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Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced


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12 Questions MCQ Test Maths 35 Years JEE Main & Advanced Past year Papers | Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced

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Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 1

Given positive integers r > 1, n >2 and that the coefficient of (3r)th and (r + 2)th terms in the binomial expansion of (1+ x)2n are equal . Then (1983 - 1 Mark)

Detailed Solution for Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 1

Given that r and n are +ve integers such that r > 1, n > 2
Also in the expansion of  (1+ x)2n coeff. of  (3r)th term = coeff. of (r + 2)th term
2nC3r–1= 2nCr+1
⇒ 3r –1 = r + 1 or 3r – 1+ r + 1 = 2n
⇒ r = 1 or 2r = n
But r  > 1      
∴       n = 2r

Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 2

The coefficient of  x4 in    is (1983 - 1 Mark)

Detailed Solution for Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 2

General term in the expansion is 

 

For coeff of x4, we should have

10 – 3r = 4  ⇒ r = 2

∴ Coeff  of  x4 =

Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 3

The expression     is a polynomial of degree (1992 -  2 Marks)

Detailed Solution for Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 3

The given expression is

We know by binomial theorem, that (x + a ) n + ( x – a ) n = 2 [ nCx n +  nC 2x n – 2a 2 + nC4xn – 4a4 + ......]

∴ The given expression is equal to 2 [5C0x5  +  5C2x3(x3 – 1) +  5C4x (x3  – 1)2]
Max. power of x involved here is 7, also only +ve  integral powers of x are involved, therefore given expression is a polynomial of degree 7.

Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 4

If in the expansion of (1 + x)m (1 – x)n, the coefficients of x and x2 are 3 and – 6 respectively, then m is  (1999 - 2 Marks)

Detailed Solution for Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 4

We have (1 + x)m (1– x)n

= 1 + (m + n)x + 

Given,  m – n = 3 ....(1)

and

⇒ m2 + n2 - 2mn - (m +n) =-12

⇒ (m -n)2 - (m +n) =-12
⇒ m +n = 9 + 12= 21 ....(2)
From (1) and (2), we get m = 12

Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 5

For             (2000S)

Detailed Solution for Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 5

NOTE THIS STEP : 

Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 6

In the binomial expansion of (a - b)n, n ≥ 5, the sum of the 5th and 6th terms is zero. Then a/b equals (2001S)

Detailed Solution for Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 6

(a – b)n, n ≥ 5 In binomial expansion of above T5+ T6= 0
nC4 an–4 b4 + nC5 an–5 b5 = 0

 

Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 7

The sum  (where is maximum when m is (2002S)

Detailed Solution for Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 7

= Coeff of xm in the expansion of product (1+ x)10 (1 + x)20 =  Coeff of xm  in the expansion of (1+ x)30 = 30Cm To get  max. value of given sum, 30Cm should be max. which is so when m = 30/2 = 15.

 

Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 8

Coefficient of t24 in (1 +t2)12 (1+t12) (1 + t24) is (2003S)

Detailed Solution for Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 8

(1 + t2)12  (1 + t12) (1 + t24) = (1 + t12 + t24 + t36) (1 + t2)12
∴ Coeff. of t24 = 1× Coeff. of  t24 in (1+ t2)12 + 1 × Coeff. of t12 in (1 + t2)12 + 1 × constant term in (1 + t2)12 = 12C12 + 12C6 + 12C0 = 1+12C6 + 1=12C6 + 2

Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 9

If n–1Cr = (k2 – 3) nCr +1, then k ∈ (2004S) 

Detailed Solution for Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 9

Since  0 ≤ r ≤ n – 1

Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 10

The value of   is where  (2005S)

Detailed Solution for Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 10

To find 30C030C1030C130 C11 + 30 C230C12 – .... + 30C2030C30
We know that (1 + x)30 = 30C0 + 30C1x + 30C2x2 + .... + 30C20x20 + ....30C30x30 ....(1)
(x – 1)30 = 30C0x30 –  30C1x29 +....+ 30C10x2030C11x19 + 30C12x18 +.... 30C30x0 ....(2)
Multiplying eqn (1) and (2), we get (x2 – 1)30 = (   ) × (   )
Equating the coefficients of x20 on both sides,
we get 30C10 = 30C030 C1030 C130C11  + 30C230C12– ....
+ 30C20  30C30
∴ Req. value is 30C10

Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 11

For r = 0, 1, …, 10, let Ar, Band Cr  denote, respectively, the coefficient of xr in the expansions of (1 + x)10 , (2010)

  (1 + x)20 and (1 + x)30.  Then   is equal to

Detailed Solution for Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 11

Clearly  


Now expanding (1 + x )10 and (1 + x )20 by bin om ial theorem and comparing the coefficients of x20 in their product, on both sides, we get

 
= coeff of x20 in (1 + x )30 = 30 C20= 30C10

Again expending (1 + x )10 and ( x + 1)10 by bin omial theorem  and comparing the coefficients of x10 in their

product on both sides, we get

coeff of x10 in (1 + x) 20 = 20C10

Substituting these values in equation (1), we get

Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 12

Coefficient of x11 in the expansion of (1 + x2)4(1 + x3)7 (1 + x4)12 is (JEE Adv. 2014)

Detailed Solution for Test: Single Correct MCQs: Mathematical Induction and Binomial Theorem | JEE Advanced - Question 12

Coeff. of x11 in exp. of 

= (Coeff. of xa ) × (Coeff. of xb ) × (Coeff. of xc )
Such that a + b + c = 11 Here a = 2m, b = 3n, c = 4p
∴ 2m + 3n + 4p = 11
Case I : m = 0, n = 1, p = 2
Case II : m = 1, n = 3, p = 0
Case III : m = 2, n = 1, p = 1
Case IV : m = 4, n = 1, p = 0
∴ Required coeff.

= 462 + 140 + 504 + 7 = 1113

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