If the coefficients of $x^{7}$ in $\left(a x^{2}+\frac{1}{2 b x}\right)^{11}$ and $x^{-7}$ in $\left(a x-\frac{1}{3 b x^{2}}\right)^{11}$ are equal, then

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Binomial Theorem

If the coefficients of

x^{7}

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

and

x^{- 7}

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

are equal, then

Answer: (B)

For

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

, the general term is

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

. For

x^{7}

, we need

22 - 3 r = 7

, giving

r = 5

. The coefficient is

(5 11) \frac{a ^{6}}{32 b ^{5}}

. For

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

, the general term is

T_{r + 1} = (r 11) a^{11 - r} (\frac{- 1}{3 b})^{r} x^{11 - 3 r}

. For

x^{- 7}

, we need

11 - 3 r = - 7

, giving

r = 6

. The coefficient is

(6 11) \frac{a ^{5}}{729 b ^{6}}

. Equating:

\frac{a ^{6}}{32 b ^{5}} = \frac{a ^{5}}{729 b ^{6}}

, which simplifies to

729 ab = 32

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Single correct option