of x² and x³ in the expansion of (3 + ax)⁹ are equal, the value of a.
Solution:

Question 9.
Find the coefficient of x⁵ in the expansion of (x + 3)⁸
Solution:
Suppose x⁵ appears in (r + 1)^th term T_r+1 = nC₁x^n-ra^r
Here n = 8, x = x, a = 3
T_r+1 = ⁸C_r(x)^{8 – r}(3)^r
For the coefficient of x⁵,
8 – r = 5
=> r = 3
T₃₊₁ = ⁸C₃(3)³

= 1512 × x⁵
Hence coefficient of x⁵ is 1512.

Question 10.
Find the coefficient of a⁵b⁷ in the expansion of (a – 2b)¹².
Solution:

Question 11.
If the 17^th and 18^th terms in the expansion of (2 + a)⁵⁰ are equal, then find the value of a. (NCERT)
Solution:
In the expansion of (x + a)ⁿ
T_r+1 = ⁿC_r x^{n – r} a^r
Here n = 50, x = 2, a = a
T₁₇ = T_{16 + 1} = ⁵⁰C₁₆ (2)^{50 – 16} (a)¹⁶
⇒ T₁₇ = ⁵⁰C₁₆ (2)³⁴ (a)¹⁶
and T₁₈ = T_{17 + 1} = ⁵⁰C₁₇ (2)^{50 – 17} (a)¹⁷
= ⁵⁰C₁₇ (2)³³ (a)¹⁷

Question 12.
Prove that the value of the middle term in the expansion of (1+x)²ⁿ is 2ⁿ xⁿ
Solution:

Question 13.
In the expansion of (x + 1)ⁿ, the coefficient of the (r – 1)^th, r^th and (r + 1)^th terms are in the ratio 1 : 3 : 5, then find the value of n and r.
Solution:
In the expansion of (x + 1)ⁿ,
T_{r + 1} = ⁿC_rx^{n – r}(1)^r
T_{r – 1} = T_{r – 2 + 1} = ⁿC_{r – 2}(x)^{n – (r – 2)}(1)^{r – 2}
Coefficient of T_{r – 1}^th term = ⁿC_{r – 2}
T_r = T_{r – 1 + 1} = ⁿC_{r – 1}(x)^{n – (r – 1)}(1)^{r – 1}
Coefficient of T_r^thterm = ⁿC_{r – 1}
T_{r + 1} = ⁿC_r x^{n – r} (1)^r
Coefficient of T_r+1^th term = ⁿC_r

Put n = 4r – 5 from equation (1) in equation (2),
3(4r – 5) – 8r = – 3
⇒ 12r – 15 – 8r = – 3
⇒ 4r = 12
∴ r = 3
Put r = 3 in equation (2),
n – 4 x 3 = – 5
⇒ n = 12 – 5
⇒ n = l
n = 7, r = 3

Question 14.
Prove that the coefficient of xⁿ in the expansion of (1 + x)²ⁿ of in the expansion of (1 + x)^{2n – 1}.
Solution:
In the expansion of (x + a)ⁿ
T_r+1 = ⁿC_r x^{n – r} a^r
Here x = 1, a = x, n = 2n
T_r+1 = ²ⁿC_r(1)^{2n – r}(x)^r
For the coefficient of xⁿ, put r = n,
T_n+1 = ²ⁿC_n(a)^{2n – n}(x)ⁿ
and T₁₈ = T_{17 + 1} = ⁵⁰C₁₇ (2)^{50 – 17} (a)¹⁷
= (²ⁿC_n) xⁿ
∴ In the expansion of (1 + x)²ⁿ, the coefficient of xⁿ = ²ⁿC_n …. (1)
and in the expansion of (1 + x)^{2n – 1}, x = 1, a = x, n = 2n – 1
∴ T_r+1 = ²ⁿC_r (1)^{2n – 1  -r} (x)ⁿ
For the coefficient of xⁿ, put r = n, ‘
We get T_n+1 = ^{2n – 1}C_n xⁿ
The coefficient of xⁿ in the expansion of (1 + x)^{2n – 1} = ^{n – 1}C_n

∴ The coefficient of xⁿ in the expansion of (1 + x)²ⁿ
= 2 x The coefficient of xⁿ in the expansion of (1 + x)²ⁿ, [from equation (1) and (2)]

Question 15.