Answer:
m can only be equal to 1
Step-by-step explanation:
Range of f(x) = Range of g(x)
[tex]\sqrt{mx} =m\sqrt{x}[/tex]
Squaring both sides,
[tex](\sqrt{mx}) ^{2} =(m\sqrt{x} )^{2}[/tex]
[tex]mx=m^{2} x[/tex]
[tex]m=m^{2}[/tex]
[tex]m^{2} -m=0[/tex]
m(m-1) = 0
m cannot be 0.
So, m - 1 = 0 and m = 1.
So, m can boly be 1.
The statement which is true about the value of m is:
We are given two functions f(x) and g(x) as:
[tex]f(x)=\sqrt{mx}[/tex]
and g(x) is given by:
[tex]g(x)=m\sqrt{x}[/tex]
Now, we know that:
The domain of a square root function is always greater than or equal to zero.
By function f(x) we have:
[tex]mx\geq 0[/tex]
so, the range is the set of all the positive real numbers.
Now, by looking at the function g(x) we have:
[tex]x\geq 0[/tex]
but m could be positive or negative,
This means that the range of the function g(x) is the set of all the real numbers.
But it is given that:
Range of both the functions are equal.
This means that:
[tex]g(x)\geq 0[/tex] for all the values of x.
This means that m has to be positive.
i.e. m≥0
Hence, the value of m has to be the set of all the positive real numbers.
