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 :
Solution
Given, $$f\left( x \right) = \left\{ {\matrix{
{{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr
{{k_2}\cos x,} & {x > \pi } \cr
} } \right.$$
<br><br>Differentiating one time,
<br><br>$$f'\left( x \right) = \left\{ {\matrix{
{2{k_1}\left( {x - \pi } \right),} & {x \le \pi } \cr
{ - {k_2}\sin x,} & {x > \pi } \cr
} } \right.$$
<br><br>Differentiating one more time,
<br><br>$$f''\left( x \right) = \left\{ {\matrix{
{2{k_1},} & {x \le \pi } \cr
{ - {k_2}\cos x,} & {x > \pi } \cr
} } \right.$$
<br><br>As f''(x) is differentiable so
<br><br>f''($\pi$<sup>+</sup>) = f''($\pi$<sup>-</sup>)
<br><br>$\Rightarrow$ -k<sub>2</sub>(-1) = 2k<sub>1</sub>
<br><br>$\Rightarrow$ 2k<sub>1</sub> = k<sub>2</sub>
<br><br>$\therefore$ (k<sub>1</sub>, k<sub>2</sub>) = $\left( {{1 \over 2},1} \right)$
About this question
Subject: Mathematics · Chapter: Limits, Continuity and Differentiability · Topic: Limits and Standard Results
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