Question

Assertion (A ): For $x<0, \frac{d^{2}}{d x^{2}}(\log |x|)=\frac{1}{|x|^{2}}$
Reason (R) : For $x<0,|x|=-x$

• A. (A) is true, (R) is true and (R) is the correct explanation to (A)
• B. (A) is true, (R) is true but (R) is not a correct explanation to (A)
• C. (A) is true, (R) is false
• D. (A) is false, (R) is true

Answer 1

Answer: (D)

Let $f(x)=\log |x|$
$=\begin{cases}\log (-x), & x<0 \\ \log x, & x \geq 0 .\end{cases}$
$\therefore f'(x)=\begin{cases}\frac{-1}{x}(-1), & x<0 \\ \frac{1}{x}, & x \geq 0\end{cases}=\begin{cases}\frac{1}{x}, & x<0 \\ \frac{1}{x}, & x \geq 0 .\end{cases}$
and $f''(x)=\begin{cases}\frac{-1}{x^{2}}, & x<0 \\ \frac{-1}{x^{2}}, & x \geq 0\end{cases}$
$\therefore f''(x)=\frac{-1}{x^{2}}=\frac{-1}{|x|^{2}}$
So, $A$ is false.
We know that, $|x|=-x$,
when $x< 0$
So, $R$ is true.