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 side PQ and a curve of the form y=xn,n1 . If the area of the region taken away by the farmer F2 is exactly 30 % of the area of Δ PQR, then the value of n is

Question

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 side 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 PQR, then the value of n is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-0zx8xfbwlifantmf.jpg" /> Area =∫ limits 01 (x - xn) d x = (3/10) [(x2/2) - (xn + 1/n + 1)]01=(3/10) (1/2)-(1/n + 1)=(3/10) ∴ n+1=5 ⇒ n=4

Area =0∫1​(x−xn)dx=103​ [2x2​−n+1xn+1​]01​=103​ 21​−n+11​=103​ ∴n+1=5 ⇒n=4

Area $= 0 \int 1 (x - x^{n}) d x = \frac{3}{10}$
$[\frac{x ^{2}}{2} - \frac{x ^{n + 1}}{n + 1}]_{0}^{1} = \frac{3}{10}$
$\frac{1}{2} - \frac{1}{n + 1} = \frac{3}{10}$ $∴ n + 1 = 5$
$\Rightarrow n = 4$