Mathematics Study Guide

Rolle's theorem

Rolle's theorem is a foundational pillar of differential calculus guaranteeing the existence of at least one critical point with a horizontal tangent under smooth boundary conditions.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

📖 Topic progress

○ Mark as complete

Definition and statement

Rolle's Theorem (Theorem of the Vanishing Derivative)

Let $f: [a, b] \to \mathbb{R}$ be a real-valued function satisfying three hypotheses: (1) $f$ is continuous on the closed interval $[a, b]$, (2) $f$ is differentiable on the open interval $(a, b)$, and (3) $f(a) = f(b)$. Then there exists at least one real number $c \in (a, b)$ such that $$f'(c) = 0$$.

📐 Geometric Interpretation

If the graph of a continuous, smooth curve starts and ends at the exact same height ($f(a) = f(b)$), the curve must peak or trough at least once in the interior. Geometrically, this means there is at least one point $c$ in the open interval $(a, b)$ where the tangent line to the curve is perfectly parallel to the $x$-axis, i.e., its slope is zero.

⚙️ Algebraic & Structural Meaning

Algebraically, Rolle's theorem bridges the gap between function values and derivative values, acting as the stepping stone for Lagrange's Mean Value Theorem and Taylor's Theorem. It implies that between any two distinct real roots of a differentiable function, there lies at least one real root of its derivative.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Continuity on $[a, b]$	`$[a, b]$ (Closed interval)`	Strictly Mandatory / Necessary	If the function has a jump or removable discontinuity in the interval, the Extreme Value Theorem fails, and the curve may not attain local extrema.
Differentiability on $(a, b)$	`$(a, b)$ (Open interval)`	Strictly Mandatory	A sharp corner or cusp (like $f(x) = \|x\|$ at $x=0$) breaks the smooth flow, making the derivative undefined at that point, violating the conclusion.
Boundary Equality $f(a) = f(b)$	`Boundary values $a$ and $b$`	Essential for horizontal tangent	If $f(a) \neq f(b)$, the secant line is tilted, which shifts the theorem into Lagrange's Mean Value Theorem rather than guaranteeing a horizontal tangent.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Root Isolation & Polynomial Derivatives

$$ \text{If } f(x) \text{ has } n \text{ real roots, } f'(x) \text{ has at least } n-1 \text{ real roots in between.} $$

Rule: Between any two consecutive roots of a differentiable function, there is at least one root of its derivative. Use this to count roots instantly without solving higher-degree equations.

Application: Frequently asked in UP PGT to find the minimum number of roots for derivatives of polynomials like $x^n + ax + b = 0$.

Symmetric Intervals & Parity

$$ \text{For } f(x) \text{ even on } [-a, a], \; f'(0) = 0 \text{ automatically by Rolle's / Fermat's principle.} $$

Rule: If a function is even and differentiable on a symmetric interval, its derivative at the origin is strictly zero. If odd, examine symmetry properties for rapid root spotting.

Application: Standard appearance in TGT/PGT matching columns for function properties and critical points.

Algebraic Derivative Roots

$$ \text{If } f(x) = (x-a)^m (x-b)^n, \text{ then } c = \frac{mb + na}{m+n} \in (a, b) $$

Rule: For factored algebraic polynomials, the value of $c$ where $f'(c) = 0$ is simply the weighted arithmetic mean of the roots.

Trigonometric & Exponential Bounds

$$ \text{For } f(x) = e^x \sin x \text{ on } [0, \pi], \; c = \frac{3\pi}{4} $$

Rule: Recognize standard product combinations of trigonometric and exponential functions where $f(0) = f(\pi) = 0$, leading to $c$ at standard angle offsets like $\frac{3\pi}{4}$.

Determinant / Wronskian / Matrix Shortcut

$$ \text{If } \begin{vmatrix} a & b \\ c & d \end{vmatrix} \text{ structures arise, apply auxiliary function } F(x) = f(x)e^{\lambda x}. $$

Rule: Construct auxiliary function $F(x) = f(x)g(x)$ or exponential shifts to satisfy $F(a) = F(b)$ and apply Rolle's theorem to prove existence of roots for linear combinations.

Rapid Elimination Trick

$$ c \in (a, b) \implies \text{Eliminate options where } c \le a \text{ or } c \ge b. $$

Rule: Instantly discard any option where the value of $c$ lies outside the open interval $(a, b)$ or equals the endpoints.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

The Sharp Corner / Modulus Fallacy

Common Pitfall: Candidates apply the theorem to piecewise or absolute value functions without verifying differentiability at interior points.

Counterexample: $f(x) = |x|$ on $[-1, 1]$

Why it fails: Continuous everywhere and $f(-1) = f(1) = 1$, but $f'(0)$ does not exist, so no $c \in (-1, 1)$ satisfies $f'(c) = 0$.

🛡️ Exam Defense: Always test derivative from left and right ($f'_-(x_0) = f'_+(x_0)$) at points where argument changes sign.

⚠️ COMMON ERROR #02

Open vs. Closed Interval Confusion

Common Pitfall: Confusing where continuity vs differentiability is required, or asserting that $c$ can be an endpoint $a$ or $b$.

Counterexample: $c \in (a, b)$ strictly, never $c = a$ or $c = b$

Why it fails: If $f'(x) = 0$ only at endpoints, the theorem is violated because $c$ must be in the open interior $(a, b)$.

🛡️ Exam Defense: Instantly discard any option that offers $c = a$ or $c = b$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Rolle's Theorem	Continuity on $[a,b]$, Differentiability on $(a,b)$, $f(a) = f(b)$	$f'(c) = 0$ for $c \in (a,b)$	Horizontal tangent line exists in the interior	Mentions $f(a) = f(b)$ and horizontal slope
Lagrange's MVT	Continuity on $[a,b]$, Differentiability on $(a,b)$	$f'(c) = \frac{f(b)-f(a)}{b-a}$	Tangent is parallel to secant chord joining endpoints	Mentions general chord slope without $f(a)=f(b)$
Cauchy's MVT	Continuity on $[a,b]$, Differentiability on $(a,b)$, $g'(x) \neq 0$	$\frac{f'(c)}{g'(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}$	Parametric curve slope ratio equivalence	Involves two functions $f(x)$ and $g(x)$
Intermediate Value Theorem	Continuity on $[a,b]$	$f(a) < k < f(b) \implies f(c) = k$	Graph crosses every horizontal line between endpoints	Asks for root existence between $a$ and $b$ without derivatives

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If Rolle's theorem is applicable to $f(x) = x^3 - 6x^2 + 11x - 6$ on $[1, 3]$, find the value of $c \in (1, 3)$ such that $f'(c) = 0$.

(A) $\frac{3 + \sqrt{3}}{3}$(B) $2 - \frac{1}{\sqrt{3}}$(C) $2 + \sqrt{3}$(D) $\frac{3 - \sqrt{3}}{2}$

📚 Method A: Formal Textbook Derivation

Step 1: Verify $f(1) = 1 - 6 + 11 - 6 = 0$ and $f(3) = 27 - 54 + 33 - 6 = 0$, so $f(1) = f(3)$. Polynomial is continuous and differentiable everywhere. Step 2: Compute $f'(x) = 3x^2 - 12x + 11$. Step 3: Set $f'(c) = 0 \implies 3c^2 - 12c + 11 = 0$. Step 4: Solve using quadratic formula $c = \frac{12 \pm \sqrt{144 - 132}}{6} = \frac{12 \pm \sqrt{12}}{6} = 2 \pm \frac{1}{\sqrt{3}}$. Step 5: Since $2 + \frac{1}{\sqrt{3}} \approx 2.577 \in (1,3)$, both roots are in range, but check specific option mapping ($2 - \frac{1}{\sqrt{3}}$ is also valid; option B matches standard printing).

⚡ Method B: 25-Second Exam Shortcut

Use shortcut $c = \frac{a+b}{2} \pm \frac{b-a}{2\sqrt{3}}$ for cubic expansions, yielding $2 \pm \frac{1}{\sqrt{3}}$.

💡 Tip: Check interval membership immediately before locking the option.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Let $f(x) = (x-a)^m (x-b)^n$ where $m, n > 0$ on $[a, b]$. The value of $c$ where $f'(c) = 0$ is:

(A) $\frac{ma + nb}{m+n}$(B) $\frac{mb + na}{m+n}$(C) $\frac{a+b}{2}$(D) $\sqrt{ab}$

📚 Method A: Formal Textbook Derivation

Step 1: $f(a) = 0$ and $f(b) = 0$, satisfying Rolle's theorem. Step 2: Apply logarithmic differentiation: $\ln f(x) = m \ln(x-a) + n \ln(x-b)$. Step 3: Differentiate: $\frac{f'(x)}{f(x)} = \frac{m}{x-a} + \frac{n}{x-b}$. Step 4: Set $f'(c) = 0 \implies \frac{m}{c-a} + \frac{n}{c-b} = 0 \implies m(c-b) + n(c-a) = 0 \implies c(m+n) = mb + na \implies c = \frac{mb + na}{m+n}$.

⚡ Method B: 25-Second Exam Shortcut

Direct formula application from Shortcut 3.

💡 Tip: Memorize this weighted mean result as it appears directly in UP PGT exams without numerical changes.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: Which of the following functions satisfies all conditions of Rolle's theorem on $[-1, 1]$?

A $f(x) = |x|$

B $f(x) = x^{2/3}$

C $f(x) = x^2 - 1$

D $f(x) = \tan x$

Derivation & Analysis:

$f(x) = x^2 - 1$ is continuous on $[-1, 1]$, differentiable on $(-1, 1)$, and $f(-1) = f(1) = 0$. $f(x) = |x|$ and $x^{2/3}$ fail differentiability at $0$, and $\tan x$ is undefined at $\pm \pi/2$ or has singularities.

MCQ 2: If $f(x) = e^x(\sin x - \cos x)$ satisfies Rolle's theorem on $[\pi/4, 5\pi/4]$, then the value of $c$ is:

A $\pi$

B $\frac{\pi}{2}$

C $\frac{3\pi}{4}$

D $\frac{3\pi}{2}$

Derivation & Analysis:

Compute $f(\pi/4) = e^{\pi/4}(0) = 0$ and $f(5\pi/4) = e^{5\pi/4}(0) = 0$. Differentiating gives $f'(x) = e^x(\sin x - \cos x) + e^x(\cos x + \sin x) = 2e^x \sin x$. Setting $f'(c) = 0 \implies \sin c = 0 \implies c = \pi \in (\pi/4, 5\pi/4)$.

MCQ 3: If $f(x) = a x^2 + b x + c$ satisfies the conditions of Rolle's theorem on $[x_1, x_2]$, then the value of $c$ is:

A $x_1 + x_2$

B $\frac{x_1 + x_2}{2}$

C $\sqrt{x_1 x_2}$

D $\frac{2x_1 x_2}{x_1 + x_2}$

Derivation & Analysis:

For any quadratic polynomial satisfying Rolle's theorem, the critical point $c$ is precisely the midpoint of the interval: $c = -b / (2a) = (x_1 + x_2)/2$.

MCQ 4: Between two consecutive roots of the equation $e^x \sin x = 1$, the derivative $e^x(\sin x + \cos x)$ has:

A exactly one root

B no root

C at least two roots

D infinite roots

Derivation & Analysis:

By Rolle's theorem, between any two roots of a differentiable function, there lies at least one root of its derivative.

MCQ 5: The function $f(x) = x(x-1)(x-2)$ on $[0, 2]$ has how many values of $c$ satisfying Rolle's theorem?

A 0

B 1

C 2

D 3

Derivation & Analysis:

$f(0) = f(2) = 0$. $f(x) = x^3 - 3x^2 + 2x \implies f'(x) = 3x^2 - 6x + 2 = 0 \implies c = \frac{6 \pm \sqrt{36 - 24}}{6} = 1 \pm \frac{1}{\sqrt{3}}$, both of which lie in $(0, 2)$. Thus there are 2 values.

MCQ 6: Which hypothesis of Rolle's theorem fails for $f(x) = \frac{1}{x}$ on $[-1, 1]$?

A $f(a) = f(b)$

B Continuity on $[-1, 1]$

C Differentiability on $(-1, 1)$

D Both continuity and differentiability

Derivation & Analysis:

The function $\frac{1}{x}$ is discontinuous at $x = 0$, violating the closed-interval continuity requirement.

MCQ 7: If $f(x)$ is differentiable on $\mathbb{R}$ and $f'(x) \neq 1$ for all $x$, then $f(1) - f(0)$ can be:

A 1

B 2

C any real number

D not equal to 1

Derivation & Analysis:

By Lagrange's MVT (consequence of Rolle's), $\frac{f(1)-f(0)}{1-0} = f'(c) \neq 1$, hence $f(1) - f(0) \neq 1$.

MCQ 8: Assertion (A): Rolle's theorem cannot be applied to $f(x) = \tan x$ on $[0, \pi]$. Reason (R): $f(x)$ is not continuous at $x = \pi/2$.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Discontinuity at $x = \pi/2$ violates the closed interval continuity requirement, making Rolle's theorem inapplicable.

MCQ 9: If $f(x) = (x-1)(x-2)(x-3)$, then roots of $f'(x) = 0$ lie in:

A $(1, 2)$ and $(2, 3)$

B $(0, 1)$ and $(1, 2)$

C $(2, 3)$ and $(3, 4)$

D outside $(1, 3)$

Derivation & Analysis:

Roots of $f(x)$ are $1, 2, 3$. By Rolle's theorem, one root of $f'(x)$ lies in $(1, 2)$ and another in $(2, 3)$.

Q10

MCQ 10: If $f(x) = x^n$ on $[a, b]$ where $0 < a < b$, then the value of $c$ given by Rolle's theorem does not exist because:

A $f(a) \neq f(b)$

B $f$ is not continuous

C $f$ is not differentiable

D $f'(x) = 0$ has no root

Derivation & Analysis:

Since $a > 0$ and $b > 0$, $f(a) = a^n \neq b^n = f(b)$, violating the boundary equality condition $f(a) = f(b)$.

Common questions

No. If $f(a) \neq f(b)$, the fundamental hypothesis $f(a) = f(b)$ fails. However, this scenario leads directly to Lagrange's Mean Value Theorem where the secant line is tilted.

Candidates frequently fail to check differentiability at interior points containing absolute values, cube roots, or trigonometric singularities, leading to invalid selections of $c$.

Memorize standard weighted mean formulas for factored polynomials (e.g., Shortcut 3) and utilize interval elimination tricks to bypass lengthy quadratic or cubic root calculations.