Q. The angle between the curves $y = a^{x}$ and $y = b^{x}$ is equal to
A) $\tan^{- 1}\left( \left| \frac{a - b}{1 + ab} \right| \right)$
B) $\tan^{- 1}\left( \left| \frac{a + b}{1 - ab} \right| \right)$
C) $\tan^{- 1}\left( \left| \frac{\log{b + \log a}}{1 + \log{a\log b}} \right|\ \right)\$
D) $\tan^{- 1}\left( \left| \frac{\log{a - \log b}}{1 + \log{a\log b}} \right|\ \right)\$
Explanation
(d)
The point of intersection of given curve is (0, 1). On differentiating given curves, we get
$\frac{dy}{dx} = a^{x}$log $a,\frac{dy}{dx} = b^{x}$ log $b$
$\Rightarrow \ \ \ \ \ m_{1} = a^{x}$ log $a,m_{2} = b^{x}$log $b$
At (0,1) $m_{1} =$log $a,m_{2} =$log $b$
$\therefore\ \ \ \$tan$\theta = \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}}$
$\Rightarrow \ \ \ \ \ \ \theta = \tan^{- 1}\left| \frac{\log a - \log b}{1 + \log a\log b} \right|$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.