JEE Mains Maths · Application of Derivatives

Rolle's and Mean Value Theorems

The two theorems that guarantee a point where the derivative takes a given value, and how they count the zeros that f′ and f″ must have.

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Why this matters

Six PYQs, four of them multiple choice. Three apply Rolle's theorem or the mean value theorem directly — find the point, or a parameter that makes the theorem hold; three count how many zeros f′ or f″ must have, given where f vanishes or takes equal values. Two ideas cover the page.

Concept 1 of 2: Rolle's theorem and the mean value theorem

f

is continuous on

[a,b]

, differentiable on

(a,b)

, and

f(a)=f(b)

, the graph must turn somewhere between:

f'(c)=0

for some

c

(a,b)

. Tilt the picture and you get the mean value theorem: some tangent is parallel to the chord,

f'(c)=\frac{f(b)-f(a)}{b-a}

Definition

Rolle: continuous on $[a,b]$ , differentiable on $(a,b)$ , $f(a)=f(b)$ gives $f'(c)=0$ .
MVT: under the first two conditions, $f'(c)=\frac{f(b)-f(a)}{b-a}$ .
Both give at least one such $c$ , strictly inside.

Mean value theorem

f'(c)=\frac{f(b)-f(a)}{b-a},\quad c\in(a,b)

Worked example

Find

c

for the mean value theorem for

x^2

[1,3]

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 3 · Q74Moderate

Example 1 · Application of Derivatives · Rolle's and Mean Value Theorems

If Rolle's theorem holds for the function

f(x) =x^{3}- ax^{2}+ bx - 4,x \in \lbrack 1,2\rbrack

with

f^{'}\left( \frac{4}{3} \right)= 0

, then ordered pair

(a,b)

is equal to :

Check the hypotheses

Rolle's theorem needs differentiability inside the interval.

|x|

[-1,1]

has equal end values but no point with

f'=0

Drill 2 more on rolle's theorem and the mean value theorem

Concept 2 of 2: Zeros of f′ and f″

Between two zeros of

f

lies a zero of

f'

, so

n

distinct zeros of

f

give at least

n-1

f'

and

n-2

f''

. When

f

is not zero but meets a line or curve several times, subtract it:

h=f-(\text{line})

has zeros there, and

h''=f''

. Symmetry supplies zeros too: an odd function vanishes at 0.

Definition

$n$ distinct zeros of $f$ : at least $n-1$ of $f'$ , $n-2$ of $f''$ .
$f$ meets $y=mx+c$ at $n$ points: $h=f-mx-c$ has $n$ zeros, and $h''=f''$ .
An odd differentiable function has $f(0)=0$ ; the derivative of an even one is odd.

Rolle, repeated

n\ \text{zeros of }f\ \Rightarrow\ \ge n-1\ \text{zeros of }f'

Worked example

A twice differentiable

f

has

f(0)=f(1)=f(2)=0

. How many zeros must

f'

and

f''

have?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 23 · Q78Moderate

Example 2 · Application of Derivatives · Rolle's and Mean Value Theorems

Let

f

be any continuous function on

\lbrack 0,2\rbrack

and twice differentiable on

(0,2)

. If

f(0) = 0, f(1) = 1

and

f(2) = 2

, then

The zeros must be distinct

Rolle needs two different points with equal values. A repeated root counts once for this argument, so count the distinct zeros before subtracting one.

Drill 2 more on zeros of f′ and f″

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Rolle's theorem and the mean value theorem
Mean value theorem
$f'(c)=\frac{f(b)-f(a)}{b-a},\quad c\in(a,b)$
Zeros of f′ and f″
Rolle, repeated
$n\ \text{zeros of }f\ \Rightarrow\ \ge n-1\ \text{zeros of }f'$

Watch out for (2)

Check the hypotheses
→ Rolle's theorem and the mean value theorem
The zeros must be distinct
→ Zeros of f′ and f″

Test yourself on Application of Derivatives

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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