一条曲线的方程为 $y = x^2 - x - 12$。点 A 和 B 在该曲线上面，其 x 坐标分别为 $k$ 和 $2k$。

1. Express the y-coordinates of A and B in terms of k

For point A, $x = k$, so $y_A = k^2 - k - 12$.
For point B, $x = 2k$, so $y_B = (2k)^2 - (2k) - 12 = 4k^2 - 2k - 12$.

2. Form the inequality $y_B < y_A$ and simplify

$4k^2 - 2k - 12 < k^2 - k - 12$
$3k^2 - k < 0$

3. Solve the quadratic inequality

$k(3k - 1) < 0$
$0 < k < \frac{1}{3}$

1. Find the gradient function and the gradients at A and B

$\dfrac{dy}{dx}=2x-1$

Gradient at $A$ (where $x=k$): $2k-1$

Gradient at $B$ (where $x=2k$): $2(2k)-1=4k-1$

2. Compare the two gradients to justify the inequality

Gradient at $B$ $-$ Gradient at $A$ $=(4k-1)-(2k-1)=2k$.

Since $k>0$ (the points are valid for $0<k<\tfrac{1}{3}$ from part (a)), $2k>0$.

Therefore gradient at $B$ > gradient at $A$, i.e. the gradient at $A$ is always less than the gradient at $B$.