Applications Of Derivatives Question 136
Question: If the curve $ y=a^{x} $ and $ y=b^{x} $ intersect at angle $ \alpha $ then, $ \tan \alpha = $
[MP PET 2001]
Options:
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} $
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Answer:
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 $
therefore $ {{( \frac{dy}{dx} )} _{(0,,1)}}=m_1=\log a $
Slope of tangent of second curve $ m_2=\frac{dy}{dx}=b^{x}\log b $
therefore $ m_2={{( \frac{dy}{dx} )} _{(0,,1)}}=\log b $ \ $ \tan \alpha =\frac{m_1-m_2}{1+m_1m_2}=\frac{\log a-\log b}{1+\log a\log b} $ .
BETA
