JEE MAIN - Mathematics (2020 - 9th January Evening Slot - No. 12)
Given : $$f(x) = \left\{ {\matrix{
{x\,\,\,\,\,,} & {0 \le x < {1 \over 2}} \cr
{{1 \over 2}\,\,\,\,,} & {x = {1 \over 2}} \cr
{1 - x\,\,\,,} & {{1 \over 2} < x \le 1} \cr
} } \right.$$
and $$g(x) = \left( {x - {1 \over 2}} \right)^2,x \in R$$
Then the area (in sq. units) of the region bounded by the curves, y = ƒ(x) and y = g(x) between the lines, 2x = 1 and 2x = $$\sqrt 3 $$, is :
and $$g(x) = \left( {x - {1 \over 2}} \right)^2,x \in R$$
Then the area (in sq. units) of the region bounded by the curves, y = ƒ(x) and y = g(x) between the lines, 2x = 1 and 2x = $$\sqrt 3 $$, is :
$${1 \over 2} + {{\sqrt 3 } \over 4}$$
$${1 \over 2} - {{\sqrt 3 } \over 4}$$
$${1 \over 3} + {{\sqrt 3 } \over 4}$$
$${{\sqrt 3 } \over 4} - {1 \over 3}$$
Explanation
_9th_January_Evening_Slot_en_12_1.png)
Required area = Area of trepezium ABCD – Area of parabola between x = $${1 \over 2}$$ and x = $${{\sqrt 3 } \over 2}$$
= Area of trepezium ABCD - $$\int\limits_{{1 \over 2}}^{{{\sqrt 3 } \over 2}} {{{\left( {x - {1 \over 2}} \right)}^2}dx} $$
= $${1 \over 2}\left( {{{\sqrt 3 } \over 2} - {1 \over 2}} \right)\left( {{1 \over 2} + 1 - {{\sqrt 3 } \over 2}} \right)$$ - $${1 \over 3}\left[ {{{\left( {x - {1 \over 2}} \right)}^3}} \right]_{{1 \over 2}}^{{{\sqrt 3 } \over 2}}$$
= $${1 \over 2}\left( {{{\sqrt 3 - 1} \over 2}} \right)\left( {{{3 - \sqrt 3 } \over 2}} \right)$$ $$ - {1 \over 3}\left[ {{{\left( {{{\sqrt 3 - 1} \over 2}} \right)}^3} - 0} \right]$$
= $${{\sqrt 3 } \over 4} - {1 \over 3}$$
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