Question:medium
If \[ f(x)= \begin{cases} x^2, & \text{if } x\leq 2,\\[4pt] 4x-\alpha, & \text{if } x>2, \end{cases} \] is continuous at \(x=2\), then the value of \(\alpha\) is equal to:
If \[ f(x)= \begin{cases} x^2, & \text{if } x\leq 2,\\[4pt] 4x-\alpha, & \text{if } x>2, \end{cases} \] is continuous at \(x=2\), then the value of \(\alpha\) is equal to:
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For piecewise linear or quadratic functions, continuity simply means "the two parts must meet at the boundary". Just plug the boundary value into both pieces and set them equal.
Updated On: Jun 25, 2026
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The Correct Option is D
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