<p><strong>136.</strong> Let \(f:R\to R\) be given as \(f(x)=\begin{cases}2x+\alpha^2, & x\geq 2\\ \dfrac{\alpha x}{2}+10, & x<2\end{cases}\). If \(f(x)\) is into function then least integral positive value of \(\alpha\) is:</p>
Step-by-Step Solution
Key Concept: For f to be continuous at x=2 (the transition point), the left-hand limit must equal the right-hand limit: lim(x→2⁻) f(x) = lim(x→2⁺) f(x). This equality gives us an equation to solve for α.
<p><strong>Step 1:</strong> Identify the transition point. The function changes definition at x=2.</p><p><strong>Step 2:</strong> For continuity at x=2, set left-hand limit equal to right-hand limit.</p><p>From the left (x→2⁻, use second piece): f(2⁻) = (α·2)/2 + 10 = α + 10</p><p>From the right (x→2⁺, use first piece): f(2⁺) = 2(2) + α² = 4 + α²</p><p><strong>Step 3:</strong> Apply continuity condition: α + 10 = 4 + α²</p><p><strong>Step 4:</strong> Rearrange: α² - α - 6 = 0</p><p><strong>Step 5:</strong> Factor: (α - 3)(α + 2) = 0</p><p><strong>Step 6:</strong> Solve: α = 3 or α = -2</p><p>∴ Answer: B</p>
Correct Answer: B