For every pair of continuous functions $f, g:[0,1] \rightarrow R$ such that $\max \{f(x): x \in[0,1]\}$ $=\max \{g(x): x \in[0,1]\}$, the correct statement (s) is (are)
(a) $(f(c))^2+3 f(c)=(g(c))^2+3 g(c)$ for some $c \in[0,1]$
(b) $(f(c))^2+f(c)=(g(c))^2+3 g(c)$ for some $c \in[0,1]$
(c) $(f(c))^2+3 f(c)=(g(c))^2+g(c)$ for some $c \in(0,1]$
(d) $(f(c))^2=(g(c))^2$ for some $c \in[0,1]$
Answer & explanation
Correct answer: option 1
Since f(x) and g(x) are continuous on [0, 1]. So, they attain their maximum and minimum values in [0, 1]. Suppose f(x) and g(x) attain their maximum values at $x_1$ and $x_2$ respectively. It is given that $f\left(x_1\right)=g\left(x_2\right)$.
Let h(x) = f(x) - g(x). Then, h(x) is continuous on [0, 1] such that
$h\left(x_1\right)=f\left(x_1\right)-g\left(x_1\right) \geq 0$ $\left[\begin{array}{l}∵ f\left(x_1\right)=g\left(x_2\right) \geq g\left(x_1\right) \\ ∴ f\left(x_1\right)-g\left(x_1\right) \geq 0\end{array}\right]$
and, $h\left(x_2\right)=f\left(x_2\right)-g\left(x_2\right) \leq 0$ $\left[\begin{array}{l}∵ g\left(x_2\right)=f\left(x_1\right) \geq f\left(x_2\right) \\ ∴ f\left(x_2\right)-g\left(x_2\right) \leq 0\end{array}\right]$
Therefore, there exists $c \in(0,1)$ such that
$h(c)=0 \Rightarrow f(c)=g(c)$
Clearly, f(c) = g(c) satisfy options (a) and (d).