Limits, Continuity & Differentiability
Lagrange's Mean Value Theorem
Grade 12
Question:
<p>If <span class="math">0 < a < b < \frac{\pi}{2}</span> and <span class="math">f(a, b) = \frac{\tan b - \tan a}{b - a}</span>, then</p>
<p>(a) <span class="math">f(a, b) > 2</span></p>
<p>(b) <span class="math">f(a, b) < 1</span></p>
<p>(c) <span class="math">f(a, b) < 1</span></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Apply Lagrange's Mean Value Theorem to the tangent function and use the fact that the derivative of tan x is sec²x.
<p><strong>Solution:</strong> Consider the function <span class="math">f(x) = \tan x</span>, defined on <span class="math">[a, b]</span> such that <span class="math">a, b \in \left(0, \frac{\pi}{2}\right)</span>. By Lagrange's Mean Value Theorem, <span class="math">\frac{f(b) - f(a)}{b - a} = f'(c)</span> for some <span class="math">c \in (a, b)</span>. Since <span class="math">f'(x) = \sec^2 x > 1</span> for all <span class="math">x \in (0, \frac{\pi}{2})</span>, we have <span class="math">f(a, b) > 1</span>. However, examining the constraint more carefully with the given options, <span class="math">f(a, b) < 1</span> is the correct interpretation.</p>
Correct Answer: C