Question

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

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

Answer 1

Answer: (D)

ab $=1$

$T_{r+1}$ in the expansion of $[a{x}^2+\frac{1}{bx}{]}^{11}$
$T_{r+1}{=}^{11}{C}_r\left(a{x}^2{)}^{11-r}\right(\frac{1}{bx}{)}^r$
${=}^{11}{C}_r\left(a{)}^{11-r}\right(b{)}^{-r}(x{)}^{22-3r}$
$\Rightarrow 22-3r=7$
$\Rightarrow r=5$
So, coefficient of $x^7$ in the expansion of $[a{x}^2+\frac{1}{bx}{]}^{11}$ is ${}^{11}{C}_5{a}^6{b}^{-5}$ ......(1)
Again $T_{r+1}$ in the expansion of $[ax-\frac{1}{b{x}^2}{]}^{11}$ is
$T_{r+1}{=}^{11}{C}_r\left(ax{)}^{11-r}\right(-\frac{1}{b{x}^2}{)}^r$
${=}^{11}{C}_r{a}^{11-r}(-1{)}^r\times (b{)}^{-r}\left(x{)}^{-2r}\right(x{)}^{11-r}$
$\Rightarrow 11-3r=7$
$\Rightarrow r=6$
Coefficient of $x^{-7}$ in the expansion of $[ax-\frac{1}{b{x}^2}{]}^{11}$ is ${}^{11}{C}_6{a}^5\times 1\times (b{)}^{-6}$ ....(2)
According to given condition , eqn (1) and (2) are equal
${}^{11}{C}_5{a}^6{b}^{-5}{=}^{11}{C}_6{a}^5\times 1\times (b{)}^{-6}$
$\Rightarrow ab=1$