Area Ques 16
- 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 $F_2$ takes away the region which lies between the sides $P Q$ 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
(2018 Adv.)
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Answer:
Correct Answer: 16.$(4)$
Solution: We have, $y=x^\pi, n>1$
$\because \quad P(0,0) Q(1,1)$ and $R(2,0)$ are vertices of $\triangle P Q R$.
$\therefore \quad $ Area of shaded region $=30 \%$ of area of $\triangle P Q R$
$\Rightarrow \quad \int_0^1\left(x-x^n\right) d x=\frac{30}{100} \times \frac{1}{2} \times 2 \times 1$
$\Rightarrow \quad \left[\frac{x^2}{2}-\frac{x^{n+1}}{n+1}\right]_0^1=\frac{3}{10} \Rightarrow\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{3}{10}$
$\Rightarrow \quad \frac{1}{n+1}=\frac{1}{2}-\frac{3}{10}=\frac{2}{10}=\frac{1}{5} \Rightarrow n+1=5 \Rightarrow n=4$
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