Question

The value of a and b such that $\underset{x\to 0}{lim}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$, are

• A. $\frac{5}{2},\frac{3}{2}$
• B. $\frac{5}{2},-\frac{3}{2}$
• C. $-\frac{5}{2},-\frac{3}{2}$
• D. none of these

Answer 1

Answer: (C)

$-\frac{5}{2},-\frac{3}{2}$

$\underset{x\to 0}{lim}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$
Consider, ${lim}_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{x^3}$
It is of the form $\frac{0}{0}$, so applying L-Hospital's rule
$=\underset{x\to 0}{lim}\frac{-ax\sin x+1+a\cos x-b\cos x}{3x^2}$
Again applying L-Hospital's rule,

$=\underset{x\to 0}{lim}\frac{-ax\cos x-2a\sin x+b\sin x}{6x}$
$=\underset{x\to 0}{lim}\frac{-a(\cos x-x\sin x)-2a\cos x+b\cos x}{6x}$
$\frac{-3a+b}{6}$
$\Rightarrow -3a+b=6$
Option C satisfies.