JEE Main 2020 — Application Of Derivatives Question with Solution
From: JEE Main 2020 (Online) 8th January Morning Slot
Question
Let ƒ(x) = xcos–1(–sin|x|), , then
which of the following is true?
Choose an option
Show full solutionCorrect option: A
Correct answer
Aƒ' is decreasing in and increasing
in
Step-by-step explanation
We know, cos-1(-x) = - cos-1x
ƒ(x) = x( - cos–1(sin|x|))
= x( - + sin–1(sin|x|))
= x( - + sin–1(sin|x|))
= x + x|x|
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.
ƒ' is decreasing in and increasing in
ƒ(x) = x( - cos–1(sin|x|))
= x( - + sin–1(sin|x|))
= x( - + sin–1(sin|x|))
= x + x|x|
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.
ƒ' is decreasing in and increasing in
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