Question

The value of \(c\) satisfied by Rolle's theorem for the function \(f(x)=\log\left(\frac{x^2+6}{5x}\right)\) in the interval \([2,3]\) is

• A. \(\sqrt{6}\)
• B. \(-\sqrt{6}\)
• C. \(5\sqrt{6}\)
• D. \(-5\sqrt{6}\)

Hint:

In Rolle's theorem questions, first verify \(f(a)=f(b)\), then solve \(f'(c)=0\) and choose \(c\in(a,b)\).

Answer 1

The Correct Option is A

Concept:
Rolle's theorem says that if a function \(f(x)\) is continuous on \([a,b]\), differentiable on \((a,b)\), and: \[ f(a)=f(b) \] then there exists at least one \(c\in(a,b)\) such that: \[ f'(c)=0 \] Here: \[ f(x)=\log\left(\frac{x^2+6}{5x}\right) \] and the interval is: \[ [2,3] \]

Step 1: Check Rolle's theorem condition.
\[ f(2)=\log\left(\frac{2^2+6}{5(2)}\right) \] \[ f(2)=\log\left(\frac{4+6}{10}\right) \] \[ f(2)=\log(1)=0 \] Now: \[ f(3)=\log\left(\frac{3^2+6}{5(3)}\right) \] \[ f(3)=\log\left(\frac{9+6}{15}\right) \] \[ f(3)=\log(1)=0 \] So: \[ f(2)=f(3) \]

Step 2: Differentiate the function.
\[ f(x)=\log(x^2+6)-\log(5x) \] Differentiate: \[ f'(x)=\frac{2x}{x^2+6}-\frac{1}{x} \]

Step 3: Apply Rolle's theorem.
By Rolle's theorem: \[ f'(c)=0 \] So: \[ \frac{2c}{c^2+6}-\frac{1}{c}=0 \] \[ \frac{2c}{c^2+6}=\frac{1}{c} \] Cross multiply: \[ 2c^2=c^2+6 \] \[ c^2=6 \] \[ c=\pm\sqrt{6} \]

Step 4: Choose the value lying in the interval.
Since: \[ c\in(2,3) \] and: \[ \sqrt{6}\approx2.449 \] So: \[ \sqrt{6}\in(2,3) \] But: \[ -\sqrt{6}\notin(2,3) \] Therefore: \[ c=\sqrt{6} \] Hence, the correct answer is: \[ \boxed{(A)\ \sqrt{6}} \]