Question

If $f\left(x\right)=\left|\log x\right|$, then

• A. $y^{\prime }\left(1^{+}\right)=1$
• B. $y^{\prime }\left(1^{-}\right)=-1$
• C. $y^{\prime }\left(1\right)=2$
• D. $y^{\prime }\left(1\right)=-2$

Answer 1

Answer: (A), (B)

The correct options are
A $y^{\prime }\left(1^{+}\right)=1$
B $y^{\prime }\left(1^{-}\right)=-1$

We have, $f\left(x\right)=|\log x|=\{\begin{array}{c}-\log x,0<x\le 1\\ \log x,x>1\end{array}$
$\Rightarrow f^{\prime }\left(x\right)=\{\begin{array}{c}-\frac{1}{x},0<x\le 1\\ \frac{1}{x},>1\end{array}$
Hence $y^{\prime }\left(1^{+}\right)=1$ and $y^{\prime }\left(1^{-}\right)=-1$