JEE Main 2020 (Online) 5th September Morning ShiftQuestion 1
If the function $f(x)=\begin{cases}k_1(x-\pi)^2-1, & x\le\pi\\ k_2\cos x, & x>\pi\end{cases}$ is twice differentiable, then the ordered pair $(k_1,k_2)$ is equal to :
Differentiation and its Application
Solution
Concept. Twice differentiable at $\pi$: values, first derivatives and second derivatives must match. Formulas used. (1) $\cos\pi=-1$, $\sin\pi=0$Step 1 - Values. $-1=k_2\cos\pi=-k_2$, so $k_2=1$. Step 2 - First derivatives. Left $2k_1(x-\pi)\to0$; right $-k_2\sin\pi=0$. Equal for any $k_1$. Step 3 - Second derivatives. Left $2k_1$; right $-k_2\cos\pi=1$. So $k_1=\frac12$. Answer: (A)