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

(π / 3, π)

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(0, π)

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(π / 2, π)

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(π / 3, 5 π / 3)

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Solution

The correct options are
A $(π / 3, π)$
D $(π / 3, 5 π / 3)$
$f (x) = x - 2 s i n x$
$f^{'} (x) = 1 - 2 c o s x$
For $f (x)$ to be increasing
$f^{'} (x) > 0 \Rightarrow 1 - 2 c o s x > 0$
$c o s x < 1 / 2$
$\Rightarrow x ϵ (\frac{π}{3}, \frac{5 π}{3})$
Hence option A,D are correct

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An interval of increase of the function y=x−2sinx if 0≤x≤2π, is

The correct options are A (π/3,π) D (π/3,5π/3)f(x)=x−2sinxf′(x)=1−2cosxFor f(x) to be increasingf′(x)>0⇒1−2cosx>0cosx<1/2⇒xϵ(π3,5π3)Hence option A,D are correct

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

The correct options are
A $(π / 3, π)$
D $(π / 3, 5 π / 3)$
$f (x) = x - 2 s i n x$
$f^{'} (x) = 1 - 2 c o s x$
For $f (x)$ to be increasing
$f^{'} (x) > 0 \Rightarrow 1 - 2 c o s x > 0$
$c o s x < 1 / 2$
$\Rightarrow x ϵ (\frac{π}{3}, \frac{5 π}{3})$
Hence option A,D are correct