<p><strong>Paragraph for Question nos. 626 and 627</strong><br>Consider, \(f(x) = \lim_{n \to \infty} \dfrac{\text{sgn}(\sqrt{ac}-b)e^{nx} + x^2 + f}{2e^{nx+x} + x + d}\) where \(a > b > c > 0\) and \(d, f \in R\).<br>[Note: sgn\((y)\) denotes the signum function of \(y\).]<br><br>If \(a\), \(b\) and \(c\) are in A.P. and \(f(x)\) is continuous for all \(x \in R\), then the value of \((2f + d + 1)\) is equal to:</p>
Step-by-Step Solution
Key Concept: For the limit to exist and be continuous for all x ∈ ℝ, the dominant exponential terms must cancel or be controlled. Since a, b, c are in A.P., we have √(ac) = b, making sgn(√ac - b) = 0, which eliminates the e^(nx) term in numerator and forces the denominator's exponential term to vanish too.
<p><strong>Step 1:</strong> Since a, b, c are in A.P., we have: 2b = a + c, which implies √(ac) = b (by AM-GM equality when terms are in A.P.).</p><p><strong>Step 2:</strong> Therefore, sgn(√(ac) - b) = sgn(0) = 0. The limit becomes:</p><p>f(x) = lim_{n→∞} [0·e^(nx) + x² + f] / [2e^(nx+x) + x + d] = lim_{n→∞} (x² + f) / [2e^(nx+x) + x + d]</p><p><strong>Step 3:</strong> For x > 0: e^(nx+x) → ∞, so f(x) = 0.</p><p>For x < 0: e^(nx+x) → 0, so f(x) = (x² + f) / (x + d).</p><p>For x = 0: f(0) = f / (2 + d).</p><p><strong>Step 4:</strong> For continuity at x = 0, we need f(0⁺) = f(0⁻) = f(0):</p><p>• From right: 0 = f/(2 + d), so f = 0</p><p>• From left: (0 + f)/(0 + d) = f/d must equal 0, confirming f = 0</p><p>• At x = 0: f(0) = 0/(2 + d) = 0 (satisfied for any d ≠ -2)</p><p><strong>Step 5:</strong> For continuity as x → 0⁻, (x² + 0)/(x + d) must approach 0, requiring d = 0.</p><p><strong>Step 6:</strong> Therefore, f = 0 and d = 0.</p><p>∴ (2f + d + 1) = 2(0) + 0 + 1 = <strong>1</strong></p>
Correct Answer: A