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

Hint:
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 \[c \in \left[ {0,1} \right]\].