A) \[\frac{a-b}{1+ab}\]
B) \[\frac{\log a-\log b}{1+\log a\log b}\]
C) \[\frac{a+b}{1-ab}\]
D) \[\frac{\log a+\log b}{1-\log a\log b}\]
Correct Answer: B
Solution :
Clearly the point of intersection of curves is (0, 1). Now, slope of tangent of first curve \[{{m}_{1}}=\frac{dy}{dx}={{a}^{x}}\log a\] Þ \[{{\left( \frac{dy}{dx} \right)}_{(0,\,1)}}={{m}_{1}}=\log a\] Slope of tangent of second curve \[{{m}_{2}}=\frac{dy}{dx}={{b}^{x}}\log b\] Þ \[{{m}_{2}}={{\left( \frac{dy}{dx} \right)}_{(0,\,1)}}=\log b\] \ \[\tan \alpha =\frac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}}=\frac{\log a-\log b}{1+\log a\log b}\].
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