JEE Challenger
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MathHardAdv 2018 • Paper 2

Evaluating Properties of Twice Differentiable Function from Limit Relation

Assume f:(0,π)→Rf: (0, \pi) \rightarrow \mathbb{R} is a twice differentiable function satisfying lim⁡t→xf(x)sin⁡t−f(t)sin⁡xt−x=sin⁡2xfor every x∈(0,π).\lim_{t \to x} \frac{f(x) \sin t - f(t) \sin x}{t-x} = \sin^2 x \quad \text{for every } x \in (0, \pi). Given that f(π6)=−π12f\left(\frac{\pi}{6}\right) = -\frac{\pi}{12}, identify which of the following assertion(s) is (are) CORRECT:

Options

A

f(π4)=π42f\left(\frac{\pi}{4}\right) = \frac{\pi}{4\sqrt{2}}

B

f(x)<x46−x2f(x) < \frac{x^4}{6} - x^2 for all x∈(0,π)x \in (0, \pi)

Correct
C

There exists some α∈(0,π)\alpha \in (0, \pi) such that f′(α)=0f'(\alpha) = 0

Correct
D

f′′(π2)+f(π2)=0f''\left(\frac{\pi}{2}\right) + f\left(\frac{\pi}{2}\right) = 0

Correct

Topics & Concepts

Differential EquationsLinear Differential EquationsDifferentiationLimits

Step-by-Step Solution

To determine the correct options, we first find the explicit form of the function f(x)f(x).

Step 1: Simplify the Given Limit

We are given: lim⁡t→xf(x)sin⁡t−f(t)sin⁡xt−x=sin⁡2xfor every x∈(0,π)\lim_{t \to x} \frac{f(x) \sin t - f(t) \sin x}{t-x} = \sin^2 x \quad \text{for every } x \in (0, \pi)

Since this limit is of the indeterminate form 00\frac{0}{0}, we apply L'Hôpital's Rule with respect to tt: lim⁡t→xddt[f(x)sin⁡t−f(t)sin⁡x]ddt[t−x]=lim⁡t→xf(x)cos⁡t−f′(t)sin⁡x1=f(x)cos⁡x−f′(x)sin⁡x\lim_{t \to x} \frac{\frac{d}{dt}[f(x) \sin t - f(t) \sin x]}{\frac{d}{dt}[t-x]} = \lim_{t \to x} \frac{f(x) \cos t - f'(t) \sin x}{1} = f(x) \cos x - f'(x) \sin x

Thus, the differential equation is: f(x)cos⁡x−f′(x)sin⁡x=sin⁡2xf(x) \cos x - f'(x) \sin x = \sin^2 x

Step 2: Solve the Differential Equation

Divide both sides by sin⁡2x\sin^2 x (since x∈(0,π)x \in (0, \pi), sin⁡x≠0\sin x \neq 0): f′(x)sin⁡x−f(x)cos⁡xsin⁡2x=−1\frac{f'(x) \sin x - f(x) \cos x}{\sin^2 x} = -1

Recognizing the left-hand side as the derivative of f(x)sin⁡x\frac{f(x)}{\sin x}: ddx(f(x)sin⁡x)=−1\frac{d}{dx} \left( \frac{f(x)}{\sin x} \right) = -1

Integrating both sides with respect to xx: f(x)sin⁡x=−x+C  ⟹  f(x)=(C−x)sin⁡x\frac{f(x)}{\sin x} = -x + C \implies f(x) = (C - x) \sin x

Using the initial condition f(π6)=−π12f\left(\frac{\pi}{6}\right) = -\frac{\pi}{12}: f(π6)=(C−π6)sin⁡(π6)=12(C−π6)=−π12f\left(\frac{\pi}{6}\right) = \left(C - \frac{\pi}{6}\right) \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \left(C - \frac{\pi}{6}\right) = -\frac{\pi}{12} C−π6=−π6  ⟹  C=0C - \frac{\pi}{6} = -\frac{\pi}{6} \implies C = 0

Therefore, the function is: f(x)=−xsin⁡xfor x∈(0,π)f(x) = -x \sin x \quad \text{for } x \in (0, \pi)

Step 3: Analyze the Options

  • Option A: f(π4)=−π4sin⁡(π4)=−π42f\left(\frac{\pi}{4}\right) = -\frac{\pi}{4} \sin\left(\frac{\pi}{4}\right) = -\frac{\pi}{4\sqrt{2}} Option A states f(π4)=π42f\left(\frac{\pi}{4}\right) = \frac{\pi}{4\sqrt{2}}, so Option A is incorrect.

  • Option B: For x∈(0,π)x \in (0, \pi), consider the expansion inequality of sin⁡x\sin x: sin⁡x>x−x36\sin x > x - \frac{x^3}{6} Multiplying by −x<0-x < 0 (since x>0x > 0): −xsin⁡x<−x(x−x36)-x \sin x < -x \left(x - \frac{x^3}{6}\right) f(x)<x46−x2f(x) < \frac{x^4}{6} - x^2 Thus, Option B is correct.

  • Option C: The first derivative of f(x)f(x) is: f′(x)=ddx(−xsin⁡x)=−sin⁡x−xcos⁡x=−(sin⁡x+xcos⁡x)f'(x) = \frac{d}{dx}(-x \sin x) = -\sin x - x \cos x = -(\sin x + x \cos x) Let h(x)=sin⁡x+xcos⁡xh(x) = \sin x + x \cos x. The function h(x)h(x) is continuous on [π2,π]\left[\frac{\pi}{2}, \pi\right]: h(π2)=sin⁡(π2)+π2cos⁡(π2)=1>0h\left(\frac{\pi}{2}\right) = \sin\left(\frac{\pi}{2}\right) + \frac{\pi}{2} \cos\left(\frac{\pi}{2}\right) = 1 > 0 h(π)=sin⁡(π)+πcos⁡(π)=−π<0h(\pi) = \sin(\pi) + \pi \cos(\pi) = -\pi < 0 By the Intermediate Value Theorem, there exists at least one α∈(π2,π)⊂(0,π)\alpha \in \left(\frac{\pi}{2}, \pi\right) \subset (0, \pi) such that h(α)=0h(\alpha) = 0, which implies f′(α)=0f'(\alpha) = 0. Thus, Option C is correct.

  • Option D: The second derivative of f(x)f(x) is: f′′(x)=ddx(−sin⁡x−xcos⁡x)=−cos⁡x−(cos⁡x−xsin⁡x)=−2cos⁡x+xsin⁡xf''(x) = \frac{d}{dx}(-\sin x - x \cos x) = -\cos x - (\cos x - x \sin x) = -2\cos x + x \sin x Adding f(x)=−xsin⁡xf(x) = -x \sin x: f′′(x)+f(x)=(−2cos⁡x+xsin⁡x)+(−xsin⁡x)=−2cos⁡xf''(x) + f(x) = (-2\cos x + x \sin x) + (-x \sin x) = -2\cos x Evaluating at x=π2x = \frac{\pi}{2}: f′′(π2)+f(π2)=−2cos⁡(π2)=0f''\left(\frac{\pi}{2}\right) + f\left(\frac{\pi}{2}\right) = -2 \cos\left(\frac{\pi}{2}\right) = 0 Thus, Option D is correct.

Conclusion

The correct options are B, C, and D.

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