Mean-value theorems (Rolle, Lagrange, Cauchy)

At a Glance

• Frequency: 3 sub-parts across 3 of 14 years (2014, 2021, 2025)
• Priority tier: T3
• Marks (count): 10 (2), 15 (1)
• Average solve time: ~9 min
• Difficulty mix: medium 2, easy 1
• Section: A | Dominant type: proof

Why This Chapter Matters

Mean-value theorem questions are among the most proof-heavy items in Section A, offering 10–15 marks for a clean three- or four-step argument. Two recurring patterns cover every past question: using Rolle’s theorem on a cleverly crafted auxiliary function to locate roots of a related equation, and using Lagrange’s MVT to sandwich a specific value between computable bounds. Both patterns reduce to the same structure — choose the right function, verify the hypotheses, invoke the theorem, translate back. Mastering the auxiliary-function trick (multiply by $e^{-x}$ to shift from $e^x\cos x$ to $\cos x+e^{-x}$) and the derivative-bounding strategy for MVT inequalities covers every question in the corpus.

Minimum Theory

Rolle’s Theorem. If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then there exists $c\in(a,b)$ with $f'(c)=0$.

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

$f(b)-f(a)=f'(c)(b-a).$

Rearranged: $f(b)=f(a)+f'(c)(b-a)$. To prove an inequality $L<f(b)<U$, bound $f'(c)$ between its minimum and maximum on $(a,b)$ using monotonicity.

Cauchy’s MVT. If $f,g$ are continuous on $[a,b]$, differentiable on $(a,b)$, and $g'(x)\ne 0$ on $(a,b)$, then there exists $c\in(a,b)$ with $\dfrac{f'(c)}{g'(c)}=\dfrac{f(b)-f(a)}{g(b)-g(a)}$. (L’Hôpital’s rule is an application.)

Question Archetypes

rolle-root-location (2 question(s); 2014, 2021)

Recognition Cues

• “Between two real roots of [equation involving $e^x\cos x$], a root of [equation involving $e^x\sin x$] lies.”
• The two equations are related by differentiation after an appropriate auxiliary transform.
• Both equations have the form "$e^x\times(\text{trig})\pm 1=0$": the $e^{-x}$ multiplier reduces them to clean trig-plus-exponential expressions.

Solution Template

1. Choose auxiliary function $\phi$. Multiply the original equation’s LHS by $e^{-x}$ (or take $e^{-x}-\cos x$, etc.) so that $\phi(\alpha)=\phi(\beta)=0$ at the two given roots.
2. Verify Rolle’s hypotheses. $\phi$ is continuous on $[\alpha,\beta]$ and differentiable on $(\alpha,\beta)$ (standard for elementary functions).
3. Apply Rolle’s theorem. Conclude $\exists c\in(\alpha,\beta)$ with $\phi'(c)=0$.
4. Compute $\phi'$ and translate. Show that $\phi'(c)=0$ is equivalent to the target equation evaluated at $c$.
5. Conclude. $c$ is a root of the target equation lying strictly between the two given roots.

Worked Example

2014 Paper 1, 2014-P1-Q1c (10 marks)

Prove that between two real roots of $e^x\cos x+1=0$, a real root of $e^x\sin x+1=0$ lies.

Step 1 — Define auxiliary function. Let $\alpha<\beta$ be roots of $e^x\cos x+1=0$. Define

$\phi(x)=e^{-x}(e^x\cos x+1)=\cos x+e^{-x}.$

Since $e^{-x}>0$, $\phi(x)=0\iff e^x\cos x+1=0$, so $\phi(\alpha)=\phi(\beta)=0$.

Step 2 — Rolle’s hypotheses. $\phi=\cos x+e^{-x}$ is $C^\infty$ on $\mathbb R$; in particular continuous on $[\alpha,\beta]$ and differentiable on $(\alpha,\beta)$.

Step 3 — Apply Rolle’s theorem. There exists $\xi\in(\alpha,\beta)$ with $\phi'(\xi)=0$.

Step 4 — Compute $\phi'$ and translate.

$\phi'(x)=-\sin x-e^{-x}.$

So $\phi'(\xi)=0$ gives $-\sin\xi-e^{-\xi}=0$, i.e.\ $e^{-\xi}=-\sin\xi$, i.e.\ $e^\xi\sin\xi+1=0$.

Conclusion. $\xi\in(\alpha,\beta)$ is a root of $e^x\sin x+1=0$.

$\boxed{\;\xi\in(\alpha,\beta):\quad e^\xi\sin\xi+1=0.\;}\qquad\blacksquare$

2021 Paper 1, 2021-P1-Q1d (10 marks)

Show that between any two roots of $e^x\cos x=1$, there exists at least one root of $e^x\sin x-1=0$.

Step 1 — Auxiliary function. Let $a<b$ be roots of $e^x\cos x=1$. Rewrite: $\cos x=e^{-x}$, equivalently $\phi(x)=e^{-x}-\cos x=0$ at both $a$ and $b$.

Step 2 — Rolle’s hypotheses. $\phi$ is $C^\infty$; $\phi(a)=\phi(b)=0$.

Step 3 — Apply Rolle. $\exists c\in(a,b)$ with $\phi'(c)=0$.

Step 4 — Translate. $\phi'(x)=-e^{-x}+\sin x$, so $\phi'(c)=0\Rightarrow\sin c=e^{-c}\Rightarrow e^c\sin c=1$, i.e.\ $e^c\sin c-1=0$.

$\boxed{\;c\in(a,b):\quad e^c\sin c-1=0.\;}\qquad\blacksquare$

Common Traps

• Wrong auxiliary function: applying Rolle directly to $f(x)=e^x\cos x+1$ gives $f'(\xi)=e^\xi(\cos\xi-\sin\xi)=0$, i.e.\ $\tan\xi=1$ — not the desired conclusion. The trick is to multiply by $e^{-x}$ first so that differentiating $e^{-x}$ produces the needed $-e^{-x}$ term alongside $-\sin x$.
• For 2021: the auxiliary is $\phi=e^{-x}-\cos x$ (not $e^{-x}\cos x$); $\phi'=\sin x-e^{-x}$, which vanishes where $e^x\sin x=1$.
• Rolle requires $\phi(a)=\phi(b)=0$ — verify this step explicitly; don’t just assert it.
• After applying Rolle, always compute $\phi'$ explicitly before translating, to make the equivalence visible to the examiner.

mvt-inequality (1 question(s); 2025)

Recognition Cues

• “Using the Mean Value Theorem, prove $L<f(b)<U$” where $f$ is a standard function and $b$ is a specific value.
• The function is $\sin^{-1}$, $\ln$, $\sqrt{\cdot}$, or similar; the interval $[a,b]$ has a known value $f(a)$ and a derivative that can be bounded on $(a,b)$.
• The bounds $L$ and $U$ are obtained by evaluating $f'$ at the endpoints of the interval (monotone derivative argument).

Solution Template

1. Choose $f$ and interval $[a,b]$. Pick $a$ where $f(a)$ is known; the value to bound is $f(b)$.
2. Apply MVT. Write $f(b)=f(a)+f'(c)(b-a)$ for some $c\in(a,b)$.
3. Bound $f'(c)$. Use monotonicity of $f'$ on $(a,b)$: $f'(a)<f'(c)<f'(b)$ (if $f'$ increasing), so $f'(a)(b-a)<f(b)-f(a)<f'(b)(b-a)$.
4. Add $f(a)$ throughout. Obtain $f(a)+f'(a)(b-a)<f(b)<f(a)+f'(b)(b-a)$.
5. Simplify each bound to the stated numerical values.

Worked Example

2025 Paper 1, 2025-P1-Q2b (15 marks)

Using the Mean Value Theorem, prove $\dfrac{\pi}{6}+\dfrac{\sqrt{3}}{15}<\sin^{-1}\!\left(\dfrac{3}{5}\right)<\dfrac{\pi}{6}+\dfrac{1}{8}$.

Step 1 — Choose $f$ and interval. Let $f(x)=\sin^{-1}x$, continuous on $[1/2,\,3/5]$ and differentiable on $(1/2,\,3/5)$, with $f'(x)=\dfrac{1}{\sqrt{1-x^2}}$.

Note $f(1/2)=\pi/6$, and $3/5-1/2=1/10$.

Step 2 — Apply MVT. There exists $c\in(1/2,\,3/5)$ with

$\sin^{-1}\!\left(\tfrac35\right)=\frac{\pi}{6}+\frac{1}{10}\cdot\frac{1}{\sqrt{1-c^2}}.\qquad(\star)$

Step 3 — Bound $f'(c)$. Since $f'(x)=1/\sqrt{1-x^2}$ is strictly increasing and $1/2<c<3/5$,

$f'\!\left(\tfrac12\right)=\frac{1}{\sqrt{1-1/4}}=\frac{2}{\sqrt{3}}$

$f'\!\left(\tfrac35\right)=\frac{1}{\sqrt{1-9/25}}=\frac{1}{\sqrt{16/25}}=\frac{5}{4}.$

So $\dfrac{2}{\sqrt3}<\dfrac{1}{\sqrt{1-c^2}}<\dfrac{5}{4}$.

Step 4 — Multiply by $1/10$ and add $\pi/6$.

$\frac{\pi}{6}+\frac{1}{10}\cdot\frac{2}{\sqrt3}<\sin^{-1}\!\left(\tfrac35\right)<\frac{\pi}{6}+\frac{1}{10}\cdot\frac{5}{4}.$

Step 5 — Simplify.

$\frac{1}{10}\cdot\frac{2}{\sqrt{3}}=\frac{2}{10\sqrt{3}}=\frac{1}{5\sqrt{3}}=\frac{\sqrt{3}}{15},\qquad\frac{1}{10}\cdot\frac{5}{4}=\frac{5}{40}=\frac{1}{8}.$

$\boxed{\;\frac{\pi}{6}+\frac{\sqrt{3}}{15}<\sin^{-1}\!\left(\frac{3}{5}\right)<\frac{\pi}{6}+\frac{1}{8}.\;}\qquad\blacksquare$

Common Traps

• Verify that $f'$ is monotone on $(a,b)$ before bounding by endpoint values; if $f'$ is not monotone, the endpoint bound fails.
• $f'(1/2)=2/\sqrt{3}$ gives the lower bound (since $f'$ is increasing, $c>1/2$ means $f'(c)>f'(1/2)$). Students who confuse which endpoint gives which bound invert the inequality.
• Simplify $1/(10\sqrt{3})=\sqrt{3}/15$ by rationalising: $1/(5\sqrt3)=\sqrt3/15$.
• State the MVT explicitly (“by Lagrange’s MVT, $\exists c\in(a,b)$ such that …”) — don’t just write the equation without naming the theorem.

Marks-Aware Writing

10-mark questions (2014, 2021): The three-step structure — define auxiliary function (identifying it correctly earns 4 marks), verify hypotheses briefly, apply Rolle and translate — covers all marks. Missing the step where $\phi'(c)=0$ is translated back into the target equation loses 3 marks. An answer that applies Rolle to the wrong function earns at most 2 marks.

15-mark question (2025): Step 1 (choose the right interval and recognise $f(1/2)=\pi/6$) is worth 3 marks; the MVT application in Step 2 is worth 4 marks; bounding $f'(c)$ by endpoint values (Step 3) is worth 5 marks; simplifying to the stated closed-form bounds (Step 5) is worth 3 marks. An answer that invokes the MVT but fails to bound the derivative earns 7 marks.

Practice Set

(No additional practice items in the corpus beyond the three worked examples.)