JEE Main 2020 (Online) 4th September Morning ShiftQuestion 1
Let $f(x)=|x-2|$ and $g(x)=f(f(x))$, $x\in[0,4]$. Then $\int_0^3(g(x)-f(x))\,dx$ is equal to :
Definite Integration
Solution
Concept. Write $f$ and $g=f(f(x))$ without modulus on $[0,2]$ and on $[2,3]$. Formulas used. (1) $|t|=t$ for $t\ge0$ and $|t|=-t$ for $t<0$Step 1 - On $[0,2]$. $f(x)=2-x\in[0,2]$, so $g(x)=|(2-x)-2|=|x|=x$. Then $$\int_0^2(g-f)\,dx=\int_0^2(2x-2)\,dx=\big[x^2-2x\big]_0^2=0.$$ Step 2 - On $[2,3]$. $f(x)=x-2\in[0,1]$, so $g(x)=|x-4|=4-x$. Then $$\int_2^3(g-f)\,dx=\int_2^3(6-2x)\,dx=\big[6x-x^2\big]_2^3=9-8=1.$$ Step 3 - Add. $0+1=1$. Answer: (C)