Question

Let $f:(0,\pi )\to R$ be a twice differentiable function such that $\underset{t\to x}{lim}\frac{f\left(x\right)\sin t-f\left(t\right)\sin x}{t-x}={\sin }^{2}x$ for all $x\in (0,\pi )$.
If $f\left(\frac{\pi }{6}\right)=-\frac{\pi }{12}$, then which of the following statement(s) is (are) TRUE?

• A. $f\left(\frac{\pi }{4}\right)=\frac{\pi }{4\sqrt{2}}$
• B. $f\left(x\right)<\frac{x^4}{6}-x^2$ for all $x\in (0,\pi )$
• C. There exists $\alpha \in (0,\pi )$ such that $f^{\prime }\left(\alpha \right)=0$
• D. $f^{\prime \prime }\left(\frac{\pi }{2}\right)+f\left(\frac{\pi }{2}\right)=0$

Answer 1

Answer: (B), (C), (D)

$f^{\prime \prime }\left(\frac{\pi }{2}\right)+f\left(\frac{\pi }{2}\right)=0$

$\underset{t\to x}{lim}\frac{f\left(x\right)\sin t-f\left(t\right)\sin x}{t-x}={\sin }^{2}x$
By using L Hopital's rule on LHS, we get
$\underset{t\to x}{lim}\frac{f\left(x\right)\cos t-f^{\prime }\left(t\right)\sin x}{1}={\sin }^{2}x$
$\Rightarrow \frac{f\left(x\right)\cos x-f^{\prime }\left(x\right)\sin x}{{\sin }^{2}x}=1$
$\Rightarrow -\frac{d}{dx}\left(\frac{f\left(x\right)}{\sin x}\right)=1$
$\Rightarrow \frac{f\left(x\right)}{\sin x}=-x+C$
$\text{Given},f\left(\frac{\pi }{6}\right)=-\frac{\pi }{12}\Rightarrow C=0$
$\therefore f\left(x\right)=-x\sin x$

$f\left(\frac{\pi }{4}\right)=-\frac{\pi }{4\sqrt{2}}$

We know $\sin x>x-\frac{x^3}{6}$
$\Rightarrow f\left(x\right)=-x\sin x<x^2-\frac{x^4}{6}$

$f\left(x\right)=-x\sin x$ is continuous on $[0,\pi ]$ and differentiable on $(0,\pi )$.
Also, $f\left(0\right)=f\left(\pi \right)=0$
$\therefore$ By Rolle's theorem, there exists $\alpha \in (0,\pi )$ such that $f^{\prime }\left(\alpha \right)=0$

$f^{\prime }\left(x\right)=-\sin x-x\cos x$
$f^{\prime \prime }\left(x\right)=-2\cos x+x\sin x$
$\Rightarrow f^{\prime \prime }\left(\frac{\pi }{2}\right)+f\left(\frac{\pi }{2}\right)=0$