fog(x)=f(g(x)) =f(x2+1x2)
=x2+1x2−1 =x2+1x2−x2−1 =x2+1−1
We know that, 0≤x2<∞,∀x∈R
⇒1≤x2+1<∞,∀x∈R ⇒1≥x2+11>0,∀x∈R ⇒−1≤x2+1−1<0,∀x∈R
So, range of fog(x) is [−1,0)⊂R.
Hence, the function fog(x) is into function and fog(−x)=f(g(−x))=(−x)2+1−1=x2+1−1=f(g(x))
∴fog(x) is an even function. So, it is a many one function.
Hence, fog(x) is neither one-one nor onto function.