JEE Mains · Maths · STD 12 - 5. continuity and differentiation
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 & Solution
Correct Answer
(A) \(\left(\frac{1}{2}, 1\right)\)
Step-by-step Solution
Detailed explanation
\(f ( x )\) is continuous and differentiable \(f \left(\pi^{-}\right)= f (\pi)= f \left(\pi^{+}\right)\) \(-1=-k_{2}\) \(k _{2}=1\) \(f^{\prime}(x)=\left\{\begin{array}{l}2 k_{1}(x-\pi) ; x \leq \pi \\ -k_{2} \sin x \quad ; x>\pi\end{array}\right.\)…
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