Q. If $I_{1} = \int_{0}^{1}{\frac{1}{|x|}\ dx}$ and $I_{2} = \int_{0}^{1}{\frac{1}{\sqrt{1 + x^{2}}}\ dx,}$ then
A) $I_{1} = I_{2}$
B) $I_{1} < I_{2}$
C) $I_{1} > I_{2}$
D) $\frac{I_{1}}{I_{2}} = 2$
Explanation
(c)
$|x| < \sqrt{1 + x^{2}}\ \ \Rightarrow \ \frac{1}{|x|} > \frac{1}{\sqrt{1 + x^{2}}}$
$\therefore\ \int_{0}^{1}{\frac{1}{|x|}dx >}\int_{0}^{1}{\frac{1}{\sqrt{1 + x^{2}}}dx}$
$\Rightarrow \ I_{1} > I_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.