Statement I: f(x)=1+∣x∣x. For x≥0, f(x)=1+xx is strictly increasing. For x<0, f(x)=1−xx with f′(x)=(1−x)21>0, also strictly increasing. The function is continuous at x=0, so f is strictly increasing on R and hence one-one. Statement I is true.
Statement II: f(x)=x2−8x+18x2+4x−30. The denominator (x−4)2+2>0 for all x.
f′(x)=(x2−8x+18)2−12(x2−8x+14)
f′(x)=0⇒x=4±2. Since f′ changes sign, f is not monotonic and hence many-one. Statement II is true.
Both statements are true.