JEE MATHEMATICS MCQ: DIFFERENTIATION

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Analyzing +4 | -1

f(x)=\left|\begin{array}{ccc}\cos x & x & 1 \\ 2 \sin x & x^{2} & 2 x \\ \tan x & x & 1\end{array}\right|

, then

\lim _{x \rightarrow 0} \frac{f^{\prime}(x)}{x}

View answer and explanation

Correct option: A

Exists and is equal to -2 [Online April 15, 2018]

Option A (Exists and is equal to -2) is correct because the limit of f'(x)/x as x approaches 0 is -2, obtained by applying L'Hopital's rule or by simplifying the expression for f'(x).

More Questions from DIFFERENTIATION

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

, then

f^{\prime}\left(\frac{-\pi}{6}\right)

is equal to Let

x

and

y

be two real numbers such that

x>0

and

x y=1

. The minimum value of

x+y

is Let

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

where

[x]

denotes the integer part of

x

. Then at

x=k, k

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],

then

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

is equal to

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

is equal to If

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

where

v = \left( {\frac{x}{y}} \right),

then

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