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QJEEAdv18Mac753: JEE Advanced 2018 Mathematics - Class 12 JEEAdv18Mac753 Solution: A farmer F 1 has a land in the shape of a triangle with…

A farmer F1 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 F2 takes away the region which lies between the sides PQ and a curve of the form y = xn (n > 1). If the area of the region taken away by the farmer F2 is exactly 30% of the area of Δ\Delta PQR, then the value of n is .................

Step-by-Step Verified Solution

We have,

y = xn, n > 1

∵\because P(0, 0) Q(1, 1) and R(2, 0) are vertices of Δ\Delta PQR.

JEE Advanced 2018 Paper 1 Offline Mathematics - Application of Integration Question 12 English Explanation

∴\therefore Area of shaded region = 30% of area of Δ\Delta PQR

⇒∫01(x−xn)dx=30100×12×2×1\Rightarrow \int_0^1 {(x - {x^n})dx = {{30} \over {100}} \times {1 \over 2} \times 2 \times 1}

⇒[x22−xn+1n+1]01=310\Rightarrow \left[ {{{{x^2}} \over 2} - {{{x^{n + 1}}} \over {n + 1}}} \right]_0^1 = {3 \over {10}}

⇒(12−1n+1)=310\Rightarrow \left( {{1 \over 2} - {1 \over {n + 1}}} \right) = {3 \over {10}}

⇒1n+1=12−310=210=15\Rightarrow {1 \over {n + 1}} = {1 \over 2} - {3 \over {10}} = {2 \over {10}} = {1 \over 5}

⇒n+1=5⇒n=4\Rightarrow n + 1 = 5 \Rightarrow n = 4

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