Question

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

$T_{r+1}$ in the expansion
${\left[a{x}^2-\left(\frac{1}{bx}\right)\right]}^{11}{=}^{11}{C}_r{\left(a{x}^2\right)}^{11-r}{\left(\frac{1}{bx}\right)}^r$
${=}^{11}{C}_r{\left(a\right)}^{11-r}{\left(b\right)}^{-r}{\left(x\right)}^{22-2r-r}$
For the Coefficient of $x^7$ , we have
22 - 3r = 7 $\Rightarrow$ r = 5
$\therefore$ Coefficient of $x^7$
${=}^{11}{C}_5{\left(a\right)}^6{\left(b\right)}^{-5}...(1)$
Again $T_{r+1}$ in the expansion
${\left[ax-\frac{1}{b{x}^2}\right]}^{11}{=}^{11}{C}_r{\left(a{x}^2\right)}^{11-r}{\left(-\frac{1}{b{x}^2}\right)}^r$
${=}^{11}{C}_r{\left(a\right)}^{11-r}{\left(-1\right)}^r\times {\left(b\right)}^{-r}{\left(x\right)}^{-2r}{\left(x\right)}^{11-r}$
For the Coefficient of $x^{-7}$ , we have
Now 11 - 3r = - 7 $\Rightarrow$ 3r = 18 $\Rightarrow$ r = 6
$\therefore$ Coefficient of $x^{-7}$
$={}^{11}{C}_6{a}^5\times 1\times (b{)}^{-6}$
$\therefore$ Coefficient of $x^7$ = Coefficient of $x^{-7}$
$\Rightarrow {}^{11}{C}_5(a{)}^6(b{)}^{-5}={}^{11}{C}_6{a}^5\times (b{)}^{-6}$
$\Rightarrow ab=1$.