If \(f\) is a function defined by \(f(x) = \begin{cases} 5x^2 - x + 3, & x \lt 1\\ 3x + 4, & x \ge 1 \end{cases}\), then, at x = 1, \(f\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → continuous but not differentiable
For continuity at x = 1, the left-hand limit, the right-hand limit and f(1) must be equal:
\(\text{LHL} = 5(1)^2 - 1 + 3 = 7\)
\(\text{RHL} = f(1) = 3(1) + 4 = 7\)
All three are 7, so f is continuous at x = 1.
For differentiability, compare the left-hand and right-hand derivatives:
\(x \lt 1:\ f'(x) = 10x - 1,\quad Lf'(1) = 9\)
\(x \gt 1:\ f'(x) = 3,\quad Rf'(1) = 3\)
Since \(9 \ne 3\), f is not differentiable at x = 1. So f is continuous but not differentiable at x = 1.
Option (3), \(\displaystyle\lim_{x \to 1^-} f'(x) = 10\), is wrong: for x < 1, f’(x) = 10x − 1, which tends to 9.
Option (4), \(f''(1) = 10\), is wrong: f’’(1) does not exist, because f’(1) itself does not exist.