Mathematics - Functions Question with Solution | TestHub

For $x\in [0,1]$, we have $x<2$, so $f(x)=x^2-4x+6=(x-2{)}^2+2$. For $x\in [0,1]$, $x+1\in [1,2]$. We need to evaluate $f(x+1)$ based on its definition. Case 1: $x\in [0,1)$. Here, $x+1\in [1,2)$. Since $x+1<2$, $f(x+1)=((x+1)-2{)}^2+2=(x-1{)}^2+2$. So, $g(x)=f(x)+f(x+1)=(x^2-4x+6)+((x-1{)}^2+2)=(x^2-4x+6)+(x^2-2x+1+2)=2x^2-6x+9$.

This is a parabola opening upwards with vertex at $x=-(-6)/(2\cdot 2)=3/2=1.5$. Since $x\in [0,1)$, $g(x)$ is decreasing on this interval.$g(0)=2(0{)}^2-6(0)+9=9$. As $x\to 1^{-}$, $g(x)\to 2(1{)}^2-6(1)+9=5$. So for $x\in [0,1)$, the range of $g(x)$ is $(5,9]$. Case 2: $x=1$. Here, $x+1=2$. So $g(1)=f(1)+f(2)$.$f(1)=(1-2{)}^2+2=1+2=3$.$f(2)=|2-4|+2=|-2|+2=2+2=4$.$g(1)=3+4=7$. Combining both cases, the range of $g(x)$ for $x\in [0,1]$ is $(5,9]$. The integers in this range are $6,7,8,9$. There are 4 such integers.