JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

Choose the correct answer:

Explanation

In the expansion of $(a x - \frac{1}{b x ^{2}})^{13}$

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

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

For the coefficient of $x^{7}$ , putting $13 - 3 r = 7$

$\Rightarrow r = 2$

So,

$T_{2 + 1} =^{13} C_{2} a^{11} \frac{1}{b ^{2}} x^{7}$

$\Rightarrow$ Coefficient of $x^{7}$ in $(a x - \frac{1}{b x ^{2}})^{13}$ is
$^{13} C_{2} \frac{a ^{11}}{b ^{2}}$

Similarly, in the expansion of $(a x + \frac{1}{b x ^{2}})^{13}$ , we have

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

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

For the coefficient of $x^{- 5}$ , putting $13 - 3 r = - 5$

$\Rightarrow r = 6$

So,

$T_{6 + 1} =^{13} C_{6} a^{7} \frac{1}{b ^{6}} x^{- 5}$

$\Rightarrow$ Coefficient of $x^{- 5}$ in $(a x + \frac{1}{b x ^{2}})^{13}$ is
$^{13} C_{6} \frac{a ^{7}}{b ^{6}}$

\binom{13}{2} \frac{a^{11}}{b^2} = \binom{13}{6} \frac{a^7}{b^6}

\binom{13}{2} \frac{a^{11}}{b^2} = \binom{13}{6} \frac{a^7}{b^6}