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Differentiation question

2025 · 2 Apr · Shift 1 · Q40
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  5. /2025 · 2 Apr · Shift 1 · Q40

Differentiation question

2025 · 2 Apr · Shift 1 · Q40

JEE MainMathematicsDifferentiationMCQ+4 / −1
Let f:R→Rf: \mathbf{R} \rightarrow \mathbf{R}f:R→R be a twice differentiable function such that (sin⁡xcos⁡y)(f(2x+2y)−f(2x−2y))=(cos⁡xsin⁡y)(f(2x+2y)+f(2x−2y))(\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))(sinxcosy)(f(2x+2y)−f(2x−2y))=(cosxsiny)(f(2x+2y)+f(2x−2y)), for all x,y∈Rx, y \in \mathbf{R}x,y∈R. If f′(0)=12f^{\prime}(0)=\frac{1}{2}f′(0)=21​, then the value of 24f′′(5π3)24 f^{\prime \prime}\left(\frac{5 \pi}{3}\right)24f′′(35π​) is :
  1. A
    2
  2. B
    3
  3. C
    −-− 3
  4. D
    −-− 2
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