Question

Let $C$ be the curve $y=x^3$ (where $x$ takes all real values.) The tangent at $A$ meets the curve again at $B$. If the gradient of the curve at $B$ is $K$ times the gradient at $A$ then $K$ is equal to

• A. $4$
• B. $2$
• C. $-2$
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$4$

Given equation of curve is $y=x^3$
Let the coordinates of A be $(t,t^3)$ and coordinates of B be $(T,T^3)$
$\frac{dy}{dx}=3x^2$
Slope of tangent to curve at A $=3t^2$.

Now , slope of AB $=\frac{T^3-t^3}{T-t}$

$\Rightarrow \frac{T^3-t^3}{T-t}=3t^2$

$\Rightarrow T^3-t^3=3t^2(T-t)$

$\Rightarrow (T-t)(T^2+tT+t^2)-3t^2(T-t)=0$

$\Rightarrow (T-t)(T^2+tT-2t^2)=0$

$\Rightarrow T=t,T=-2t$

$T=t$ is not possible. Hence, $T=-2t$

Given, $m_B=K\cdot m_A$
$\Rightarrow T^2=K{t}^2$
$\Rightarrow 4t^2=K{t}^2$
$\Rightarrow K=4$