ExamBro
ExamBro
JEE Mains · Maths · STD 12 - 5. continuity and differentiation

मान लीजिए कि \(f: \mathbf{R} \rightarrow \mathbf{R}\) एक ऐसा द्वि-अवकलनीय फलन है कि सभी \(\mathrm{x}, \mathrm{y} \in \mathbf{R}\) के लिए \((\sin x \cos y)(f(2 x+2 y)-f(2 x-2 y))=(\cos x\) \(\sin \mathrm{y})(f(2 \mathrm{x}+2 \mathrm{y})+f(2 \mathrm{x}-2 \mathrm{y}))\) है। यदि \(f^{\prime}(0)=\frac{1}{2}\) है, तो \(24 f^{\prime \prime}\left(\frac{5 \pi}{3}\right)\) का मान क्या है?

  1. A 2
  2. B \(-3\)
  3. C 3
  4. D \(-2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(-3\)

Step-by-step Solution

Detailed explanation

\begin{aligned} & (\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-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)} \\ & \text { Put } 2 x+2 y=m, 2 x-2 y=n \\ & \frac{f(m)}{\sin…

Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app