JEE MAIN - Mathematics (2020 - 8th January Morning Slot - No. 4)
Let ƒ(x) = xcos–1(–sin|x|), $$x \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$$, then
which of the following is true?
ƒ' is decreasing in $$\left( { - {\pi \over 2},0} \right)$$ and increasing
in $$\left( {0,{\pi \over 2}} \right)$$
ƒ '(0) = $${ - {\pi \over 2}}$$
ƒ is not differentiable at x = 0
ƒ' is increasing in $$\left( { - {\pi \over 2},0} \right)$$ and decreasing
in $$\left( {0,{\pi \over 2}} \right)$$
Explanation
We know, cos-1(-x) = $$\pi $$ - cos-1x
$$ \therefore $$ ƒ(x) = x($$\pi $$ - cos–1(sin|x|))
= x($$\pi $$ - $${\pi \over 2}$$ + sin–1(sin|x|))
= x($$\pi $$ - $${\pi \over 2}$$ + sin–1(sin|x|))
= x$${\pi \over 2}$$ + x|x|
$$ \therefore $$ f(x) = $$\left\{ {\matrix{ {x{\pi \over 2} - {x^2},} & {x < 0} \cr {x{\pi \over 2} + {x^2},} & {x \ge 0} \cr } } \right.$$
Now f'(x) = $$\left\{ {\matrix{ {{\pi \over 2} - 2x,} & {x < 0} \cr {{\pi \over 2} + 2x,} & {x \ge 0} \cr } } \right.$$
and f''(x) = $$\left\{ {\matrix{ { - 2,} & {x < 0} \cr {2,} & {x \ge 0} \cr } } \right.$$
$$ \therefore $$ ƒ' is decreasing in $$\left( { - {\pi \over 2},0} \right)$$ and increasing in $$\left( {0,{\pi \over 2}} \right)$$
$$ \therefore $$ ƒ(x) = x($$\pi $$ - cos–1(sin|x|))
= x($$\pi $$ - $${\pi \over 2}$$ + sin–1(sin|x|))
= x($$\pi $$ - $${\pi \over 2}$$ + sin–1(sin|x|))
= x$${\pi \over 2}$$ + x|x|
$$ \therefore $$ f(x) = $$\left\{ {\matrix{ {x{\pi \over 2} - {x^2},} & {x < 0} \cr {x{\pi \over 2} + {x^2},} & {x \ge 0} \cr } } \right.$$
Now f'(x) = $$\left\{ {\matrix{ {{\pi \over 2} - 2x,} & {x < 0} \cr {{\pi \over 2} + 2x,} & {x \ge 0} \cr } } \right.$$
and f''(x) = $$\left\{ {\matrix{ { - 2,} & {x < 0} \cr {2,} & {x \ge 0} \cr } } \right.$$
$$ \therefore $$ ƒ' is decreasing in $$\left( { - {\pi \over 2},0} \right)$$ and increasing in $$\left( {0,{\pi \over 2}} \right)$$
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