The point at which the tangent to the curve ;$\displaystyle y=x^{3}+5$ is perpendicular to the line $x + 3y = 2$ are

Let point be $\displaystyle \left ( x_{1},y_{1} \right )$

$y = \displaystyle x^{3}+5$

$\Rightarrow \displaystyle \dfrac{dy}{dx}=3x^{2}$

Now, the slope of tangent at this point is $=\displaystyle \left ( \frac{dy}{dx} \right )_{x_{1}y_{1}}=3x_{1}^{2}$

It is& ;$\displaystyle \perp$ to $x + 3y - 2 = 0$

So $\displaystyle 3x_{1}^{2}\times -\frac{1}{3}=-1\Rightarrow x_1=\pm 1$

Thus,

at $\displaystyle x_{1}=1$,             $\Rightarrow$                $\displaystyle y_{1}=6$

and at $\displaystyle x_{1}=-1$      $\Rightarrow$              $\displaystyle y_{1}=4$

Thus, the points are $(1, 6)$ and $(-1, 4)$

Hence, option 'D' is correct.