JEE Advance - Mathematics (2018 - Paper 1 Offline - No. 14)
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 $$\Delta $$PQR, then the value of n is .................
Answer
4
Explanation
We have,
y = xn, n > 1
$$ \because $$ P(0, 0) Q(1, 1) and R(2, 0) are vertices of $$\Delta $$PQR.
$$ \therefore $$ Area of shaded region = 30% of area of $$\Delta $$PQR
$$ \Rightarrow \int_0^1 {(x - {x^n})dx = {{30} \over {100}} \times {1 \over 2} \times 2 \times 1} $$
$$ \Rightarrow \left[ {{{{x^2}} \over 2} - {{{x^{n + 1}}} \over {n + 1}}} \right]_0^1 = {3 \over {10}}$$
$$ \Rightarrow \left( {{1 \over 2} - {1 \over {n + 1}}} \right) = {3 \over {10}}$$
$$ \Rightarrow {1 \over {n + 1}} = {1 \over 2} - {3 \over {10}} = {2 \over {10}} = {1 \over 5}$$
$$ \Rightarrow n + 1 = 5 \Rightarrow n = 4$$
y = xn, n > 1
$$ \because $$ P(0, 0) Q(1, 1) and R(2, 0) are vertices of $$\Delta $$PQR.
$$ \therefore $$ Area of shaded region = 30% of area of $$\Delta $$PQR
$$ \Rightarrow \int_0^1 {(x - {x^n})dx = {{30} \over {100}} \times {1 \over 2} \times 2 \times 1} $$
$$ \Rightarrow \left[ {{{{x^2}} \over 2} - {{{x^{n + 1}}} \over {n + 1}}} \right]_0^1 = {3 \over {10}}$$
$$ \Rightarrow \left( {{1 \over 2} - {1 \over {n + 1}}} \right) = {3 \over {10}}$$
$$ \Rightarrow {1 \over {n + 1}} = {1 \over 2} - {3 \over {10}} = {2 \over {10}} = {1 \over 5}$$
$$ \Rightarrow n + 1 = 5 \Rightarrow n = 4$$
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