JEE Advance - Mathematics (2019 - Paper 1 Offline - No. 4)
The area of the region
{(x, y) : xy $$ \le $$ 8, 1 $$ \le $$ y $$ \le $$ x2} is
{(x, y) : xy $$ \le $$ 8, 1 $$ \le $$ y $$ \le $$ x2} is
$$8{\log _e}2 - {{14} \over 3}$$
$$8{\log _e}2 - {{7} \over 3}$$
$$16{\log _e}2 - {{14} \over 3}$$
$$16{\log _e}2 - 6$$
Explanation
The given region
{(x, y) : xy $$ \le $$ 8, 1 $$ \le $$ y $$ \le $$ x2}.
From the figure, region A and B satisfy the given region, but only A is bounded region, so area of bounded region
$$A = \int_1^2 {({x^2} - 1)} dx + \int_2^8 {\left( {{8 \over x} - 1} \right)} dx$$
[$$ \therefore $$ Points P(1, 1), Q(2, 4) and R(8, 1)]
$$ = \left[ {{{{x^3}} \over 3} - x} \right]_1^2 + [8\log |x| - x]_2^8$$
$$ = \left( {{8 \over 3} - 2 - {1 \over 3} + 1} \right) + 8\log 8 - 8 - 8\log 2 + 2$$
$$ = - {{14} \over 3} + 16\log 2$$
$$ = 16\log 2 - {{14} \over 3}$$
{(x, y) : xy $$ \le $$ 8, 1 $$ \le $$ y $$ \le $$ x2}.
From the figure, region A and B satisfy the given region, but only A is bounded region, so area of bounded region
$$A = \int_1^2 {({x^2} - 1)} dx + \int_2^8 {\left( {{8 \over x} - 1} \right)} dx$$
[$$ \therefore $$ Points P(1, 1), Q(2, 4) and R(8, 1)]
$$ = \left[ {{{{x^3}} \over 3} - x} \right]_1^2 + [8\log |x| - x]_2^8$$
$$ = \left( {{8 \over 3} - 2 - {1 \over 3} + 1} \right) + 8\log 8 - 8 - 8\log 2 + 2$$
$$ = - {{14} \over 3} + 16\log 2$$
$$ = 16\log 2 - {{14} \over 3}$$
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