Let ƒ(x) = xcos–1(–sin|x|), $x \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$, then which of the following is true?
Solution
We know, cos<sup>-1</sup>(-x) = $\pi$ - cos<sup>-1</sup>x
<br><br>$\therefore$ ƒ(x) = x($\pi$ - cos<sup>–1</sup>(sin|x|))
<br><br>= x($\pi$ - ${\pi \over 2}$ + sin<sup>–1</sup>(sin|x|))
<br><br>= x($\pi$ - ${\pi \over 2}$ + sin<sup>–1</sup>(sin|x|))
<br><br>= x${\pi \over 2}$ + x|x|
<br><br>$\therefore$ f(x) = $$\left\{ {\matrix{
{x{\pi \over 2} - {x^2},} & {x < 0} \cr
{x{\pi \over 2} + {x^2},} & {x \ge 0} \cr
} } \right.$$
<br><br>Now f'(x) = $$\left\{ {\matrix{
{{\pi \over 2} - 2x,} & {x < 0} \cr
{{\pi \over 2} + 2x,} & {x \ge 0} \cr
} } \right.$$
<br><br>and f''(x) = $$\left\{ {\matrix{
{ - 2,} & {x < 0} \cr
{2,} & {x \ge 0} \cr
} } \right.$$
<br><br>$\therefore$ ƒ' is decreasing in $\left( { - {\pi \over 2},0} \right)$ and increasing
in $\left( {0,{\pi \over 2}} \right)$
About this question
Subject: Mathematics · Chapter: Application of Derivatives · Topic: Tangents and Normals
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