Question

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

The Correct Option is D

$T_{r+1 }$ in the expansion $\left[ax^2 - \left(\frac{1}{bx}\right)\right]^{11} = ^{11}C_{r} \left(ax^{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}{bx^{2}}\right]^{11} =^{11}C_{r} \left(ax^{2}\right)^{11-r} \left(- \frac{1}{bx^{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} $ $\Rightarrow \, ab = 1$.