Question

If sum of the coefficients of x⁷ and x⁴ in the expansion of ${(\frac{x^2}{a}-\frac{b}{x})}^{11}$ is zero, then

• (A) ab = 1
• (B) a = b
• (C) ab = -1
• (D) a+b = 0

Answer 1

The Correct Option is A

Explanation:
$$\begin{array}{rl}{T}_{r+1}& {=}^{11}{C}_r{\left(\frac{x^2}{a}\right)}^{11-r}{(-\frac{b}{x})}^r\\ & {=}^{11}{C}_r(-1{)}^r{a}^{r-11}{b}^r{x}^{22-3r}\end{array}$$For coefficient of $x^4$ and $x^7,$ we put $22-3r=4$ and $22-3r=7$$\Rightarrow r=6$ and $r=5$We are given$$\begin{array}{l}{}^{11}{C}_6(-1{)}^6{a}^{-5}{b}^6{+}^{11}{C}_5(-1{)}^5{a}^{-6}{b}^5=0\\ \Rightarrow \frac{b^6}{a^5}=\frac{b^5}{a^6}\Rightarrow ab=1\end{array}$$