GATE CS 2015 Set 2 Q.36 (CS Q.26): Calculus

Let $f(x) = x^{-(1/3)}$ and $A$ denote the area of the region bounded by $f(x)$ and the X-axis, when $x$ varies from $-1$ to 1. Which of the following statements is/are TRUE?

I) $f$ is continuous in $[-1, 1]$
II) $f$ is not bounded in $[-1, 1]$
III) $A$ is nonzero and finite

II only

III only

II and III only

I, II and III

Official answer and solution

Official answer: C
From: Official GATE 2015 answer key

Solution

Investigate continuity, boundedness, and improper integrability of $f(x) = x^{-1/3}$ over $[-1, 1]$ .

Continuity: The function $f(x) = x^{-1/3} = \dfrac{1}{x^{1/3}}$ is undefined at $x = 0$ . In fact, $\lim_{x \to 0^+} f(x) = +\infty$ and $\lim_{x \to 0^-} f(x) = -\infty$ . Hence, $f(x)$ has an infinite discontinuity at $x = 0$ , so Statement I is false.

Boundedness: Because $\lim_{x \to 0} |f(x)| = +\infty$ , $f(x)$ is unbounded on $[-1, 1]$ . Hence, Statement II is true.

Area $A$ : The geometric area between $f(x)$ and the X-axis is the improper integral of $|f(x)|$ over $[-1, 1]$ :

A = \int_{-1}^1 |f(x)|\, dx = \int_{-1}^0 (-x^{-1/3})\, dx + \int_0^1 x^{-1/3}\, dx.

By symmetry around $x = 0$ :

A = 2 \int_0^1 x^{-1/3}\, dx = 2 \lim_{\epsilon \to 0^+} \left[ \frac{x^{2/3}}{2/3} \right]_\epsilon^1 = 2 \times \frac{3}{2} [1 - 0] = 3.

Since $A = 3$ , the area is non-zero and finite. Hence, Statement III is true.

Thus, Statements II and III only are true (option C).

Final answer

Common trap

Computing the signed integral $\int_{-1}^1 x^{-1/3}\,dx = 0$ instead of the geometric area $A = \int_{-1}^1 |f(x)|\,dx = 3$ .

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2015 Computer Science and Information Technology (CS) Session 2
Question: 36 of 65, printed as CS Q.26
Section: Computer Science and Information Technology

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