I want to understand my mistake, other answers are welcome but I'll only accept the one which the one which points out where and why I went wrong.
For $y=x^3$, tangent at A meets the curve again at B. Gradient at B is $k$ times the gradient at A.Then the number of integral values of k is:
Let the point $A$ be $(a, a^3)$ and $B~(b, b^3)$.
$$y'=3x^2$$
So equation of tangent at $A$.
$$\frac{b^3-a^3}{b-a}=3a^2$$
$$\frac{(b-a)(b^2 + a^2 + ab)}{b-a}=3a^2$$
$$b^2-2a^2+ab=0$$
Therefore $b=-2a$ I ignored the repeated root $b=a$ as it is the same point.
So we get one integral value of $k=4$. However the number of integral values of $k$ has been given as 3.
What's wrong with my method?
Best Answer
Your working is fine.
We find that $b=-2a$, of which the gradient at the point is $3(-2a)^2=4(3a^2)$.
For a sanity check, consider the point $(1,1)$, the tangent is $y=\color{blue}3x-2$ which intersect the graph at $(-2,-8)$. The gradient at that point is $3(-2)^2=4(\color{blue}3)$. The answer is $4$.