A) \[\frac{\log a-\log b}{1+\log a\log b}\]
B) \[\frac{\log a+\log b}{1-\log a\log b}\]
C) \[\frac{\log a-\log b}{1-\log a\log b}\]
D) None of these
Correct Answer: A
Solution :
The angle between the two curves is the angle between their tangents at their point of contact. \[\therefore \]The point of intersection of two curves is\[{{a}^{x}}={{b}^{x}}\] For distinct values of a and b, the equality is true, if\[x=0\] Now, if\[x=0\], then\[y={{a}^{0}}=1\] \[\therefore \]The two curves intersect at (0, 1). Let\[{{m}_{1}}\]and\[{{m}_{2}}\]be the slope of the tangent to the curve \[y={{a}^{x}}\] and\[y={{b}^{x}}\] respectively at\[P(0,\,\,1)\]. \[\therefore \] \[{{m}_{1}}={{\left( \frac{dy}{dx} \right)}_{p}}={{({{a}^{x}}\log a)}_{(0,\,\,1)}}=\log a\] \[{{m}_{2}}={{\left( \frac{dy}{dx} \right)}_{p}}={{({{b}^{x}}\log b)}_{(0,\,\,1)}}=\log b\] If\[\alpha \]be the angle of intersection of the tangents. Then, \[\tan \alpha =\left| \frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|\] \[\Rightarrow \] \[\tan \alpha =\left| \frac{\log a-\log b}{1+\log a\log b} \right|\]
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