Let f : 0 ,   π → R be a twice differentiable function such that lim t → x f x sin t -…

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

Solution

$\underset{t\to x}{\lim }\frac{f\left(x\right)\sin t-f\left(t\right)\sin x}{t-x}={\sin }^{2}x$
By using L'Hopital rule
$\underset{t\to x}{\lim }\frac{f\left(x\right)\cos t-f^{\prime }\left(t\right)\sin x}{1}={\sin }^{2}x$
$\Rightarrow f\left(x\right)\cos x-f^{\prime }\left(x\right)\sin x={\sin }^{2}x$
$\Rightarrow -\left(\frac{f^{\prime }\left(x\right)\sin x-f\left(x\right)\cos x}{{\sin }^{2}x}\right)=1$
$\Rightarrow -\frac{d\frac{f\left(x\right)}{\sin x}}{dx}=1$
$\Rightarrow \frac{f\left(x\right)}{\sin x}=-x+c$
Put $x=\frac{\pi }{6}$ and $f\left(\frac{\pi }{6}\right)=-\frac{\pi }{12}$
$\therefore$ $c=0\Rightarrow f\left(x\right)=-x\sin x$
$\left(A\right)$ $f\left(\frac{\pi }{4}\right)=-\frac{\pi }{4}\frac{1}{\sqrt{2}}$
$\left(B\right)$ $f\left(x\right)=-x\sin x$
As $\sin x>x-\frac{x^3}{6}$ , $-x\sin x<-x^2+\frac{x^4}{6}$
$\therefore f\left(x\right)<-x^2+\frac{x^4}{6} \forall x\in \left(0,\pi \right)$
$\left(C\right)$ $f^{\prime }\left(x\right)=-\sin x-x\cos x$
$f^{\prime }\left(x\right)=0\Rightarrow \tan x=-x\Rightarrow$ there exist $\alpha \in \left(0,\pi \right)$ for which $f^{\prime }\left(\alpha \right)=0$

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

Asked in: JEE Advanced 2018 (Paper 2)

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