Application of DerivativeshardFree

f(x)/f(y) <= 2^((x - y)^2) and constant functions | JEE Advanced

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Question
The function ff satisfies f(x)(f(y))−1≤2(x−y)2f\left(x\right)\left(f\left(y\right)\right)^{-1} \le 2^{\left(x-y\right)^2} ∀ x,y∈Df\forall\,x,y \in D_f, where DfD_f is the domain set of ff. Then f(x)f\left(x\right) will be
A(x+x3)\left(\sqrt x+x^3\right)
B∫0x2t3 dt\displaystyle\int_0^x 2t^3\,dt
C∫0sin⁡2xsin⁡−1t dt+∫0cos⁡2xcos⁡−1t dt\displaystyle\int_0^{\sin^2x}\sin^{-1}\sqrt t\,dt+\int_0^{\cos^2x}\cos^{-1}\sqrt t\,dtcorrect
D−∣x∣\sqrt{-\left|x\right|}correct
Solution
Step 1: The ratio needs f(y)≠0f\left(y\right) \ne 0, and where f>0f>0 we may take log⁡2\log_2:
f(x)f(y)≤2(x−y)2⟹log⁡2f(x)−log⁡2f(y)≤(x−y)2\frac{f\left(x\right)}{f\left(y\right)} \le 2^{\left(x-y\right)^2} \quad\Longrightarrow\quad \log_2f\left(x\right)-\log_2f\left(y\right) \le \left(x-y\right)^2
 Step 2: The condition holds for all pairs, so swapping xx and yy,
log⁡2f(y)−log⁡2f(x)≤(y−x)2=(x−y)2\log_2f\left(y\right)-\log_2f\left(x\right) \le \left(y-x\right)^2 = \left(x-y\right)^2
 so, writing g=log⁡2fg = \log_2f,
∣g(x)−g(y)∣≤(x−y)2∀ x,y∈Df\left|g\left(x\right)-g\left(y\right)\right| \le \left(x-y\right)^2 \qquad\forall\,x,y \in D_f
 Step 3: For x≠yx \ne y,
∣g(x)−g(y)x−y∣≤∣x−y∣→y→x0\left|\frac{g\left(x\right)-g\left(y\right)}{x-y}\right| \le \left|x-y\right| \xrightarrow[y\to x]{} 0
 so g′(x)g'\left(x\right) exists and equals 00 at every interior point of DfD_f, hence gg is constant on each interval of DfD_f and
f(x)=2g is a constant function.f\left(x\right) = 2^{g} \ \text{is a constant function.}
 So the question is: which of the four options is constant on its domain? Step 4: x+x3\sqrt x+x^3 is strictly increasing on [0,∞)\left[0,\infty\right), not constant
∫0x2t3dt=x42\int_0^x2t^3dt = \frac{x^4}{2}
 also not constant Both vanish at x=0x = 0, a point of their own domains, where (f(y))−1\left(f\left(y\right)\right)^{-1} is undefined, so they fail the hypothesis as well. Step 5: For (3), by the fundamental theorem and the chain rule, for x∈(0,π2)x \in \left(0,\tfrac\pi2\right),
ddx[∫0sin⁡2xsin⁡−1t dt]=sin⁡−1(sin⁡2x)⋅2sin⁡xcos⁡x=xsin⁡2x\frac{d}{dx}\left[\int_0^{\sin^2x}\sin^{-1}\sqrt t\,dt\right] = \sin^{-1}\left(\sqrt{\sin^2x}\right)\cdot2\sin x\cos x = x\sin2x
ddx[∫0cos⁡2xcos⁡−1t dt]=cos⁡−1(cos⁡2x)⋅(−2cos⁡xsin⁡x)=−xsin⁡2x\frac{d}{dx}\left[\int_0^{\cos^2x}\cos^{-1}\sqrt t\,dt\right] = \cos^{-1}\left(\sqrt{\cos^2x}\right)\cdot\left(-2\cos x\sin x\right) = -x\sin2x
 The two derivatives cancel, so the sum is constant, its value being π4\tfrac\pi4 Step 6: For (4), −∣x∣\sqrt{-\left|x\right|} is real only when −∣x∣≥0-\left|x\right| \ge 0, i.e. x=0x = 0, so
Df={0},f(0)=0D_f = \left\{0\right\},\qquad f\left(0\right) = 0
 On a one-point domain the condition only ever involves x=yx = y, where it reads 1≤11 \le 1, true. 1 function on a single point is constant Answer: (3) and (4)
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