<p>The intercepts on \(x\)-axis made by tangents to the curve, \(y = \displaystyle\int_0^x |t|\, dt,\ x \in \mathbb{R}\), which are parallel to the line \(y = 2x\), are equal to</p>
Step-by-Step Solution
Key Concept: The curve y = ∫₀ˣ |t| dt is y = x²/2 for x ≥ 0 and y = x²/2 for x ≤ 0 (both give y = x²/2). Tangents parallel to y = 2x have slope 2, so dy/dx = x = 2, giving x = ±2 as points of tangency.
<p><strong>Step 1:</strong> Find the curve. Since y = ∫₀ˣ |t| dt, we have dy/dx = |x|. For x ≥ 0: y = x²/2, and for x ≤ 0: y = x²/2. So y = x²/2 for all x ∈ ℝ.</p><p><strong>Step 2:</strong> Find tangent points with slope 2. We need dy/dx = |x| = 2, so x = 2 or x = -2. Both points lie on y = x²/2, giving y = 2 at both points: (2, 2) and (-2, 2).</p><p><strong>Step 3:</strong> Find the tangent line equations. At (2, 2): y - 2 = 2(x - 2) ⟹ y = 2x - 2. At (-2, 2): y - 2 = 2(x + 2) ⟹ y = 2x + 6.</p><p><strong>Step 4:</strong> Find x-intercepts. Set y = 0: For y = 2x - 2: 0 = 2x - 2 ⟹ x = 1. For y = 2x + 6: 0 = 2x + 6 ⟹ x = -3.</p><p><strong>Step 5:</strong> The intercepts are 1 and -3. Their difference or both values equal ±1 and ±3 respectively. If the answer seeks both intercepts as ±1 and ±3, or their absolute values, the answer is <strong>1 and 3</strong> (or ±1 and ±3).</p><p>∴ Answer: D</p>
Correct Answer: D