If the function


If the function $f(x)\left\{\begin{array}{ll}k_{1}(x-\pi)^{2}-1, & x \leq \pi \\ k_{2} \cos x, & x>\pi\end{array}\right.$ is twice differentiable, then the ordered pair $\left(k_{1}, k_{2}\right)$ is equal to:

  1. (1) $\left(\frac{1}{2}, 1\right)$

  2. (2) $(1,0)$

  3. (3) $\left(\frac{1}{2},-1\right)$

  4. (4) $(1,1)$

Correct Option: 1


$f(x)$ is differentiable then, $f(x)$ is also continuous.

$\therefore \lim _{x \rightarrow \pi^{+}} f(x)=\lim _{x \rightarrow \pi^{-}} f(x)=f(\pi)$

$\Rightarrow-1=-K_{2} \Rightarrow K_{2}=1$

$\therefore f^{\prime}(x)=\left\{\begin{array}{rl}2 K_{1}(x-\pi): & x \leq \pi \\ -K_{2} \sin x & x>\pi\end{array}\right.$

Then, $\lim _{x \rightarrow \pi^{+}} f(x)=\lim _{x \rightarrow \pi^{-}} f(x)=0$

$f^{\prime \prime}(x)=\left\{\begin{array}{ccc}2 K_{1} & ; & x \leq \pi \\ -K_{2} \cos x & ; & x>\pi\end{array}\right.$

Then, $\lim _{x \rightarrow \pi^{+}} f(x)=\lim _{x \rightarrow \pi^{-}} f(x)$

$\Rightarrow 2 K_{1}=K_{2} \Rightarrow K_{1}=\frac{1}{2}$

$\operatorname{So},\left(K_{1}, K_{2}\right)=\left(\frac{1}{2}, 1\right)$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now