Mathematics · Area under the Curves

JEE Advanced 2018 — Paper 1 — Question 40

A farmer $F_{1}$ has a land in the shape of a triangle with vertices at $P(0,0), Q(1,1)$ and $R(2,0)$ . From this land, a neighbouring farmer $\mathrm{F}_{2}$ takes away the region which lies between the side PQ and a curve of the form $y=x^{n}(n>1)$ . If the area of the region taken away by the farmer $F_{2}$ is exactly $30 \%$ of the area of $\triangle P Q R$ , then the value of $n$ is $\_\_\_\_\_$ .

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

Area of $\triangle PQR = \frac{1}{2} |0(1-0) + 1(0-0) + 2(0-1)| = 1$ sq. unit. 30% of this area is $0.3 = \frac{3}{10}$ . The side PQ has equation $y = x$ (from (0,0) to (1,1)). The region taken is between $y = x$ and $y = x^n$ from $x=0$ to $x=1$ . Area = $\int_{0}^{1} (x - x^n) \, dx = \left[ \frac{x^2}{2} - \frac{x^{n+1}}{n+1} \right]_{0}^{1} = \frac{1}{2} - \frac{1}{n+1}$ . Set equal to $\frac{3}{10}$ : $\frac{1}{2} - \frac{1}{n+1} = \frac{3}{10}$ . Solve: $\frac{1}{n+1} = \frac{1}{2} - \frac{3}{10} = \frac{2}{10} = \frac{1}{5}$ . Thus $n+1 = 5$ , so $n = 4$ .

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 1
Subject: Mathematics
Chapter: Area under the Curves
Topic: Area under the Curves