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

The Lagrange's Mean Value Theorem (LMVT) states that if a function \( f(x) \) is continuous on the closed interval [\( a, b \)] and differentiable on the open interval (\( a, b \)), there exists at least one \( c \) in the interval (\( a, b \)) such that: 

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

Let's apply this theorem to the function \( f(x) = \sqrt{x^2 - 4} \) with \( a = 2 \) and \( b = 3 \).

1. First, check the conditions of the theorem:
   • Continuity: The function \( f(x) = \sqrt{x^2 - 4} \) is continuous for \( x \geq 2 \) because the expression under the square root is positive or zero (i.e., \( x^2 - 4 \geq 0 \)).
   • Differentiability: The function is differentiable for \( x > 2 \). Thus, it satisfies all conditions of LMVT on this interval.
2. Calculate \( f(a) \) and \( f(b) \) :
   • \(f(2) = \sqrt{2^2 - 4} = \sqrt{0} = 0\)
   • \(f(3) = \sqrt{3^2 - 4} = \sqrt{9 - 4} = \sqrt{5}\)
3. Find the derivative \( f'(x) \) :
   • Using the chain rule, \(f'(x) = \frac{d}{dx}\left(\sqrt{x^2-4}\right) = \frac{1}{2\sqrt{x^2-4}} \cdot 2x = \frac{x}{\sqrt{x^2-4}}\)
4. Apply LMVT:

According to LMVT, there exists some \( c \) in (2, 3) such that:

\(\frac{f(3) - f(2)}{3 - 2} = f'(c)\)

• This becomes \(\frac{\sqrt{5} - 0}{1} = \frac{c}{\sqrt{c^2 - 4}}\)
• Simplifying gives: \(\sqrt{5} = \frac{c}{\sqrt{c^2 - 4}}\)
• Let us solve for \( c \):
  • Cross-multiply to get \(c = \sqrt{5} \cdot \sqrt{c^2 - 4}\)
  • Square both sides: \(c^2 = 5(c^2 - 4)\)
  • Expanding: \(c^2 = 5c^2 - 20\)
  • Rearrange to find: \(4c^2 = 20 \Rightarrow c^2 = 5 \Rightarrow c = \sqrt{5}\) (Only consider positive \( c \) because \( c \) is in (2, 3))

1. Conclusion: The value of \( c \) prescribed by the Lagrange's Mean Value Theorem is \(\sqrt{5}\).