Question

$f\left(x\right)=[x^2+1],$ ( [ x] is the greatest integer less than or equal to x) . Then $f$ is continuous.

• A. on [1, 3]
• B. for all $x$ in $[1,3]$ except four points
• C. for all $x$ in $[1,3]$ except seven points
• D. for all $x$ in $[1,3]$ except 9 points

Answer 1

The correct option is D for all $x$ in $[1,3]$ except 9 points

$f\left(x\right)=[x^2+1]$
In the interval $xϵ[1,3]$,
$f\left(1\right)=[1^2+1]=2f\left(3\right)=[3^2+1]=10$
Clearly, all possible values of $f\left(x\right)$ in the interval $[1,3]$ must be $2,3,4,5,6,7,8,9,10$.
The greatest integer function will become discontinuous at the eight points where $[x^2+1]=2,3,4,5,6,7,8,9,10$.
Explanation: If $[x^2+1]=n$, this means
$n\le x^2+1<n+1\Rightarrow x^2-(n-1)\ge 0and{x}^2-n<0$
For positive $x$, the solution is:
$\Rightarrow xϵ(\sqrt{n-1},\sqrt{n})$
The function becomes discontinuous every time the value of $n$ changes. This value changes for $n=2,3,4,5,6,7,8,9,10$.
So, the function is continuous everywhere in the interval $[1,3]$ except at nine points.