Question

For every pair of continuous function $\mathrm{f},\mathrm{g}:\left[0,1\right]\to \mathrm{ℝ}$ such that $\max \mathrm{f}\left(x\right):x\in \left[0,1\right]=\max \mathrm{g}\left(x\right):x\in \left[0,1\right]$, the correct statement(s) is (are)

• A. ${\left[\mathrm{f}\left(c\right)\right]}^2+3\mathrm{f}\left(c\right)={\left[\mathrm{g}\left(\mathrm{c}\right)\right]}^2+3\mathrm{g}\left(\mathrm{c}\right)$ for some $\mathrm{c}\in \left[0,1\right]$
• B. ${\left[\mathrm{f}\left(c\right)\right]}^2+\mathrm{f}\left(c\right)={\left[\mathrm{g}\left(c\right)\right]}^2+3\mathrm{g}\left(c\right)$ for some $\mathrm{c}\in \left[0,1\right]$
• C. ${\left[\mathrm{f}\left(c\right)\right]}^2+3\mathrm{f}\left(c\right)={\left[\mathrm{g}\left(\mathrm{c}\right)\right]}^2+\mathrm{g}\left(\mathrm{c}\right)$ for some $\mathrm{c}\in \left[0,1\right]$
• D. ${\left[\mathrm{f}\left(c\right)\right]}^2={\left[\mathrm{g}\left(c\right)\right]}^2$ for some $\mathrm{c}\in \left[0,1\right]$

Answer 1

Answer: (A), (D)

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

Explanation for the correct option:

It is given that $\mathrm{f}$ and $\mathrm{g}$ are continuous functions such that $\mathrm{f},\mathrm{g}:\left[0,1\right]\to \mathrm{ℝ}$.

And, $\max \mathrm{f}\left(x\right):x\in \left[0,1\right]=\max \mathrm{g}\left(x\right):x\in \left[0,1\right]$

The graphs of both functions will intersect at one point given $\mathrm{x}\in \left[0,1\right]$.

Let that point be $c$.

So, $\mathrm{f}\left(c\right)=\mathrm{g}\left(c\right)$ …(i)

Option (A): Using equation (i),

$\because$ $\mathrm{f}\left(c\right)=\mathrm{g}\left(c\right)$

$\Rightarrow$ $3\mathrm{f}\left(c\right)=3\mathrm{g}\left(c\right)$ [multiplying by $3$ on both sides]

$\Rightarrow$ ${\left[\mathrm{f}\left(c\right)\right]}^2+3\mathrm{f}\left(c\right)={\left[\mathrm{f}\left(c\right)\right]}^2+3\mathrm{g}\left(c\right)$ [adding ${\left[\mathrm{f}\left(c\right)\right]}^2$ on both sides]

$\Rightarrow$ ${\left[\mathrm{f}\left(c\right)\right]}^2+3\mathrm{f}\left(c\right)={\left[\mathrm{g}\left(c\right)\right]}^2+3\mathrm{g}\left(c\right)$ $\left[\because f\left(c\right)=g\left(c\right)\right]$

Where, $\mathrm{c}\in \left[0,1\right]$

Hence, option (A) is the correct option.

Option (D): Using equation (i),

$\because$ $\mathrm{f}\left(c\right)=\mathrm{g}\left(c\right)$

$\Rightarrow$ ${\left[\mathrm{f}\left(c\right)\right]}^2={\left[\mathrm{g}\left(c\right)\right]}^2$ [squaring on both sides]

Where, $\mathrm{c}\in \left[0,1\right]$

Hence, option (D) is the correct option.

Explanation for the incorrect options:

Option (B): The option states that,

${\left[\mathrm{f}\left(c\right)\right]}^2+\mathrm{f}\left(c\right)={\left[\mathrm{g}\left(c\right)\right]}^2+3\mathrm{g}\left(c\right)$

On substituting $\mathrm{f}\left(c\right)=\mathrm{g}\left(c\right)$ in above, we get,

${\left[\mathrm{f}\left(c\right)\right]}^2+\mathrm{f}\left(c\right)={\left[\mathrm{f}\left(c\right)\right]}^2+3\mathrm{f}\left(\mathrm{c}\right)$ which is not true.

$\Rightarrow$ ${\left[\mathrm{f}\left(c\right)\right]}^2+\mathrm{f}\left(c\right)\ne {\left[\mathrm{f}\left(c\right)\right]}^2+3\mathrm{f}\left(\mathrm{c}\right)$

Where, $\mathrm{c}\in \left[0,1\right]$

Hence, option (B) is an incorrect option.

Option (C): The option states that,

${\left[\mathrm{f}\left(c\right)\right]}^2+3\mathrm{f}\left(c\right)={\left[\mathrm{g}\left(\mathrm{c}\right)\right]}^2+\mathrm{g}\left(\mathrm{c}\right)$

On substituting $\mathrm{f}\left(c\right)=\mathrm{g}\left(c\right)$ in above, we get,

${\left[\mathrm{f}\left(c\right)\right]}^2+3\mathrm{f}\left(c\right)={\left[\mathrm{f}\left(\mathrm{c}\right)\right]}^2+\mathrm{f}\left(\mathrm{c}\right)$ which is not true.

$\Rightarrow$ ${\left[\mathrm{f}\left(c\right)\right]}^2+3\mathrm{f}\left(c\right)\ne {\left[\mathrm{f}\left(\mathrm{c}\right)\right]}^2+\mathrm{f}\left(\mathrm{c}\right)$

Where, $\mathrm{c}\in \left[0,1\right]$

Hence, option (C) is an incorrect option.

Therefore, options (A) and (D) are the correct options.