Exams Joint Entrance Examination Mathematics Differentiation Question 4 Question 4 of 67 Medium MCQ
Let f : R → R f:R\to R f : R → R be a twice differentiable function such that ( sin x cos y ) ( f ( 2 x + 2 y ) − f ( 2 x − 2 y ) ) = ( cos x sin y ) ( f ( 2 x + 2 y ) + f ( 2 x − 2 y ) ) (\sin x\cos y)(f(2x+2y)-f(2x-2y))=(\cos x\sin y)(f(2x+2y)+f(2x-2y)) ( sin x cos y ) ( f ( 2 x + 2 y ) − f ( 2 x − 2 y )) = ( cos x sin y ) ( f ( 2 x + 2 y ) + f ( 2 x − 2 y )) , for all x , y ∈ R x,y\in R x , y ∈ R .
If f ′ ( 0 ) = 1 2 {f}^{'}(0)=\frac{1}{2} f ′ ( 0 ) = 2 1 , then the value of 24 f ′ ′ ( 5 π 3 ) 24{f}^{''}(\frac{5\pi }{3}) 24 f ′′ ( 3 5 π ) is :
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sin ( x − y ) f ( 2 x + 2 y ) = f ( 2 x − 2 y ) sin ( x + y ) f ( 2 x + 2 y ) sin ( x + y ) = f ( 2 x − 2 y ) sin ( x − y ) = k ( ) f ( 2 x + 2 y ) = k sin ( x + y ) f ( 2 x ) = 5 sin x ( ∵ y = 0 ) f ( x ) = k sin x 2 f ′ ( x ) = k 2 cos x 2 f ′ ( 0 ) = 1 2 ⇒ k = 1 f ( x ) = sin x 2 ⇒ f ′ ( x ) = 1 2 cos x 2 f ′ ′ ( x ) = − 1 4 sin x 2 24 f ′ ′ ( 5 π 3 ) = − 3 \begin{gathered} \sin (x-y)f(2x+2y)=f(2x-2y)\sin (x+y) \\ \frac{f(2x+2y)}{\sin (x+y)}=\frac{f(2x-2y)}{\sin (x-y)}=k() \\ f(2x+2y)=k\sin (x+y) \\ f(2x)=5\sin x\, (\because y=0) \\ f(x)=k\sin \frac{x}{2} \\ {f}^{'}(x)=\frac{k}{2}\cos \frac{x}{2} \\ {f}^{'}(0)=\frac{1}{2}\Rightarrow k=1 \\ f(x)=\sin \frac{x}{2}\Rightarrow {f}^{'}(x)=\frac{1}{2}\cos \frac{x}{2} \\ {f}^{''}(x)=-\frac{1}{4}\sin \frac{x}{2} \\ 24{f}^{''}(\frac{5\pi }{3})=-3 \end{gathered} sin ( x − y ) f ( 2 x + 2 y ) = f ( 2 x − 2 y ) sin ( x + y ) sin ( x + y ) f ( 2 x + 2 y ) = sin ( x − y ) f ( 2 x − 2 y ) = k ( ) f ( 2 x + 2 y ) = k sin ( x + y ) f ( 2 x ) = 5 sin x ( ∵ y = 0 ) f ( x ) = k sin 2 x f ′ ( x ) = 2 k cos 2 x f ′ ( 0 ) = 2 1 ⇒ k = 1 f ( x ) = sin 2 x ⇒ f ′ ( x ) = 2 1 cos 2 x f ′′ ( x ) = − 4 1 sin 2 x 24 f ′′ ( 3 5 π ) = − 3 Answer the question to unlock