SAT1000 - P158

Take $f(x)=\sqrt {|x+1|}$ .

Solution to this problem :

The position coordinates of $A, B, F, P, Q, M$ as $(\frac{y_1^2}{2},y_1),(\frac{y_2^2}{2},y_2), (\frac{1}{2},0),(-\frac{1}{2},y_1),(-\frac{1}{2},y_2),(\frac{y_1^2+y_2^2}{4},\frac{y_1+y_2}{2})$ .

The condition imposed over the triangle areas yield $y_1y_2=-2$ .

So we can write $x_M=\dfrac {y_1^2+y_2^2}{4}, y_M=\dfrac {y_1+y_2}{2}$

$\implies 4y_M^2=4x_M-4\implies x_M=1+y_M^2$

At $y_M=50,x_M=2501$ .

A function has the property that an input cannot have more than one output.

Option A cannot be correct because I am able to find an input to $f(\sin{2x})$ that has two different outputs. If x = 0 then $f(0) = 0$ but if x = $\pi/2$ then $f(0) = 1$ .

Option B cannot be correct for the same reason that Option A is incorrect. If x = 0 then $f(0) = 0$ but if x = $\pi/2$ then $f(0) = \pi^{2}/4 + \pi/2$ .

Option C also violates the function property. If x = 1 then $f(2) = 2$ but if x = -1 then $f(2) = 0$ .

By the process of elimination, we can conclude that Option D is the correct answer. We can think of $|x + 1|$ as the horizontal deviation from the line x = -1 which happens to be the vertical line separating $x^{2} + 2x$ into two halves. The function $f(x^{2} + 2x) = |x + 1|$ therefore represents the horizontal deviation from the center of the quadratic y = x^{2} + 2x given its y-value (i.e: f(3) = 2 since the point (1, 3) is a difference of 2 units between x = -1 and x = 1). Since the y-values are symmetric across the line x = -1 and each point on y = x^{2} + 2x has the same horizontal deviation from x = -1 as the point symmetric to it, there does not exist any input to the function with more than one output.