Question

The angle which the tangent to a curve at any point $(x,y)$ on it makes with axis of $x$ is ${\tan }^{-1}(x^2-2x)$ for all values of $x$ and it passes through the point $(2,0)$.Determine the point on it whose ordinate is maximum.

• A. $(2,8/3)$
• B. $(0,4/3)$
• C. $(1,2/3)$
• D. $(-1,4/3)$

Answer 1

Answer: (B)

$(0,4/3)$

$\frac{dy}{dx}=tan\left(ta{n}^{-1}\right(x^2-2x\left)\right)$
Or
$\frac{dy}{dx}=x^2-2x$
Or
$dy=(x^2-2x).dx$
Integrating both sides we get
$y=\frac{x^3}{3}-x^2+c$.
Now $y=0$ at $x=2$
Substituting in the above expression, we get
$0=\frac{8}{3}-4+c$
Or
$c-\frac{4}{3}=0$
Or
$c=\frac{4}{3}$.
Hence
$f\left(x\right)=\frac{x^3}{3}-x^2+\frac{4}{3}$ is the required equation of the curve.
Now
$\frac{dy}{dx}=0$ gives the critical points.
Hence
$x^2-2x=0$
$x(x-2)=0$
Or
$x=0$ and $x=2$
$f^{\prime \prime }\left(x\right)=2x-2$
$f^{\prime \prime }\left(2\right)>0$ ... (minimum ordinate)
$f^{\prime \prime }\left(0\right)<0$ ... (maximum ordinate).
$f\left(0\right)=\frac{4}{3}$.
Thus the point with maximum ordinate is $(0,\frac{4}{3})$.