Q71 / 170 · Manipal (MET) 2020 · Mathematics
If $\lim _{x \rightarrow 0} \frac{\sin (\sin x)-\sin x}{a x^3+b x^5+c}=\frac{-1}{12}$, then
- A$a=2, b \in R, c=0$
- Correct answerB$a=-2, b \in R, c=0$
- C$a=1, b \in R, c=0$
- D$a=-1, b \in R, c=0$
Solution
Given, $\lim _{x \rightarrow 0} \frac{\sin (\sin x)-\sin x}{a x^3+b x^5+c}=-\frac{1}{12}$ ....(i)
$\Rightarrow \quad \frac{\sin \sin 0-\sin 0}{a(0)^3+b(0)^5+c}=-\frac{1}{12}$
$\Rightarrow \quad \frac{0}{c}=-\frac{1}{12} \Rightarrow c=0$
Applying L' Hospital's rule in Eq.(i),we get
$\lim _{x \rightarrow 0} \frac{\cos (\sin x) \cos x-\cos x}{3 a x^2+5 b x^4+0}=-\frac{1}{12}$
$\Rightarrow \lim _{x \rightarrow 0} \frac{\cos x(\cos (\sin x)-1)}{x^2(3 a+5 b x)}=-\frac{1}{12}$
$\Rightarrow \lim _{x \rightarrow 0} \cos x \frac{2 \sin ^2\left(\frac{\sin x}{2}\right)}{x^2(3 a+5 b x)}=-\frac{1}{12}$
$\Rightarrow \frac{\cos 0}{3 a+5 b \times 0} \times \lim _{x \rightarrow 0} \frac{2 \sin ^2\left(\frac{\sin x}{2}\right)}{4\left(\frac{\sin x}{2}\right)^2 \times\left(\frac{x}{\sin x}\right)^2}=-\frac{1}{12}$
$\Rightarrow \frac{1}{3 a+0} \times \frac{1}{2} \times \frac{1}{1}=-\frac{1}{12}\left(\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right)$
$\Rightarrow \quad \frac{1}{6 a}=-\frac{1}{12} \Rightarrow 6 a=-12$
$\therefore \quad a=-2$
Hence, $a=-2, b \in R$ and $c=0$
Hence, the correct answer is option: 'B'.