Question

Range of f (x) = 1/(1-2 cos ⁡x )is *


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Answer 1

Answer:

æ.......I represent this as infinity

If a>b then 1/a<1/b

Step-by-step explanation:

Range of cosx[-1,1]

So, -1≤cosx≤1

-2≤-2cosx≤2

1-2≤1-2cosx≤1+2

-1≤1-2cosx≤3

-1≤1-2cosx<0. 0≤1-2cosx≤3

1/-1≥1/(1-2cosx)>1/0. 1/0≥1/(1-2cosx)≥1/3

-1≥1/(1-2cosx)>æ. æ≥1/(1-2cosx)≥1/3

x€(-æ,-1]u[1/3,æ)

Answer 2

Let,

$\longrightarrow y=\dfrac{1}{1-2\cos x}$

$\longrightarrow 1-2\cos x=\dfrac{1}{y}$

$\longrightarrow 2\cos x=1-\dfrac{1}{y}$

$\longrightarrow 2\cos x=\dfrac{y-1}{y}$

$\longrightarrow \cos x=\dfrac{y-1}{2y}\quad\quad\dots(1)$

We know the range of $\cos x.$

$\longrightarrow \cos x\in[-1,\ 1]$

From (1),

$\longrightarrow\dfrac{y-1}{2y}\in\left[-1,\ 1]$

Multiply by 2.

$\longrightarrow\dfrac{y-1}{y}\in\left[-2,\ 2]$

$\longrightarrow1-\dfrac{1}{y}\in\left[-2,\ 2]$

Subtract 1.

$\longrightarrow-\dfrac{1}{y}\in\left[-3,\ 1]$

Multiply by -1. [Note the interval limit change]

$\longrightarrow\dfrac{1}{y}\in\left[-1,\ 3]$

Or,

$\longrightarrow\dfrac{1}{y}\in\left[-1,\ 0]\cup[0,\ 3]$

Take the reciprocal. [Note the interval limit change]

$\longrightarrow y\in\left(\dfrac{1}{0},\ \dfrac{1}{-1}\right]\cup\left[\dfrac{1}{3},\ \dfrac{1}{0}\right)$

$\longrightarrow \underline{\underline{f(x)\in\left(-\infty,\ -1\right]\cup\left[\dfrac{1}{3},\ \infty\right)}}$

This is the range of our function.