Question

Let f, g and h be real valued functions defined on the interval [0,1] by $f\left( x \right) = {e^{{x^2}}} + {e^{ - {x^2}}},g\left( x \right) = x{e^{{x^2}}} + {e^{ - {x^2}}}$ and $h\left( x \right) = {x^2}{e^{{x^2}}} + {e^{ - {x^2}}}$ If a, b and c denote, respectively the absolute maximum value of f, g and h on [0,1] respectively, then
A. a = b and ${\text{b}} \ne {\text{c}}$
B. a = c and ${\text{a}} \ne {\text{b}}$
C. ${\text{a}} \ne {\text{b}}$ and ${\text{c}} \ne {\text{b}}$
D. a = b = c.

Answer 1

To solve this question, we will use the concept of maxima and minima of application of derivatives. If $y = f\left( x \right)$ be a function defined on [a,b], then we will use the following algorithm for finding the maximum and minimum values on closed interval [a,b]:
Step I: find $f'\left( x \right)$
Step II: find $f'\left( x \right) = 0$ and find values of x. let ${c_1},{c_2},{c_3},........,{c_n}$ be the values of x.
Step III: take the maximum and minimum values obtained in step III are respectively the largest (or absolute maximum) and the smallest (or absolute minimum) values of the function.

Complete step-by-step answer :
Given that,

$$\Rightarrow f\left( x \right) = {e^{{x^2}}} + {e^{ - {x^2}}} \\\ \Rightarrow g\left( x \right) = x{e^{{x^2}}} + {e^{ - {x^2}}} \\\ \Rightarrow h\left( x \right) = {x^2}{e^{{x^2}}} + {e^{ - {x^2}}} \\\$$

Where $x \in \left[ {0,1} \right]$
Let us find the absolute maximum values of f, g and h in [0,1].
1. $f\left( x \right) = {e^{{x^2}}} + {e^{ - {x^2}}}$
Differentiate both sides with respect to x,
$\Rightarrow f'\left( x \right) = {e^{{x^2}}}\left( {2x} \right) + {e^{ - {x^2}}}\left( { - 2x} \right)$
$\Rightarrow f'\left( x \right) = \left( {2x} \right)\left( {{e^{{x^2}}} - {e^{ - {x^2}}}} \right)$
Here we can see that,
$f'\left( x \right) \geqslant 0,$ for $0 \leqslant x \leqslant 1$, it means it is an increasing function.
Thus, we will get the maximum value of $f\left( x \right)$ at $f\left( 1 \right)$.
So,
$\Rightarrow f\left( 1 \right) = e + \dfrac{1}{e}$
And according to the question, it is denoted as a, i.e.
$e + \dfrac{1}{e} = a$
Similarly,
$\Rightarrow g\left( x \right) = x{e^{{x^2}}} + {e^{ - {x^2}}}$
Differentiate both sides with respect to x,

$$\Rightarrow g'\left( x \right) = x\left( {{e^{{x^2}}} \times 2x} \right) + {e^{{x^2}}} + {e^{ - {x^2}}}\left( { - 2x} \right) \\\ \Rightarrow g'\left( x \right) = {e^{{x^2}}} + 2x\left( {x{e^{{x^2}}} - {e^{ - {x^2}}}} \right) \\\$$

Here,
$g'\left( x \right) \geqslant 0,$ for$0 \leqslant x \leqslant 1$, hence it is an increasing function.
We will get the maximum value of $g\left( x \right)$ at $g\left( 1 \right)$
So,
$\Rightarrow g\left( 1 \right) = e + \dfrac{1}{e}$
and it will be denoted as b, i.e. $e + \dfrac{1}{e} = b$
Now,
$h\left( x \right) = {x^2}{e^{{x^2}}} + {e^{ - {x^2}}}$
Differentiate both sides with respect to x,

$$\Rightarrow h'\left( x \right) = {x^2}\left( {{e^{{x^2}}} \times 2x} \right) + 2x{e^{{x^2}}} + {e^{ - {x^2}}}\left( { - 2x} \right) \\\ \Rightarrow h'\left( x \right) = 2x\left( {{x^2}{e^{{x^2}}} + {e^{{x^2}}} - {e^{ - {x^2}}}} \right) \\\$$

Here, we can see that
$h'\left( x \right) \geqslant 0,$ for$0 \leqslant x \leqslant 1$, it means it is also an increasing function.
We will get the maximum value of $h\left( x \right)$ at $h\left( 1 \right)$
So,
$\Rightarrow h\left( 1 \right) = e + \dfrac{1}{e}$
And it will be denoted as c.
So, now we can clearly see that $f\left( 1 \right) = g\left( 1 \right) = h\left( 1 \right)$
i.e. a = b = c.
Therefore, the correct answer is option (D).

Note : Whenever we are asked such types of questions, we have to remember that if $f\left( x \right)$ be a real valued function defined on an interval [a, b]. Then, $f\left( x \right)$ is said to have the maximum value in [a, b], if there exists a point c in [a, b] such that $f\left( x \right) \leqslant f\left( c \right)$ for all $x \in \left[ {a,b} \right]$, in such a case $f\left( c \right)$ is called the absolute maximum value.