For what value of \(\alpha\), the function \(f\) defined by \(f(x) = \begin{cases}\alpha(x^2 - 2x + 1), & \text{if } x \le 0\\2x + 1, & \text{if } x \gt 0\end{cases}\) is continuous at \(x = 0\)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\alpha = 1\)
f is continuous at x = 0 if the left-hand limit, the right-hand limit and \(f(0)\) are equal.
\(f(0) = \alpha(0 - 0 + 1) = \alpha\)
\( \displaystyle \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \alpha(x^2 - 2x + 1) = \alpha \)
\( \displaystyle \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2x + 1) = 1 \)
Equating them gives \(\alpha = 1\).