Mathematics · Methods of Differentiation

JEE Main 2025 — 2 April, Morning Shift — Question 27

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ 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 $x, y \in \mathbb{R}$ . If $\quad f^{\prime}(0)=\frac{1}{2}$ , then the value of $24 f^{\prime \prime}\left(\frac{5 \pi}{3}\right)$ is :

Option A:
$3$
Option B:
$2$
Option C:
$-3$
Correct
Option D:
$-2$

Answer: C

Step-by-step solution

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

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

$f(2 x)=5 \sin x \quad(\because y=0)$

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

$f^{\prime}(x)=\frac{k}{2} \cos \frac{x}{2}$

$f^{\prime}(0)=\frac{1}{2} \Rightarrow k=1$

$f(x)=\sin \frac{x}{2} \Rightarrow f^{\prime}(x)=\frac{1}{2} \cos \frac{x}{2}$

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

$24 f^{\prime \prime}\left(\frac{5 \pi}{3}\right)=-3$

Answer key and solution verified before publishing.

Practise Methods of Differentiation

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2025
Paper: 2 April, Morning Shift
Subject: Mathematics
Chapter: Methods of Differentiation
Topic: First principle Rule & its application in functional equations