Question

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

Answer 1

Hint: We solve this question by directly using Mean Value Theorem which states that suppose that a function f(x) satisfies the two given condition, that is, it is continuous on closed interval [a, b] and it is differentiable on open interval $ \left( {a,b} \right) $ then, there exist a number c such that $ a < c < b $ and $ f'\left( c \right) = \left( {\dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}} \right) $ .

Complete step-by-step answer:
We are given $ f\left( x \right) = {e^x} $ and $ a = 0 $ , $ b = 1 $ which means that the interval is $ \left[ {0,1} \right] $ . We have to calculate the value of c in mean value theorem, this implies that, the given function satisfies both the conditions of Mean Value Theorem, such that, there exist a number c such that $ a < c < b $ and $ f'\left( c \right) = \left( {\dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}} \right) $ .
Now directly using the above formula, we get
 $ f'\left( x \right) = {e^x} $
 $ f'\left( c \right) = {e^c} $
Substituting values of $ f'\left( c \right) $ in the formula we get,
 $ {e^c} = \dfrac{{{e^b} - {e^a}}}{{b - a}} - - - - \left( 1 \right) $
We know that $ a = 0 $ and $ b = 1 $ , putting value of a, b in equation $ \left( 1 \right) $ , we get,
 $ \Rightarrow {e^c} = \dfrac{{{e^1} - {e^0}}}{{1 - 0}} $
We know that any number raised to power zero is equal to one. Simplifying the expression, we get,
 $ \Rightarrow {e^c} = e - 1 $
Taking natural logarithm on both sides of the equation, we get,
 $ \Rightarrow \ln \left( {{e^c}} \right) = \ln \left( {e - 1} \right) $
Now, we know the law of logarithm $ \ln {x^n} = n\log x $ . So, we get,
 $ \Rightarrow c\left( {\ln e} \right) = \ln \left( {e - 1} \right) $
We know the value of $ \ln e $ is one. So, we get,
 $ \Rightarrow c = \ln \left( {e - 1} \right) $
So, the value of c is $ \ln \left( {e - 1} \right) $ .
Hence, option (B) is the correct answer.
So, the correct answer is “Option B”.

Note: The main thing to keep in mind while doing this type of question is that Mean Value Theorem is applicable only if both the conditions are satisfied. We should also remember the basic properties of Log. We should also keep in mind the direct result of the Mean Value Theorem, that is, $ f'\left( c \right) = \left( {\dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}} \right) $ .