JEE MAIN - Mathematics (2020 - 5th September Morning Slot - No. 15)
If the function
$$f\left( x \right) = \left\{ {\matrix{ {{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr {{k_2}\cos x,} & {x > \pi } \cr } } \right.$$ is
twice differentiable, then the ordered pair (k1, k2) is equal to :
$$f\left( x \right) = \left\{ {\matrix{ {{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr {{k_2}\cos x,} & {x > \pi } \cr } } \right.$$ is
twice differentiable, then the ordered pair (k1, k2) is equal to :
$$\left( {{1 \over 2},-1} \right)$$
(1, 1)
(1, 0)
$$\left( {{1 \over 2},1} \right)$$
Explanation
Given, $$f\left( x \right) = \left\{ {\matrix{
{{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr
{{k_2}\cos x,} & {x > \pi } \cr
} } \right.$$
Differentiating one time,
$$f'\left( x \right) = \left\{ {\matrix{ {2{k_1}\left( {x - \pi } \right),} & {x \le \pi } \cr { - {k_2}\sin x,} & {x > \pi } \cr } } \right.$$
Differentiating one more time,
$$f''\left( x \right) = \left\{ {\matrix{ {2{k_1},} & {x \le \pi } \cr { - {k_2}\cos x,} & {x > \pi } \cr } } \right.$$
As f''(x) is differentiable so
f''($$\pi $$+) = f''($$\pi $$-)
$$ \Rightarrow $$ -k2(-1) = 2k1
$$ \Rightarrow $$ 2k1 = k2
$$ \therefore $$ (k1, k2) = $$\left( {{1 \over 2},1} \right)$$
Differentiating one time,
$$f'\left( x \right) = \left\{ {\matrix{ {2{k_1}\left( {x - \pi } \right),} & {x \le \pi } \cr { - {k_2}\sin x,} & {x > \pi } \cr } } \right.$$
Differentiating one more time,
$$f''\left( x \right) = \left\{ {\matrix{ {2{k_1},} & {x \le \pi } \cr { - {k_2}\cos x,} & {x > \pi } \cr } } \right.$$
As f''(x) is differentiable so
f''($$\pi $$+) = f''($$\pi $$-)
$$ \Rightarrow $$ -k2(-1) = 2k1
$$ \Rightarrow $$ 2k1 = k2
$$ \therefore $$ (k1, k2) = $$\left( {{1 \over 2},1} \right)$$
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