Assertion: For x

Question

(A) Note: Assuming the base of the logarithm is $e$.Given:Assertion: For $x < 0$,$\frac{d^2}{d x^2}(\log |x|) = \frac{1}{|x|^2}$.Reason: For $x < 0$,$

Vedclass · Accepted Answer

Assertion is false but Reason is true.

Vedclass · Answer

(A) Note: Assuming the base of the logarithm is $e$.Given:Assertion: For $x Reason: For $x Let $f(x) = \log |x|$.For $x Now,differentiating with respect to $x$:$\frac{d}{d x}(\log(-x)) = \frac{1}{-x} \cdot \frac{d}{d x}(-x) = \frac{1}{-x} \cdot (-1) = \frac{1}{x}$.Now,differentiating again with respect to $x$:$\frac{d^2}{d x^2}(\log |x|) = \frac{d}{d x}(\frac{1}{x}) = -\frac{1}{x^2}$.Since $|x|^2 = (-x)^2 = x^2$,the Assertion states $\frac{d^2}{d x^2}(\log |x|) = \frac{1}{x^2}$,but we found $-\frac{1}{x^2}$.Therefore,the Assertion is false and the Reason is true.

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

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