NIMCET 2020 — Mathematics PYQ

- 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 $y=f(x)fromx=atox=b$ 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 ${\int }_a^{b}f(x)dx=[g(x){]}_a^b=g(b)-g(a)$.
- $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$.

$\underline{\text{Calculation:}}$

Let's say that the two curves are f( x, y) = x+ y- 3= 0 and g( x, y) = $x^2$- y- 9= 0. 

The points of their intersection are the points where f(x, y)=g(x, y).

$\Rightarrow \mathbf{x}+\mathbf{y}-3={\mathbf{x}}^2-\mathbf{y}-9=0$

$\Rightarrow -x-y+3=x^2-y-9=0$

$\Rightarrow {\mathbf{x}}^2+\mathbf{x}-12=0$

$\Rightarrow {\mathbf{x}}^2+4\mathbf{x}-3\mathbf{x}-12=0$

$\Rightarrow \mathbf{x}(\mathbf{x}+4)-3(\mathbf{x}+4)=0$

$\Rightarrow (\mathbf{x}+4)(\mathbf{x}-3)=0$

$\Rightarrow \mathbf{x}+4=0$ OR x-3=0

$\Rightarrow \mathbf{x}=-4$ OR x=3.

And,  y= 3- ( - 4) = 7 OR y= 3- 3= 0. 

Hence, the curves intersect at the points $\mathcal{B}(3,0)$ and C(4,7) as shown in the diagram below:

The points where y = x^2 - 9 cuts the x-axis (y = 0) are A(-3, 0) and B(3, 0).
The required area is the shaded part ABC = Area of BDC - Area of ADC.
$${\int }_{-4}^3(3-x)dx-{\int }_{-4}^3(x^2-9)dx$$
$$<br>=3[x{]}_{-4}^3-{\left[\frac{x^2}{2}\right]}_{-4}^3-{\left[\frac{x^3}{3}\right]}_{-4}^3+9[x{]}_{-4}^3$$
$$=3[3-(-4)]-\frac{1}{2}[3^2-(-4{)}^2]-\frac{1}{3}[(-3{)}^3-(-4{)}^3]+9[-3-(-4)]$$
$$<br>=3(7)-\frac{1}{2}(7)-\frac{1}{3}(37)+9(1)$$
$$=21+\frac{7}{2}-\frac{37}{3}+9$$
$$<br>=\frac{127}{6}$$.
The answer is None of these.