Question

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

• A. ${\tan }^{-1}\left(\left|\frac{a-b}{1+ab}\right|\right)$
• B. ${\tan }^{-1}\left(\mid \frac{a+b}{1-ab}\right)$
• 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
$\frac{dy}{dx}=a^x\log a,\frac{dy}{dx}=b^x\log b$
$\Rightarrow m_1=a^x\log a,m_2=b^x\log b$
At $(0,1),m_1=\log a,m_2=\log b$
$\therefore \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\left|\frac{\log a-\log b}{1+\log a\log b}\right|$
$\Rightarrow \theta ={\tan }^{-1}\left|\frac{\log a-\log b}{1+\log a\log b}\right|$