Question

If the function $f(x)=\{\begin{array}{ll}{k}_1(x-\pi {)}^2-1,& \text{ x≤π }\\ k_{2}cosx,& \text{ x>π }\end{array}$
is twice differentiable, then the ordered pair $(k_1,k_2)$ is equal to :

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

Answer 1

Answer: (A)

$f(x)$ is continuous and differentiable
$f\left({\pi }^{-}\right)=f(\pi )=f\left({\pi }^{+}\right)$
$-1=-k_2$
$k_2=1$
$f^{\prime }(x)=\{\begin{array}{ll}2k_1(x-\pi );& \text{ x≤π }\\ -k_{2}sinx,;& \text{ x>π }\end{array}$
$f^{\prime }\left({\pi }^{-}\right)=f^{\prime }\left({\pi }^{+}\right)$
$0=0$
so, differentiable at $x=0$
$f^{\prime \prime }(x)=\{\begin{array}{ll}2k_1;& \text{if x≤π }\\ -k_{2}cosx;& \text{if x>π}\end{array}$
$f^{\prime \prime }\left({\pi }^{-}\right)=f^{\prime \prime }\left({\pi }^{+}\right)$
$2k_1=k_2$
$k_1=\frac{1}{2}$
$\left(k_1,k_2\right)=\left(\frac{1}{2},1\right)$