NIMCET 2020 — Mathematics PYQ

Concept:
- Two curves $f(x,y)=0$ and $g(x,y)=0$ cut/touch at a point (a, b) if f(a, b) = g(a, b) = 0.
- The area under the function $y=f(x)$ from $x=a$ to $x=b$ and the x-axis is given by the definite integral $\left|{\int }_a^{b}f(x)dx\right|$, for curves which are entirely on the same side of the x-axis in the given range.
- If the curves are on both the sides of the x-axis, then we calculate the areas of both the sides separately and add them.
- Definite integral: If $\int f(x)dx=g(x)+C$, then ${\int }_a^{b}f(x)dx=[g(x){]}_a^b=g(b)-g(a)$.
- $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$.

Calculations:
Let's first find the points of intersection/touching of the two curves $y^2=x$ and $y=|x|$.
Using the condition for intersecting, we must have:
$$y^2-x=y-|x|=0$$
CASE 1: For $x\ge 0$, $|x|=x$.
$$<br>\Rightarrow y^2-x=y-x=0$$
$$\Rightarrow y^2-y=0$$
$$<br>\Rightarrow y(y-1)=0$$
$$\Rightarrow y=0\text{ OR }y-1=0$$
$$<br>\Rightarrow y=0\text{ OR }y=1.$$
And, $x=0^2=0\text{ OR }x=1^2=1.$
The points of intersection are $(0,0)$ and $(1,1)$.
CASE 2: For x &lt; 0, $|x|=-x$.
$$\Rightarrow y^2-x=y-(-x)=0$$
$$<br>\Rightarrow y^2-x=-y-x=0$$
$$\Rightarrow y^2+y=0$$
$$<br>\Rightarrow y(y+1)=0$$

⇒ $y=0ORy+1=0$
⇒ $y=0ORy=-1$.
∵ $y=|x|$, y cannot be negative.
∴ The only solution in this case $y=0$ and $x=0^2=0$.
Finally, the solutions (points of intersection/touching) of $y^2=x$ and $y=|x|$ are (0, 0) and (1, 1).
The graph of the curves is shown below:

\begin{aligned}
 & \mathrm{The~required~(shaded)~area~is:} \\
 & \mathrm{(Area~under~y^{2}=x~from~x=0~to~x=1.)-(Area~under~y=|x|~from~x=0~to~x=1.)} \\
 & =\left|\int_{0}^{1}\sqrt{\mathbf{x}}\mathrm{dx}\right|-\left|\int_{0}^{1}\mathbf{x}\mathrm{dx}\right| \\
 & =\left|
\begin{array}
{c}2 \\
3
\end{array}\left[\mathbf{x}^3\right]_0^1\right|-\left|
\begin{array}
{c}1 \\
2
\end{array}\left[\mathbf{x}^2\right]_0^1\right| \\
 & =
\begin{vmatrix}
2 & 3 & 3 \\
3 & [1 & 2 & -0 & 2
\end{vmatrix}-
\begin{vmatrix}
1 & 1 \\
2 & [1^2-0^2]
\end{vmatrix} \\
 & =
\begin{array}
{cc}{2} & {1} \\
{3} & {-} & {2}
\end{array} \\
 & =
\begin{array}
{c}{1} \\
{6}
\end{array}\mathrm{sq.~units.}
\end{aligned}