Question

Let C be the curve $y=x^3\forall x\in R.$ The tangent at $A$ meets the curve again at $B$. If the gradient at $B$ is $k$ times the gradient at $A$, then the number of integral values of $k$ is

• A. $0$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

tangent at $pt(1,1)$ which lies on tje given curve has curve has slope $\frac{dy}{dx}=3x$

$\frac{dy}{dx}{|}_{(1,1)}={3.1}^2=3$

$(y-1)=3(x-1)\Rightarrow y=3x-2$

$yu=x^{3}3x-2=y^3$

$x^3-3x+2=0$

factors $\pm 1,\pm 2$

$x=1,x=-2$ are the solution

so $y=(2{)}^3=8$

other coordinates would be $C(-2,8)$

Gradient at $(-2,8)=3x^2=3.4=12$

Gradient at $C$ is given to be $K$ times

gradient at $A=K$. gradient at $C$

$K=12/3=4\Rightarrow K=4$