Question

Let x belongs to (0, 1), what is the relationship between x and sinx.

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Answer 1

$\large\underline{\sf{Solution-}}$

As given that

$\rm :\longmapsto\:x \: \in \: (0, \: 1)$

We know,

$\red{\boxed{ \rm{ \: 1 \: radian \: \approx \: 57 \degree \: \: }}}$

$\bf\implies \:x \: \in \: {1}^{st} \: quadrant$

Now, to find the relationship between x and sinx, Let assume that,

$\rm :\longmapsto\:f(x) = x - sinx$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:f'(x) = 1 - cosx$

$\rm :\longmapsto\:f'(x) = {2cos}^{2}\bigg[\dfrac{x}{2} \bigg]$

$\rm :\longmapsto\:f'(x) \geqslant 0$

$\bf\implies \:f(x) \: is \: increasing$

By definition of increasing function,

$\rm :\longmapsto\:x > 0$

$\bf\implies \:f(x) > f(0)$

$\bf\implies \:x - sinx > 0 - sin0$

$\bf\implies \:x - sinx > 0$

$\bf\implies \:x > sinx$

Basic Concepts

1. For decreasing function,

$\red{\boxed{ \rm{ \: x \geqslant y \: \implies \: f(x) \leqslant f(y)}}}$

$\red{\boxed{ \rm{ \: x \leqslant y \: \implies \: f(x) \geqslant f(y)}}}$

$\red{\boxed{ \rm{ \: x < y \: \implies \: f(x) > f(y)}}}$

$\red{\boxed{ \rm{ \: x > y \: \implies \: f(x) < f(y)}}}$

2. For increasing function

$\red{\boxed{ \rm{ \: x > y \: \implies \: f(x) > f(y)}}}$

$\red{\boxed{ \rm{ \: x <y \: \implies \: f(x) < f(y)}}}$

$\red{\boxed{ \rm{ \: x \leqslant y \: \implies \: f(x) \leqslant f(y)}}}$

$\red{\boxed{ \rm{ \: x \geqslant y \: \implies \: f(x) \geqslant f(y)}}}$