Question

Let a function, \(f:(-1,3) \rightarrow \mathbf{R}\) be defined as \(f(x)=\min \{x[x],|x[x]-2|+2\}\), where \([x]\) denotes the greatest integer \(\leq x\). Then \(f\) is :

• A. not continuous at only one point and not differentiable at three points.
• B. neither continuous nor differentiable at exactly 3 points.
• C. neither continuous nor differentiable at exactly two points.
• D. not continuous at two points and not differentiable at 3 points.

Hint:

Split \((-1,3)\) at the integers, simplify \(f\) on each piece, and check continuity and differentiability at \(0,1,2\).

Answer 1

The Correct Option is D

On \((-1,0)\): \([x]=-1\), \(x[x]=-x\in(0,1)\), and \(|x[x]-2|+2=x+4\). The minimum is \(-x\), so \(f=-x\).
On \([0,1)\): \(x[x]=0\) and the other term is \(4\). So \(f=0\).
On \([1,2)\): \(x[x]=x\) and the other term is \(|x-2|+2=4-x\). Since \(x<4-x\) for \(x<2\), \(f=x\).
On \([2,3)\): \(x[x]=2x\) and the other term is \(|2x-2|+2=2x\). So \(f=2x\).
At \(x=0\): left limit \(0\), \(f(0)=0\), right value \(0\), so \(f\) is continuous; but the slopes are \(-1\) (left) and \(0\) (right), so it is not differentiable.
At \(x=1\): left limit \(0\) but \(f(1)=1\), so it is discontinuous and hence not differentiable.
At \(x=2\): left limit \(2\) but \(f(2)=4\), so it is discontinuous and hence not differentiable.
So \(f\) is discontinuous at 2 points (\(x=1,2\)) and not differentiable at 3 points (\(x=0,1,2\)).
Answer: (4) not continuous at two points and not differentiable at 3 points.