Question

Let f be a function defined by $y = f(x)$ where $x = 2t - |t|$ & $y = t^2 + t/|t|$ for $t \in R$, then

• A. f(x) is both continuous & differentiable at x = 0
• B. f(x) is non-differentiable at x = 0
• C. f(x) is discontinuous at x = 0
• D. f(x) is neither continuous nor differentiable at x = 0

Answer 1

Answer: (B), (C), (D)

B, C, D

Answer 2

For $t > 0$, $x = 2t - t = t$ and $y = t^2 + t/t = t^2 + 1$. So, $f(x) = x^2 + 1$ for $x > 0$. For $t < 0$, $x = 2t - (-t) = 3t$ and $y = t^2 + t/(-t) = t^2 - 1$. So, $t = x/3$, which gives $f(x) = (x/3)^2 - 1 = x^2/9 - 1$ for $x < 0$. For $t=0$, $x = 2(0) - |0| = 0$, but $y = 0^2 + 0/|0|$ is undefined. Thus $f(0)$ is undefined.

Continuity at $x=0$: Right-hand limit: $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 + 1) = 1$. Left-hand limit: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x^2/9 - 1) = -1$. Since the left-hand limit $\neq$ right-hand limit, the limit does not exist. Also, $f(0)$ is undefined. Thus, $f(x)$ is discontinuous at $x=0$.

Differentiability at $x=0$: A function must be continuous at a point to be differentiable at that point. Since $f(x)$ is discontinuous at $x=0$, it is also non-differentiable at $x=0$.

Therefore, $f(x)$ is discontinuous at $x=0$ and non-differentiable at $x=0$. This means options B, C, and D are all correct.