Question

A function f : R → R satisfies

sin x cos y (ƒ(2x + 2y) – ƒ (2x – 2y))

= cos x sin y (ƒ(2x + 2y) + ƒ(2x – 2y)) If ƒ′(0) = $\frac{1}{2}$, then –

• A. ƒ′′(x) = ƒ(x) = 0
• B. 4ƒ′′(x) + ƒ(x) = 0
• C. ƒ′′(x) + ƒ(x) = 0
• D. 4ƒ′′(x) – ƒ(x) = 0

Answer 1

Answer: (B)

4ƒ′′(x) + ƒ(x) = 0

Answer 2

We have $\frac{ƒ(2x + 2y)}{ƒ(2x - 2y)}$ = $\frac{\sin(x + y)}{\sin(x - y)}$

⇒ $\frac{ƒ(\alpha)}{\sin\frac{\alpha}{2}}$ = $\frac{ƒ(\beta)}{\sin\frac{\beta}{2}}$ = K

⇒ ƒ(x) = K sin$\frac{x}{2}$

⇒ ƒ′(x) = $\frac{K}{2}$ cos $\frac{x}{2}$

and ƒ′′ (x) = $\frac{–K}{4}$ sin $\frac{x}{2}$

⇒ 4ƒ′′ (x) + ƒ(x) = 0

Hence (2) is the correct answer.