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STD 12 - 5. continuity and differentiation — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 5. continuity and differentiation1 Mark
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
  • ✓
    $\left(\frac{1}{2}, 1\right)$
  • B
    $(1,1)$
  • C
    $\left(\frac{1}{2},-1\right)$
  • D
    $(1,0)$
✓

Answer

Correct option: A.
$\left(\frac{1}{2}, 1\right)$
a
$f ( x )$ is continuous and differentiable

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

$-1=-k_{2}$

$k _{2}=1$

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

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

$0=0$

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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