JEE 2025 — Mathematics PYQ

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Question

JEE 2025 — Mathematics PYQ

JEE | Mathematics | 2025

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 $x, y \in R$ and $f^{'} (0) = \frac{1}{2}$ . If $f (x) = \frac{1}{2}$ , then the value of $24 f^{''} (\frac{5 π}{3})$ is

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Explanation

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

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

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

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

$Put 2 x + 2 y = m, 2 x - 2 y = n$

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

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

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

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

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

$Put x = 0 : f^{'} (0) = \frac{K}{2} = \frac{1}{2} \Rightarrow K = 1$

$f^{'} (x) = \frac{1}{2} cos (\frac{x}{2})$

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

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

$= - \frac{24}{4} sin (\frac{5 π}{6})$

$= - 6 sin (π - \frac{π}{6})$

$= - 6 sin (\frac{π}{6})$

$= - 6 \times \frac{1}{2} = - 3$

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