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 equal to

• A. 0
• B. $\pi$
• C. $2\pi$
• D. None of these

Answer 1

The correct option is C $2\pi$

Clearly $-\pi ,\pi$ are the roots of quadratics
$\Rightarrow f\left(x\right)=a(x-\pi )(x+\pi )$
Also $f\left(\frac{\pi }{2}\right)=a(-\frac{3{\pi }^2}{4})=-\frac{3{\pi }^2}{4}\Rightarrow a=1$
$\therefore \underset{x\to -\pi }{lim}\frac{f\left(x\right)}{\sin \left(\sin x\right)}=\underset{x\to -\pi }{lim}\frac{(x-\pi )(x+\pi )}{\frac{\sin \left(\sin x\right)}{\sin x}\cdot \sin \left(x\right)}$
$=\underset{x\to -\pi }{lim}\frac{(x-\pi )}{\frac{\sin \left(\sin x\right)}{\sin x}\cdot \frac{-\sin (x+\pi )}{x+\pi }}=\frac{-\pi -\pi }{1(-1)}=2\pi$