If sum of the coefficients of x 7 and x 4 in the expansion of ( x 2 a − b x ) 11 is zero, then

Explanation: T r + 1 = 11 C r ( x 2 a ) 11 − r ( − b x ) r = 11 C r ( − 1 ) r a r − 11 b r x 22 − 3 r For coefficient of x 4 and x 7 , we put 22 − 3 r = 4 and 22 − 3 r = 7 ⇒ r = 6 and r = 5 We are given 11 C 6 ( − 1 ) 6 a − 5 b 6 + 11 C 5 ( − 1 ) 5 a − 6 b 5 = 0 ⇒ b 6 a 5 = b 5 a 6 ⇒ a b = 1

Question:

If sum of the coefficients of x⁷ and x⁴ in the expansion of ${(\frac{x^{2}}{a} - \frac{b}{x})}^{11}$ is zero, then

WBJEE

Updated On: Apr 27, 2024

(A) ab = 1
(B) a = b
(C) ab = -1
(D) a+b = 0

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The Correct Option is A

Solution and Explanation

Explanation:

\begin{aligned} T_{r + 1} & =^{11} C_{r} {(\frac{x^{2}}{a})}^{11 - r} {(- \frac{b}{x})}^{r} \\ =^{11} C_{r} (- 1)^{r} a^{r - 11} b^{r} x^{22 - 3 r} \end{aligned}

For coefficient of

x^{4}

and

x^{7},

we put

22 - 3 r = 4

and

22 - 3 r = 7

\Rightarrow r = 6

and

r = 5

We are given

\begin{array}{l} ^{11} C_{6} (- 1)^{6} a^{- 5} b^{6} +^{11} C_{5} (- 1)^{5} a^{- 6} b^{5} = 0 \\ \Rightarrow \frac{b^{6}}{a^{5}} = \frac{b^{5}}{a^{6}} \Rightarrow a b = 1 \end{array}

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If sum of the coefficients of x7 and x4 in the expansion of (x2a−bx)11 is zero, then

The Correct Option is A

Solution and Explanation

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Questions Asked in WBJEE exam

If sum of the coefficients of x⁷ and x⁴ in the expansion of ${(\frac{x^{2}}{a} - \frac{b}{x})}^{11}$ is zero, then