Question

For the function $f\left( x \right)={{e}^{x}}$, $a=0$,$b=1$, the value of $c$ in mean value theorem will be
A) 0
B) b$\log \left( e-1 \right)$
C) $\log x$
D) 1

Answer 1

Here we have to find the value of the given variable. For that, we will use the mean value theorem for the given function. The given function will satisfy all the conditions of the mean value theorem i.e. the given function will be continuous in the given range and it will be differentiable in the given range and then we will use the formula in mean value theorem as the last step to find the value of the required variable.

Complete step by step solution:
The given function is $f\left( x \right)={{e}^{x}}$ and it is given that $x$ lies in $\left[ 0,1 \right]$
The given function satisfies all the conditions of mean value theorem i.e. the given function is continuous in $\left[ 0,1 \right]$ and also the given function is differentiable in $\left[ 0,1 \right]$.
Thus, there exists $c\in \left( 0,1 \right)$ such that,
$\Rightarrow \dfrac{f\left( 1 \right)-f\left( 0 \right)}{1-0}=f'\left( c \right)$ …….. $\left( 1 \right)$
First we will find the value of $f\left( 1 \right)$ by putting the value of $x$ as 1 in the given function.
$\Rightarrow f\left( 1 \right)={{e}^{1}}=e$
Now, we will find the value of $f\left( 0 \right)$ by putting the value of $x$ as 0 in the given function
$\Rightarrow f\left( 0 \right)={{e}^{0}}=1$
To find the value of $f'\left( c \right)$, we will first differentiate the given function with respect to $x$ and then we will substitute the value of $x$ as $c$
$\begin{aligned} & \Rightarrow \dfrac{df\left( x \right)}{dx}=\dfrac{d{{e}^{x}}}{dx} \\\ & \Rightarrow f'\left( x \right)={{e}^{x}} \\\ \end{aligned}$
Now, we will find the value of $f'\left( c \right)$ by substituting the value of $x$ as $c$.
$\Rightarrow f'\left( c \right)={{e}^{c}}$
Now, we will
Now, we will substitute the value of $f\left( 1 \right)$, $f'\left( c \right)$ and $f\left( 0 \right)$ in equation 1, we get
$\begin{aligned} & \Rightarrow \dfrac{e-1}{1-0}={{e}^{c}} \\\ & \Rightarrow e-1={{e}^{c}} \\\ \end{aligned}$
Taking logarithm on both sides, we get
$\begin{aligned} & \Rightarrow \log \left( e-1 \right)=\log \left( {{e}^{c}} \right) \\\ & \Rightarrow c=\log \left( e-1 \right) \\\ \end{aligned}$

Thus, the correct option is option B.

Note:
Let’s understand the mean value theorem as we have used this here. Suppose $f\left( x \right)$ is a function such that it is continuous in the closed interval $\left[ a,b \right]$ and differentiable in the closed interval $\left[ a,b \right]$, then there exists a number $c$ such that $a\text{ }<\text{ }c\text{ }<\text{ }b$ then,
$f\prime (c)=\dfrac{f(b)-f(a)}{b-a}$ .