\lim_{x\to 1^-} f'(x)=-2 

\lim_{x\to 1^+} f(x)=1

f is right continuous at 1.

\lim_{x\to 1^+} f'(x)=3 

\lim_{x\to 1^-} f(x)=(-1)^+

\lim_{x\to 1^+} f(x)=1^+

\lim_{x\to 1^-} f'(x)=(-2)^-

f is left continuous at 1.

\lim_{x\to 1} f(x) does not exist

f is right differentiable at 1 with f'_{+}(1)=3

f is left differentiable at 1 with f'_{-}(1)=-2

f is differentiable at 1 

f is continuous at 1.

There exists c \in (a,b) such that f(c) = 0 and c= \sup A .

If \sup A=c \in (a,b) and f(c) > 0 then there are upper bounds of A to the left of c and that is a contradiction.

If \sup A=c \in (a,b) and f(c) > 0 then there are elements of A to the left of c and that is a contradiction.

If \sup A=c \in (a,b) and f(c) < 0 then there are elements of A to the right of c and that is a contradiction.

If \sup A=a then there are elements of is A to the right of a and that is a contradiction.

f(a) < 0, therefore a \in A and A is a non-empty subset of \mathbb{R}. 

If \sup A=c \in (a,b) and f(c) < 0 then there are upper bounds of A to the right of c and that is a contradiction.

If \sup A=a then there are upper bounds of A to the right of a and that is a contradiction.

If A is a non-empty subset of \mathbb{R} then it has a supremum, \sup A\in \mathbb{R}.

If

If

None of these.

\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}} 

\sum_{n=1}^{\infty}\frac{1}{n} 

\sum_{n=1}^{\infty}\frac{1}{n^2(n+1)^2} 

\sum_{n=1}^{\infty}\frac{1}{n^2} 

\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}\sqrt{n+1}}