JEE Main 2023 — Mathematics Calculus

We have,

f(x)=\frac{x+|x|}{2}={\begin{matrix}\begin{matrix}x; & x\geq 0\end{matrix} \\ 0;x<0\end{matrix}

g(x)={\begin{matrix}\begin{matrix}{x}^{2}; & x\geq 0\end{matrix} \\ x;x<0\end{matrix}

So,

$$fog(x)=f{g(x)}=[\begin{matrix}g(x); & g(x)\geq 0 \\ 0; & g(x)<0\end{matrix}$$fog(x)=fg(x)=[g(x);0;​g(x)≥0g(x)<0​

\Rightarrow fog(x)={\begin{matrix}\begin{matrix}{x}^{2}; & x\geq 0\end{matrix} \\ 0;x<0\end{matrix}

And,

$$2y-x=15$$2y−x=15

Solving $2y-x=15$2y−x=15 and $y={x}^{2}$y=x2, we get

$$2{x}^{2}-x-15=0$$2x2−x−15=0

$$\Rightarrow x=\frac{1\pm 11}{4}$$⇒x=41±11​

$$\Rightarrow x=3,-\frac{5}{2}$$⇒x=3,−25​

Required area

$$A={\int }_{0}^{3}(\frac{x+15}{2}-{x}^{2})\mathrm{dx}+\frac{1}{2}\times \frac{15}{2}\times 15$$A=∫03​(2x+15​−x2)dx+21​×215​×15

$$\Rightarrow A={[\frac{{x}^{2}}{4}+(\frac{15}{2})x-\frac{{x}^{3}}{3}]}_{0}^{3}+\frac{225}{4}$$⇒A=[4x2​+(215​)x−3x3​]03​+4225​

$$\Rightarrow A=\frac{9}{4}+\frac{45}{2}-9+\frac{225}{4}$$⇒A=49​+245​−9+4225​

$$\Rightarrow A=\frac{99-36+225}{4}$$⇒A=499−36+225​

$$\Rightarrow A=72$$⇒A=72