Question

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

The correct options are
A $f\left(x\right)$ is a many one and into function
B $f\left(x\right)=0$ for infinite number of values of $x$

Given that
$f\left(x\right)={\left[x\right]}^2+[x+1]-3\Rightarrow f\left(x\right)={\left[x\right]}^2+\left[x\right]+1-3\Rightarrow f\left(x\right)={\left[x\right]}^2+\left[x\right]-2\Rightarrow f\left(x\right)=(\left[x\right]+2)(\left[x\right]-1)$
Now $f\left(x\right)=0\Rightarrow (\left[x\right]+2)(\left[x\right]-1)=0$
$\Rightarrow$either $\left[x\right]=-2$ or $\left[x\right]=1$
$\Rightarrow$ either $-2\le x<-1$ or $1\le x<2$
i.e. $f\left(x\right)=0$ for infinite number of value of $x$.
Also, $f\left(x\right)$ will not be one-to-one
Since, $f\left(x\right)=(\left[x\right]+2)(\left[x\right]-1)=($ integer $\left)\right($ integer $)=$ integer
Hence, $f\left(x\right)$ is into on the set real numbers $R$ as codomain.