Question

Let $I={\int }_0^1\frac{sinx}{\sqrt{x}}$ and $J={\int }_0^1\frac{cosx}{\sqrt{x}}dx$. Then which one of the following is true

• A. $I>\frac{2}{3}$ and $J>2$
• B. $I<\frac{2}{3}$ and $J<2$
• C. $I<\frac{2}{3}$ and $J>2$
• D. $I>\frac{2}{3}$ and $J<2$

Answer 1

The correct option is B $I<\frac{2}{3}$ and $J<2$

We have,

$sinx<x$ for $0<x<1$

$\Rightarrow \frac{sinx}{\sqrt{x}}<\frac{x}{\sqrt{x}}$ for $0<x<1$

$\Rightarrow {\int }_0^1\frac{sinx}{\sqrt{x}}dx<{\int }_0^1\sqrt{x}dx\Rightarrow I<\frac{2}{3}{\left[x^{3/2}\right]}_0^1\Rightarrow I<\frac{2}{3}$

Also,

$cosx<1$ for all x $\in$ (0, 1)

$\Rightarrow \frac{cosx}{\sqrt{x}}<\frac{1}{\sqrt{x}}$ for x $\in$ (0, 1)

$\Rightarrow {\int }_0^1\frac{cosx}{\sqrt{x}}dx<{\int }_0^1\frac{1}{\sqrt{x}}dx\Rightarrow J<2{\left[x^{1/2}\right]}_0^1\Rightarrow J<2$.