Question

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

Answer 1

Answer: (A)

$ab=1$

Given that:

Sum of coefficients of$x^7$ and $x^4$ in the expansion ${(\frac{x^2}{a}-\frac{b}{x})}^{11}$ is zero.

Solution:

Coefficients of$x^7$ in the expansion ${(\frac{x^2}{a}-\frac{b}{x})}^{11}={}^{11}{C}_6\left(\frac{1}{a^6}\right)(-b{)}^5$

Coefficients of$x^4$ in the expansion ${(\frac{x^2}{a}-\frac{b}{x})}^{11}={}^{11}{C}_5\left(\frac{1}{a^5}\right)(-b{)}^6$

Sum of coefficients of$x^7$ and $x^4$ in the expansion ${(\frac{x^2}{a}-\frac{b}{x})}^{11}$ $={}^{11}{C}_6\left(\frac{1}{a^6}\right)(-b{)}^5+{}^{11}{C}_5\left(\frac{1}{a^5}\right)(-b{)}^6=0$

or, $-{}^{11}{C}_6\left(\frac{1}{a^6}\right)b^5+{}^{11}{C}_5\left(\frac{1}{a^5}\right)b^6=0$

or, ${}^{11}{C}_6\left(\frac{1}{a^6}\right)b^5={}^{11}{C}_5\left(\frac{1}{a^5}\right)b^6$ $(\because {}^{11}{C}_6={}^{11}{C}_5)$

or, $\frac{1}{a}=b$

or, $ab=1$

Hence, A is the correct option.