If the coefficient of $x^{7}$ in $\left(ax-\frac{1}{bx^{2}}\right)^{13}$ and the coefficient of $x^{-5}$ in $\left(ax+\frac{1}{bx^{2}}\right)^{13}$ are equal, then $a^{4}b^{4}$ is equal to :

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Binomial Theorem

If the coefficient of

x^{7}

(a x - \frac{1}{b x ^{2}})^{13}

and the coefficient of

x^{- 5}

(a x + \frac{1}{b x ^{2}})^{13}

are equal, then

a^{4} b^{4}

is equal to :

Answer: (B)

For the first expansion, the general term is

T_{r + 1} =^{13} C_{r} (a x)^{13 - r} (- \frac{1}{b x ^{2}})^{r} =^{13} C_{r} a^{13 - r} (- 1)^{r} b^{- r} x^{13 - 3 r}

.

For

x^{7}

13 - 3 r = 7 \Rightarrow r = 2

. Coefficient:

^{13} C_{2} a^{11} b^{- 2}

.

For the second expansion with

x^{- 5}

13 - 3 r = - 5 \Rightarrow r = 6

. Coefficient:

^{13} C_{6} a^{7} b^{- 6}

.

Equating:

^{13} C_{2} a^{11} b^{- 2} =^{13} C_{6} a^{7} b^{- 6}

, so

a^{4} b^{4} = \frac{^{13} C _{6}}{^{13} C _{2}} = \frac{1716}{78} = 22

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Single correct option