JEE Main 2022 (Online) 25th June Morning Shift | Differentiation Question 36 | Mathematics

Let f : R $$\to$$ R be defined as $$f(x) = {x^3} + x - 5$$. If g(x) is a function such that $$f(g(x)) = x,\forall 'x' \in R$$, then g'(63) is equal to ________________.

$${1 \over {49}}$$

$${3 \over {49}}$$

$${43 \over {49}}$$

$${91 \over {49}}$$

JEE Main 2022 (Online) 24th June Evening Shift

MCQ (Single Correct Answer)

-1

If $$y = {\tan ^{ - 1}}\left( {\sec {x^3} - \tan {x^3}} \right),{\pi \over 2} < {x^3} < {{3\pi } \over 2}$$, then

$$xy'' + 2y' = 0$$

$${x^2}y'' - 6y + {{3\pi } \over 2} = 0$$

$${x^2}y'' - 6y + 3\pi = 0$$

$$xy'' - 4y' = 0$$

JEE Main 2021 (Online) 27th August Evening Shift

MCQ (Single Correct Answer)

-1

If $$y(x) = {\cot ^{ - 1}}\left( {{{\sqrt {1 + \sin x} + \sqrt {1 - \sin x} } \over {\sqrt {1 + \sin x} - \sqrt {1 - \sin x} }}} \right),x \in \left( {{\pi \over 2},\pi } \right)$$, then $${{dy} \over {dx}}$$ at $$x = {{5\pi } \over 6}$$ is :

$$ - {1 \over 2}$$

$$-$$1

$${1 \over 2}$$

JEE Main 2021 (Online) 26th August Morning Shift

MCQ (Single Correct Answer)

-1

Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1. Then :

$${(1 - x)^2}f'(x) - 2{(f(x))^2} = 0$$

$${(1 + x)^2}f'(x) + 2{(f(x))^2} = 0$$

$${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$$

$${(1 + x)^2}f'(x) - 2{(f(x))^2} = 0$$

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