Question

Let $f\left(x\right)=[x{]}^2+[x+1]-3,$ where $\left[x\right]=$ the greatest integer less than or equal $x$.Then

• A. $f\left(x\right)$ is a many-one and into function
• B. $f\left(x\right)=0$ for infinite number of values of $x$
• C. $f\left(x\right)=0$ for only two real values
• D. none of these

Answer 1

Answer: (A), (B)

$f\left(x\right)=0$ for infinite number of values of $x$
$f\left(x\right)$ is a many-one and into function

$For0\le x<1,f\left(x\right)=0+1-3=-2For1\le x<2,f\left(x\right)=1+2-3=0Forx>2,f\left(x\right)>2^2+3-3=4For-1\le x<0,f\left(x\right)=1+0-3=-4$

From this data we can say that $f\left(x\right)$ is zero for infinte numbers between

$1$ & $2$, also $f\left(x\right)$ can only take some integer values and therefore it is not onto but into