Question

If $ab\ne 0$ and the sum of the coefficients of $x^7$ and $x^4$ in the expansion of ${(\frac{x^2}{a}-\frac{b}{x})}^{11}$ is zero, then

• A. $a=b$
• B. $a+b=0$
• C. $ab=-1$
• D. $ab=1$

Answer 1

Answer: (D)

$ab=1$

We know the general term in the expansion is $(\frac{11}{r})(-1{)}^r{x}^{22-3r}\frac{b^r}{a^{11-r}}$

Now for the coefficient of $x^7$ put $r=5$

$\Rightarrow -(\frac{11}{5})\frac{b^5}{a^6}$

And for $x^4$ put $r=6$

$\Rightarrow (\frac{11}{6})\frac{b^6}{a^5}$

Since their sum is zero, $\frac{b^5}{a^6}=\frac{b^6}{a^5}$ as $(\frac{11}{5})=(\frac{11}{6})$

$\frac{1}{a}=\frac{b}{1}\Rightarrow ab=1$

So, Option (D)