Limits, Continuity & Differentiability
Continuity and Differentiability of Composite Functions
Grade 12
<p><strong>Example 31:</strong> Let \(f(x) = \begin{cases} [x], & -2 \le x < 0 \\ |x|, & 0 \le x \le 2 \end{cases}\) and \(g(x) = \sec x, x \in \mathbb{R} - \{(2n-1)\frac{\pi}{2}\}\).</p><p>Match the following statements in Column I with their values in Column II:</p><p><strong>Column I:</strong></p><p>(A) Limit of fog exists at</p><p>(B) Limit of gof doesn't exist at</p><p>(C) Points of discontinuity of fog is/are</p><p>(D) Points of differentiability of fog is/are</p><p><strong>Column II:</strong></p><p>(p) \(-1\)</p><p>(q) \(\pi\)</p><p>(r) \(\frac{5\pi}{6}\)</p><p>(s) \(-\pi\)</p>
Step-by-Step Solution
Key Concept: Understanding composition of functions and analyzing limits, continuity, and differentiability at specific points using properties of greatest integer function and secant function.
<p><strong>Solution:</strong> By analyzing the composition fog and gof:</p><p>For fog: The composition involves applying f first, then g(sec x).</p><p>For gof: The composition involves applying g first (sec x), then f.</p><p><strong>Analysis of each statement:</strong></p><p>(A) Limit of fog exists at: $x = -1, \pi, -\pi$ → matches (p, q, s)</p><p>(B) Limit of gof doesn't exist at: $x = -1$ → matches (p)</p><p>(C) Points of discontinuity of fog are: $\pi, -\pi$ → matches (q, s)</p><p>(D) Points of differentiability of fog are: $x = -1, \frac{5\pi}{6}$ → matches (p, r)</p>
Correct Answer: (A)-(p,q,s), (B)-(p), (C)-(q,s), (D)-(p,r)