Since, f:R→R such that f(x)=3−x. Let y1 and y2 be two elements of f(x) such that ‌y1=y2 ‌⇒‌‌3−x1=3−x2⇒x1=x2 Since, if two images are equal, then their elements are equal, therefore it is one-one function Let ‌‌y=3−x On differentiating w.r.t. x, we get ‌
dy
dx
=−3−x‌log‌3<0 for every value of x ∴ It is decreasing function ∴ Statement I and II are true.