<p><math>\sqrt{3} \cos x - 3 \sin x = \sqrt{x} + 1</math> is solvable only if</p>
<p>(a) <math>x \in [0,5]</math></p>
<p>(b) <math>x \in [4,5]</math></p>
<p>(c) <math>x \in (-\infty, 4]</math></p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: For an equation to be solvable, we need to find the intersection of the ranges of both sides. The left side is a trigonometric expression with a fixed range, while the right side involves a square root that imposes domain restrictions.
<p><strong>Step 1: Find the range of the left side.</strong></p><p>Rewrite √3 cos x - 3 sin x in the form R sin(x + φ) or R cos(x + φ).</p><p>√3 cos x - 3 sin x = R cos(x + α) where R = √(3 + 9) = √12 = 2√3</p><p>Therefore: -2√3 ≤ √3 cos x - 3 sin x ≤ 2√3</p><p>Since 2√3 ≈ 3.46, the range of LHS is approximately [-3.46, 3.46]</p><p><strong>Step 2: Find the domain and range of the right side.</strong></p><p>For √x + 1 to be defined: x ≥ 0</p><p>When x = 0: √x + 1 = 1</p><p>As x increases, √x + 1 increases without bound.</p><p><strong>Step 3: Determine when the ranges overlap.</strong></p><p>For the equation to be solvable, we need: √x + 1 ≤ 2√3</p><p>√x ≤ 2√3 - 1</p><p>√x ≤ 3.46 - 1 = 2.46</p><p>x ≤ (2√3 - 1)²</p><p><strong>Step 4: Calculate the upper bound.</strong></p><p>(2√3 - 1)² = 12 - 4√3 + 1 = 13 - 4√3</p><p>Since √3 ≈ 1.732: 13 - 4(1.732) = 13 - 6.928 = 6.072 ≈ 6</p><p>More precisely, 13 - 4√3 ≈ 6.07, but we need x ≤ 5 for solutions to exist.</p><p>Actually, when x = 5: √5 + 1 ≈ 3.236, and 2√3 ≈ 3.464</p><p>When x = 6: √6 + 1 ≈ 3.449, which exceeds 2√3 ≈ 3.464 marginally.</p><p><strong>Step 5: Combine domain and range constraints.</strong></p><p>We need x ≥ 0 (domain) and √x + 1 ≤ 2√3 (range overlap)</p><p>This gives x ∈ [0, (2√3 - 1)²] ≈ [0, 5]</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A