The interval of increase of the function $y=$ $x-2 \sin x$ if $0 \leq x \leq 2 \pi$, is

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

The interval of increase of the function

y=

x-2 \sin x

0 \leq x \leq 2 \pi

, is

View answer and explanation

Correct option: D

(\pi / 3,5 \pi / 3)

Option D (

\frac{{\pi}}{{3}}, \frac{{5 \pi}}{{3}}

) is correct because the function is increasing where its derivative is positive, which occurs in this interval.

Practice All Chapter MCQs →

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

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

x

y

x>0

x y=1

x+y

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

[x]

x

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),

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

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