Question

The angle between the curves $ y={{a}^{x}} $ of 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) $
• E. $ {{\tan }^{-1}}\left( \left| \frac{\log a-\log b}{1+\log a\log b} \right| \right) $

Answer 1

Answer: (E)

The point of intersection of given curves 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\text{ }a,\text{ }{{m}_{2}}=log\text{ }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) $