Wht cannot this term be equal to zero?

Question

I came across the following problem:

   For y=x³, 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:

I proceeded as:-

Let A (a,y1) and B (b,y2) for all real x

The slope of tangent at B = (a³-b³)/(a-b)

And we know that slope of tangent at A, that is, 3a² = K times slope at B

So, process this information, I got the following quadratic equation:

      (3-K)(b/a)² - K(b/a) - K = 0

Since x is real, the discriminant of the above equation has to be greater than or equal to zero. Processing it I got:

        K belongs to the interval [0,4]

So the answer came out to be 5, since we need the integral values. But in the answer key, it is given that k cannot be zero, thus making the answer 3.

My question is why can't K be zero? Am I missing something?

Please note that a similar question has been asked by someone, but the answer is not at all satisfactory. Thus, please refrain from providing any links to that answer

Answer 1

The equation of the tangent line at $A = (a,a^3)$ is

$$ y - a^3 = 3a^2 (x - a).$$

Since this passes through $B = (b,b^3)$ we have

$$b^3 - a^3 = 3a^2 (b - a).$$

This equation can be rewritten as

$$(b -a)^2(b + 2a) = 0.$$

Since $b \ne a$, we have $b = -2a$, whenever $a \ne 0$. If $a = 0$, then according to the terms of the question, there is no corresponding value of $b$.

We have $3b^2 = 4(3a^2)$. Thus $k$ can take only one value, namely $4$.

Answer 2

One reason $k$ cannot be zero is that "zero times the gradient at $A$" is zero, that is, $k=0$ implies the gradient at $B$ is zero, which can be true only if $B = (0,0).$ But there is no tangent to the curve $y = x^3$ that passes through $(0,0),$ so $(0,0)$ cannot be the point at which the tangent at $A$ meets the curve again.

Another reason $k$ cannot be zero is that $k = 4,$ a fact that has been proved already in another answer to your question and in For $y=x^3$, tangent at A meets the curve again at B. Gradient at B is $k$ times the gradient at A. This fact also proves that the answer key you read the answer "$3$" from is wrong. The number of integral values of $k$ is $1$. (That is, there is only one possible value of $k,$ and that value happens to be an integer.)

By the way, I count at least three errors in your working. It is remarkable that you got a result as close to the answer key as you did.