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  4. If the function f(x) = begincases k1(x-π)2-1, text x leπ [2ex] k2 cos x, text x > π endcases is twice differentiable, then the ordered pair (k1, k2) is equal to :

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Q. If the function $f(x) = \begin{cases} k_{1}(x-\pi)^{2}-1, & \text{ $x \le\pi$ } \\[2ex] k_{2}\,cos\,x, & \text{ $x >\,\pi$ } \end{cases}$
is twice differentiable, then the ordered pair $(k_{1}, k_{2})$ is equal to :

JEE MainJEE Main 2020Continuity and Differentiability

Solution:

$f ( x )$ is continuous and differentiable
$f\left(\pi^{-}\right)=f(\pi)=f\left(\pi^{+}\right)$
$-1=-k_{2}$
$k _{2}=1$
$f'(x) = \begin{cases} 2k_{1}(x-\pi); & \text{ $x \le\pi$ } \\[2ex] - k_{2}\,sin\,x, ;& \text{ $x >\,\pi$ } \end{cases}$
$f^{\prime}\left(\pi^{-}\right)=f^{\prime}\left(\pi^{+}\right)$
$0=0$
so, differentiable at $x=0$
$f''(x) = \begin{cases} 2k_{1} ;& \text{if $x \le\pi$ } \\[2ex] -k_{2}\,cos\,x ;& \text{if $x >\,\pi$} \end{cases}$
$f ^{\prime \prime}\left(\pi^{-}\right)= f ^{\prime \prime}\left(\pi^{+}\right)$
$2 k _{1}= k _{2}$
$k _{1}=\frac{1}{2}$
$\left( k _{1}, k _{2}\right)=\left(\frac{1}{2}, 1\right)$