If the function f\left( x ight) = \left\{ {\begin{matrix} {{k_1}{{\left( {x - \pi } ight)}^2} - 1,} & {x \le \pi } \ {{k_2}\cos x,} & {x \pi } \ \end{matrix} } ight. is twice differentiable, then the ordered pair (k1, k2) is equal to :

Question

If the function f\left( x ight) = \left\{ {\begin{matrix} {{k_1}{{\left( {x - \pi } ight)}^2} - 1,} & {x \le \pi } \ {{k_2}\cos x,} & {x > \pi } \ \end{matrix} } ight. is twice differentiable, then the ordered pair (k1, k2) is equal to :

Accepted Answer

Answer: D. \left( {{1 \over 2},1} ight) Given, f\left( x ight) = \left\{ {\begin{matrix} {{k_1}{{\left( {x - \pi } ight)}^2} - 1,} & {x \le \pi } \ {{k_2}\cos x,} & {x > \pi } \ \end{matrix} } ight. Differentiating one time, f'\left( x ight) = \left\{ {\begin{matrix} {2{k_1}\left( {x - \pi } ight),} & {x \le \pi } \ { - {k_2}\sin x,} & {x > \pi } \ \end{matrix} } ight. Differentiating one more time, f''\left( x ight) = \left\{ {\begin{matrix} {2{k_1},} & {x \le \pi } \ { - {k_2}\cos x,} & {x > \pi } \ \end{matrix} } ight. As f''(x) is differentiable so f''(\pi +) = f''(\pi -) \Rightarrow -k2(-1) = 2k1 \Rightarrow 2k1 = k2 	herefore (k1, k2) = \left( {{1 \over 2},1} ight)

If the function $$f\left( x \right) = \left\{ {\begin{matrix} {{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \\ {{k_2}\cos x,} & {x > \pi } \\ \end{matrix} } \right.$$ is twice differentiable, then the ordered pair (k1, k2) is equal to :

Answer

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