Let \(f(x) = \begin{cases}\displaystyle\frac{x^{2}}{4} - 2x + 5, & x \lt 2\\ |x - 4|, & x \ge 2\end{cases}\). The set of all points at which f is not differentiable is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\{4\}\)
Each piece is differentiable inside its own interval, except |x − 4| at its corner x = 4. Only x = 2 and x = 4 need checking.
For \(2 \le x \lt 4\), \(|x - 4| = 4 - x\).
At x = 2 (the join). Both pieces give the value 2:
\( \displaystyle \frac{4}{4} - 4 + 5 = 2, \quad 4 - 2 = 2 \)
So f is continuous at 2. The one-sided derivatives are:
\( \displaystyle Lf'(2) = \frac{2}{2} - 2 = -1 \)
\( Rf'(2) = \) derivative of \(4 - x\) \(= -1\)
They are equal, so f is differentiable at 2.
At x = 4 (the corner).
\(Lf'(4) = -1\) (from \(4 - x\)), \(\ Rf'(4) = 1\) (from \(x - 4\))
They are unequal, so f is not differentiable at 4.
So the set is {4}.