STATEMENT-1: The curve $y = \frac{-x^{2}}{2} + x + 1$ is symmetric with respect to the line $x = 1$. because STATEMENT-2: A parabola is symmetric about its axis. (1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1 (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False (4) Statement-1 is False, Statement-2 is True.

STATEMENT-1: The curve y = (-x²)/2 + x + 1 is symmetric with respect to the line x = 1. because STATEMENT-2: A parabola is symmetric about its axis.

$\text{Given: } y = -\dfrac{x^2}{2}+x+1 \quad (\text{a parabola})$

$\text{S-1: this curve is symmetric about } x = 1$

$\text{S-2: every parabola is symmetric about its axis}$

$\text{To find: truth of S-1, truth of S-2, and whether S-2 explains S-1}$

$y = -\dfrac{x^2}{2}+x+1 \quad (\text{quadratic})$

$\Rightarrow \text{graph is a parabola}$

$y = -\dfrac{1}{2}\,(x^2-2x)+1$

$x^2-2x = (x-1)^2 - 1$

$y = -\dfrac{1}{2}\,[\,(x-1)^2-1\,]+1$

$y = -\dfrac{1}{2}(x-1)^2+\dfrac{3}{2}$

$\text{vertex } \left(1,\dfrac{3}{2}\right), \quad \text{axis } x=1$

$x_{\text{axis}} = -\dfrac{b}{2a}, \quad a=-\dfrac{1}{2},\ b=1$

$x_{\text{axis}} = -\dfrac{1}{2(-1/2)} = 1$

$f(1+t) = -\dfrac{1}{2}t^2+\dfrac{3}{2}$

$f(1-t) = -\dfrac{1}{2}t^2+\dfrac{3}{2}$

$f(1+t) = f(1-t) \;\Rightarrow\; \text{symmetric about } x=1$

$\therefore \text{S-1 is True}$

$\text{S-2: a parabola is symmetric about its axis}$

$\text{parabola}: PF = \text{distance from } P \text{ to directrix}$

$\text{focus lies on the axis; directrix } \perp \text{ axis}$

$\Rightarrow \text{reflection in the axis preserves the locus}$

$y = a(x-h)^2+k \;\Rightarrow\; f(h+t)=f(h-t)=at^2+k$

$\therefore \text{every parabola is symmetric about } x=h$

$\therefore \text{S-2 is True}$

$\text{curve is a parabola with axis } x=1$

$\text{S-2} \Rightarrow \text{symmetry about } x=1$

$\boxed{\text{S-1 True, S-2 True; S-2 explains S-1}}$

$\text{Answer: option (1)}$