7.Binomial Theorem
hard

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:

A

$-1$

B

$2$

C

$-2$

D

$1$

(JEE MAIN-2021)

Solution

Coefficient of $x^{7} \operatorname{in}\left(x^{2}+\frac{1}{b x}\right)^{11}$

${ }^{11} \mathrm{C}_{\mathrm{r}}\left(\mathrm{x}^{2}\right)^{11-\mathrm{r}} \cdot\left(\frac{1}{\mathrm{bx}}\right)^{\mathrm{r}}$

${ }^{11} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{22-3 \mathrm{r}} \cdot \frac{1}{\mathrm{~b}^{r}}$

$22-3 \mathrm{r}=7$

$r=5$

$\therefore{ }^{11} \mathrm{C}_{5} \cdot \frac{1}{\mathrm{~b}^{5}} \cdot \mathrm{x}^{7}$

Coefficient of $x^{-7}$ in $\left(x-\frac{b}{b x^{2}}\right)^{11}$

${ }^{11} \mathrm{C}_{\mathrm{r}}(\mathrm{x})^{11-\mathrm{r}} \cdot\left(-\frac{1}{\mathrm{bx}^{2}}\right)^{\mathrm{r}}$

${ }^{11} \mathrm{C}_{\mathrm{r}} \mathrm{x}^{11-3 r} \cdot \frac{(-1)^{r}}{\mathrm{~b}^{r}}$

$11-3 \mathrm{r}=-7 \therefore \mathrm{r}=6$

${ }^{11} \mathrm{C}_{6} \cdot \frac{1}{b^{6}} \mathrm{x}^{-7}$

${ }^{11} \mathrm{C}_{5} \cdot \frac{1}{\mathrm{~b}^{5}}={ }^{11} \mathrm{C}_{6} \cdot \frac{1}{\mathrm{~b}^{6}}$

Since $b \neq 0 \therefore b=1$

Standard 11
Mathematics

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