Let $$S$$ denote the locus of the mid-points of those chords of the parabola $$y^2 = x$$, such that the area of the region enclosed between the parabola and the chord is $$\frac{4}{3}$$. Let $$\mathcal{R}$$ denote the region lying in the first quadrant, enclosed by the parabola $$y^2 = x$$, the curve $$S$$, and the lines $$x = 1$$ and $$x = 4$$.

Then which of the following statements is (are) TRUE?

Let the parabola be $$x = y^{2}$$  (opening towards the right).
Take any two points on it with parametric values $$t_{1},\,t_{2}$$:
$$P\,(t_{1}^{2},\,t_{1}),\;Q\,(t_{2}^{2},\,t_{2}).$$

Mid-point $$M(h,k)$$ of the chord $$PQ$$ is therefore
$$h = \dfrac{t_{1}^{2}+t_{2}^{2}}{2},\qquad k = \dfrac{t_{1}+t_{2}}{2}.$$ We must find the locus of $$M$$ under the condition that the area of the parabolic segment cut off by $$PQ$$ equals $$\dfrac{4}{3}$$.

1. Equation of the chord through $$P,Q$$
Using two-point form, $$(y-t_{1})(t_{2}^{2}-t_{1}^{2})=(x-t_{1}^{2})(t_{2}-t_{1}).$$ Because $$t_{2}^{2}-t_{1}^{2}=(t_{2}-t_{1})(t_{2}+t_{1})$$, the chord simplifies to $$x=(t_{1}+t_{2})y-t_{1}t_{2}\;.\tag{-1}$$

2. Area between the parabola and the chord
For a fixed ordinate $$y$$ (between $$t_{1}$$ and $$t_{2}$$), $$x_{\text{chord}}-(x_{\text{parabola}})=\bigl((t_{1}+t_{2})y-t_{1}t_{2}\bigr)-y^{2} =-(y-t_{1})(y-t_{2})=(t_{2}-y)(y-t_{1}).$$ Hence the required area is $$A=\int_{y=t_{1}}^{t_{2}}\!(t_{2}-y)(y-t_{1})\,dy.$$ Put $$y=t_{1}+u,\;u\in[0,d]$$ where $$d=t_{2}-t_{1}$$: $$A=\int_{0}^{d}\!\bigl(d-u\bigr)u\,du =\int_{0}^{d}\!(du-u^{2})\,du =d\left[\frac{u^{2}}{2}\right]_{0}^{d}-\left[\frac{u^{3}}{3}\right]_{0}^{d} =\frac{d^{3}}{2}-\frac{d^{3}}{3} =\frac{d^{3}}{6}.\tag{-2}$$

The question states that $$A=\dfrac{4}{3}$$. Using $$(\text{-2})$$: $$\frac{d^{3}}{6}=\frac{4}{3}\;\Longrightarrow\;d^{3}=8\; \Longrightarrow\;d=t_{2}-t_{1}=2.\tag{-3}$$

3. Locus of the mid-point
From $$(\text{-3})$$, set $$t_{1}=k-1,\;t_{2}=k+1.$$ Then $$h=\frac{(k-1)^{2}+(k+1)^{2}}{2} =\frac{k^{2}-2k+1+k^{2}+2k+1}{2} =k^{2}+1.$$ Thus the locus $$S$$ of all such mid-points is $$x=y^{2}+1\qquad\bigl(\text{or }y^{2}=x-1\bigr).$$

4. Verification of the given points
(i) For $$(4,\sqrt{3})$$:  $$y^{2}+1=3+1=4=x$$ ⇒ lies on $$S$$.
(ii) For $$(5,\sqrt{2})$$:  $$y^{2}+1=2+1=3\neq5$$ ⇒ does not lie on $$S$$.

5. Region $$\mathcal{R}$$ in the first quadrant
Within $$x=1$$ to $$x=4$$ we have
• lower curve (from $$S$$):  $$y=\sqrt{x-1}$$,
• upper curve (original parabola):  $$y=\sqrt{x}$$.
Therefore the required area is $$\text{Area}=\int_{x=1}^{4}\!\bigl(\sqrt{x}-\sqrt{x-1}\bigr)\,dx.$$

$$\int_{1}^{4}\!\sqrt{x}\,dx =\left[\frac{2}{3}x^{3/2}\right]_{1}^{4} =\frac{2}{3}(8-1)=\frac{14}{3},$$ $$\int_{1}^{4}\!\sqrt{x-1}\,dx =\int_{0}^{3}\!\sqrt{u}\,du =\left[\frac{2}{3}u^{3/2}\right]_{0}^{3} =\frac{2}{3}\,(3\sqrt{3})=2\sqrt{3}.$$ Hence $$\text{Area}(\mathcal{R})=\frac{14}{3}-2\sqrt{3}.$$

6. Conclusions
Option A is correct because $$(4,\sqrt{3})\in S$$.
Option B is incorrect.
Option C is correct since $$\text{Area}(\mathcal{R})=\dfrac{14}{3}-2\sqrt{3}$$.
Option D is incorrect.

Therefore the TRUE statements are:
Option A  and  Option C.