Let `C` be the curve `y=x^3` (where `x` takes all real values). The tangent at `A` meets the curve again at `Bdot` If the gradient at `B` is `K` times the gradient at `A ,` then `K` is equal to (a) 4 (b) 2 (c) `-2` (d) `1/4`

Options

A. 4
B. 2
C. `-2`
D. `1/4`

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the curve and the point of tangency
The curve is given by the equation \( y = x^3 \). Let the point \( A \) on the curve be represented by the coordinates \( (t, t^3) \).

### Step 2: Find the gradient of the curve at point \( A \)
To find the gradient (slope) of the curve at point \( A \), we need to calculate the derivative of \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 3x^2 \]
At point \( A \) where \( x = t \):
\[ \text{Gradient at } A = 3t^2 \]

### Step 3: Write the equation of the tangent line at point \( A \)
The equation of the tangent line at point \( A(t, t^3) \) can be expressed using the point-slope form:
\[ y - t^3 = 3t^2(x - t) \]
Rearranging this gives:
\[ y = 3t^2x - 3t^3 + t^3 = 3t^2x - 2t^3 \]

### Step 4: Find the point \( B \) where the tangent meets the curve again
To find point \( B \), we set the equation of the tangent equal to the curve:
\[ 3t^2x - 2t^3 = x^3 \]
Rearranging gives:
\[ x^3 - 3t^2x + 2t^3 = 0 \]
This is a cubic equation in \( x \). Since \( x = t \) is one root (point \( A \)), we can factor the cubic equation:
\[ (x - t)(x^2 + ax + b) = 0 \]
Using polynomial long division or synthetic division, we find:
\[ x^3 - 3t^2x + 2t^3 = (x - t)(x^2 + tx - 2t^2) \]
Setting the quadratic to zero to find the other intersection point \( B \):
\[ x^2 + tx - 2t^2 = 0 \]
Using the quadratic formula:
\[ x = \frac{-t \pm \sqrt{t^2 + 8t^2}}{2} = \frac{-t \pm 3t}{2} \]
This gives us the roots:
\[ x = t \quad \text{or} \quad x = -2t \]
Thus, point \( B \) is \( (-2t, (-2t)^3) = (-2t, -8t^3) \).

### Step 5: Find the gradient at point \( B \)
Now, we find the gradient at point \( B(-2t, -8t^3) \):
\[ \text{Gradient at } B = 3(-2t)^2 = 3 \cdot 4t^2 = 12t^2 \]

### Step 6: Relate the gradients at points \( A \) and \( B \)
We know from the problem statement that the gradient at \( B \) is \( K \) times the gradient at \( A \):
\[ 12t^2 = K \cdot 3t^2 \]
Dividing both sides by \( 3t^2 \) (assuming \( t \neq 0 \)):
\[ K = \frac{12t^2}{3t^2} = 4 \]

### Conclusion
Thus, the value of \( K \) is \( 4 \).