Question

Assertion : $\int_{-2}^{2} \log \left(\frac{1+x}{1-x}\right) d x=0$
Reason : If $f$ is an odd function, then $\int_{-a}^{a} f(x) d x=0$.

• A. Assertion is correct, Reason is correct; Reason is a correct explanation for assertion
• B. Assertion is correct, Reason is correct; Reason is not a correct explanation for Assertion
• C. Assertion is correct, Reason is incorrect
• D. Assertion is incorrect, Reason is correct

Answer 1

Answer: (A)

$
\begin{array}{l}
\int_{-2}^{2} \log \left(\frac{1+x}{1-x}\right) d x=0 \\
f(x)=\log \left(\frac{1+x}{1-x}\right) \\
f(-x)=\log \left(\frac{1-x}{1+x}\right)=-\log \left(\frac{1+x}{1-x}\right)=-f(x) \\
f \text { is an odd function } \Rightarrow \int_{-a}^{a} f(x) d x=0
\end{array}
$
Both are true and Reason is correct explanation of Assertion.