GATE CS 2015 Set 2 Q26 — Calculus: Limits, Continuity, Differentiability

Let us analyze each statement for the function $f(x) = x^{-1/3} = \frac{1}{x^{1/3}}$ in the interval $[-1, 1]$:

1. Statement I: $f$ is continuous in $[-1, 1]$
   At $x = 0$, which lies within $[-1, 1]$, the function is undefined because: $$\lim_{x \to 0^+} f(x) = \infty \quad \text{and} \quad \lim_{x \to 0^-} f(x) = -\infty$$\ Since the limit does not exist at $x = 0$, $f(x)$ is discontinuous at $x = 0$. Thus, Statement I is FALSE.


2. Statement II: $f$ is not bounded in $[-1, 1]$
   As $x \to 0$, $|f(x)| \to \infty$. Since the function values can be arbitrarily large in magnitude near $x = 0$, the function is unbounded on $[-1, 1]$. Thus, Statement II is TRUE.


3. Statement III: $A$ is nonzero and finite
   The area $A$ bounded by $f(x)$ and the X-axis from $x = -1$ to $1$ is given by the integral of the absolute value of $f(x)$: $$A = \int_{-1}^{1} |f(x)| \, dx = \int_{-1}^{1} |x^{-1/3}| \, dx$$ Since $|x^{-1/3}|$ is an even function, we can simplify this improper integral as: $$A = 2 \int_{0}^{1} x^{-1/3} \, dx$$ Evaluating the limit: $$A = 2 \lim_{t \to 0^+} \left[ \frac{x^{2/3}}{2/3} \right]_t^1 = 2 \cdot \frac{3}{2} \lim_{t \to 0^+} (1 - t^{2/3}) = 3$$\ Since $A = 3$, the area is nonzero and finite. Thus, Statement III is TRUE.

Therefore, only statements II and III are true.