Question

What is/are the tangents to $y=(x^3-1)(x-2)$ at the points where the curve cuts the x-axis

• A. $y+3x=3$
• B. $y+2x=3$
• C. $y-7x+14=0$
• D. $y-5x-14=0$

Answer 1

Answer: (A), (C)

The correct options are
A $y+3x=3$
C $y-7x+14=0$

Given curve is, $y=(x^3-1)(x-2)$
For intersection of this curve with x-axis put $y=0$
$\Rightarrow (x^3-1)(x-2)=0\Rightarrow x=1,2$
Thus points are $(1,0)$ and $(2,0)$
Now $\frac{dy}{dx}=(x^3-1)+(x-2).3x^2=4x^3-6x^2-1$
So slope of the tangent are $m_1={\left(\frac{dy}{dx}\right)}_{x=1}=-3$ and $m_2={\left(\frac{dy}{dx}\right)}_{x=2}=7$
Hence required tangents are, $y+3x=3,y-7x+14=0$