JEE Main 2013 23 April Online
Exam JEE Main
Year 2013
Date 23 April
Mode Online
Subject Mathematics
Chapter Differentiation JEE Main 2013 · 23 April · Online · Mathematics · Differentiation
If f(x)= ( x) and f^ (x)+ x f^ (x)+g(x) =0 , then g(x) is :
If
f ( x ) = sin ( sin x ) f(x)=\sin (\sin x) f ( x ) = sin ( sin x ) and
f ′ ′ ( x ) + tan x f ′ ( x ) + g ( x ) f^{\prime \prime}(x)+\tan x f^{\prime}(x)+g(x) f ′′ ( x ) + tan x f ′ ( x ) + g ( x ) = 0 =0 = 0 , then
g ( x ) g(x) g ( x ) is :
Options A cos 2 x cos ( sin x ) \cos ^2 x \cos (\sin x) cos 2 x cos ( sin x ) B sin 2 x cos ( cos x ) \sin ^2 x \cos (\cos x) sin 2 x cos ( cos x ) C sin 2 x sin ( cos x ) \sin ^2 x \sin (\cos x) sin 2 x sin ( cos x ) D cos 2 x sin ( sin x ) \cos ^2 x \sin (\sin x) cos 2 x sin ( sin x ) Correct answer D. cos 2 x sin ( sin x ) \cos ^2 x \sin (\sin x) cos 2 x sin ( sin x )
Step-by-step solution
f ( x ) = sin ( sin x ) ⇒ f ′ ( x ) = cos ( sin x ) ⋅ \cosx ⇒ f ′ ′ ( x ) = − sin ( sin x ) ⋅ cos 2 x + cos ( sin x ) . ( − sin x ) = − cos 2 x ⋅ sin ( \sinx ) − sin x ⋅ cos ( \sinx ) Now f ′ ′ ( x ) + tan x . f ′ ( x ) + g ( x ) = 0 ⇒ g ( x ) = cos 2 x ⋅ sin ( \sinx ) + sin x ⋅ cos ( \sinx ) − tan x ⋅ \cosx ⋅ cos ( \sinx ) ⇒ g ( x ) = cos 2 x ⋅ sin ( \sinx ) . \begin{aligned}& f(x)=\sin (\sin x) \\
& \Rightarrow f^{\prime}(x)=\cos (\sin x) \cdot \cosx \\
& \Rightarrow f^{\prime \prime}(x)=-\sin (\sin x) \cdot \cos ^2 x+\cos (\sin x) \text{. }\\
& (-\sin x) \\
& =-\cos ^2 x \cdot \sin (\sinx)-\sin x \cdot \cos (\sinx) \\
& \text{ Now }f^{\prime \prime}(x)+\tan x . f^{\prime}(x)+g(x)=0 \\
& \Rightarrow g(x)=\cos ^2 x \cdot \sin (\sinx)+\sin x \cdot \cos (\sinx) \\
& -\tan x \cdot \cosx \cdot \cos (\sinx) \\
& \Rightarrow g(x)=\cos ^2 x \cdot \sin (\sinx) \text{. }\\
&
\end{aligned} f ( x ) = sin ( sin x ) ⇒ f ′ ( x ) = cos ( sin x ) ⋅ \cosx ⇒ f ′′ ( x ) = − sin ( sin x ) ⋅ cos 2 x + cos ( sin x ) . ( − sin x ) = − cos 2 x ⋅ sin ( \sinx ) − sin x ⋅ cos ( \sinx ) Now f ′′ ( x ) + tan x . f ′ ( x ) + g ( x ) = 0 ⇒ g ( x ) = cos 2 x ⋅ sin ( \sinx ) + sin x ⋅ cos ( \sinx ) − tan x ⋅ \cosx ⋅ cos ( \sinx ) ⇒ g ( x ) = cos 2 x ⋅ sin ( \sinx ) .
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