Question

If $f\left(x\right)=0$ is a quadratic equation such that $f(-\pi )=f\left(\pi \right)=0$ and $f\left(\frac{\pi }{2}\right)=-\frac{3{\pi }^2}{4},$ then $\underset{x\to -\pi }{lim}\frac{f\left(x\right)}{\sin \left(\sin x\right)}$ is

• A. $0$
• B. $\pi$
• C. $2\pi$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is C $2\pi$

Let, $f\left(x\right)=a(x-\pi )(x+\pi )$
we have,
$f\left(\frac{\pi }{2}\right)=-\frac{3{\pi }^2}{4}$

$-\frac{3a{\pi }^2}{4}=-\frac{3{\pi }^2}{4}$
$\Rightarrow a=1$
$\therefore f\left(x\right)=x^2-{\pi }^2$
$\Rightarrow \underset{x\to -\pi }{lim}\frac{x^2-{\pi }^2}{\sin \left(\sin x\right)}=\frac{0}{0}$ form
Using L' Hospital rule:
$=\underset{x\to -\pi }{lim}\frac{2x}{\cos \left(\sin x\right)\cdot \cos x}$
$=\frac{-2\pi }{-1}=2\pi$

Alternate:
$\underset{x\to -\pi }{lim}\frac{x^2-{\pi }^2}{\sin \left(\sin x\right)}=\underset{x\to -\pi }{lim}\frac{(x-\pi )(x+\pi )}{\sin \left(\sin x\right)}$
Let, $x+\pi =t$
$\Rightarrow t\to 0$
$=\underset{t\to 0}{lim}\frac{(t-2\pi )t}{\sin \left(\sin \right(t-\pi \left)\right)}$

As $t\to 0\Rightarrow t-2\pi \to -2\pi$
$=\underset{t\to 0}{lim}\frac{t(-2\pi )}{\sin \left(\sin \right(t-\pi \left)\right)}$
$=\underset{t\to 0}{lim}\frac{-2t\pi }{-\sin \left(\sin t\right)}$
$=\underset{t\to 0}{lim}2\pi \times \frac{\sin t}{\sin \left(\sin t\right)}\times \frac{t}{\sin t}=2\pi$