Question

Let $C(m,n)$ be a point on the curve $y=x^3,$ where the tangent is parallel to the chord connecting the points $O(0,0)$ and $A(2,8)$. Then the value of $m\cdot n$ is:

• A. $\frac{16}{9}$
• B. $\frac{12}{5}$
• C. $\frac{16}{27}$
• D. $\frac{32}{9}$

Answer 1

The correct option is A $\frac{16}{9}$

$y=x^3$ is continuous and differentiable for all $x\in R$. Therefore, we can use LMVT.
Let $f\left(x\right)=x^3$
$f^{\prime }\left(c\right)=\frac{f\left(2\right)-f\left(0\right)}{2-0}=\frac{8-0}{2}$
$\Rightarrow 3c^2=4\Rightarrow c=\pm \frac{2}{\sqrt{3}}$
The only value of $c$ possible within the given end points is
$c=\frac{2}{\sqrt{3}}$
$\therefore m=\frac{2}{\sqrt{3}}$
So, $n=m^3={\left(\frac{2}{\sqrt{3}}\right)}^3=\frac{8}{3\sqrt{3}}$
$\therefore m\cdot n=\frac{16}{9}$