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) \[{{\tan }^{-1}}\left( \left| \frac{\log a-\log b}{1+\log a\log b} \right| \right)\]
Correct Answer: E
Solution :
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), \[{{m}_{1}}=log\text{ }a,\text{ }{{m}_{2}}=log\text{ }b\] \[\therefore \] \[\tan \theta =\frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}\] \[\Rightarrow \] \[\theta ={{\tan }^{-1}}\left( \frac{\log a-\log b}{1+\log a\log b} \right)\]
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