If the function defined by \(f(x) = \begin{cases}kx^2 + 1 & ,\text{ if } x \le 1\\2 & ,\text{ if } x \gt 1\end{cases}\) is continuous at \(x = 1\), then \(k\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 1
For continuity at x = 1, the left-hand limit, the right-hand limit and f(1) must be equal.
\(\displaystyle \lim_{x \to 1^-} f(x) = f(1) = k + 1\)
\(\displaystyle \lim_{x \to 1^+} f(x) = 2\)
Equating:
\(k + 1 = 2,\quad k = 1\)