Question

If the binomial expansion of $(a+bx{)}^{-2}$ is $\frac{1}{4}-3x+...$, then the value of $a$ and $b$ are

• A. $2,12$
• B. $2,10$
• C. $1,12$
• D. $1,15$

Answer 1

Answer: (A)

We have, ${\left(a+bx\right)}^{-2}=\frac{1}{4}-3x+\dots$ (Given)
Also, ${\left(a+bx\right)}^{-2}=a^{-2}{\left(1+\frac{b}{a}x\right)}^{-2}$
$=a^{-2}\left(1-\frac{2b}{a}x\right)$
[Neglecting $x^2$ and higher powers of $x$]
$=\frac{1}{a^2}-\frac{2bx}{a^3}$
$\therefore \frac{1}{a^2}-\frac{2bx}{a^3}=\frac{1}{4}-3x$
$\Rightarrow \frac{1}{a^2}=\frac{1}{4}$ and $\frac{2b}{a^3}=3$
$\Rightarrow a=2$ and $b=\frac{3}{2}{a}^3$
$=\frac{3}{2}\times {\left(2\right)}^3=\frac{3}{2}\times 8=12$