The curve $y=f\left(x\right)$ which satisfies the condition $f\text{'}\left(x\right)>0$ and $f^{\text{'}\text{'}}\left(x\right)<0$ for all real $x$, is

Given: $f^{\text{'}}\left(\mathrm{x}\right)>0$ and $f^{\text{'}\text{'}}\left(x\right)<0$ for all real $x$.

To find: Curve $y=f\left(x\right)$ which satisfy given conditions.

Step-$1$

Choose $x_1,x_2$ such that $x_2>x_1$and $f\left(x_2\right)>f\left(x_1\right)$,

example: 

                                                                                                                                OR

which means that function is increasing in interval $\left(x_1,x_2\right)$.

Step-$2$

Consider, $f^{\text{'}}\left(x\right)>0$ then function is strictly increasing in its domain. 

Hence, option $\left(1\right) , \left(2\right)$are ruled out since they both decrease at some interval. 

As it can be observed that the slope of the curve is not strictly increasing it is increasing in some interval and decreasing in other intervals same is the case for option-$2$ as , In option-$2$

As it can easily seen in the graph that $f\left(x\right)$ is strictly decreasing because its slope is always negative.

Now , ${\mathrm{f}}^{\text{'}\text{'}}\left(\mathrm{x}\right)<0 \forall \mathrm{x}\in \mathrm{D}$ then graph of $f\left(x\right)$ is concave downward like,

Thus curve $y=f\left(x\right)$ that satisfies the condition $f^{\text{'}}\left(x\right)>0,f^{\text{'}\text{'}}\left(x\right)<0\forall x\in R$ is concave downward . 

Hence option-$4$ is the correct answer.