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Official PYQ · 2022Q2MCQ1 mark

GATE CS 2022 Q2 — General Aptitude

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

y(x)={2if0≤x<133if13≤x<341if34≤x≤1y(x) = \left\{\begin{matrix} 2& \text{if} & 0 \leq x < \frac{1}{3} \\ 3& \text{if}& \frac{1}{3} \leq x < \frac{3}{4} & \\ 1 & \text{if} & \frac{3}{4} \leq x \leq 1& \end{matrix}\right.

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

Choose one option
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Answer: C

The area under the curve y(x)y(x) over the interval [0,1][0, 1] can be calculated by integrating the function over its sub-intervals, which is equivalent to summing the areas of the three rectangular regions formed by the step function:

Area=∫01y(x) dx=∫01/32 dx+∫1/33/43 dx+∫3/411 dx\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

Let us compute the area for each sub-interval:

  1. First Interval [0,1/3)[0, 1/3): Area1=2×(13−0)=23\text{Area}_1 = 2 \times \left(\frac{1}{3} - 0\right) = \frac{2}{3}
  2. Second Interval [1/3,3/4)[1/3, 3/4): Area2=3×(34−13)=3×9−412=3×512=54\text{Area}_2 = 3 \times \left(\frac{3}{4} - \frac{1}{3}\right) = 3 \times \frac{9 - 4}{12} = 3 \times \frac{5}{12} = \frac{5}{4}
  3. Third Interval [3/4,1][3/4, 1]: Area3=1×(1−34)=14\text{Area}_3 = 1 \times \left(1 - \frac{3}{4}\right) = \frac{1}{4}

Summing these individual areas gives the total area under the curve:

Total Area=23+54+14\text{Total Area} = \frac{2}{3} + \frac{5}{4} + \frac{1}{4} Total Area=23+64=23+32\text{Total Area} = \frac{2}{3} + \frac{6}{4} = \frac{2}{3} + \frac{3}{2} Total Area=4+96=136\text{Total Area} = \frac{4 + 9}{6} = \frac{13}{6}

Thus, the correct option is C.

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