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Try this problem from ISI-MSQMS 2018 which involves the concept of Real numbers, sequence and series and Definite integral.

## DEFINITE INTEGRAL | ISI 2018| MSQMS | PART A | PROBLEM 22

Let $I=\int_{0}^{1} \frac{\sin x}{\sqrt{x}} d x$ and $J=\int_{0}^{1} \frac{\cos x}{\sqrt{x}} d x,$ then which 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 $J<2$

### Key Concepts

REAL NUMBERS

REIMANN INTEGRATION

SEQUENCE AND SERIES

But try the problem first…

Answer:(d) $I<\frac{2}{3}$ and $J<2$

Source

ISI 2018|MSQMS |QMA|PROBLEM 22

INTRODUCTION TO REAL ANALYSIS :BARTLE SHERBERT

## Try with Hints

First hint

We know when $f(x)$>$g(x)$

$\int \limits_a^bf(x)$>$\int \limits_a^bg(x)$

We know for $0<x<1$, $\cos x <1$

Second hint

$\frac{\cos x}{\sqrt x}$< $\frac{1}{\sqrt x}$ implies $\int \limits_0^1\frac{\cos x}{\sqrt x}\mathrm dx$<$\int \limits_0^1\frac{1}{\sqrt x}\mathrm dx$

$\int \limits_0^1\frac{1}{\sqrt x}\mathrm dx = 2$

$\int \limits_0^1\frac{\cos x}{\sqrt x}\mathrm dx$<$2$

$J$<$2$

Third hint

Again we claim $x-s\sin x$>$0$ for $0 \leq x\leq 1$

Let $f(x)=x-\sin x$

$f'(x)=1-\cos x\geq 0$

hence $f(x)$ is monotonic increasing.

Therefore $x-\sin x$> $0$, $x\epsilon [0,1]$

So,$x$>$sinx$

$\sqrt x$ > $\frac{\sin x}{\sqrt x}$ $x\epsilon [0,1]$

integrating both sides with limits $0$ to $1$ we get;

$\int \limits_0^1\frac{\sin x}{\sqrt x} \mathrm dx$<$\frac{2}{3}$

$I$<$\frac{2}{3}$

Final Step

Therefore,$I<\frac{2}{3}$ and $J<2$