---
title: "Verify Rolle's theorem for a log function"
subject: "Mathematics I"
chapter: 1
key: "semester_1/mathematics_i:rolle-verify-log-function"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/rolle-verify-log-function
site: BCT Engineering
---

# Verify Rolle's theorem for a log function

Mathematics I · Chapter 1: Derivatives and their Applications

- Asked in: 2073 Shrawan
- Video: [verify Rolle’s theorem for the functions f(x)=log{x^2+ab/x(a+b) }in [a, b]|verify rolle's theorem](https://www.youtube.com/watch?v=p16SqkZJZEc) — RS ACADEMY
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/rolle-verify-log-function
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-1

## Question

Verify Rolle's Theorem for f(x) = log((x²+ab)/((a+b)x)); x ∈ [a,b]. How does Rolle's Theorem differ from Lagrange's Mean value theorem?

## Short answer

f(x) = log(x²+ab) − log(a+b) − log x. f(a)=0, f(b)=0. f'(x) = (x²−ab)/(x(x²+ab)). Setting f'(c)=0 gives c = √(ab) which lies in (a,b). Thus Rolle's theorem is verified. Rolle's theorem requires f(a)=f(b) and yields f'(c)=0; Lagrange's MVT does not require equal endpoints and gives f'(c)=(f(b)−f(a))/(b−a).

## Answer

#### Verification of Rolle's theorem
f(x) = log[(x² + ab)/((a + b)x)] = log(x² + ab) − log(a + b) − log x on [a, b], with 0 < a < b.

1. **Continuity/differentiability:** x² + ab > 0 and x > 0 on [a, b], so f is continuous on [a, b] and differentiable on (a, b).
2. **End values:**
   - f(a) = log[(a² + ab)/((a+b)a)] = log[a(a+b)/((a+b)a)] = log 1 = 0
   - f(b) = log[(b² + ab)/((a+b)b)] = log[b(a+b)/((a+b)b)] = log 1 = 0

   So f(a) = f(b).
3. **Derivative:**

```text
f′(x) = 2x/(x² + ab) − 1/x = (2x² − x² − ab)/(x(x² + ab))
      = (x² − ab)/(x(x² + ab))
```
f′(c) = 0 gives c² = ab, so c = √(ab). Since a < √(ab) < b (geometric mean), c lies in (a, b).

**Rolle's theorem is verified with c = √(ab).**

#### Rolle's vs Lagrange's theorem
Rolle's theorem needs f(a) = f(b) and gives f′(c) = 0 (tangent parallel to the x-axis). Lagrange's mean value theorem drops the condition f(a) = f(b) and gives f′(c) = (f(b) − f(a))/(b − a) (tangent parallel to the chord). Rolle's theorem is the special case of Lagrange's theorem when f(a) = f(b).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. f(a) and f(b) for f = log[(x^2+ab)/((a+b)x)] are:
   A. log a and log b
   B. both 0
   C. both 1
   D. a and b
   Answer: B. Both arguments equal 1. (Key point: End values)

2. f'(x) =
   A. (2x^2 + ab)/x
   B. 2x/(x^2 + ab)
   C. (x^2 - ab)/(x(x^2 + ab))
   D. (x^2 + ab)/(x(x^2 - ab))
   Answer: C. 2x/(x^2+ab) - 1/x. (Key point: Derivative)

3. c = sqrt(ab) lies in (a, b) because:
   A. a = b
   B. ab > b
   C. it equals (a+b)/2
   D. it is the geometric mean
   Answer: D. a < sqrt(ab) < b for 0 < a < b. (Key point: Value of c)

4. Rolle's theorem differs from Lagrange's in that it:
   A. needs f' continuous
   B. needs f(a) = f(b) and gives f'(c) = 0
   C. gives a chord slope
   D. drops continuity
   Answer: B. Rolle is the special case of Lagrange. (Key point: Difference)

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