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A. \[{\left[ {f\left( c \right)} \right]^2} + {\text{3}}f\left( c \right) = {\left[ {g\left( c \right)} \right]^2} + 3g\left( c \right)\]for some\[c \in \left[ {0,{\text{ }}1} \right]\].

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

C. \[{\left[ {f\left( c \right)} \right]^2} + {\text{3}}f\left( c \right) = {\left[ {g\left( c \right)} \right]^2} + g\left( c \right)\] for some\[c \in \left[ {0,{\text{ }}1} \right]\].

D.\[{\left[ {f\left( c \right)} \right]^2}{\text{ }} = {\text{ }}{\left[ {g\left( c \right)} \right]^2}\]for some\[c \in \left[ {0,{\text{ }}1} \right]\].

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We are given that for every pair of continuous function \[f,g:\left[ {0,{\text{ }}1} \right] \to R\] such that\[max\{ f\left( x \right):x \in \left[ {0,1} \right]\} = max\{ g\left( x \right):x \in \left[ {0,1} \right]\} \].

Let us take one function at a time. First we take \[f\left( x \right)\].

Let the value of which the function \[f\left( x \right)\] is maximum by ${c_1}$. This is shown as –

\[max\{ f\left( x \right):x \in \left[ {0,1} \right]\} \]=\[f\left( {{c_1}} \right)\]

Similarly, we take the second function \[g\left( x \right)\] and we get,

Let the value of which the function \[g\left( x \right)\] is maximum by ${c_2}$. This is shown as –

\[max\{ g\left( x \right):x \in \left[ {0,1} \right]\} = g({c_2})\]

Now let us take one other function \[h\left( x \right)\] as shown –

\[h\left( x \right)\]=\[f\left( x \right)\]-\[g\left( x \right)\]

Let us find Function \[h\left( x \right)\] for $x = {c_1}$. We get,

\[h\left( {{c_1}} \right)\]=\[f\left( {{c_1}} \right)\]-\[g\left( {{c_1}} \right)\]

Since, \[f\left( {{c_1}} \right)\] is greater than \[g\left( {{c_1}} \right)\] because at $x = {c_1}$ we get maximum value of \[f\left( x \right)\] whereas in case of \[g\left( x \right)\] at $x = {c_1}$ we do not get the maximum value. Therefore,

\[h\left( {{c_1}} \right) > 0\]

Let us find Function \[h\left( x \right)\] for $x = {c_2}$. We get,

\[h\left( {{c_2}} \right)\]=\[f\left( {{c_2}} \right)\]-\[g\left( {{c_2}} \right)\]

Since, \[g\left( {{c_2}} \right)\] is greater than \[f\left( {{c_2}} \right)\] because at $x = {c_2}$ we get maximum value of \[g\left( x \right)\] whereas in case of \[f\left( x \right)\] at \[x = {c_2}\] we do not get the maximum value. Therefore,

\[h\left( {{c_2}} \right) < 0\]

Therefore, for any arbitrary \[c\] which lie in between ${c_1}$ and ${c_2}$ \[h\left( x \right)\] has a root that is –

\[h\left( c \right) = 0\] For \[c \in \left[ {0,{\text{ }}1} \right]\].

Which implies,

\[f(c) - g\left( c \right) = 0\] For \[c \in \left[ {0,{\text{ }}1} \right]\]

\[ \Rightarrow f(c) = g\left( c \right)\]…………. (1)

Now, squaring both sides we get –

\[ \Rightarrow {[f(c)]^2} = {[g\left( c \right)]^2}\]…….. (2)

Multiplying equation (1) by 3 we get,

\[ \Rightarrow 3f(c) = 3g\left( c \right)\]………… (3)

Adding both the equations (2) and (3) we get,

\[{\left[ {f\left( c \right)} \right]^2} + {\text{3}}f\left( c \right) = {\left[ {g\left( c \right)} \right]^2} + 3g\left( c \right)\]…….. (4)

Therefore, from equations (2) and (4) we get that options (A) and (D) are correct options.