JEE Corner (Mains +Advanced) # 2

Geometry Level 3

Let $f: A \rightarrow B$ and is given that $f(x) =\log_2 \left (\sqrt{(\cos(\pi x) - \sin(\pi x) )(\cos(\pi x) + \sin(\pi x) ) - 1 } + 2 \right)$ is real and $g: A \rightarrow C$ such that $g(x) = \lfloor x \rfloor ^{\{ x \} }$ . Find the range of $g(x)$ .

Note that $\{x\}$ denote the fractional part of $x$ .

{Set Of Integers} {1} Function f(x) Is Not Defined {0,1} Function g(x) Is Not Defined

1 solution

Devarsh Wali
Jun 27, 2015

Since f(x) is defined from Set A to Set B and g(x) is defined from Set A to Set C which implies that the domain of f(x) and g(x) is same.So,if we find out the domain of f(x) then we could use that domain to find the range of g(x). Now Domain of f(x) is :-

But just because a function has a point where it is undefined doesn't mean it doesn't have range, right? Like the range of $f(x)=\frac{1}{x}$ would be $(-\infty,0) \cup (0,\infty)$ even though it is undefined at $x=0$

Jason Zou - 5 years, 11 months ago

Just a point, $\frac{1}{0}$ is not undefined. It's infinity. $0^0$ is one of the undefined forms, and thus, it can not be included in the range.

Kushashwa Ravi Shrimali - 5 years, 11 months ago

I am not saying that $0^0$ should be included in the range though. However, we also have $1^0, 2^0, \ldots=1$ which are defined and thus should be included in the range.

Jason Zou - 5 years, 11 months ago

@Jason Zou – @Kushashwa Ravi Shrimali @Jason Zou guys, in response to both of you, a function is defined when all the elements of its domain has an image or range but here the domain of g(x) is set of integers in which when we put 0 from the domain of g(x) in the function g(x) it doesn't gives any image as 0^0 is undetermined hence g(x) is not a defined function since it is discontinuous at x=0