For every pair of continuous functions f, g: | vedclass

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(D) Let $\max \{f(x): x \in [0, 1] \} = \max \{g(x): x \in [0, 1] \} = \lambda$. Since $f$ and $g$ are continuous on $[0, 1]$,there exist $a, b \in [0

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$(A, D)$

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(D) Let $\max \{f(x): x \in [0, 1] \} = \max \{g(x): x \in [0, 1] \} = \lambda$. Since $f$ and $g$ are continuous on $[0, 1]$,there exist $a, b \in [0, 1]$ such that $f(a) = \lambda$ and $g(b) = \lambda$. Define $h(x) = f(x) - g(x)$. Then $h(a) = f(a) - g(a) = \lambda - g(a) \ge 0$ and $h(b) = f(b) - g(b) = f(b) - \lambda \le 0$. By the Intermediate Value Theorem,there exists $c \in [0, 1]$ such that $h(c) = 0$,which implies $f(c) = g(c)$. For option $(A)$: $(f(c))^2 + 3f(c) = (g(c))^2 + 3g(c)$. Since $f(c) = g(c)$,this holds true. For option $(D)$: $(f(c))^2 = (g(c))^2$. Since $f(c) = g(c)$,this holds true. For options $(B)$ and $(C)$,consider $f(x) = g(x) = \lambda$ where $\lambda 
eq 0$. Then $(B)$ becomes $\lambda^2 + \lambda = \lambda^2 + 3\lambda$,which implies $\lambda = 3\lambda$,or $\lambda = 0$,which contradicts $\lambda 
eq 0$. Similarly for $(C)$. Thus,$(A)$ and $(D)$ are correct.

For every pair of continuous functions $f, g: [0, 1] \rightarrow \mathbb{R}$ such that $\max \{f(x): x \in [0, 1] \} = \max \{g(x): x \in [0, 1] \} = \lambda$,the correct statement$(s)$ is (are):

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