Let be a twice differentiale function such that (sin x cos y)(f(2x + 2y) – f(2x – 2y)) = (cos x sin y)(f(2x + 2y) + f(2x – 2y)), for all x, y R. If , then value of is : [2025]

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Solution

Q.
Let $f : R \to R$ be a twice differentiale function such that (sin x cos y)(f(2x + 2y) – f(2x – 2y)) = (cos x sin y)(f(2x + 2y) + f(2x – 2y)), for all x, y $\in$ R.

If $f' (0) = \frac{1}{2}$ , then the value of $24 f'' (\frac{5 π}{3})$ is : [2025]

1 –3

2 –2

3 3

4 2

Ans.
(1)

We have,

(sin x cos y)(f(2x + 2y) – f(2x – 2y)) = (cos x sin y)(f(2x + 2y) + f(2x – 2y))

$\Rightarrow$ f(2x + 2y) sin (x – y) = f(2x – 2y) sin (x + y)

$\Rightarrow \frac{f (2 x + 2 y)}{\sin (x + y)} = \frac{f (2 x - 2 y)}{\sin (x - y)}$

Put 2x + 2y = m and 2x – 2y = n, we get

$\frac{f (m)}{\sin (\frac{m}{2})} = \frac{f (n)}{\sin (\frac{n}{2})} = K$

$\Rightarrow f (m) = K \sin (\frac{m}{2}) and f (n) = K \sin (\frac{n}{2})$

$∴$ $f (x) = K \sin (\frac{x}{2}) \Rightarrow f' (x) = \frac{K}{2} \cos (\frac{x}{2})$

Now, $f' (0) = \frac{1}{2} \Rightarrow \frac{1}{2} = \frac{K}{2} \Rightarrow K = 1$

$∴ f' (x) = \frac{1}{2} \cos \frac{x}{2} \Rightarrow f'' (x) = - \frac{1}{4} \sin \frac{x}{2}$

$∴ 24 f'' (\frac{5 π}{3}) = 24 (- \frac{1}{4} \sin (\frac{5 π}{6})) = \frac{- 24}{8} = - 3$ .

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