If the function f(x)= ll k_ 1 (x- )^ 2 -1, & x k_ 2 x, & x> . is … - Vidyadip

MCQ

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

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

✓

Answer

$f ( x )$ is continuous and differentiable$f \left(\pi^{-}\right)= f (\pi)= f \left(\pi^{+}\right)$
$-1=-k_{2}$
$k _{2}=1$
$f^{\prime}(x)=\left\{\begin{array}{l}2 k_{1}(x-\pi) ; x \leq \pi \\ -k_{2} \sin x \quad ; x>\pi\end{array}\right.$
$f^{\prime}\left(\pi^{-}\right)=f^{\prime}\left(\pi^{+}\right)$
$0=0$
so, differentiable at $x=0$
$f ^{\prime \prime}( x )=\left\{\begin{array}{cc}2 k _{1} & ; x \leq \pi \\ - k _{2} \cos x ; x >\pi\end{array}\right.$
$f ^{\prime \prime}\left(\pi^{-}\right)= f ^{\prime \prime}\left(\pi^{+}\right)$
$2 k _{1}= k _{2}$
$k_{1}=\frac{1}{2}$
$\left( k _{1}, k _{2}\right)=\left(\frac{1}{2}, 1\right)$

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