Let S be the set of all twice differentiable functions f from ℝ to ℝ such that d 2 f d x 2 x…

Given,

$f^{"}\left(x\right)>0 \& f\left(x\right)-x=0$

Now let,

$g\left(x\right)=f\left(x\right)-x$

Now differentiating above equation we get,

$\Rightarrow g^{\text{'}}\left(x\right)=f^{\text{'}}\left(x\right)-1$

Again differentiating we get,

$g^{"}\left(x\right)=f^{"}\left(x\right)>0\Rightarrow \text{ concave up }$

So, possible graphs will be,

Here in above graph $X_f=0$ as there is no point for which $f\left(x\right)=x$

Now here $X_f\le 2$ as there will be minimum two points when $y=x$ will cut the given graph,

Now in above graph $X_f=1$ so there will minimum one intersection between line $y=x$ and the given graph so option $\left(D\right)$ is wrong.