Question

If the coefficient of $x^7$ in ${[a{x}^2+\left(\frac{1}{bx}\right)]}^{11}$ equals the coefficient of $x^{-7}$ in ${[ax-\left(\frac{1}{b{x}^2}\right)]}^{11}$, then $a$ and $b$ satisfy the relation

• A. $ab=1$
• B. $\frac{a}{b}=1$
• C. $a+b=1$
• D. $a-b=1$

Answer 1

The correct option is A $ab=1$

Let $x^7$ is contained in $(r+1)$th term in the expansion of ${(a{x}^2+\frac{1}{bx})}^{11}$

Thus $T_{r+1}{=}^{11}{C}_r{\left(a{x}^2\right)}^{11-r}{\left(\frac{1}{bx}\right)}^r$

${=}^{11}{C}_r\frac{a^{11-r}}{b^r}\cdot x^{22-3r}$
For coefficient of $x^7$,

$\Rightarrow 22-3r=7$

$\Rightarrow 3r=15$

$\Rightarrow r=5$

Thus $T_6{=}^{11}{C}_5\frac{a^6}{b^5}{x}^7$
Therefore, coefficient of $x^7$ in the expansion of

${(a{x}^2+\frac{1}{bx})}^{11}{=}^{11}{C}_5\frac{a^6}{b^5}$

Similarly, coefficient of $x^{-7}$ in the expansion of

${(ax-\frac{1}{b{x}^2})}^{11}{=}^{11}{C}_6\frac{a^5}{b^6}$

Given, ${}^{11}{C}_5\frac{a^6}{b^5}{=}^{11}{C}_6\frac{a^5}{b^6}$

$\Rightarrow \frac{a^6}{b^5}=\frac{a^5}{b^6}$ $(\because {}^{11}{C}_5{=}^{11}{C}_6)$

$\Rightarrow ab=1$