Functions Question 334
Question: The function $ f(x)={{
[x]}^{2}}-[x^{2}] $ , (where [y] is the greatest integer less than or equal to y),is discontinuous at [IIT 1999]
Options:
A) All integers
B) All integers except 0 and 1
C) All integers except 0
D) All integers except 1
Show Answer
Answer:
Correct Answer: D
Solution:
Given $ f(x)={{[x]}^{2}}-[x^{2}] $
$ -1<x<0,f(x)={{(-1)}^{2}}-0=1 $
$ x=0,f(x)=0^{2}-0=0 $
$ 0<x<1,f(x)=0^{2}-0=0 $
$ x=1,f(x)=1^{2}-1=0 $
$ 1<x<\sqrt{2},f(x)=1^{2}-1=0 $
$ x=\sqrt{2},f(x)=1^{2}-2=-1 $
$ \sqrt{2}<x<\sqrt{3},f(x)=1^{2}-2=-1 $
$ x=\sqrt{3},f(x)=1^{2}-3=-2 $
$ \sqrt{3}<x<2,f(x)=1^{2}-3=-2 $
$ x=2,f(x)=4-4=0 $ ; $ 2<x<\sqrt{5},f(x)=4-4=0 $
$ x=\sqrt{5},f(x)=4-5=-1 $
Hence function is discontinuous at all integers except 1.
BETA
