JEE Main 2020 — Continuity and Differentiability Question with Solution

$\mathrm{f}\left(\mathrm{x}\right)$ is differentiable then will also continuous then $f\left(\pi \right)=k_1{\left(\pi -\pi \right)}^2-1=-1$

$f\left({\pi }^{+}\right)=\underset{h\to 0}{\lim }{k}_2\cos \left(\pi +h\right)=-k_2$

$f\left({\pi }^{+}\right)=f\left(\pi \right)\Rightarrow k_2=1\dots \left(1\right)$,$f$ is continuous.

Now, $f\text{'}\left(x\right)=\left\{\begin{array}{ll}\frac{d}{dx}\left\{k_1(x-\pi {)}^2\right\},& x\le \pi \\ \frac{d}{dx}\left\{k_2\cos x\right\},& x>\pi \end{array}\right.$

 $f\text{'}\left(x\right)=\left\{\begin{array}{ll}2k_1(x-\pi ),& x\le \pi \\ -k_2\sin x,& x>\pi \end{array}\right.$

then, $f\text{'}\left({\pi }^{-}\right)=2k_1\left(\pi -\pi \right)=0$

$f\text{'}\left({\pi }^{+}\right)=-k_2\sin \left(\pi \right)=0$

 $\Rightarrow f\text{'}\left({\pi }^{-}\right)=f\text{'}\left({\pi }^{+}\right)=0$

Similarly, we get

$f\text{'}\text{'}\left(x\right)=\left\{\begin{array}{cc}2k_1,& x\le \pi \\ -k_2\cos x,& x>\pi \end{array}\right.$

As the function is twice differentiable, the second-order derivatives, $LHD=RHD$ 

$\Rightarrow f\text{'}\text{'}\left({\pi }^{-}\right)=f\text{'}\text{'}\left({\pi }^{+}\right)\Rightarrow 2k_1=-k_2\mathrm{cos\pi }$

$\Rightarrow 2k_1=k_2$

$\Rightarrow k_1=\frac{1}{2}$ (from $\left(1\right)$)