JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

f\left( x \right) = {\sin ^{ - 1}}\left( {\frac{{2x}}{{1 + {x^2}}}} \right)

, then

f\left( x \right)

is differentiable on

View answer and explanation

Correct option: B

R - \left\{ { - 1,1} \right\}

Option B (

R - \\{-1, 1\\}

) is correct because the derivative of

f(x)

exists for all real numbers except

x = -1

and

x = 1

, where the function is not differentiable.

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

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x

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x y=1

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[x]

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x=k, k

f(x)

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],

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

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

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

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v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}