Question

The value of \(c\) prescribed by Lagrange's mean value theorem, when \(f(x) = \sqrt{x^2 - 4}\), \(a = 2\) and \(b = 3\), is

• A. 2.5
• B. \(\sqrt{5}\)
• C. \(\sqrt{3}\)
• D. \(\sqrt{3} + 1\)

Hint:

LMVT requires \(f\) to be continuous on \([a,b]\) and differentiable on \((a,b)\).

Answer 1

The Correct Option is B

To find the value of \(c\) prescribed by Lagrange's Mean Value Theorem (LMVT) for the function \(f(x) = \sqrt{x^2 - 4}\) in the interval \( [2, 3] \), we need to follow these steps:

Check the conditions of LMVT: The function must be continuous on \([a, b]\) and differentiable on \((a, b)\).

• \(f(x) = \sqrt{x^2 - 4}\) is continuous for \(x \geq 2\).
• It is also differentiable for \(x > 2\).

Apply Lagrange's Mean Value Theorem:

According to LMVT, there exists at least one \(c\) in \((a, b)\) such that:

\(f'(c) = \frac{f(b) - f(a)}{b - a}\)

Calculate \(f(a)\) and \(f(b)\):

• \(f(2) = \sqrt{2^2 - 4} = \sqrt{0} = 0\)
• \(f(3) = \sqrt{3^2 - 4} = \sqrt{9 - 4} = \sqrt{5}\)

Calculate the average rate of change:

\(\frac{f(3) - f(2)}{3 - 2} = \frac{\sqrt{5} - 0}{1} = \sqrt{5}\)

Determine \(f'(x)\):

Using the chain rule and differentiating \(f(x) = \sqrt{x^2 - 4}\), we get:

\(f'(x) = \frac{1}{2\sqrt{x^2 - 4}} \cdot 2x = \frac{x}{\sqrt{x^2 - 4}}\)

Find \(c\) such that \(f'(c) = \sqrt{5}\):

Equate the derivative to the average rate of change:

\(\frac{c}{\sqrt{c^2 - 4}} = \sqrt{5}\)

Squaring both sides:

\(\frac{c^2}{c^2 - 4} = 5\)

Solving for \(c\):

\(c^2 = 5(c^2 - 4)\)

\(c^2 = 5c^2 - 20\)

\(4c^2 = 20\)

\(c^2 = 5\)

\(c = \sqrt{5}\) (since \(c\) is in the interval \((2, 3)\))

Therefore, the value of \(c\) that satisfies Lagrange's Mean Value Theorem for \(f(x) = \sqrt{x^2 - 4}\) on \([2, 3]\) is \(\sqrt{5}\).