JEE MAIN - Mathematics (2020 - 7th January Evening Slot - No. 16)
The area (in sq. units) of the region
{(x, y) $$ \in $$ R2 | 4x2 $$ \le $$ y $$ \le $$ 8x + 12} is :
{(x, y) $$ \in $$ R2 | 4x2 $$ \le $$ y $$ \le $$ 8x + 12} is :
$${{125} \over 3}$$
$${{128} \over 3}$$
$${{127} \over 3}$$
$${{124} \over 3}$$
Explanation
For point of intersection
4x2
= 8x + 12
$$ \Rightarrow $$ x2 - 2x - 3 = 0
$$ \Rightarrow $$ x = –1, 3_7th_January_Evening_Slot_en_16_1.png)
Required area = area of the shaded region
= $$\int\limits_{ - 1}^3 {\left( {8x + 12 - 4{x^2}} \right)dx} $$
= $$4\left[ {2.{{{x^2}} \over 2} + 3x - {{{x^3}} \over 3}} \right]_{ - 1}^3$$
= (36 + 36 – 36) – (4 – 12 + $${4 \over 3}$$)
= $${{128} \over 3}$$
$$ \Rightarrow $$ x2 - 2x - 3 = 0
$$ \Rightarrow $$ x = –1, 3
Required area = area of the shaded region
= $$\int\limits_{ - 1}^3 {\left( {8x + 12 - 4{x^2}} \right)dx} $$
= $$4\left[ {2.{{{x^2}} \over 2} + 3x - {{{x^3}} \over 3}} \right]_{ - 1}^3$$
= (36 + 36 – 36) – (4 – 12 + $${4 \over 3}$$)
= $${{128} \over 3}$$
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