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