Question

For the function $f(x)=\begin{cases} -x, & x<0 \\ x^{2}, & x\ge 0 \end{cases}$ the CORRECT statement(s) is/are
 

• A. $f(x)$ is continuous at $x=1$
• B. $f(x)$ is differentiable at $x=1$
• C. $f(x)$ is continuous at $x=0$
• D. $f(x)$ is differentiable at $x=0$

Hint:

For piecewise functions, continuity requires matching limits; differentiability additionally requires matching left and right derivatives.

Answer 1

The Correct Option is A, B, C

For $x=1$: Since $1 \ge 0$, the function uses $f(x)=x^{2}$. Polynomials are continuous and differentiable everywhere. Thus, (A) and (B) are correct.
[6pt] For $x=0$: Left limit: $\lim_{x\to 0^-} f(x) = \lim_{x\to 0} (-x) = 0$.
Right limit: $\lim_{x\to 0^+} f(x) = \lim_{x\to 0} x^{2} = 0$.
Function value: $f(0)=0^{2}=0$.
Hence $f$ is continuous at $0$ → (C) correct.
[6pt] Differentiability at $0$: Left derivative: $\frac{d}{dx}(-x)=-1$.
Right derivative: $\frac{d}{dx}(x^{2})=2x $\Rightarrow$ 0$.
Left and right derivatives differ → not differentiable → (D) false.