If the function f(x) = { k1 (x-π)^2 – 1 ,
Question:

If the function $\mathrm{f}(\mathrm{x})= \begin{cases}\mathrm{k}_{1}(\mathrm{x}-\pi)^{2}-1, & \mathrm{x} \leq \pi \\ \mathrm{k}_{2} \cos \mathrm{x}, & \mathrm{x}>\pi\end{cases}$

is twice differentiable, then the ordered pair $\left(\mathrm{k}_{1}, \mathrm{k}_{2}\right)$ is equal to :

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

2. $(1,1)$

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

4. $(1,0)$

Correct Option: 1

Solution:

$f(x)$ is continuous and differentiable

$f(\pi)=f(\pi)=f\left(\pi^{+}\right)$

$-1=-k_{2}$

$\mathrm{k}_{2}=1$

$f^{\prime}(x)= \begin{cases}2 k_{1}(x-\pi) ; & x \leq \pi \\ -k_{2} \sin x & ; x>\pi\end{cases}$

$f^{\prime}(\pi)=f^{\prime}\left(\pi^{+}\right)$

$0=0$

so, differentiable at $x=0$

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

$f^{\prime \prime}\left(\pi^{-}\right)=f^{\prime \prime}\left(\pi^{+}\right)$

$2 \mathrm{k}_{1}=\mathrm{k}_{2}$

$\mathrm{k}_{1}=\frac{1}{2}$

$\left(\mathrm{k}_{1}, \mathrm{k}_{2}\right)=\left(\frac{1}{2}, 1\right)$