Question

If the coefficient of $x^7$ in $(a{x}^2+\frac{1}{bx}{)}^{11}$ is equal to the coefficient of $x^{-7}$ in $(ax-\frac{1}{b{x}^2}{)}^{11}$ then

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

Answer 1

Answer: (C)

$ab=1$

General term $T_{r+1}$ in the expansion of $(a+b{)}^n$ is given by

$T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$

Applying to ${(a{x}^2+\frac{1}{bx})}^{11}$, we get

$T_{r+1}={}^{11}{C}_r{x}^{22-3r}{a}^{11}(ab{)}^{-r}$ ...(i)

Therefore $22-3r=7$ for coefficient of $x^7$

$\Rightarrow r=5$

Thus, $\left(i\right)$ reduces to

$T_6={}^{11}{C}_5{x}^7{a}^{11}(ab{)}^{-5}$

Similarly applying to ${(ax-\frac{1}{b{x}^2})}^{11}$, we get

$T_{r+1}=(-1{)}^r{}^{11}{C}_r{x}^{11-3r}{a}^{11}(ab{)}^{-r}$ ...(ii)

Therefore $11-3r=-7$ for coefficient of $x^{-7}$

$\Rightarrow r=6$

Equation $\left(ii\right)$ reduces to

$T_7={}^{11}{C}_6{x}^{-7}{a}^{11}{ab}^{-6}$ ...(b)

Hence applying the given condition we get

${}^{11}{C}_6{a}^{11}(ab{)}^{-6}={}^{11}{C}_5{a}^{11}(ab{)}^{-5}$ from(a,b)

$ab=1$

Hence answer is C