Question

If the function $f\left(x\right)=\{\begin{array}{ccc}{k}_1{(x-\pi )}^2-1,& x\le \pi & \\ k_2\cos x,& x>\pi & \end{array}$ is twice differentiable, then the ordered pair $(k_1,k_2)$ is equal to:

• A. $(1,1)$
• B. $(1,0)$
• C. $(\frac{1}{2},-1)$
• D. $(\frac{1}{2},1)$

Answer 1

The correct option is D $(\frac{1}{2},1)$

$f\left(x\right)$ is continuous and differentiable
$f\left({\pi }^{-}\right)=f\left(\pi \right)=f\left({\pi }^{+}\right)$
$\Rightarrow k_2\cos \pi =k_1\left(0\right)-1$
$\Rightarrow k_2(-1)=-1$
$\Rightarrow k_2=1$

$f^{\prime }\left(x\right)=\{\begin{array}{ccc}2k_1(x-\pi );& x<\pi & \\ -k_2\sin x;& x>\pi & \end{array}$
$f^{\prime }\left({\pi }^{-}\right)=f^{\prime }\left({\pi }^{+}\right)$
$\Rightarrow 0=0$
So, differentiable at $x=0$

$f^{\prime \prime }\left(x\right)=\{\begin{array}{ccc}2k_1;& x<\pi & \\ -k_2\cos x;& x>\pi & \end{array}$
$\Rightarrow f^{\prime \prime }\left({\pi }^{-}\right)=f^{\prime \prime }\left({\pi }^{+}\right)$
$\Rightarrow 2k_1=k_2$
$\Rightarrow k_1=\frac{1}{2}$