# Solve this

Question:

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:

1. 2

2. –1

3. 1

4. –2

Correct Option: , 3

Solution:

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

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

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

$22-3 r=7$

$r=5$

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

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

${ }^{11} C_{r}(x)^{11-r} \cdot\left(-\frac{1}{b x^{2}}\right)^{r}$

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

$11-3 r=-7 \quad \therefore r=6$

${ }^{11} C_{6} \cdot \frac{1}{b^{6}} x^{-7}$

${ }^{11} C_{5} \cdot \frac{1}{b^{5}}={ }^{11} C_{6} \cdot \frac{1}{b^{6}}$

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

Leave a comment

Click here to get exam-ready with eSaral