Question:
If the function
\( f(x)=\begin{cases} k_1(x-\pi)^2-1, & x \le \pi \\ k_2\cos x, & x \gt \pi \end{cases} \)
is twice differentiable, then the ordered pair \((k_1, k_2)\) is equal to :
If the function
\( f(x)=\begin{cases} k_1(x-\pi)^2-1, & x \le \pi \\ k_2\cos x, & x \gt \pi \end{cases} \)
is twice differentiable, then the ordered pair \((k_1, k_2)\) is equal to :
\( f(x)=\begin{cases} k_1(x-\pi)^2-1, & x \le \pi \\ k_2\cos x, & x \gt \pi \end{cases} \)
is twice differentiable, then the ordered pair \((k_1, k_2)\) is equal to :
Show Hint
Impose continuity of \(f\), \(f'\) and \(f''\) at \(x=\pi\) and compare left and right values.
Updated On: Nov 5, 2026
- \(\left(\frac{1}{2}, -1\right)\)
- \((1, 1)\)
- \((1, 0)\)
- \(\left(\frac{1}{2}, 1\right)\)
Show Solution
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The Correct Option is D
Solution and Explanation
Continuity at \(x=\pi\): left value \(=k_1\cdot 0-1=-1\); right value \(=k_2\cos\pi=-k_2\). So \(k_2=1\).
First derivative: \(f'(x)=2k_1(x-\pi)\) for \(x\lt\pi\) and \(-k_2\sin x\) for \(x\gt\pi\). At \(x=\pi\): left \(=0\), right \(=-k_2\sin\pi=0\). Equal for every \(k_1,k_2\).
Second derivative: \(f''(x)=2k_1\) for \(x\lt\pi\) and \(-k_2\cos x\) for \(x\gt\pi\). At \(x=\pi\): left \(=2k_1\), right \(=-k_2\cos\pi=k_2\).
So \(2k_1=k_2=1\), giving \(k_1=\frac{1}{2}\), \(k_2=1\).
Answer: (4) \(\left(\frac{1}{2}, 1\right)\)
First derivative: \(f'(x)=2k_1(x-\pi)\) for \(x\lt\pi\) and \(-k_2\sin x\) for \(x\gt\pi\). At \(x=\pi\): left \(=0\), right \(=-k_2\sin\pi=0\). Equal for every \(k_1,k_2\).
Second derivative: \(f''(x)=2k_1\) for \(x\lt\pi\) and \(-k_2\cos x\) for \(x\gt\pi\). At \(x=\pi\): left \(=2k_1\), right \(=-k_2\cos\pi=k_2\).
So \(2k_1=k_2=1\), giving \(k_1=\frac{1}{2}\), \(k_2=1\).
Answer: (4) \(\left(\frac{1}{2}, 1\right)\)
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