JEE Main 2022 · 25 Jul · Shift 1 · Mathematics Q8

MathematicsSingle correct+4 / -1Question 8 of 90

The area of the region given by $A=\left\{(x,y):x^2\le y\le\min\{x+2,4-3x\}\right\}$ is :

\dfrac{31}8

\dfrac{17}6

\dfrac{19}6

\dfrac{27}8

Answer and worked solution

Official answer: B

Region. $y=x^2$ meets $y=x+2$ at $x=-1$ and $y=4-3x$ at $x=1$ ; the two lines cross at $x=\frac12$ .

Area. $\displaystyle\int_{-1}^{1/2}(x+2-x^2)dx+\int_{1/2}^1(4-3x-x^2)dx=\frac{17}6$ .

(B)

Topics:Area between curves

JEE Main 2022 · 25 Jul · Shift 1, Mathematics question 8