Method of DifferentiationmediumFree

Method of Differentiation: Function Satisfies Value

JEE Maths question with a full step-by-step solution.

Question
A function f:R→Rf:\mathbb{R}\to\mathbb{R} satisfies sin⁡xcos⁡y (f(2x+2y)−f(2x−2y))=cos⁡xsin⁡y (f(2x+2y)+f(2x−2y))\sin x\cos y\,(f(2x+2y)-f(2x-2y)) = \cos x\sin y\,(f(2x+2y)+f(2x-2y)). If f′(0)=12f'(0) = \dfrac{1}{2}, then the value of 4f′′(x)+f(x)4f''(x)+f(x) is.
Solution
Answer: 0
Step 1: Simplify the functional equation Rearranging:
f(2x+2y)sin⁡(x−y)=f(2x−2y)sin⁡(x+y)f(2x+2y)\sin(x-y) = f(2x-2y)\sin(x+y)
f(2x+2y)f(2x−2y)=sin⁡(x+y)sin⁡(x−y)\frac{f(2x+2y)}{f(2x-2y)} = \frac{\sin(x+y)}{\sin(x-y)}
 This identifies f(x)=ksin⁡x2f(x) = k\sin\dfrac{x}{2}. Step 2: Apply f′(0)=12f'(0) = \dfrac{1}{2}
f′(x)=k2cos⁡x2  ⟹  f′(0)=k2=12  ⟹  k=1f'(x) = \frac{k}{2}\cos\frac{x}{2} \implies f'(0) = \frac{k}{2} = \frac{1}{2} \implies k = 1
 So f(x)=sin⁡x2f(x) = \sin\dfrac{x}{2}. Step 3: Evaluate 4f′′(x)+f(x)4f''(x)+f(x)
f′′(x)=−14sin⁡x2f''(x) = -\frac{1}{4}\sin\frac{x}{2}
4f′′(x)+f(x)=−sin⁡x2+sin⁡x2=04f''(x)+f(x) = -\sin\frac{x}{2}+\sin\frac{x}{2} = 0
 Answer: 0
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