JEE Main 2020 — Mathematics Calculus

$f(x)$f(x) is differentiable then will also continuous then $f(\pi )={k}_{1}{(\pi -\pi )}^{2}-1=-1$f(π)=k1​(π−π)2−1=−1

$$f({\pi }^{+})=\underset{h\rightarrow 0}{\mathrm{lim}}{k}_{2}\mathrm{cos}(\pi +h)=-{k}_{2}$$f(π+)=h→0lim​k2​cos(π+h)=−k2​

$f({\pi }^{+})=f(\pi )\Rightarrow {k}_{2}=1\ldots (1)$f(π+)=f(π)⇒k2​=1…(1),$f$f is continuous.

Now, f'(x)={\begin{matrix}\frac{d}{dx}{{k}_{1}(x-\pi {)}^{2}}, & x\leq \pi \\ \frac{d}{dx}{{k}_{2}\mathrm{cos}x}, & x>\pi \end{matrix}

f'(x)={\begin{matrix}2{k}_{1}(x-\pi ), & x\leq \pi \\ -{k}_{2}\mathrm{sin}x, & x>\pi \end{matrix}

then, $f'({\pi }^{-})=2{k}_{1}(\pi -\pi )=0$f′(π−)=2k1​(π−π)=0

$$f'({\pi }^{+})=-{k}_{2}\mathrm{sin}(\pi )=0$$f′(π+)=−k2​sin(π)=0

$$\Rightarrow f'({\pi }^{-})=f'({\pi }^{+})=0$$⇒f′(π−)=f′(π+)=0

Similarly, we get

f''(x)={\begin{matrix}2{k}_{1}, & x\leq \pi \\ -{k}_{2}\mathrm{cos}x, & x>\pi \end{matrix}

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

$$\Rightarrow f''({\pi }^{-})=f''({\pi }^{+})\Rightarrow 2{k}_{1}=-{k}_{2}cos\pi$$⇒f′′(π−)=f′′(π+)⇒2k1​=−k2​cosπ

$$\Rightarrow 2{k}_{1}={k}_{2}$$⇒2k1​=k2​

$\Rightarrow {k}_{1}=\frac{1}{2}$⇒k1​=21​ (from $(1)$(1))