Question

A curve passes through the point $(2,0)$ and the slope of the tangent at any point $(x,y)$ is $x^2-2x$ for all values of $x$. The point of maximum ordinate on the curve is :

• A. $(0,\frac{2}{3})$
• B. $(0,1)$
• C. $(0,\frac{4}{3})$
• D. $(\frac{4}{3},0)$

Answer 1

The correct option is C

$(0,\frac{4}{3})$

Explanation for the correct option:

Step 1: Find the equation of the curve.

We have been given that, a curve whose slope of the tangent at any point $(x,y)$ is $x^2-2x$ for all values of $x$.

We need to find the point of maximum ordinate on the curve.

The slope of the tangent be,

$\frac{dy}{dx}=x^2-2x$

To find the equation of curve we integrate the above equation:

$$\begin{gathered} \int dy=\int (x^2-2x)dx \\ y=\frac{x^3}{3}-x^2+c \end{gathered}$$
Step 2: Find the value of integration constant ‘c’.

Since the curve passes through $(2,0)$ , equation of the curve would be,

$$\begin{gathered} \Rightarrow 0=\frac{2^3}{3}-2^2+c \\ \Rightarrow c=-(\frac{8}{3}-4) \\ \Rightarrow c=\frac{4}{3} \end{gathered}$$

So, the equation of the curve would be,

$y=\frac{x^3}{3}-x^2+\frac{4}{3}$

Step 3: Find the critical point.

We know, to find the critical point, $\frac{dy}{dx}=0$

$$\begin{gathered} \Rightarrow (x^2-2x)=0 \\ \Rightarrow x(x-2)=0 \\ \Rightarrow x=0,x-2=0 \\ \Rightarrow x=0,x=2 \end{gathered}$$

Step 4: To decide maximum and minimum point find $\frac{d^{2}y}{d{x}^2}$.

$$\begin{gathered} \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(x^2-2x) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=2x-2 \end{gathered}$$

Step 5: Find the point of maxima.

$$\begin{gathered} Forx=0, \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=2\left(0\right)-2 \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=-2 \\ \Rightarrow \frac{d^{2}y}{d{x}^2}<0 \end{gathered}$$

This means the curve has the maximum value at $x=0$.

Step 6: Find the point of maximum ordinate on the curve.

At $x=0$,

$$\begin{gathered} \Rightarrow y_{max}=\frac{{\left(0\right)}^3}{3}-{\left(0\right)}^2+\frac{4}{3} \\ \Rightarrow y_{max}=\frac{4}{3} \end{gathered}$$

The required point is $\left(0,\frac{4}{3}\right)$

Therefore, option (C) is the correct answer.