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MathMedAdv 2010 • Paper 1

Compare Absolute Maxima of Real-Valued Exponential Functions on Unit Interval

Let f\mathrm{f}, g\mathrm{g} and h\mathrm{h} be real-valued functions defined on the interval [0,1][0, 1] by f(x)=ex2+e−x2\mathrm{f}(\mathrm{x}) = \mathrm{e}^{\mathrm{x}^2} + \mathrm{e}^{-\mathrm{x}^2}, g(x)=xex2+e−x2\mathrm{g}(\mathrm{x}) = \mathrm{x}\mathrm{e}^{\mathrm{x}^2} + \mathrm{e}^{-\mathrm{x}^2} and h(x)=x2ex2+e−x2\mathrm{h}(\mathrm{x}) = \mathrm{x}^2\mathrm{e}^{\mathrm{x}^2} + \mathrm{e}^{-\mathrm{x}^2}. If a,b\mathrm{a}, \mathrm{b} and c\mathrm{c} denote, respectively, the absolute maximum of f,g\mathrm{f}, \mathrm{g} and h\mathrm{h} on [0,1][0, 1], then

Options

A

a=b\mathrm{a} = \mathrm{b} and c≠b\mathrm{c} \neq \mathrm{b}

B

a=c\mathrm{a} = \mathrm{c} and a≠b\mathrm{a} \neq \mathrm{b}

C

a≠b\mathrm{a} \neq \mathrm{b} and c≠b\mathrm{c} \neq \mathrm{b}

D

a=b=c\mathrm{a} = \mathrm{b} = \mathrm{c}

Correct

Topics & Concepts

Application of DerivativesMaxima and MinimaMonotonicity

Step-by-Step Solution

To find the absolute maximum of the functions f(x)\mathrm{f}(x), g(x)\mathrm{g}(x), and h(x)\mathrm{h}(x) on the interval [0,1][0, 1], we analyze their definitions:

f(x)=ex2+e−x2\mathrm{f}(x) = e^{x^2} + e^{-x^2}
g(x)=xex2+e−x2\mathrm{g}(x) = x e^{x^2} + e^{-x^2}
h(x)=x2ex2+e−x2\mathrm{h}(x) = x^2 e^{x^2} + e^{-x^2}

Step 1: Maximize f(x)\mathrm{f}(x) on [0,1][0, 1]

Differentiating f(x)\mathrm{f}(x) with respect to xx:
f′(x)=2xex2−2xe−x2=2x(ex2−e−x2)\mathrm{f}'(x) = 2x e^{x^2} - 2x e^{-x^2} = 2x \left(e^{x^2} - e^{-x^2}\right)

For all x∈(0,1]x \in (0, 1]:

  • x>0x > 0
  • x2>0  ⟹  ex2>1x^2 > 0 \implies e^{x^2} > 1 and e−x2<1e^{-x^2} < 1, which gives ex2−e−x2>0e^{x^2} - e^{-x^2} > 0

Thus, f′(x)>0\mathrm{f}'(x) > 0 for all x∈(0,1]x \in (0, 1], meaning f(x)\mathrm{f}(x) is strictly increasing on [0,1][0, 1].
Consequently, the absolute maximum of f(x)\mathrm{f}(x) on [0,1][0, 1] occurs at the right endpoint x=1x = 1:
a=f(1)=e1+e−1=e+1ea = \mathrm{f}(1) = e^1 + e^{-1} = e + \frac{1}{e}

Step 2: Compare g(x)\mathrm{g}(x) and h(x)\mathrm{h}(x) with f(x)\mathrm{f}(x)

For any x∈[0,1]x \in [0, 1], we have the inequalities:
0≤x2≤x≤10 \le x^2 \le x \le 1

Multiplying by ex2>0e^{x^2} > 0 and adding e−x2e^{-x^2}:
x2ex2+e−x2≤xex2+e−x2≤1⋅ex2+e−x2x^2 e^{x^2} + e^{-x^2} \le x e^{x^2} + e^{-x^2} \le 1 \cdot e^{x^2} + e^{-x^2}

Hence, for all x∈[0,1]x \in [0, 1]:
h(x)≤g(x)≤f(x)\mathrm{h}(x) \le \mathrm{g}(x) \le \mathrm{f}(x)

Since a=e+1ea = e + \frac{1}{e} is the absolute maximum of f(x)\mathrm{f}(x) on [0,1][0, 1], we have for all x∈[0,1]x \in [0, 1]:
h(x)≤g(x)≤f(x)≤e+1e\mathrm{h}(x) \le \mathrm{g}(x) \le \mathrm{f}(x) \le e + \frac{1}{e}

Step 3: Evaluate g(1)\mathrm{g}(1) and h(1)\mathrm{h}(1)

Evaluating g(x)\mathrm{g}(x) and h(x)\mathrm{h}(x) at x=1x = 1:
g(1)=1⋅e12+e−12=e+1e\mathrm{g}(1) = 1 \cdot e^{1^2} + e^{-1^2} = e + \frac{1}{e}
h(1)=12⋅e12+e−12=e+1e\mathrm{h}(1) = 1^2 \cdot e^{1^2} + e^{-1^2} = e + \frac{1}{e}

Since g(x)≤e+1e\mathrm{g}(x) \le e + \frac{1}{e} and attains this value at x=1x = 1, the absolute maximum of g(x)\mathrm{g}(x) on [0,1][0, 1] is:
b=e+1eb = e + \frac{1}{e}

Similarly, since h(x)≤e+1e\mathrm{h}(x) \le e + \frac{1}{e} and attains this value at x=1x = 1, the absolute maximum of h(x)\mathrm{h}(x) on [0,1][0, 1] is:
c=e+1ec = e + \frac{1}{e}

Conclusion

Therefore, we have:
a=b=c=e+1ea = b = c = e + \frac{1}{e}

This matches option D.

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