Let for x \in \mathbbR , f(x)=\fracx+|x|2 and g(x)= beginarraycc x, & x\lt 0 x^2, & x \geq 0 \endarray. . Then area bounded by the curve y=(f \circ g)(x) and the lines y=0, 2y-x=15 is equal to \underline\hspace2cm .

Area Under the Curves PYQ — JEE Main 2023 · 31 Jan · Shift 1 · Q46 with Solution

View written solutionFree

Correct answer: 72

Simplify the functions

Given $f(x)=\frac{x+|x|}{2}$ This is the standard function $f(x)=\begin{cases}0,&x<0\\ x,&x\ge 0\end{cases}$ because:

if $x<0$ , then $|x|=-x$ , so $f(x)=\frac{x-x}{2}=0$
if $x\ge 0$ , then $|x|=x$ , so $f(x)=\frac{x+x}{2}=x$

Also, $g(x)=\begin{cases}x,&x<0\\ x^2,&x\ge 0\end{cases}$

So $(f\circ g)(x)=f(g(x))$

Find $f(g(x))$ piecewise

For $x<0$ : $g(x)=x<0$ , hence $f(g(x))=f(x)=0$
For $x\ge 0$ : $g(x)=x^2\ge 0$ , hence $f(g(x))=f(x^2)=x^2$

Therefore, $(f\circ g)(x)=\begin{cases}0,&x<0\\ x^2,&x\ge 0\end{cases}$

So the curve is:

the $x$ -axis for $x<0$
the parabola $y=x^2$ for $x\ge 0$

Write the line in standard form

Given line: $2y-x=15$ $\Rightarrow 2y=x+15$ $\Rightarrow y=\frac{x+15}{2}$

This line meets the $x$ -axis at $0=\frac{x+15}{2}\Rightarrow x=-15$ So one intersection with $y=0$ is $(-15,0)$ .

Find intersection of the line with the curve $y=x^2$

For $x\ge 0$ , set $x^2=\frac{x+15}{2}$ $2x^2-x-15=0$ $2x^2-6x+5x-15=0$ $2x(x-3)+5(x-3)=0$ $(2x+5)(x-3)=0$ So $x=3\quad \text{or} \quad x=-\frac52$ Since we need $x\ge 0$ , only $x=3$ Thus intersection point is $(3,9)$

Understand the bounded region

The three boundaries are:

$y=0$
$y=(f\circ g)(x)$
$y=\frac{x+15}{2}$

The closed region is formed by:

along $y=0$ from $x=-15$ to $x=0$
along $y=x^2$ from $(0,0)$ to $(3,9)$
along the line from $(3,9)$ back to $(-15,0)$

Hence area must be computed in two parts:

From $x=-15$ to $x=0$ : between line and $x$ -axis
From $x=0$ to $x=3$ : between line and parabola

So, $A=\int_{-15}^{0}\left(\frac{x+15}{2}-0\right)dx+\int_0^3\left(\frac{x+15}{2}-x^2\right)dx$

Compute the first integral

=\frac12\int_{-15}^{0}(x+15)\,dx$$ $$=\frac12\left[\frac{x^2}{2}+15x\right]_{-15}^{0}$$ At $x=0$: $$\frac{0}{2}+0=0$$ At $x=-15$: $$\frac{225}{2}-225=-\frac{225}{2}$$ Thus, $$A_1=\frac12\left(0-\left(-\frac{225}{2}\right)\right)=\frac{225}{4}$$ --- 7. **Compute the second integral** $$A_2=\int_0^3\left(\frac{x+15}{2}-x^2\right)dx$$ $$=\left[\frac{x^2}{4}+\frac{15x}{2}-\frac{x^3}{3}\right]_0^3$$ At $x=3$: $$\frac{9}{4}+\frac{45}{2}-9 =\frac{9}{4}+\frac{90}{4}-\frac{36}{4} =\frac{63}{4}$$ At $x=0$: $$0$$ So, $$A_2=\frac{63}{4}$$ --- 8. **Total area** $$A=A_1+A_2=\frac{225}{4}+\frac{63}{4}=\frac{288}{4}=72$$ Therefore, the required area is $$\boxed{72}$$

More from Area Under the Curves

View written solutionFree

Correct answer: 72

Simplify the functions

Given $f(x)=\frac{x+|x|}{2}$ This is the standard function $f(x)=\begin{cases}0,&x<0\\ x,&x\ge 0\end{cases}$ because:

if $x<0$ , then $|x|=-x$ , so $f(x)=\frac{x-x}{2}=0$
if $x\ge 0$ , then $|x|=x$ , so $f(x)=\frac{x+x}{2}=x$

Also, $g(x)=\begin{cases}x,&x<0\\ x^2,&x\ge 0\end{cases}$

So $(f\circ g)(x)=f(g(x))$

Find $f(g(x))$ piecewise

For $x<0$ : $g(x)=x<0$ , hence $f(g(x))=f(x)=0$
For $x\ge 0$ : $g(x)=x^2\ge 0$ , hence $f(g(x))=f(x^2)=x^2$

Therefore, $(f\circ g)(x)=\begin{cases}0,&x<0\\ x^2,&x\ge 0\end{cases}$

So the curve is:

the $x$ -axis for $x<0$
the parabola $y=x^2$ for $x\ge 0$

Write the line in standard form

Given line: $2y-x=15$ $\Rightarrow 2y=x+15$ $\Rightarrow y=\frac{x+15}{2}$

This line meets the $x$ -axis at $0=\frac{x+15}{2}\Rightarrow x=-15$ So one intersection with $y=0$ is $(-15,0)$ .

Find intersection of the line with the curve $y=x^2$

Understand the bounded region

The three boundaries are:

$y=0$
$y=(f\circ g)(x)$
$y=\frac{x+15}{2}$

The closed region is formed by:

along $y=0$ from $x=-15$ to $x=0$
along $y=x^2$ from $(0,0)$ to $(3,9)$
along the line from $(3,9)$ back to $(-15,0)$

Hence area must be computed in two parts:

From $x=-15$ to $x=0$ : between line and $x$ -axis
From $x=0$ to $x=3$ : between line and parabola

So, $A=\int_{-15}^{0}\left(\frac{x+15}{2}-0\right)dx+\int_0^3\left(\frac{x+15}{2}-x^2\right)dx$

Compute the first integral

=\frac12\int_{-15}^{0}(x+15)\,dx$$ $$=\frac12\left[\frac{x^2}{2}+15x\right]_{-15}^{0}$$ At $x=0$: $$\frac{0}{2}+0=0$$ At $x=-15$: $$\frac{225}{2}-225=-\frac{225}{2}$$ Thus, $$A_1=\frac12\left(0-\left(-\frac{225}{2}\right)\right)=\frac{225}{4}$$ --- 7. **Compute the second integral** $$A_2=\int_0^3\left(\frac{x+15}{2}-x^2\right)dx$$ $$=\left[\frac{x^2}{4}+\frac{15x}{2}-\frac{x^3}{3}\right]_0^3$$ At $x=3$: $$\frac{9}{4}+\frac{45}{2}-9 =\frac{9}{4}+\frac{90}{4}-\frac{36}{4} =\frac{63}{4}$$ At $x=0$: $$0$$ So, $$A_2=\frac{63}{4}$$ --- 8. **Total area** $$A=A_1+A_2=\frac{225}{4}+\frac{63}{4}=\frac{288}{4}=72$$ Therefore, the required area is $$\boxed{72}$$

Area Under the Curves question

Area Under the Curves question

More from Area Under the Curves

Area Under the Curves question

Area Under the Curves question

More from Area Under the Curves