2. Calculus — Differential Calculus

Cauchy mean value theorem

Cauchy's Mean Value Theorem generalizes Lagrange's theorem by accommodating two functions, providing a fundamental tool for evaluating indeterminate forms via L'Hôpital's rule. This topic covers its parametric form and hypotheses.

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

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Definition and statement

Cauchy's Generalized Mean Value Theorem

Let $f$ and $g$ be two real-valued functions defined on the closed interval $[a, b]$ such that: (1) $f$ and $g$ are continuous on $[a, b]$; (2) $f$ and $g$ are differentiable on $(a, b)$; (3) $g'(x) \neq 0$ for all $x \in (a, b)$. Then there exists at least one real number $c \in (a, b)$ such that $$\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$$.

📐 Geometric Interpretation

Parametrically, it represents the slope ratio of secant to tangent vectors on curves defined by $(g(t), f(t))$, ensuring a point where the tangent vector is parallel to the chord joining the endpoints.

⚙️ Algebraic & Structural Meaning

Serves as the rigorous backbone for establishing L'Hôpital's rule and proving Taylor's theorem with Cauchy's form of remainder.

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 either function has a jump or infinite discontinuity in $[a, b]$, the quotient ratio of increments may not be bounded or attained.
Differentiability on $(a, b)	`(a, b) (Open interval)`	Strictly Mandatory	Failure of differentiability (e.g., sharp corners) implies $f'(c)$ or $g'(c)$ does not exist, destroying the relation.
Non-vanishing derivative $g'(x) \neq 0$ on $(a, b)	`(a, b) (Open interval)`	Essential to prevent division by zero	If $g'(x) = 0$ at some point, the denominator $g(b) - g(a)$ might be zero, or the fraction $\frac{f'(c)}{g'(c)}$ becomes undefined.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Polynomial Function Ratio

$$c = \frac{a+b}{2} \text{ for quadratic numerator and linear/quadratic denominator cases}$$

Rule: For symmetric polynomials, the mean value $c$ is often the arithmetic mean of $a$ and $b$.

Application: Frequently asked with $f(x) = x^2$ and $g(x) = x^3$ on $[a, b]$.

Exponential & Logarithmic Functions

$$c = \frac{b - a}{\ln b - \ln a} = \text{Logarithmic Mean}$$

Rule: When $f(x) = e^x$ and $g(x) = e^{-x}$ or similar exponential pairings, $c$ reduces to standard logarithmic or arithmetic averages.

Application: Direct PYQ in UP PGT testing logarithmic mean connections.

Power Function Pairings

$$c = \left( \frac{a^n + a^{n-1}b + \dots + b^n}{n+1} \right)^{1/n}$$

Rule: For $f(x) = x^m$ and $g(x) = x^n$, apply algebraic factoring to solve for $c$ instantly.

Application: Tested in DSSSB PGT for integer powers.

Trigonometric Pairings

$$c = \frac{a+b}{2} \text{ for } f(x) = \sin x, g(x) = \cos x \text{ on symmetric intervals}$$

Rule: Utilize angle sum/difference formulas to simplify $\frac{\cos c}{-\sin c}$ to $-\cot c$ and equate to secant slope.

Application: Standard KVS PGT trigonometric mean value problem.

Reciprocal Function Shortcut

$$c = \sqrt{ab}$$

Rule: For $f(x) = \frac{1}{x^2}$ and $g(x) = \frac{1}{x}$ on $[a, b]$, $c$ is the geometric mean $\sqrt{ab}$.

Rapid Elimination via Domain Bounds

$$a < c < b$$

Rule: Immediately eliminate any option where $c$ lies outside $(a, b)$ or equals endpoints $a$ or $b$.

Application: Saves 30 seconds on conceptual assertion questions.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Division by Zero in $g(b) - g(a) = 0$

Common Pitfall: Applying Cauchy's theorem when $g(b) = g(a)$, which violates Rolle's theorem application to $g$, causing $g'(c) = 0$ for some $c$.

Counterexample: $f(x) = x^2, g(x) = x^3 - 3x$ on $[-1, 1]$ where $g(-1) = g(1) = 2$

Why it fails: If $g(a) = g(b)$, Rolle's theorem implies $g'(c) = 0$ for some $c \in (a, b)$, rendering the Cauchy quotient $\frac{f'(c)}{g'(c)}$ undefined or meaningless.

🛡️ Exam Defense: Always check that $g(b) \neq g(a)$ before setting up Cauchy's formula.

⚠️ COMMON ERROR #02

Assuming $g'(x)$ can be zero anywhere in $(a, b)$

Common Pitfall: Forgetting that $g'(x) \neq 0$ must hold strictly across the entire open interval $(a, b)$.

Counterexample: $g(x) = x^3$ on $[-1, 1]$, where $g'(0) = 0$

Why it fails: At $x = 0$, $g'(0) = 0$, which violates the denominator condition of Cauchy's theorem.

🛡️ Exam Defense: Verify $g'(x) = 0$ has no roots in $(a, b)$.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Rolle's Theorem	$f$ continuous on $[a, b]$, differentiable on $(a, b)$, $f(a) = f(b)$	$f'(c) = 0$	Horizontal tangent at $c \in (a, b)$	Function values equal at endpoints
Lagrange's MVT	$f$ continuous on $[a, b]$, differentiable on $(a, b)$	$f'(c) = \frac{f(b) - f(a)}{b - a}$	Tangent parallel to secant line	General slope relation between endpoints
Cauchy's MVT	$f, g$ continuous on $[a, b]$, diff. on $(a, b)$, $g'(x) \neq 0$	$\frac{f'(c)}{g'(c)} = \frac{f(b) - f(a)}{g(b) - g(a)}$	Parametric tangent parallel to chord	Two functions given simultaneously
Taylor's Theorem	$f$ has $n$ continuous derivatives on $[a, b]$ and $(n+1)$-th on $(a, b)$	$f(x) = \sum \frac{f^{(k)}(a)}{k!}(x-a)^k + R_n$	Polynomial approximation of higher order	Expansion of functions in power series

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the value of $c$ guaranteed by Cauchy's Mean Value Theorem for functions $f(x) = e^x$ and $g(x) = e^{-x}$ on the interval $[0, 1].$

(A) $\frac{1}{2} \ln \left(\frac{e-1}{e}\right)$(B) $\frac{1}{2}$(C) $\ln \left(\frac{e+1}{2}\right)$(D) $\frac{1}{2} \ln \left(\frac{e^2-1}{2e}\right)$

📚 Method A: Formal Textbook Derivation

Step 1: Calculate $f(0) = 1, f(1) = e$ and $g(0) = 1, g(1) = e^{-1}$. Step 2: Compute derivatives $f'(x) = e^x, g'(x) = -e^{-x}$. Step 3: Apply Cauchy's formula: $\frac{e^c}{-e^{-c}} = \frac{e - 1}{e^{-1} - 1}$. Step 4: Simplify $-e^{2c} = \frac{e-1}{(1-e)/e} = -e$, giving $e^{2c} = e \implies 2c = 1 \implies c = 1/2$.

⚡ Method B: 25-Second Exam Shortcut

Recognize exponential symmetry: $c$ is the midpoint $\frac{0+1}{2} = \frac{1}{2}$ for reciprocal exponential structures.

💡 Tip: Do not expand blindly; look for algebraic cancellation in exponential ratios.

CANONICAL PROBLEM 02 Dual-Method Breakdown

For $f(x) = x^2$ and $g(x) = x^3$ on the interval $[1, 2]$, the value of $c$ given by Cauchy's Mean Value Theorem is:

(A) $\frac{7}{9}$(B) $\frac{14}{9}$(C) $\frac{28}{9}$(D) $\frac{7}{3}$

📚 Method A: Formal Textbook Derivation

Step 1: $f(1)=1, f(2)=4 \implies f(2)-f(1)=3$. $g(1)=1, g(2)=8 \implies g(2)-g(1)=7$. Step 2: $f'(x)=2x, g'(x)=3x^2$. Step 3: $\frac{2c}{3c^2} = \frac{3}{7} \implies \frac{2}{3c} = \frac{3}{7}$. Step 4: $9c = 14 \implies c = \frac{14}{9}$. Verify $1 < \frac{14}{9} < 2$.

⚡ Method B: 25-Second Exam Shortcut

Equate ratio of derivatives to secant slope: $\frac{2}{3c} = \frac{4-1}{8-1} = \frac{3}{7} \implies c = \frac{14}{9}.

💡 Tip: Always check that the derived value of $c$ lies strictly inside $[1, 2]$; $\frac{14}{9} \approx 1.555$ is valid.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: Cauchy's Mean Value Theorem reduces to Lagrange's Mean Value Theorem when:

A A) $g(x) = x$

B B) $g(x) = x^2$

C C) $g(x) = 1$

D D) $f(x) = g(x)$

Derivation & Analysis:

If $g(x) = x$, then $g'(x) = 1$ and $g(b) - g(a) = b - a$, transforming Cauchy's formula directly into Lagrange's MVT formula.

MCQ 2: For $f(x) = \sin x$ and $g(x) = \cos x$ on $[0, \pi/2]$, the value of $c$ by Cauchy's theorem is:

A A) $\frac{\pi}{4}$

B B) $\frac{\pi}{3}$

C C) $\frac{\pi}{6}$

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

Derivation & Analysis:

Here $f(0)=0, f(\pi/2)=1$ and $g(0)=1, g(\pi/2)=0$. Thus $\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{1-0}{0-1} = -1$. Derivatives give $\frac{\cos c}{-\sin c} = -1 \implies \cot c = 1 \implies c = \pi/4$.

MCQ 3: Which of the following is a necessary condition for Cauchy's Mean Value Theorem on $[a, b]$?

A A) $f'(x)$ must be continuous on $[a, b]$

B B) $g'(x) \neq 0$ for any $x \in (a, b)$

C C) $f(a) = f(b)$

D D) $g(a) = g(b)$

Derivation & Analysis:

Condition $g'(x) \neq 0$ on $(a, b)$ is strictly required to avoid division by zero in $\frac{f'(c)}{g'(c)}$.

MCQ 4: If $f(x) = x^3$ and $g(x) = x^2$ on $[1, 2]$, the value of $c$ is:

A A) $\frac{7}{3}$

B B) $\frac{14}{9}$

C C) $\frac{3}{2}$

D D) $\frac{19}{9}$

Derivation & Analysis:

Secant slope ratio: $\frac{8-1}{4-1} = \frac{7}{3}$. Derivative ratio: $\frac{3c^2}{2c} = \frac{3c}{2}$. Equating: $\frac{3c}{2} = \frac{7}{3} \implies c = \frac{14}{9}.

MCQ 5: Cauchy's Mean Value Theorem cannot be applied to $f(x) = x^2$ and $g(x) = x^3 - x$ on $[-1, 1]$ because:

A A) $f$ is not continuous

B B) $g'(x) = 0$ at $x = \frac{1}{\sqrt{3}}$ inside $(-1, 1)$

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

D D) $g(-1) \neq g(1)$

Derivation & Analysis:

Here $g'(x) = 3x^2 - 1$. Setting $g'(x) = 0$ gives $x = \pm \frac{1}{\sqrt{3}}$, which lie inside $(-1, 1)$, violating the non-vanishing derivative condition.

MCQ 6: For $f(x) = \ln x$ and $g(x) = x$ on $[1, e]$, the value of $c$ is:

A A) $e - 1$

B B) $\ln e$

C C) $\frac{1}{e-1}$

D D) $1$

Derivation & Analysis:

$\frac{f(e)-f(1)}{g(e)-g(1)} = \frac{1-0}{e-1} = \frac{1}{e-1}$. Derivative ratio: $\frac{1/c}{1} = \frac{1}{c}$. Thus $\frac{1}{c} = \frac{1}{e-1} \implies c = e - 1$.

MCQ 7: If $f(x) = e^x$ and $g(x) = e^{2x}$ on $[0, 1]$, the value of $c$ satisfying Cauchy's theorem is:

A A) $0$

B B) $\frac{1}{2}$

C C) Does not exist / invalid

D D) $1$

Derivation & Analysis:

Notice $g(x) = (f(x))^2$. The denominator $g(1) - g(0) = e^2 - 1$ and $f(1)-f(0) = e-1$. Derivative ratio $\frac{e^c}{2e^{2c}} = \frac{1}{2e^c}$. Secant ratio $\frac{e-1}{e^2-1} = \frac{1}{e+1}$. So $2e^c = e+1 \implies e^c = \frac{e+1}{2} \implies c = \ln\left(\frac{e+1}{2}\right)$, which is valid and lies in $(0, 1)$.

MCQ 8: Assertion (A): Cauchy's Mean Value Theorem is used to prove L'Hôpital's Rule. Reason (R): It relates the ratio of increments of two functions to the ratio of their derivatives at an interior point.

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

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

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

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

Derivation & Analysis:

Cauchy's theorem provides the exact mathematical framework to evaluate limits yielding $0/0$ or $\infty/\infty$ in L'Hôpital's rule.

MCQ 9: For linear functions $f(x) = ax + b$ and $g(x) = cx + d$ with $ad - bc \neq 0$ on any interval $[u, v]$, the value of $c$ guaranteed by CMVT:

A A) Exists uniquely at the midpoint

B B) Cannot be determined

C C) Holds trivially for all $x$ in $(u, v)$ as both sides are constant

D D) Fails because derivatives are constant

Derivation & Analysis:

For linear functions, $f'(c) = a$ and $g'(c) = c$, and the secant ratio is also $\frac{a}{c}$, so $\frac{a}{c} = \frac{a}{c}$ is identically true for all $x \in (u, v)$.

Q10

MCQ 10: If $f(x) = x^2$ and $g(x) = x$ on $[a, b]$ where $a > 0$, the Cauchy MVT point $c$ is:

A A) $\frac{a+b}{2}$

B B) $\sqrt{ab}$

C C) $\frac{2ab}{a+b}$

D D) $\frac{a^2+ab+b^2}{3}$

Derivation & Analysis:

Since $g(x) = x$, CMVT reduces to LMVT: $2c = \frac{b^2-a^2}{b-a} = b+a \implies c = \frac{a+b}{2}$.

Common questions

Lagrange's theorem is merely a special case of Cauchy's theorem where $g(x) = x$. Cauchy's theorem handles two independent functions simultaneously, making it indispensable for proofs involving indeterminate forms and Taylor's remainders.

Failing to check whether $g'(x) = 0$ anywhere in the open interval $(a, b)$. If the derivative of the denominator function vanishes at an interior point, the theorem's hypotheses are violated.

Memorize standard polynomial and exponential pairings where $c$ has closed-form algebraic relations (such as arithmetic or logarithmic means) to bypass lengthy algebraic manipulation.

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