Question

The abscissa of the points of the curve $y=x(x-2)(x-4)$ where tangents are parallel to x-axis is obtained as

• A. $x=2\pm \frac{2}{\sqrt{3}}$
• B. $x=1\pm \frac{1}{\sqrt{3}}$
• C. $x=2\pm \frac{1}{\sqrt{3}}$
• D. $x=\pm 1$

Answer 1

The correct option is A

$x=2\pm \frac{2}{\sqrt{3}}$

Explanation for the correct option:

Step-1 : Simplification

The given curve is : $y=x(x-2)(x-4)$.

On simplifying, we get

$$\begin{gathered} y=x(x-2)(x-4) \\ \Rightarrow y=x(x^2–6x+8) \\ \Rightarrow y=x^3–6x^2+8x\left(1\right) \end{gathered}$$

Step-2 : Finding the slope of the tangents that are parallel to $x$-axis

We know that coordinates of every point on the curve be of the form $(x,y)$ and in $(x,y)$, $x$ is called the abscissa.

Since tangents to the curve are parallel to x-axis, therefore, the slope of tangents will be zero i.e. $\frac{dy}{dx}=0$.

Step-3 : Finding the required abscissas

Differentiating $\left(1\right)$ with respect to $x$ and equating it to zero, we get

$$\begin{gathered} \frac{dy}{dx}=0 \\ \Rightarrow 3x^2–12x+8=0 \end{gathered}$$

Now, we know that the solution of a quadratic equation $a{x}^2+bx+c=0$ is given by $\left[x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\right]$.

Thus, we get :

$$\begin{gathered} 3x^2–12x+8=0 \\ \Rightarrow x=\frac{-12\pm \sqrt{{\left(12\right)}^2-4\times 3\times 8}}{2\times 3} \\ \Rightarrow x=\frac{-12\pm \sqrt{48}}{6} \\ \Rightarrow x=\frac{-12\pm 4\sqrt{3}}{6} \\ \Rightarrow x=2\pm \frac{2}{\sqrt{3}} \end{gathered}$$

So, the required abscissas are : $x=2\pm \frac{2}{\sqrt{3}}$.

Hence, option (C) is the correct answer.