Binomial Theorem & its Simple Applications Test

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Binomial Theorem & its Simple Applications Test - 5

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Weekly Quiz Competition

Question 1
2 / -0.83

If in the expansion of (1 + x)^m (1 – x)ⁿ , the co-efficients of x and x² are 3 and – 6 respectively, then m is [JEE 99,2 ]

A
6

B
9

C
12

D
24

Solution

(1+x)^m (1 −x)ⁿ = (1 + mx + m(m −1)x² /2!)(1 −nx + n(n −1)x² /2! )
= 1 + (m −n)x + [(n² −n)/2 −mn + (m2 −m)/2] x²
Given m −n = 3 or n = m −3
Hence (n² −n)/2 −mn + (m² −m)/2 = 6
⇒[(m −3)(m −4)]/2 −m(m −3) + (m² −m)/2 = −6
⇒m2 −7m + 12 −2m2 + 6m + m² −m + 12 = 0
⇒−2m + 24 = 0
⇒m = 12.
Question 2
2 / -0.83

For 2 £r £n,

A

B

C

D
Question 3
2 / -0.83

Find the largest co-efficient in the expansion of (1 + x)ⁿ , given that the sum of co-efficients of the terms in its expansion is 4096. [REE 2000 (Mains)]

A
924

B
925

C
920

D
927

Solution

We know that, the coefficients in a binomial expansion is obtained by replacing each variable by unit in the given expression.
Therefore, sum of the coefficients in (a+b)^n
= 4096=(1+1)ⁿ
⇒4096=(2)ⁿ
⇒(2)¹² =(2)ⁿ
⇒n=12
Here n is even, so the greatest coefficient is nCn/2 i.e., ¹² C₆ = 924
Question 4
2 / -0.83

In the binomial expansion of (a - b)ⁿ , n ³5, the sum of the 5th and 6th terms is zero. Then equals.

A

B

C

D

Solution

It is given that T6 +T5 =0.
Hence nC4aⁿ −4b⁴ −nC5aⁿ −5b⁵ =0
nC4 aⁿ −4b ⁴ = nC5aⁿ −5b⁵
nC4 a = nC5b
n!b/(n −4)!4! = n!b/(n −5)!5!
= a/(n −4) = b/5
Therefore a/b=(n −4)/5
Question 5
2 / -0.83

Find the coefficient of x⁴⁹ in the polynomial [REE 2001 (Mains), 3] where C_r = ⁵⁰ C_r

A
22100

B
22200

C
22250

D
22300
Question 6
2 / -0.83

The sum , (where = O if P < q) is maximum when m is [JEE 2002 (Scr.), 3]

A
5

B
10

C
15

D
20

Solution
Question 7
2 / -0.83

(a) Coefficient of t²⁴ in the expansion of (1 + t² )¹² (1 + t¹² ) (1 + t²⁴ ) is [JEE 2003 (Scr.), 3]
(b) Prove that :
[JEE 2003 (Mains),2]

A
¹² C₆ + 2

B
¹² C₆ +1

C
¹² C₆

D
None

Solution

(1+x)¹² (1+x¹² ) (1+x²⁴ )
= [C₀ + C₁ x² + C₂ x⁴ + C₃ x⁶ + C₄ x⁸ +.......C₁₂ x²⁴ ) (1 + x¹² + x²⁴ + x³⁶ )
= x²⁴ (¹² C₀ + ¹² C₆ + ¹² C₁₂ )
= 1 + ¹² C₆ + 1
= ¹² C₆ + 2
Question 8
2 / -0.83

^n_1 C_r = (k² – 3). ⁿ C_{r + 1}^, if k Î [JEE 2004 (Scr.)]

A

B

C

D

Solution

Formula,

(n - 1)!/{r! ×(n - 1 - r)!} = (k ²- 3) ×n!/(r + 1)!(n - r - 1)!
or, (n - 1)!/r! = (k ²- 3) ×n!/(r + 1)!
or, (n - 1)!/r! = (k ²- 3) ×n(n - 1)!/(r + 1)r!
or, 1/1 = (k ²- 3) ×n/(r + 1)
or, (r + 1)/n = (k ²- 3)
we know, r and n are integers so, (r + 1)/n (0, 1]
So, 0 <(r + 1)/n ≤1
or, 0 or, 3 or, √3 Hence, k (√3, 2]
Question 9
2 / -0.83

^n-1 C_r = (k² - 3). ⁿ C_{r + 1} [JEE 2004 (Scr.)]

A

B

C

D

Solution

(n - 1)!/{r! × (n - 1 - r)!} = (k² - 3) × n!/(r + 1)!(n - r - 1)!
or, (n - 1)!/r! = (k² - 3) × n!/(r + 1)!
or, (n - 1)!/r! = (k² - 3) × n(n - 1)!/(r + 1)r!
or, 1/1 = (k² - 3) × n/(r + 1)
or, (r + 1)/n = (k² - 3)
we know, r and n are integers so, (r + 1)/n (0, 1]
so, 0 < (r + 1)/n ≤ 1
or, 0 < k² - 3 ≤ 1
or, 3 < k² ≤ 4
or, √3 < k ≤ 2 , -2 ≤ k < -√3
hence, k (√3, 2]
Question 10
2 / -0.83

The value of

is, where = ⁿ C_r [JEE 2005 (Scr.)]

A

B

C

D

Solution
Question 11
2 / -0.83

The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1, 2 and 3 only, is [JEE 2009]

A
55

B
66

C
77

D
88
Question 12
2 / -0.83

For r = 0, 1, ...., 10 let A_r , B_r , C_r denote, respectively, the coefficient of x^r in the expansions of (1 + x)¹⁰ , (1 + x)²⁰ and (1 + x)³⁰ . Then is equal to [JEE 2010]

A
B₁₀ - C₁₀

B

C
0

D
C₁₀ - B₁₀
Question 13
2 / -0.83

Let a_n denote the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let b_n = the number of such n-digit integers ending with digit 1 and c_n = the number of such n-digit integers ending with digit 0. [JEE 2012]

Which of the following is correct ?

A
a₁₇ = a₁₆ + a₁₅

B

C

D
a₁₇ = c₁₇ + b₁₆
Question 14
2 / -0.83

Let a_n denote the number of all n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let b_n = the number of such n-digit integers ending with digit 1 and c_n = the number of such n-digit integers ending with digit 0. [JEE 2012]
The value of b₆ is

A
7

B
8

C
9

D
11