Intermediate Value Property on Pair of Continuous Functions

To determine the correct statement(s), we analyze the properties of continuous functions on the closed and bounded interval $[0, 1]$.

Given that $f, g: [0, 1] \to \mathbb{R}$ are continuous functions and $\max \{f(x) : x \in [0, 1]\} = \max \{g(x) : x \in [0, 1]\} = M$

Since $[0, 1]$ is a compact interval and $f, g$ are continuous, by the Extreme Value Theorem, $f$ and $g$ attain their maximum value $M$ at some points in $[0, 1]$. Thus, there exist $x_1, x_2 \in [0, 1]$ such that: $f(x_1) = M \quad \text{and} \quad g(x_2) = M$

Now, consider the function $h: [0, 1] \to \mathbb{R}$ defined by: $h(x) = f(x) - g(x)$

Since $f$ and $g$ are continuous on $[0, 1]$, $h(x)$ is also continuous on $[0, 1]$.

Evaluating $h(x)$ at $x_1$ and $x_2$:

1. At $x = x_1$: $h(x_1) = f(x_1) - g(x_1) = M - g(x_1)$ Since $g(x) \le M$ for all $x \in [0, 1]$, we have $g(x_1) \le M$, which implies: $h(x_1) \ge 0$

2. At $x = x_2$: $h(x_2) = f(x_2) - g(x_2) = f(x_2) - M$ Since $f(x) \le M$ for all $x \in [0, 1]$, we have $f(x_2) \le M$, which implies: $h(x_2) \le 0$

Since $h(x_1) \ge 0$ and $h(x_2) \le 0$, by the Intermediate Value Theorem (IVT), there must exist at least one point $c \in [0, 1]$ such that: $h(c) = 0 \implies f(c) = g(c)$

---

Analysis of Options:

• Option (A): $(f(c))^2 + 3f(c) = (g(c))^2 + 3g(c)$ Since there exists $c \in [0, 1]$ such that $f(c) = g(c)$, substituting $g(c) = f(c)$ into the equation yields: $(f(c))^2 + 3f(c) = (f(c))^2 + 3f(c)$ This statement is always true. Thus, Option (A) is correct.

• Option (D): $(f(c))^2 = (g(c))^2$ Using the same point $c \in [0, 1]$ where $f(c) = g(c)$, squaring both sides gives: $(f(c))^2 = (g(c))^2$ This statement is always true. Thus, Option (D) is correct.

• Options (B) and (C): Consider a counterexample where $f(x) = 1$ and $g(x) = 1$ for all $x \in [0, 1]$.

  • Both functions are continuous and $\max \{f(x)\} = \max \{g(x)\} = 1$.
  • For Option (B): $(f(c))^2 + f(c) = 1^2 + 1 = 2$ $(g(c))^2 + 3g(c) = 1^2 + 3(1) = 4$ Since $2 \neq 4$, Option (B) is false.
  • For Option (C): $(f(c))^2 + 3f(c) = 1^2 + 3(1) = 4$ $(g(c))^2 + g(c) = 1^2 + 1 = 2$ Since $4 \neq 2$, Option (C) is false.

---

Conclusion:

The correct statements are (A) and (D).