Let's go through the problem step by step:

Part (a)

The problem states that the graphs of $y = x - 1$ and $y = kx^2$ intersect at two distinct points $A$ and $B$. We are asked to write down the quadratic equation satisfied by the $x$-coordinates of $A$ and $B$, and show that $k < \frac{1}{4}$.

Solution for (a):

1. Set the two equations equal to each other to find the intersection points: $x - 1 = kx^2$ Rearranging the terms: $kx^2 - x + 1 = 0$ This is the quadratic equation in $x$ for the intersection points $A$ and $B$.

2. For two distinct intersection points, the discriminant of the quadratic equation must be positive. The discriminant $\Delta$ for a quadratic equation $ax^2 + bx + c = 0$ is given by: $\Delta = b^2 - 4ac$ Here, $a = k$, $b = -1$, and $c = -1$. So, the discriminant becomes: $\Delta = (-1)^2 - 4(k)(-1) = 1 + 4k$ For two distinct points, the discriminant must be positive: $1 + 4k > 0$ Solving this inequality: $4k < 1 \quad \Rightarrow \quad k < \frac{1}{4}$

So, we've shown that for two distinct points of intersection, $k$ must be less than $\frac{1}{4}$.

Part (b)

We are asked to describe the relationship between the graphs of $y = x - 1$ and $y = kx^2$ in two cases:

1. $k = \frac{1}{4}$
2. $k > \frac{1}{4}$

Solution for (b):

1. When $k = \frac{1}{4}$:

   • The discriminant $\Delta = 1 + 4k = 1 + 4\left(\frac{1}{4}\right) = 2$, which is zero.
   • This means the two graphs touch at exactly one point, i.e., they are tangent to each other at one point.

2. When $k > \frac{1}{4}$:

   • The discriminant $\Delta = 1 + 4k$ becomes negative.
   • This means there are no real solutions, and hence the graphs do not intersect at all.

Part (c)

We need to show that for negative values of $k$, the equation $x - 1 = kx^2$ has two real roots, one of which lies between 0 and 1.

Solution for (c):

1. Consider the equation again: $x - 1 = kx^2 \quad \text{or} \quad kx^2 - x + 1 = 0$ For $k < 0$, the quadratic opens downwards since $kx^2$ is negative. This means the parabola will have two real roots (since the discriminant $1 + 4k$ is still positive for small negative values of $k$).

2. To confirm that one root lies between 0 and 1:

   • When $x = 0$, the equation becomes $-1 = 0$, which is negative.
   • When $x = 1$, the equation becomes $1 - 1 = 0$, which is a root.
   • Therefore, by the Intermediate Value Theorem, since the function is continuous, there must be another root between $0$ and $1$.

Would you like further clarification or more details on any part? Here are some follow-up questions to explore this concept more deeply:

1. How does the shape of the parabola $y = kx^2$ change as $k$ varies?
2. What does the discriminant tell us about the number and nature of the solutions?
3. Can you find the exact points of intersection for specific values of $k$?
4. How would the analysis change if $k$ were negative but less than $-1$?
5. What is the geometrical interpretation of the roots in part (c)?

Tip:

When analyzing quadratic equations in geometry problems, always check the discriminant to determine the number of intersection points!