Question

For every pair of continuous functions $f,g:\left[ {0,1} \right] \to R$ such that \max \left\\{ {f\left( x \right):x \in \left[ {0,1} \right]} \right\\} = \max \left\\{ {g\left( x \right):x \in \left[ {0,1} \right]} \right\\}, the correct statement is
A) ${\left( {f\left( c \right)} \right)^2} + 3f\left( c \right) = {\left( {g\left( c \right)} \right)^2} + 3g\left( c \right)$ for some $c \in \left[ {0,1} \right]$
B) ${\left( {f\left( c \right)} \right)^2} + f\left( c \right) = {\left( {g\left( c \right)} \right)^2} + 3g\left( c \right)$ for some $c \in \left[ {0,1} \right]$
C) ${\left( {f\left( c \right)} \right)^2} + 3f\left( c \right) = {\left( {g\left( c \right)} \right)^2} + g\left( c \right)$ for some $c \in \left[ {0,1} \right]$
D) ${\left( {f\left( c \right)} \right)^2} = {\left( {g\left( c \right)} \right)^2}$ for some $c \in \left[ {0,1} \right]$

Answer 1

We will first consider the given function. Then we will let function $f\left( x \right)$ and $g\left( x \right)$ will get their maximum value at ${x_1}$ and ${x_2}$ respectively. Now we will take the difference of both the functions and put it equal to $h\left( x \right)$. Then we will substitute ${x_1}$ and ${x_2}$ in $h\left( x \right)$ and for ${x_1}$, we will take $h\left( {{x_1}} \right)$ positive and for ${x_2}$, we will take $h\left( {{x_2}} \right)$. Thus, for some $c \in \left[ {0,1} \right]$ we will get $h\left( c \right) = 0$ which will give us $f\left( c \right) = g\left( c \right)$. Thus, we will get the desired result.

Complete step by step solution:
We will first consider the given information, that is for every pair of continuous functions, $f,g:\left[ {0,1} \right] \to R$ such that \max \left\\{ {f\left( x \right):x \in \left[ {0,1} \right]} \right\\} = \max \left\\{ {g\left( x \right):x \in \left[ {0,1} \right]} \right\\}.
We need to find the correct relation between the two functions.
Now, we will let functions $f\left( x \right)$ and $g\left( x \right)$ will get their maximum value at ${x_1}$ and ${x_2}$ respectively.
Thus, we will find the difference of two functions and put it equal to $h\left( x \right)$.
We get,
$h\left( x \right) = f\left( x \right) - g\left( x \right)$
Next, we will substitute ${x_1}$ and ${x_2}$ in the above obtained function,
Thus, we get,
$h\left( {{x_1}} \right) = f\left( {{x_1}} \right) - g\left( {{x_1}} \right) \geqslant 0$ and $h\left( {{x_2}} \right) = f\left( {{x_2}} \right) - g\left( {{x_2}} \right) \leqslant 0$
Thus, by replacing the value of $x$ with some $c$ which belongs to the interval $\left[ {0,1} \right]$.
We will put $h\left( c \right) = 0$ as $f\left( x \right)$ and $g\left( x \right)$ are continuous functions so, their difference will also be a continuous function.
Thus, Using Rolle’s theorem we get,

$$\Rightarrow h\left( c \right) = 0 \\\ \Rightarrow f\left( c \right) = g\left( c \right) \\\$$

Hence, we can conclude that ${\left( {f\left( c \right)} \right)^2} = {\left( {g\left( c \right)} \right)^2}$ and ${\left( {f\left( c \right)} \right)^2} + 3f\left( c \right) = {\left( {g\left( c \right)} \right)^2} + 3g\left( c \right)$ for some $c \in \left[ {0,1} \right]$ will be valid options.

Thus, option A and D are correct.

Note:
Firstly go through the given statement properly as \max \left\\{ {f\left( x \right):x \in \left[ {0,1} \right]} \right\\} = \max \left\\{ {g\left( x \right):x \in \left[ {0,1} \right]} \right\\} which gives us that both the functions attain their maximum value at some point so, it’s necessary to let numbers at which the functions have achieved their maximum value. As $h\left( c \right) = 0$, gives us that the equation $h\left( x \right) = f\left( x \right) - g\left( x \right)$ will also be equal to zero. As the graph of both the functions will intersect at one point $x \in \left[ {0,1} \right]$ that’s why we have let that point be $c \in \left[ {0,1} \right]$.