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 $xϵ(0,\pi )$.

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 $xϵ(0,\pi )$
• C. There exists $\alpha ϵ(0,\pi )$ such that $f^{\prime }\left(\alpha \right)=0$
• D. $f"\left(\frac{\pi }{2}\right)+f\left(\frac{\pi }{2}\right)=0$

Answer 1

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

$f\left(x\right)<\frac{x^4}{6}-x^2$ for all $xϵ(0,\pi )$
There exists $\alpha ϵ(0,\pi )$ such that $f^{\prime }\left(\alpha \right)=0$
$f"\left(\frac{\pi }{2}\right)+f\left(\frac{\pi }{2}\right)=0$

$\underset{t\to x}{lim}\frac{f\left(x\right)\sin t-f\left(t\right)\sin x}{t-x}={\sin }^{2}x$
By using L'Hospital's Rule

$\underset{t\to x}{lim}\frac{f\left(x\right)\cos t-f^{\prime }\left(t\right)\sin x}{1}={\sin }^{2}x$

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

$\Rightarrow -\left(\frac{f^{\prime }\left(x\right)\sin x-f\left(x\right)\cos x}{{\sin }^{2}x}\right)=1$

$\Rightarrow -d\left(\frac{f\left(x\right)}{\sin x}\right)=1$

$\Rightarrow \frac{f\left(x\right)}{\sin x}=x+c$

Put $x=\frac{\pi }{6}$ and $f\left(\frac{\pi }{6}\right)=-\frac{\pi }{12}$

$\therefore c=0\Rightarrow f\left(x\right)=-x\sin x$

(A) $f\left(\frac{\pi }{4}\right)=\frac{-\pi }{4}\frac{1}{\sqrt{2}}$

(B) $f\left(x\right)=-x\sin x$
As $\sin x>x-\frac{x^3}{6},-x\sin x<-x^2+\frac{x^4}{6}$

$\therefore f\left(x\right)<-x^2+\frac{x^4}{6}\forall xϵ(0,\pi )$

(C) $f^{\prime }\left(x\right)=-\sin x-x\cos x$
$f^{\prime }\left(x\right)=0\Rightarrow \tan x=-x\Rightarrow$ there exist $\alpha ϵ(0,\pi )$ for which $f^{\prime }\left(\alpha \right)=0$

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