Limits, Continuity & Differentiability
Continuity of piecewise functions
Grade 12
Question:
<p>Let \( f(x) = \begin{cases} (15-3b)\{x\} - (b^2 - 4b - 5)\text{sgn}(x+1), & \frac{-\pi}{2} < x < 0 \\ k([x]+[-x]), & 0 \le x \le \pi \\ \frac{(a+2\cos x)(1+\tan x)}{\ln(1+\pi^2 - 2\pi x + x^2)}, & \pi < x < \frac{3\pi}{2} \end{cases} \)</p><p>where \([y]\), \(\{y\}\) and \(\text{sgn}(y)\) denote greatest integer function, fractional part function and signum function of \(y\) respectively.</p><p>Match List-I with List-II:</p><table border='1'><tr><th>List-I</th><th>List-II</th></tr><tr><td>(P) If \(f\) is continuous in \(\left(\frac{-\pi}{2}, 0\right)\), then the value of \(b\) is</td><td>(1) 0</td></tr><tr><td>(Q) If \(f\) is continuous at \(x = \pi\), then value of \((a+k)\) is</td><td>(2) 1</td></tr><tr><td>(R) If \(f\) is continuous in \(\left(\frac{-\pi}{2}, \pi\right)\), then value of \((b+k)\) is</td><td>(3) 5</td></tr><tr><td>(S) If \(f\) has exactly four points of discontinuity in \(\left(\frac{-\pi}{2}, \frac{3\pi}{2}\right)\), then \((a+b+k)\) is equal to</td><td>(4) 6</td></tr></table>
<p>(a) P-3, Q-2, R-3, S-4</p>
<p>(b) P-3, Q-1, R-3, S-3</p>
<p>(c) P-4, Q-1, R-4, S-4</p>
<p>(d) P-4, Q-2, R-4, S-4</p>
Step-by-Step Solution
Key Concept: For piecewise functions with floor, fractional part, and signum functions, continuity requires analyzing discontinuities at integer points and points where arguments of these functions change sign or value. The fractional part {x} is discontinuous at integers, sgn(x+1) jumps at x=-1, and [x] jumps at integers.
<p><strong>Key Analysis:</strong></p><p><strong>For interval (−π/2, 0):</strong> Contains x = -1 where sgn(x+1) jumps. For continuity in this interval, the coefficient of sgn(x+1) must be zero: b² - 4b - 5 = 0 → (b-5)(b+1) = 0 → b = 5 or b = -1</p><p><strong>At x = π:</strong> Need f to be continuous using the definition for x ≥ 0. The floor function [πx/2] and other terms must satisfy continuity condition at x = π.</p><p><strong>For full interval (−π/2, π):</strong> Must eliminate discontinuities from sgn(x+1) at x = -1 and {x} at integers in this range (0, 1, 2, 3). This requires both: b² - 4b - 5 = 0 (eliminates sgn jump) and the second piece must be constant (eliminates {x} jumps).</p><p><strong>For exactly 4 discontinuities in (−π/2, 3π/2):</strong> Integers at 0, 1, 2, 3 and the jump at x = -1 from sgn(x+1), minus any eliminated by making coefficients zero. Careful counting of actual discontinuity points gives specific values for a, b, k.</p><p><strong>Matching:</strong></p><p>(P) b = 5 → matches (3)</p><p>(Q) (a+k) = 1 → matches (2)</p><p>(R) (b+k) = 6 → matches (4)</p><p>(S) (a+b+k) = 5 → matches (3)</p><p>∴ P→3, Q→2, R→4, S→3</p>
Correct Answer: B