Question

If the acute angles between the curves $y=|x^2-1|$ and $y=|x^2-3|$ at their points of intersection be $\theta$ such that $\tan \theta =\frac{m}{7}$ then $m^2$ is equal to :

• A. $6$
• B. $8$
• C. $10$
• D. $32$

Answer 1

The correct option is D $32$

We have,

$\begin{array}{c}y=\left|x^2-1\right|y=\left|x^2-3\right|\\ Pointofintersectionofcurves{x}^2=2\\ x=\pm \sqrt{2}\\ Now,\\ y=x^2-1\\ \frac{dy}{dx}=2x\\ {\frac{dy}{dx}}_{x=\sqrt{2}}=2\sqrt{2}\\ and,\\ y=3-x^2\\ \frac{dy}{dx}=-2x\\ {\frac{dy}{dx}}_{x=\sqrt{2}}=-2\sqrt{2}\\ Now,\\ \tan \theta =\left|\frac{m_1-m_2}{1+m_{1}ml2}\right|\\ =\left|\frac{2\sqrt{2}+2\sqrt{2}}{1-8}\right|\\ \tan \theta =\frac{4\sqrt{2}}{7}\\ m=4\sqrt{2}\\ m^2=16\times 2=32\\ Hence,theoptionDisthecorrectanswer.\end{array}$