JEE Advanced 2018 - 1

Calculus Level 3

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


The answer is 4.

Shivam Jadhav by Shivam Jadhav , 22, India

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1 solution

Akshaj Garg
Sep 9, 2019

A careful observation reveals that the points ( 0 , 0 ) (0,0) and ( 1 , 1 ) (1,1) lie on the line and the curve. Therefore area of region taken by Farmer 2 is the area between the side of triangle and the curve.

⇒ \Rightarrow \ 0.3 = 0.3 \ = ∫ 0 1 x d x \int_0^1 x\,\mathrm{d}x − - ∫ 0 1 x n d x \int_0^1 x^{n}\,\mathrm{d}x

⇒ \Rightarrow \ 0.3 = 0.3 \ = x 2 2 ∣ 0 1 \left.\dfrac{x^2}{2}\right|_0^1 − - \ x n + 1 n + 1 ∣ 0 1 \left.\dfrac{x^{n+1}}{n+1}\right|_0^1

⇒ \Rightarrow \ 0.3 = 0.3 \ = 1 2 − \dfrac{1}{2} \ - 1 n + 1 \dfrac{1}{n+1}

⇒ \Rightarrow \ 1 n + 1 = 1 2 − 3 10 \dfrac{1}{n+1} \ = \dfrac{1}{2} \ - \dfrac{3}{10}

⇒ \Rightarrow \ 1 n + 1 = 2 10 \dfrac{1}{n+1} \ = \dfrac{2}{10}

T h e r e f o r e \mathrm{Therefore}

n = 4 \boxed{n \ = 4}

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