If the coefficients of \(x^{7}\) in \(\left(x^{2}+\frac{1}{b x}\right)^{11}\) and \(x^{-7}\) in \(\left(x-\frac{1}{b x^{2}}\right)^{11}, b \neq 0\) are equal, then the value of \(b\) is equal to :

Binomial Theorem — JEE Main · 2021 · 27 Jul 2021 · Shift 1 Question with Solution

If the coefficients of

x^{7}

(x^{2} + \frac{1}{b x})^{11}

and

x^{- 7}

(x - \frac{1}{b x ^{2}})^{11}, b \neq = 0

are equal, then the value of

b

is equal to :

Answer: (B)

For the first expansion

(x^{2} + \frac{1}{b x})^{11}

, the general term is

(r 11) (x^{2})^{11 - r} (\frac{1}{b x})^{r} = (r 11) \frac{1}{b ^{r}} x^{22 - 2 r - r} = (r 11) \frac{1}{b ^{r}} x^{22 - 3 r}

. For the coefficient of

x^{7}

, we need

22 - 3 r = 7

, giving

r = 5

. The coefficient is

(5 11) \frac{1}{b ^{5}}

.

For the second expansion

(x - \frac{1}{b x ^{2}})^{11}

, the general term is

(s 11) x^{11 - s} (- \frac{1}{b x ^{2}})^{s} = (s 11) \frac{( - 1 ) ^{s}}{b ^{s}} x^{11 - s - 2 s} = (s 11) \frac{( - 1 ) ^{s}}{b ^{s}} x^{11 - 3 s}

. For the coefficient of

x^{- 7}

, we need

11 - 3 s = - 7

, giving

s = 6

. The coefficient is

(6 11) \frac{( - 1 ) ^{6}}{b ^{6}} = (6 11) \frac{1}{b ^{6}}

.

Setting them equal:

(5 11) \frac{1}{b ^{5}} = (6 11) \frac{1}{b ^{6}}

. Since

(5 11) = (6 11) = 462

, we get

\frac{1}{b ^{5}} = \frac{1}{b ^{6}}

, which gives

b = 1

JEE Main · 2021 · 27 Jul 2021 · Shift 1 · Single correct option