JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Continuity and Differentiability

Let

f : R \to 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))

, for all

x, y \in R

.
If

f^{'} (0) = \frac{1}{2}

, then the value of

24 f^{''} (\frac{5 π}{3})

is:

Answer: (B)

Solution

From the functional equation, rearranging gives:

f (2 x + 2 y) sin (x - y) = f (2 x - 2 y) sin (x + y)

This implies

\frac{f ( 2 x + 2 y )}{s i n ( x + y )} = \frac{f ( 2 x - 2 y )}{s i n ( x - y )} = K

(constant).

Substituting

m = 2 x + 2 y

and

n = 2 x - 2 y

, we get

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

.

So

f (x) = K sin (\frac{x}{2})

.

Then

f^{'} (x) = \frac{K}{2} cos (\frac{x}{2})

.

Using

f^{'} (0) = \frac{1}{2}

\frac{K}{2} = \frac{1}{2}

, so

K = 1

.

Thus

f^{''} (x) = - \frac{1}{4} sin (\frac{x}{2})

.

Therefore:

24 f^{''} (\frac{5 π}{3}) = 24 \cdot (- \frac{1}{4}) sin (\frac{5 π}{6}) = - 6 \cdot \frac{1}{2} = - 3

JEE Main · 2025 · 02 Apr 2025 · Shift 1 · Single correct option