MHT-CET Maths · Applications of Derivative

Rolle's Theorem and the Mean Value Theorem

If a function is smooth on an interval with equal endpoint values, its graph must level off somewhere (Rolle); more generally, some tangent must be parallel to the chord joining the endpoints (Lagrange). The whole subtopic is checking the three hypotheses, then solving f'(c) for the point c.

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Why this matters

One of the most reliably tested subtopics in MHT-CET calculus: 18 PYQs sit here (3 HARD, 12 MODERATE, 3 EASY). The paper recycles a small set of moves — verify Rolle and find c, count how many c work, apply LMVT and solve for c, or use 'Rolle holds' to back out unknown coefficients a and b. The recurring traps are always the same: rejecting a root of f'(c)=0 that falls outside the open interval, forgetting that MVT needs differentiability (not just continuity), and confusing Rolle's f(a)=f(b) requirement with LMVT's chord slope.

Concept 1 of 5

Rolle's Theorem — the Three Hypotheses and the Conclusion

Intuition

If a smooth curve starts and ends at the same height over [a, b], it cannot rise without coming back down — somewhere in between its tangent must be flat. That flat spot is a point c where f'(c) = 0. Rolle's theorem promises at least one such c, provided three hypotheses hold.

Definition

Rolle's theorem. If $f$ satisfies all three hypotheses:

$f$ is continuous on the closed interval $[a, b]$ ,
$f$ is differentiable on the open interval $(a, b)$ ,
$f(a) = f(b)$ (equal endpoint values),

then there exists at least one $c \in (a, b)$ with $f'(c) = 0$ .

In practice: check the hypotheses (for a polynomial, continuity and differentiability are automatic — only $f(a)=f(b)$ needs verifying), then solve $f'(x)=0$ and keep the root that lies inside $(a,b)$ .

Rolle's theorem

f \text{ cont. on }[a,b],\ \text{diff. on }(a,b),\ f(a)=f(b)\ \Rightarrow\ \exists\, c\in(a,b):\ f'(c)=0

cpoint inside (a, b) where the tangent is horizontal
f(a)=f(b)the equal-endpoint hypothesis unique to Rolle

Worked example

Verify Rolle's theorem for

f(x) = x^2 - 4x + 3

[1, 3]

and find

c

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Applications of DerivativeMODERATE

The value of

c

for which Rolle's theorem for the function

f(x)=x^{3}-3x^{2}+2x

in the interval

[0,2]

are

[Q142 · 3rd May Shift 2 · 2023]

All THREE hypotheses are required, not just f(a) = f(b)

Rolle needs continuity on

[a,b]

AND differentiability on

(a,b)

AND

f(a)=f(b)

. A function like

f(x)=|x|

[-1,1]

has

f(-1)=f(1)=1

but is NOT differentiable at

0

, so Rolle does not apply — there is no

c

with

f'(c)=0

. Never skip the differentiability check.

Rolle gives f'(c) = 0, not the chord slope

Rolle's conclusion is specifically

f'(c)=0

(a horizontal tangent), because the endpoints are equal. If the endpoints differ, use the Mean Value Theorem instead — its conclusion is

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

, which reduces to

0

only when

f(a)=f(b)

Concept 2 of 5

Finding c and Rejecting Roots Outside the Interval

Intuition

Once the hypotheses hold, finding

c

is just solving

f'(x)=0

. But

f'

often has more than one root — and Rolle only promises a

c

that lies STRICTLY between

a

and

b

. So compute every root, then discard any that fall outside the open interval

(a,b)

Definition

To find the Rolle point $c$ :

Solve $f'(x) = 0$ to get all candidate values.
Keep only the candidates that lie inside the open interval $(a,b)$ ; reject the rest.

For products/exponential-times-polynomial forms, differentiate carefully (product rule + chain rule) — the exponential factor $e^{kx}$ is never zero, so it only scales $f'$ ; the roots come entirely from the polynomial factor. Set that polynomial factor to zero and filter by the interval.

Rolle point from f'(x) = 0

f'(c) = 0,\quad c \in (a,b)\ \ \text{(reject any root outside the interval)}

Worked example

Rolle's theorem holds for

f(x) = (x-1)(x-4)

[1,4]

. Find

c

, rejecting any root outside the interval.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Applications of DerivativeMODERATE

Value of c satisfying the conditions and conclusions of Rolle's theorem for the function

f(x) = x\sqrt{x+6},\; x \in [-6,0]

[Q109 · 11th May Shift 1 · 2024]

Reject any root of f'(c) = 0 that lies OUTSIDE (a, b)

Solving

f'(x)=0

can give roots that Rolle never promised. For

f(x)=x(x+3)e^{-x/2}

[-3,0]

f'

vanishes at

x=3

and

x=-2

, but only

-2

lies in

(-3,0)

. Picking

x=3

is the single most common mistake — always filter candidates by the interval.

The exponential factor is never zero

In a form like

f'(x)=e^{-x/2}\,(\text{polynomial})

, the factor

e^{-x/2}>0

always. So

f'(c)=0

forces the POLYNOMIAL factor to be zero — never set the exponential to zero. It only affects sign/scaling, not the location of the Rolle point.

Drill 1 more on finding c and rejecting roots outside the interval

Concept 3 of 5

Counting the Number of Valid c

Intuition

Some questions do not ask for the value of

c

but for HOW MANY values of

c

satisfy the conclusion. For a trigonometric function like

\sin(k\pi x)

f'(x)=0

has many solutions — so you count all the roots of

f'

that fall inside the open interval.

Definition

To count the Rolle/MVT points:

Write $f'(x)=0$ and solve the resulting trigonometric (or polynomial) equation over the open interval.
Count every solution that lies strictly inside $(a,b)$ .

For $f(x)=\sin(k\pi x)$ , $f'(x)=k\pi\cos(k\pi x)$ ; $f'=0$ where $\cos(k\pi x)=0$ , i.e. $k\pi x = \pm\tfrac{\pi}{2}, \pm\tfrac{3\pi}{2}, \dots$ . List these, then keep the ones inside the interval and count them.

Number of Rolle points

\#\{\,c\in(a,b): f'(c)=0\,\} = \text{number of roots of } f'(x)=0 \text{ inside } (a,b)

Worked example

How many values of

c

satisfy Rolle's theorem for

f(x) = \sin \pi x

[0, 2]

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Applications of DerivativeMODERATE

The number of values of

C

that satisfy the conclusion of Rolle's theorem in case of following function

f(x) = \sin 2\pi x,\, x \in [-1,1]

[Q104 · 15th May Shift 2 · 2023]

Count roots INSIDE the open interval only — mind the endpoints

When counting

c

, include only roots strictly between

a

and

b

. For

\sin 2\pi x

[-1,1]

the roots of

f'

are

\pm\tfrac14, \pm\tfrac34

— four values, all interior. A root landing exactly on an endpoint would NOT count. Sketch or list them; don't guess the count.

Solve the full trig equation — don't stop at the first solution

\cos(k\pi x)=0

has infinitely many solutions; over a finite interval several survive. Writing

k\pi x = \pm\tfrac{\pi}{2}, \pm\tfrac{3\pi}{2}, \dots

systematically and filtering by the interval avoids under-counting (a common way to pick '02' when the answer is '04').

Concept 4 of 5

Lagrange's Mean Value Theorem — Statement and Finding c

Intuition

Rolle needs equal endpoints; Lagrange removes that restriction. Over any smooth arc from

(a, f(a))

(b, f(b))

, there is a point where the tangent is parallel to the CHORD joining the ends. The chord's slope is

\frac{f(b)-f(a)}{b-a}

, so LMVT says

f'(c)

equals that slope for some

c

inside.

Definition

Lagrange's Mean Value Theorem (LMVT). If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , then there exists $c \in (a,b)$ with

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

To find

c

: compute the chord slope

\frac{f(b)-f(a)}{b-a}

, set

f'(x)

equal to it, solve, and keep the root inside

(a,b)

. Rolle is the special case

f(a)=f(b)

, where the chord slope is

0

Lagrange's Mean Value Theorem

f'(c) = \dfrac{f(b) - f(a)}{b - a},\qquad c \in (a,b)

$\frac{f(b)-f(a)}{b-a}$ slope of the chord joining the endpoints
f'(c)slope of the tangent at the guaranteed point c

Worked example

Find the value of

c

of LMVT for

f(x) = x^2

[1, 3]

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Applications of DerivativeMODERATE

The value of

c

of Lagrange's mean value theorem for

f(x)=\sqrt{25-x^2}

[1,5]

[Q126 · 9th May Shift 1 · 2024]

MVT needs DIFFERENTIABILITY, not just continuity

Both hypotheses must hold: continuous on

[a,b]

AND differentiable on

(a,b)

. A function continuous but with a corner (non-differentiable) inside the interval can fail to have a valid

c

. Continuity alone is not enough — always confirm differentiability before applying LMVT.

Chord slope uses f(b) − f(a), not f'(a) or f'(b)

The right-hand side of LMVT is the average slope

\frac{f(b)-f(a)}{b-a}

— a difference of FUNCTION values over the interval width, not a derivative at an endpoint. A frequent slip is to compute

f'(a)

f'(b)

instead of the chord slope.

Bounding trick: f(b) − f(a) ≤ (b − a)·max f'

When

f'(x) \le M

everywhere, LMVT gives

f(b) - f(a) = f'(c)(b-a) \le M(b-a)

. So if

f(1)=1

and

f'\le 5

[1,5]

, then

f(5) \le 1 + 5\cdot 4 = 21

. Use the theorem as an INEQUALITY to bound a value.

Drill 8 more on lagrange's mean value theorem — statement and finding c

Concept 5 of 5

Solving Unknown Parameters and the Tangent-Parallel-to-Chord View

Intuition

A whole family of questions says 'Rolle (or MVT) holds for

f

at a GIVEN point

c

' and asks you to find unknown coefficients

a, b

. Each piece of the theorem is an equation:

f(a)=f(b)

gives one, and

f'(c)=0

(Rolle) or

f'(c)=\text{chord slope}

(MVT) gives another — two equations, two unknowns. Geometrically, MVT is just 'find the point where the tangent is parallel to the chord.'

Definition

Back-solving parameters. Given that the theorem holds with a specified $c$ :

Write the endpoint equation: Rolle ⇒ $f(a)=f(b)$ ; MVT ⇒ (usually) also $f(a)=f(b)$ or the given chord.
Write the interior equation: Rolle ⇒ $f'(c)=0$ ; MVT ⇒ $f'(c)=\frac{f(b)-f(a)}{b-a}$ .
Solve the resulting simultaneous equations for the unknowns.

Tangent parallel to chord (geometric MVT). 'Find the point(s) where the tangent is parallel to the chord $AB$ ' means: compute the chord slope through $A, B$ , set $f'(x)$ equal to it, and solve for $x$ — the same computation as finding the LMVT point.

Two equations from 'the theorem holds at c'

\underbrace{f(a)=f(b)}_{\text{endpoint}} \quad \text{and} \quad \underbrace{f'(c)=0}_{\text{Rolle at given }c}

Worked example

Rolle's theorem holds for

f(x) = x^3 + ax^2 + bx

[0, 3]

at the point

c = 1

. Find

a

and

b

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Applications of DerivativeMODERATE

If Rolle's theorem holds for the function

x^{3}+ax^{2}+bx,1 \leq x\leq 2

at the point

\frac{4}{3}

, then the values of

a

and

b

are respectively

[Q110 · 25 April Shift I · 2025]

You need BOTH equations — endpoint and interior

Using only

f(a)=f(b)

fixes just one relation between

a

and

b

; some options may share that value. You must ALSO use

f'(c)=0

(or the chord condition) at the given

c

to pin down both. Solving one equation and matching a partial answer is how students land on the wrong option.

Watch coefficient order — 'a and b respectively'

Options like

(11, -6)

(-6, 11)

differ only by which value is

a

and which is

b

. Track exactly which unknown you solved for; re-substitute into the original

f

to confirm the pair actually satisfies

f(a)=f(b)

and

f'(c)=0

before selecting.

Drill 4 more on solving unknown parameters and the tangent-parallel-to-chord view

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 (5)

Rolle's Theorem — the Three Hypotheses and the Conclusion
Rolle's theorem
$f \text{ cont. on }[a,b],\ \text{diff. on }(a,b),\ f(a)=f(b)\ \Rightarrow\ \exists\, c\in(a,b):\ f'(c)=0$
Finding c and Rejecting Roots Outside the Interval
Rolle point from f'(x) = 0
$f'(c) = 0,\quad c \in (a,b)\ \ \text{(reject any root outside the interval)}$
Counting the Number of Valid c
Number of Rolle points
$\#\{\,c\in(a,b): f'(c)=0\,\} = \text{number of roots of } f'(x)=0 \text{ inside } (a,b)$
Lagrange's Mean Value Theorem — Statement and Finding c
Lagrange's Mean Value Theorem
$f'(c) = \dfrac{f(b) - f(a)}{b - a},\qquad c \in (a,b)$
Solving Unknown Parameters and the Tangent-Parallel-to-Chord View
Two equations from 'the theorem holds at c'
$\underbrace{f(a)=f(b)}_{\text{endpoint}} \quad \text{and} \quad \underbrace{f'(c)=0}_{\text{Rolle at given }c}$

Watch out for (11)

All THREE hypotheses are required, not just f(a) = f(b)
→ Rolle's Theorem — the Three Hypotheses and the Conclusion
Rolle gives f'(c) = 0, not the chord slope
→ Rolle's Theorem — the Three Hypotheses and the Conclusion
Reject any root of f'(c) = 0 that lies OUTSIDE (a, b)
→ Finding c and Rejecting Roots Outside the Interval
The exponential factor is never zero
→ Finding c and Rejecting Roots Outside the Interval
Count roots INSIDE the open interval only — mind the endpoints
→ Counting the Number of Valid c
Solve the full trig equation — don't stop at the first solution
→ Counting the Number of Valid c
MVT needs DIFFERENTIABILITY, not just continuity
→ Lagrange's Mean Value Theorem — Statement and Finding c
Chord slope uses f(b) − f(a), not f'(a) or f'(b)
→ Lagrange's Mean Value Theorem — Statement and Finding c
Bounding trick: f(b) − f(a) ≤ (b − a)·max f'
→ Lagrange's Mean Value Theorem — Statement and Finding c
You need BOTH equations — endpoint and interior
→ Solving Unknown Parameters and the Tangent-Parallel-to-Chord View
Watch coefficient order — 'a and b respectively'
→ Solving Unknown Parameters and the Tangent-Parallel-to-Chord View

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