Application of DerivativesmediumFree

Maxima of g(x) = 3x/f(x) When f(x+y) = f(x)f(y) | JEE Advanced

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Question
Let a function f:R→Rf : \mathbb{R} \to \mathbb{R} satisfy the functional equation
f(x+y)=f(x)⋅f(y),f(x+y) = f(x)\cdot f(y),
 where f(0)≠0f(0) \ne 0 and f′(0)=1f'(0) = 1. Given g(x)=3xf(x),g(x) = \frac{3x}{f(x)}, then
Athe value of g′(ln⁡e)g'\left(\ln\sqrt{e}\right) is 32(e)−12\dfrac32 (e)^{-\frac12}correct
Bg(x)g(x) attains local minima at x=1x = 1
Cg(x)g(x) attains local maxima at x=0x = 0
Dif g(x)=kg(x) = k has two solutions then k∈(0, 3e)k \in \left(0,\ \dfrac{3}{e}\right)correct
Solution
Step 1: Identify ff. The equation f(x+y)=f(x)f(y)f(x+y) = f(x)f(y) with ff differentiable and f(0)≠0f(0) \ne 0 has exponential solutions f(x)=axf(x) = a^{x}. Differentiating and putting x=0x = 0 gives f′(0)=ln⁡af'(0) = \ln a, and f′(0)=1f'(0) = 1 implies a=ea = e:
f(x)=ex.f(x) = e^{x} .
 Step 2: Write gg explicitly.
g(x)=3xex=3x e−x.g(x) = \frac{3x}{e^{x}} = 3x\,e^{-x} .
 Step 3: Differentiate by the product rule.
g′(x)=3e−x+3x(−e−x)=3(1−x)e−x.g'(x) = 3e^{-x} + 3x\left(-e^{-x}\right) = 3(1-x)e^{-x} .
 Step 4: Test (1). Since ln⁡e=12\ln\sqrt e = \dfrac12,
g′(12)=3(1−12)e−1/2=32e−1/2.g'\left(\tfrac12\right) = 3\left(1-\tfrac12\right)e^{-1/2} = \frac32 e^{-1/2} .
 (1) is true. Step 5: Find the turning point. g′(x)=0g'(x) = 0 gives x=1x = 1, and since e−x>0e^{-x} > 0 the sign of g′g' is the sign of 1−x1-x:
g′>0 for x<1,g′<0 for x>1.g' > 0 \ \text{for } x < 1, \qquad g' < 0 \ \text{for } x > 1 .
 Step 6: Test (2). The derivative changes from positive to negative at x=1x = 1, so that is a local **maximum**, not a minimum. (2) is false. Step 7: Test (3). g′(0)=3(1−0)e0=3≠0g'(0) = 3(1-0)e^{0} = 3 \ne 0, so x=0x = 0 is not a critical point at all and cannot be a local maximum. (3) is false. Step 8: Test (4). Describe the whole graph. The maximum value is
g(1)=3e.g(1) = \frac{3}{e} .
 As x→−∞x \to -\infty, 3xe−x→−∞3xe^{-x} \to -\infty; as x→+∞x \to +\infty, 3xe−x→0+3xe^{-x} \to 0^{+}. So gg climbs from −∞-\infty up to 3e\dfrac3e and then decreases towards 00 without reaching it. Step 9: Read off how many times a horizontal line y=ky = k meets the graph. It meets it twice exactly when kk lies strictly between the eventual level 00 and the peak:
0<k<3e.0 < k < \frac3e .
 (4) is true. Answer: (1) and (4).
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