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JEE Main 20208 JanuaryEvening Shift

Exam: JEE Main
Year: 2020
Date: 8 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2020 · 8 January · Evening Shift · Mathematics · Continuity and Differentiability

Let S , be the set of all functions f: [0, 1] → R , which are...

Let

S

, be the set of all functions

f: \left[0, 1\right] \to R

, which are continuous on \left[0, 1\right]

, and differentiable on \left(0, 1\right)

. Then for every

f in S

, there exists

c \in \left(0, 1\right), depending on

f

, such that.

Options

A $\left|f$ \left(c\right) - f $\left(1\right)\right| < \left(1 - c\right)\left|{f}^{'}\left(c\right)\right|$
B $\dfrac{f\left(1\right) - f\left(c\right)}{1 - c} = {f}^{'}\left(c\right)$
C $\left|f$ \left(c\right) + f $\left(1\right)\right| < \left(1 + c\right)\left|{f}^{'}\left(c\right)\right|$
D $\left|f$ \left(c\right) - f $\left(1\right)\right| < \left|{f}^{'}\left(c\right)\right|$

Correct answer

B. $\dfrac{f\left(1\right) - f\left(c\right)}{1 - c} = {f}^{'}\left(c\right)$

Step-by-step solution

Let us consider $f\left(x\right)$ , a constant function, as constant functions are continuous and differentiable everywhere.

Then, $f\left(1\right) = f\left(c\right) \text{ and } f ' \left(c\right) = 0$

So, $\dfrac{f\left(1\right) - f\left(c\right)}{1 - c} = f ' \left(c\right) = 0$ .

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