Question

If $f\left( x \right)$ and $g\left( x \right)$ are differentiable functions in $\left[ {0,1} \right]$ such that $f\left( 0 \right) = 2 = g\left( 1 \right)$ , $g\left( 0 \right) = 0$ , $f\left( 1 \right) = 6$ then there exists c, $0 < c < 1$ such that $f'\left( c \right)$ =
A . $g'\left( c \right)$
B . $- g'\left( c \right)$
C . $2g'\left( c \right)$
D . $3g'\left( c \right)$

Answer 1

Hint : As mentioned there are two functions, $f\left( x \right)$ and $g\left( x \right)$ which are differentiable in $\left[ {0,1} \right]$ , hence to find $f'\left( c \right)$ , apply Mean value theorem to get the relation between the two given functions. Mean value theorem is the relationship between the derivative of a function and increasing or decreasing nature of function. It basically defines the derivative of a differential and continuous function.

Complete step-by-step answer :
Let us write the given data,
$f\left( 0 \right) = 2$
$g\left( 0 \right) = 0$
$f\left( 1 \right) = 6$
$g\left( 1 \right) = 2$
Hence, we need to find $f'\left( c \right)$ and $g'\left( c \right)$ .
Let us apply, mean value theorem to find $f'\left( c \right)$ i.e.,
$f'\left( c \right) = \dfrac{{f\left( b \right) - f\left( a \right)}}{{b - a}}$
$f'\left( c \right) = \dfrac{{f\left( 1 \right) - f\left( 0 \right)}}{{1 - 0}}$
Substitute the value of the functions, as
$\Rightarrow$ $f'\left( c \right) = \dfrac{{6 - 2}}{1}$
$\Rightarrow$ $f'\left( c \right) = 4$
Now, let us find $g'\left( c \right)$ , by mean value theorem:
$g'\left( c \right) = \dfrac{{g\left( 1 \right) - g\left( 0 \right)}}{{1 - 0}}$
$\Rightarrow$ $g'\left( c \right) = \dfrac{{2 - 0}}{1}$
$\Rightarrow$ $g'\left( c \right) = 2$
Hence,
$f'\left( c \right) = 2 \cdot g'\left( c \right)$
Therefore, option C is the right answer
So, the correct answer is “Option C”.

Note : Mean value theorem is one of the most useful tools in both differential and integral calculus. It has very important consequences in differential calculus and helps us to understand the identical behaviour of different functions.
The hypothesis and conclusion of the mean value theorem shows some similarities to those of Intermediate value theorem. Mean value theorem is also known as Lagrange’s Mean Value Theorem. This theorem is abbreviated as MVT.