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Differentiation question

2025 · 2 Apr · Shift 1 · Q40
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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
View written solutionFree

Correct answer: C

  1. Given functional equation

We are given, for all x,y∈Rx,y\in \mathbb Rx,y∈R,

(sin⁡xcos⁡y)(f(2x+2y)−f(2x−2y))=(cos⁡xsin⁡y)(f(2x+2y)+f(2x−2y)).(\sin x\cos y)\big(f(2x+2y)-f(2x-2y)\big)=(\cos x\sin y)\big(f(2x+2y)+f(2x-2y)\big).(sinxcosy)(f(2x+2y)−f(2x−2y))=(cosxsiny)(f(2x+2y)+f(2x−2y)).

Let A=f(2x+2y),B=f(2x−2y).A=f(2x+2y),\qquad B=f(2x-2y).A=f(2x+2y),B=f(2x−2y). Then the equation becomes

(sin⁡xcos⁡y)(A−B)=(cos⁡xsin⁡y)(A+B).(\sin x\cos y)(A-B)=(\cos x\sin y)(A+B).(sinxcosy)(A−B)=(cosxsiny)(A+B).

Rearrange:

A(sin⁡xcos⁡y−cos⁡xsin⁡y)=B(sin⁡xcos⁡y+cos⁡xsin⁡y).A(\sin x\cos y-\cos x\sin y)=B(\sin x\cos y+\cos x\sin y).A(sinxcosy−cosxsiny)=B(sinxcosy+cosxsiny).

Using identities,

sin⁡xcos⁡y−cos⁡xsin⁡y=sin⁡(x−y),\sin x\cos y-\cos x\sin y=\sin(x-y),sinxcosy−cosxsiny=sin(x−y), sin⁡xcos⁡y+cos⁡xsin⁡y=sin⁡(x+y).\sin x\cos y+\cos x\sin y=\sin(x+y).sinxcosy+cosxsiny=sin(x+y).

So,

Asin⁡(x−y)=Bsin⁡(x+y),A\sin(x-y)=B\sin(x+y),Asin(x−y)=Bsin(x+y),

i.e.

f(2x+2y)sin⁡(x−y)=f(2x−2y)sin⁡(x+y).f(2x+2y)\sin(x-y)=f(2x-2y)\sin(x+y).f(2x+2y)sin(x−y)=f(2x−2y)sin(x+y).
  1. Change variables

Let u=x+y,v=x−y.u=x+y,\qquad v=x-y.u=x+y,v=x−y. Then

So the equation becomes

f(2u)sin⁡v=f(2v)sin⁡ufor all u,v∈R.f(2u)\sin v=f(2v)\sin u \qquad \text{for all }u,v\in\mathbb R.f(2u)sinv=f(2v)sinufor all u,v∈R.

Thus,

f(2u)sin⁡u=f(2v)sin⁡v\frac{f(2u)}{\sin u}=\frac{f(2v)}{\sin v}sinuf(2u)​=sinvf(2v)​

whenever denominators are nonzero. Hence this ratio is a constant, say ccc. Therefore,

f(2t)=csin⁡t∀t∈R.f(2t)=c\sin t \qquad \forall t\in\mathbb R.f(2t)=csint∀t∈R.

Now let z=2tz=2tz=2t. Then t=z/2t=z/2t=z/2, so

f(z)=csin⁡(z2).f(z)=c\sin\left(\frac z2\right).f(z)=csin(2z​).

Thus the function must be of the form

f(x)=csin⁡(x2).f(x)=c\sin\left(\frac x2\right).f(x)=csin(2x​).
  1. Use the condition f′(0)=12f'(0)=\tfrac12f′(0)=21​

Differentiate:

f′(x)=c2cos⁡(x2).f'(x)=\frac c2\cos\left(\frac x2\right).f′(x)=2c​cos(2x​).

So

f′(0)=c2=12.f'(0)=\frac c2=\frac12.f′(0)=2c​=21​.

Hence,

Therefore,

f(x)=sin⁡(x2).f(x)=\sin\left(\frac x2\right).f(x)=sin(2x​).
  1. Compute second derivative

We have

f′(x)=12cos⁡(x2),f'(x)=\frac12\cos\left(\frac x2\right),f′(x)=21​cos(2x​), f′′(x)=−14sin⁡(x2).f''(x)=-\frac14\sin\left(\frac x2\right).f′′(x)=−41​sin(2x​).

Now evaluate at x=5π3x=\frac{5\pi}{3}x=35π​:

f′′(5π3)=−14sin⁡(5π6).f''\left(\frac{5\pi}{3}\right)=-\frac14\sin\left(\frac{5\pi}{6}\right).f′′(35π​)=−41​sin(65π​).

Since

sin⁡(5π6)=12,\sin\left(\frac{5\pi}{6}\right)=\frac12,sin(65π​)=21​,

we get

f′′(5π3)=−14⋅12=−18.f''\left(\frac{5\pi}{3}\right)=-\frac14\cdot \frac12=-\frac18.f′′(35π​)=−41​⋅21​=−81​.

Thus,

24f′′(5π3)=24(−18)=−3.24f''\left(\frac{5\pi}{3}\right)=24\left(-\frac18\right)=-3.24f′′(35π​)=24(−81​)=−3.
  1. Check options
  • A: 222 ❌
  • B: 333 ❌
  • C: −3-3−3 ✅
  • D: −2-2−2 ❌

So the correct option is C.

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