MHTCET 10 May 2024 Shift 2 Paper

Concept:The required area is bounded below by $y=\frac{x^2}{4}$ and above by the two curves $y=1+\sqrt{x}$ and $y=3-x$, for $x\ge0$.Explanation:The boundary curves are $x+y=3$, $x^2=4y$, and $y=1+\sqrt{x}$.So the relevant curves are $y=3-x$, $y=\frac{x^2}{4}$, and $y=1+\sqrt{x}$.Find the intersection of $y=3-x$ and $y=1+\sqrt{x}$:$3-x=1+\sqrt{x}$, which gives $x+\sqrt{x}-2=0$.Let $\sqrt{x}=t$. Then $t^2+t-2=0$, so $(t+2)(t-1)=0$.Since $t=\sqrt{x}\ge0$, we get $t=1$, hence $x=1$ and $y=2$.Find the intersection of $y=3-x$ and $y=\frac{x^2}{4}$:$3-x=\frac{x^2}{4}$, which gives $x^2+4x-12=0$.So $(x+6)(x-2)=0$. Since $x\ge0$, we get $x=2$ and $y=1$.Thus, for $0\le x\le1$ the upper curve is $y=1+\sqrt{x}$, and for $1\le x\le2$ the upper curve is $y=3-x$.The lower curve is $y=\frac{x^2}{4}$ throughout.Required area:$$A=\int_0^1\left(1+\sqrt{x}-\frac{x^2}{4}\right)dx+\int_1^2\left(3-x-\frac{x^2}{4}\right)dx.$$Combining the $\frac{x^2}{4}$ terms:$$A=\int_0^1(1+\sqrt{x})dx+\int_1^2(3-x)dx-\frac14\int_0^2x^2dx.$$Evaluating the integrals:$$A=\left[x+\frac23x^{3/2}\right]_0^1+\left[3x-\frac{x^2}{2}\right]_1^2-\frac14\left[\frac{x^3}{3}\right]_0^2.$$$$A=\left(1+\frac23\right)+\left(3-\frac32\right)-\frac14\cdot\frac83=\frac53+\frac32-\frac23=\frac52.$$Answer:The area of the region is $\frac{5}{2}$ sq. units, i.e. option D.