Question

lf $\underset{x\to 0}{lim}\frac{x(1+a\cos x)-b\sin x}{x^3}=1$, then $a,b$ are

• A. $\frac{1}{2},\frac{-3}{2}$
• B. $\frac{5}{2},\frac{3}{2}$
• C. $\frac{-5}{2},\frac{-3}{2}$
• D. $\frac{5}{2},\frac{-3}{2}$

Answer 1

The correct option is C $\frac{-5}{2},\frac{-3}{2}$

On applying the limit, since the function is of the form $\frac{0}{0}$,
We apply L-Hospital's Rule,
$li{m}_{x\to 0}\frac{1+acosx-axsinx-bcosx}{3x^2}$
To proceed further, the function should again be of the form $\frac{0}{0}$
Therefore, $1+a-b=0$
Again applying L-Hospital's Rule,
$li{m}_{x\to 0}\frac{-asinx-asinx-axcosx+bsinx}{6x}$
This is expressible in the form,
$li{m}_{x\to 0}\frac{-2asinx}{6x}-acosx+\frac{bsinx}{6x}$
$=\frac{-a}{3}-a+\frac{b}{6}=1$
Solving the 2 equations in $a$ and $b$ simultaneously, we get their values as,
$a=\frac{-5}{2}$ and $b=\frac{-3}{2}$
Hence, option 'C' is correct.