If $A(1,p^2), B(0,1)$ and $C(p,0)$ are the coordinates of three points, then the value of p for which area of ABC is minimal.
Using the formula for area $$\Delta =\frac 12[1(1)+0+p(p^2-1)]$$ Taking its derivative
$$\frac {d \Delta}{dp}=\frac 12 [3p^2-1]=0$$ $$p=\frac {1}{\sqrt 3}$$
The options were, $\frac {1}{\sqrt 3} , \frac{-1}{\sqrt 3} , \frac {1}{\sqrt 2}$ while the fourth option is ‘None of these’. Fourth option is right. Why is that so?
I checked around, and the reasoning given is that the value of the triangle will be minimum when the points are collinear ie. area of triangle is zero. It seems plausible but kinda defeats the purpose of the question. If we go with the collinearity reason, no values in the option satisfy the equation. However, I feel $\frac {1}{\sqrt 3} should be the minim value in case the pints are ‘not’ collinear.
My interpretation is as follows: Since the three vertices can be collinear (when $p = p_0$ satisfies $p_0^3 + 1 = p_0$), as $p$ approaches this $p_0$ the area of the triangle approaches zero. So the area can reach every arbitrarily small positive number, but never zero itself, since in the case of collinear vertices $ABC$ is no longer a triangle and therefore does not have a well-defined area. Hence there is no minimum area.