Question

Which of the following option(s) is/are correct about curves $y=x^3-3x+4$ and $y=3(x^2-x)$

• A. Equation of common tangent is $y=9x-12$
• B. Angle between the tangents at their point of intersection is ${\tan }^{-1}\left(9\right)$
• C. Equation of common tangent is$y-6=0$
• D. Both the curves intersect atmost at one point and touch each other at one point.

Answer 1

Answer: (A), (B), (D)

Both the curves intersect atmost at one point and touch each other at one point.

$y=x^3-3x+4,y=3(x^2-x)$
For point of intersection $x^3-3x+4=3(x^2-x)$
$\Rightarrow x^3-3x^2+4=0$
On factorising we get, $(x+1)(x-2{)}^2=0$
$\therefore$ Points of intersection are $x=-1,x=2$
$x=-1\Rightarrow y=6$
$x=2\Rightarrow y=6$

At $(-1,6)$
$y=x^3-3x+4\Rightarrow \frac{dy}{dx}=3x^2-3\Rightarrow m_1=0$
$y=3(x^2-x)\Rightarrow \frac{dy}{dx}=3(2x-1)\Rightarrow m_2=-9$
$\therefore$ Angle between the tangents at $(-1,6)$ is $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=9$
$\Rightarrow \theta ={\tan }^{-1}\left(9\right)$

At $(2,6)$
$y=x^3-3x+4\Rightarrow \frac{dy}{dx}=3x^2-3\Rightarrow m_1=9$
$y=3(x^2-x)\Rightarrow \frac{dy}{dx}=3(2x-1)\Rightarrow m_2=9$
So, both the curves touch each other at $(2,6)$ $\because m_1=m_2$
$\therefore$Equation of common tangent is $(y-6)=9(x-2)\Rightarrow y=9x-12$