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 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? (1) (4, √3) ∈ S (2) (5, √2) ∈ S (3) Area of R is $\frac{14}{3} − 2√3 (4)$ Area of R is $\frac{14}{3} − √3$

$y^2=x \ \text{(parabola)}$ $S = \text{locus of mid-points of chords with enclosed area } \dfrac43$ $R = \text{first-quadrant region bounded by } y^2=x,\ S,\ x=1,\ x=4$ $\text{Check: } (4,\sqrt3)\in S,\quad (5,\sqrt2)\in S$ $\text{Area}(R)=\dfrac{14}{3}-2\sqrt3 \quad \text{or} \quad \dfrac{14}{3}-\sqrt3$ $1.\ \text{parametrise } y^2=x \text{ as } (t^2,t)$ $2.\ \text{chord area} \Rightarrow |q-p|=2$ $3.\ \text{mid-point} \Rightarrow S, \text{ then area of } R$ $P(p^2,p), \ P(q^2,q), \quad \text{chord}: x=(p+q)y-pq$ $\text{Area}=\int_p^q[(p+q)y-pq-y^2]\,dy=\dfrac{(q-p)^3}{6}$ $\dfrac{(q-p)^3}{6}=\dfrac43 \Rightarrow (q-p)^3=8 \Rightarrow |q-p|=2$ $M=\left(\dfrac{p^2+q^2}{2},\ \dfrac{p+q}{2}\right)=(x,y)$ $|q-p|=2 \Rightarrow p=y-1,\ q=y+1$ $x=\dfrac{(y-1)^2+(y+1)^2}{2}=y^2+1 \Rightarrow S:\ y^2=x-1$ $(4,\sqrt3):\ 4-1=3=(\sqrt3)^2 \Rightarrow (4,\sqrt3)\in S\ \checkmark$ $(5,\sqrt2):\ 5-1=4\ne2=(\sqrt2)^2 \Rightarrow (5,\sqrt2)\notin S$ $\Rightarrow \text{(1) TRUE}, \ \text{(2) FALSE}$ $R:\ \sqrt{x-1}\le y\le\sqrt{x}, \quad 1\le x\le4$ $\text{Area}(R)=\int_1^4\left(\sqrt{x}-\sqrt{x-1}\right)dx$ $=\left[\dfrac23 x^{3/2}-\dfrac23(x-1)^{3/2}\right]_1^4$ $=\dfrac23(8-1)-\dfrac23(3^{3/2})=\dfrac{14}{3}-2\sqrt3$ $\Rightarrow \text{(3) TRUE}, \ \text{(4) FALSE}$ $\text{(1) TRUE}, \quad \text{(3) TRUE}$ $\boxed{\text{Answer: (1) and (3)}}$