Question

range of f(x)=1/1-2cosx is​

Answer 1

Given,

$\longrightarrow f(x)=\dfrac{1}{1-2\cos x}$

We need to find its range.

Let,

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

$\longrightarrow y-2y\cos x=1$

$\longrightarrow 2y\cos x=y-1$

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

We know that,

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

So from (1),

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

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

Subtracting 1,

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

Multiplying by -1, (note the interval limit change)

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

or,

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

Taking reciprocal, (note the interval limit change)

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

That is,

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

Hence range of our function is,

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

Answer 2

Answer:

Given, f(x)=11−2cosx

We know that, −1≤cosx≤1

⇒−2≤2cosx≤2

⇒−2≤−2cosx≤2

⇒1−2≤1−2cosx≤2+1

⇒−1≤1−2cosx≤3

So, range is [-1,3]