Let $f$ be a twice differentiable function on $\mathbb{R}$. Given that $f''(x)>0$ for all $x\in \mathbb{R}$.Then which is true? Let $f$ be a twice differentiable function on $\mathbb{R}$. Given that $f''(x)&gt;0$ for all $x\in \mathbb{R}$.Then which  of the following is true?

1.$f(x)=0$ has exactly two solutions on $\mathbb{R}$.
2.$f(x)=0$ has a positive solution  if $f(0)=0$ and $f'(0)&gt;0$.
3.$f(x)=0$ has no positive solution if $f(0)=0$ and $f'(0)&gt;0$.
4.$f(x)=0$ has no positive solution if $f(0)=0$ and $f'(0)&lt;0$.


My thoughts:-
(1) is not true as $f(x)=x^2+1$ is a counter example.
Now suppose the conditions in (2) holds.then $f'(x)$ is increasing everywhere.so $f'(x)$ is never zero for all positive $x$.so $f(x)$ can not have a positive solution otherwise $f'(x)$ have a zero between $0$ and $x$ by Rolle's Theorem.
so (3) is true.
Are my arguments right?
Your arguments are correct. For (2), a counterexample would have been enough. However, your procedure was more efficient, since it dealt simultaneously with (2) and (3). 

You left out analyzing (4), which is not always true. For example, let $f(x)=-2x+x^2$.  

But there are functions for which (4) holds. For instance, let $f(x)=\frac{1}{1+x}-1$.
Correct answer for above question is option no.3.Second option is not correct because it has a counter example.$$f(x)=e^x-1$$ f satisfy condition $f(0)=0,f'(0)&gt;0$but $0$ is the only solution of $f(x)=0$.Hence $f(x)=0$ has no positive solution.