If the points (x + 1, 2),(1, x + 2),and left( | vedclass

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(D) Let $A \equiv (x + 1, 2)$,$B \equiv (1, x + 2)$,and $C \equiv \left( \frac{1}{x + 1}, \frac{2}{x + 1} \right)$.The points $A, B, C$ are collinear

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Both $(b)$ and $(c)$

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(D) Let $A \equiv (x + 1, 2)$,$B \equiv (1, x + 2)$,and $C \equiv \left( \frac{1}{x + 1}, \frac{2}{x + 1} ight)$.The points $A, B, C$ are collinear if the area of $\Delta ABC = 0$.$\left| \begin{array}{ccc} x + 1 & 2 & 1 \ 1 & x + 2 & 1 \ \frac{1}{x + 1} & \frac{2}{x + 1} & 1 \end{array} ight| = 0$Applying $R_1 	o R_1 - R_2$:$\left| \begin{array}{ccc} x & -x & 0 \ 1 & x + 2 & 1 \ \frac{1}{x + 1} & \frac{2}{x + 1} & 1 \end{array} ight| = 0$Applying $C_2 	o C_2 + C_1$:$\left| \begin{array}{ccc} x & 0 & 0 \ 1 & x + 3 & 1 \ \frac{1}{x + 1} & \frac{3}{x + 1} & 1 \end{array} ight| = 0$Expanding along $R_1$:$x \left( (x + 3)(1) - (1)\left( \frac{3}{x + 1} ight) ight) = 0$$x \left( x + 3 - \frac{3}{x + 1} ight) = 0$$x \left( \frac{(x + 3)(x + 1) - 3}{x + 1} ight) = 0$$x \left( \frac{x^2 + 4x + 3 - 3}{x + 1} ight) = 0$$\frac{x(x^2 + 4x)}{x + 1} = 0$$\frac{x^2(x + 4)}{x + 1} = 0$Thus,$x = 0$ or $x = -4$.

If the points $(x + 1, 2)$,$(1, x + 2)$,and $\left( \frac{1}{x + 1}, \frac{2}{x + 1} \right)$ are collinear,then $x$ is

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