Question

Consider all functions given in List-I in the interval [1,3]. The List-2 has the values of 'c' obtained by applying Lagrange's mean value theorem on the functions of List-1. Match the functions and values of 'c'.

• A. A-II, B-V, C-IV, D-III
• B. A-II, B-I, C-IV, D-III
• C. A-IV, B-V, C-II, D-I
• D. A-IV, B-III, C-II, D-V

Hint:

Lagrange's Mean Value Theorem connects the average slope of a function over an interval to the instantaneous slope at some point within that interval. The core of applying it is to calculate both $f'(c)$ and $\frac{f(b)-f(a)}{b-a}$ and set them equal.

Answer 1

The Correct Option is D

Step 1: State Lagrange's Mean Value Theorem (LMVT).
For a function $f(x)$ continuous on $[a,b]$ and differentiable on $(a,b)$, there exists a $c \in (a,b)$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$. Here, our interval is $[1,3]$.  

Step 2: Match function B: $f(x) = \log x$. 
$f'(x) = 1/x$, so $f'(c)=1/c$.
The average slope is $\frac{\log 3 - \log 1}{3-1} = \frac{\log 3}{2}$.
Setting them equal: $\frac{1}{c} = \frac{\log 3}{2} \implies c = \frac{2}{\log 3} = 2\log_3 e = \log_3 (e^2)$.
This matches List-2 item III. So, B-III. 

Step 3: Match function C: $f(x) = x^2+x+1$. 
$f'(x) = 2x+1$, so $f'(c) = 2c+1$.
The average slope is $\frac{f(3)-f(1)}{3-1} = \frac{(9+3+1)-(1+1+1)}{2} = \frac{13-3}{2} = 5$.
Setting them equal: $2c+1 = 5 \implies 2c=4 \implies c=2$.
This matches List-2 item II. So, C-II. 

Step 4: Match function D: $f(x) = e^x$. 
$f'(x) = e^x$, so $f'(c) = e^c$. The average slope is $\frac{f(3)-f(1)}{3-1} = \frac{e^3-e^1}{2}$.
Setting them equal: $e^c = \frac{e^3-e}{2} \implies c = \ln\left(\frac{e^3-e}{2}\right) = \log\left(\frac{e^3-e}{2}\right)$.
This matches List-2 item V. So, D-V. 

Step 5: Match function A by elimination. 
We have the matches B-III, C-II, and D-V. Looking at the options, only option (D) has this combination.
Option (D) is A-IV, B-III, C-II, D-V.
This implies A must match with IV, which is $c=\sqrt{2}$. Let's verify.
For $f(x)=|x-1|$ on $[1,3]$, we can write $f(x)=x-1$. $f'(x)=1$. The average slope is $\frac{f(3)-f(1)}{2} = \frac{2-0}{2}=1$.
So $f'(c)=1$ must be true. Since $f'(x)=1$ for all $x>1$, any $c \in (1,3)$ is a valid solution. $\sqrt{2}$ is in $(1,3)$, so it is a valid, though not unique, value for c. 

Step 6: Final Matching. 
The correct matching is: A-IV, B-III, C-II, D-V. This corresponds to option (D).