JEE Main 2017 · 2 Apr · Shift 1 · Mathematics Q76

MathematicsSingle correct+4 / -1Question 76 of 90

The area (in sq. units) of the region $\{(x,y):x\ge0,\ x+y\le3,\ x^2\le4y\text{ and }y\le1+\sqrt x\}$ is :

\dfrac32

\dfrac73

\dfrac52

\dfrac{59}{12}

Answer and worked solution

Official answer: C

The boundaries meet as follows:

$\int_0^1\left(1+\sqrt x-\frac{x^2}4\right)dx+\int_1^2\left(3-x-\frac{x^2}4\right)dx=\frac{19}{12}+\frac{11}{12}=\frac52.$

JEE Main 2017 · 2 Apr · Shift 1, Mathematics question 76