Question

The angle between the curves $y=a^x$ and $y=b^x$ is equal to

• A. $\tan ^{-1}\left(\left|\frac{a-b}{1+a b}\right|\right)$
• B. $\tan ^{-1}\left(\mid \frac{a+b}{1-a b}\right)$
• C. $\tan ^{-1}\left(\left|\frac{\log b+\log a}{1+\log a \log b}\right|\right)$
• D. $\tan ^{-1}\left(|\frac{\log a-\log b}{1+\log a \log b}|\right)$

Answer 1

Answer: (D)

The point of intersection of given curves is $(0,1)$.
On differentiating given curves, we get
$\frac{d y}{d x} =a^x \log a, \frac{d y}{d x}=b^x \log b $
$\Rightarrow m_1 =a^x \log a, m_2=b^x \log b $
At $ (0,1), m_1 =\log a, m_2=\log b $
$\therefore \tan \theta =\left|\frac{m_1-m_2}{1+m_1 m_2}\right|=\left|\frac{\log a-\log b}{1+\log a \log b}\right| $
$\Rightarrow \theta =\tan ^{-1}\left|\frac{\log a-\log b}{1+\log a \log b}\right|$