Question

Assertion :If $f\left(x\right)={\int }_1^x\frac{lntdt}{1+t+t^2}(x>0)$ then $f\left(x\right)=-f\left(\frac{1}{x}\right)$ Reason: If $f\left(x\right)={\int }_1^x\frac{lntdt}{t+1},$ then $f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{1}{2}{\left(lnx\right)}^2$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is incorrect but Reason is correct
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is incorrect but Reason is correct

$f\left(\frac{1}{x}\right)={\int }_1^{\frac{1}{x}}\frac{\log t}{1+t+t^2}dt$
Substitute $t=\frac{1}{z}\Rightarrow dt=-\frac{1}{z^2}dz$
$\therefore f\left(\frac{1}{x}\right)={\int }_1^x\frac{\log \frac{1}{z}}{1+\frac{1}{z}+\frac{1}{z^2}}\left(\frac{-dz}{z^2}\right)$
$={\int }_1^x\frac{\log zdz}{z^2+z+1}={\int }_1^x\frac{\log tdt}{1+t+t^2}=f\left(x\right)$