If $f\left( x \right) = 3{e^{{x^2}}}$, then $f'\left( x \right) - 2x{\rm{ }}f\left( x \right) + \frac{1}{3}f\left( 0 \right) - f'\left( 0 \right)$ is equal to

$\left( {7/3} \right){e^{{x^2}}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Application +4 | -1

f\left( x \right) = 3{e^{{x^2}}}

, then

f'\left( x \right) - 2x\mathrm{ }f\left( x \right) + \frac{1}{3}f\left( 0 \right) - f'\left( 0 \right)

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because when the given expression is simplified using the derivative of the function, all terms cancel out except for a constant which evaluates to 1.

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f(x)=|x|^{|\tan x|}

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and

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. The minimum value of

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is Let

f(x)=[x] \sin (\pi x)

where

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x

. Then at

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being an integer, the left derivative of

f(x)

is If

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

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\frac{{dy}}{{dx}}

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\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

is equal to If

z = y + f\left( v \right),

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