If the function {f}( {x})=\{ \table k_1(x-\\pi)^2-1{,}, {x} \leq{ \\pi} ;k_2\cosx{,},x >\\pi is twice differentiable, then the ordered pair ({k}_{1},{ k}_{2} ) is equal to : [5 Sep 2020 Shift 1]

\\pi f"(\\pi^{-})=f"(\\pi^{+}) 2 {k}_{1}= {k}_{2} [{k}_{1}= \frac{1}{2}] ({k}_{1}, {k}_{2} )=(\frac{1}{2}, 1 )" >

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Limits, Continuity and Differentiability Part 1

Section: Mathematics

Question : 4

Total: 100

k1(x−π)2−1,	x≤π
k2‌cosx‌,	x>π

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2k1(x−π)−1	;	x≤π
−k2‌cosx	;	x>π

2k1	;	x≤π
−k2‌cos‌x	;	x>π