GATE CS 2022 Q.2: Quantitative Aptitude

A function $y(x)$ is defined in the interval $[0, 1]$ on the $x$ -axis as

y(x) = \begin{cases} 2 & \text{if } 0 \le x < \frac{1}{3} \\ 3 & \text{if } \frac{1}{3} \le x < \frac{3}{4} \\ 1 & \text{if } \frac{3}{4} \le x \le 1 \end{cases}

Which one of the following is the area under the curve for the interval $[0, 1]$ on the $x$ -axis?

$\frac{5}{6}$

$\frac{6}{5}$

$\frac{13}{6}$

$\frac{6}{13}$

Official answer and solution

Solution

Integrate the piecewise constant function over $[0, 1]$ by summing the areas of the rectangular regions under each interval.

The area under the curve is given by the definite integral:

\text{Area} = \int_0^1 y(x)\,dx = \int_0^{1/3} 2\,dx + \int_{1/3}^{3/4} 3\,dx + \int_{3/4}^1 1\,dx

Evaluating each component:

$\int_0^{1/3} 2\,dx = 2 \times \left(\frac{1}{3} - 0\right) = \frac{2}{3}$
$\int_{1/3}^{3/4} 3\,dx = 3 \times \left(\frac{3}{4} - \frac{1}{3}\right) = 3 \times \frac{9 - 4}{12} = 3 \times \frac{5}{12} = \frac{5}{4}$
$\int_{3/4}^1 1\,dx = 1 \times \left(1 - \frac{3}{4}\right) = \frac{1}{4}$

Summing the three areas:

\text{Area} = \frac{2}{3} + \frac{5}{4} + \frac{1}{4} = \frac{2}{3} + \frac{6}{4} = \frac{2}{3} + \frac{3}{2} = \frac{4 + 9}{6} = \frac{13}{6}

Final answer

Option (C)

Common trap

Incorrectly computing the interval width $\frac{3}{4} - \frac{1}{3} = \frac{5}{12}$ .

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