Question

# For every pair of continuous function $f,g:\left[ {0,{\text{ }}1} \right] \to R$ such that $max\{ f\left( x \right):x \in \left[ {0,1} \right]\} = max\{ g\left( x \right):x \in \left[ {0,1} \right]\}$, the correct statement(s) is (are): 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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Hint:For any function $f\left( x \right)$ if at some point p function is positive and any other point q function is negative that is $f\left( p \right) > 0$ and $f\left( q \right) < 0$ then there is at least one root of $f\left( x \right)$ between p and q.

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.

So, the correct answer is “Option A and D”.

Note:When we use the property that for any function $f\left( x \right)$ if at some point p function is positive and any other point q function is negative that is $f\left( p \right) > 0$ and $f\left( q \right) < 0$ then there is at least one root of $f\left( x \right)$ between p and q. It should be noted that the number of roots lying between p and q are atleast one and not equal to one.