Question

Find the range of:
$f(x) = cos x + cos (x+\frac{\pi}{6} )+3$
Use closed and open brackets for representing the range.

Answer 1

Given:

↬ Function $\bf{f(x)=cosx+cos\Big(x+\dfrac{\pi }{6}\Big)+3}$

To Find:

• Range of given function

Things to know before solving Question

➺ cos(A+B)= cosAcosB - sinAsinB

➺ Range of function asinθ ± bcosθ

✏ $\Big[-\sqrt{a^{2}+b^{2}} , \sqrt{a^{2}+b^{2}}\:\Big]$

Solution:

Given function,

$\longrightarrow\sf{f(x)=cosx+cos\Big(x+\dfrac{\pi }{6}\Big)+3}$

$\longrightarrow\sf{f(x)=cosx+\Big(\cos x\cos\dfrac{\pi }{6}-\sin x\sin\dfrac{\pi }{6}\Big)+3}$

$\longrightarrow\sf{f(x)=cosx+\Big(\cos x\dfrac{\sqrt{3}}{2}-\sin x\dfrac{1}{2}\Big)+3}$

$\longrightarrow\sf{f(x)=cosx+\cos x\dfrac{\sqrt{3}}{2}-\sin x\dfrac{1}{2}+3}$

$\longrightarrow\sf{f(x)=\bigg(1+\dfrac{\sqrt{3}}{2}\bigg)\cos x-\dfrac{1}{2}\sin x+3}$

$\longrightarrow\sf{f(x)=\bigg(\dfrac{2+\sqrt{3}}{2}\bigg)\cos x-\dfrac{1}{2}\sin x+3}$

Now,

$\sf{Let\:g(x)=-\dfrac{1}{2}\sin x+\bigg(\dfrac{2+\sqrt{3}}{2}\bigg)\cos x}$

So,

$\sf{Range\:of\:g(x)=\Big[-\sqrt{\bigg(\dfrac{-1}{2}\bigg)^{2}+\bigg(\dfrac{2+\sqrt{3}}{2}\bigg)^{2}} , \sqrt{\bigg(\dfrac{-1}{2}\bigg)^{2}+\bigg(\dfrac{2+\sqrt{3}}{2}\bigg)^{2}}\:\Big]}$

$\sf{=\Big[-\sqrt{\dfrac{1}{4}+\dfrac{(2+\sqrt{3})^{2}}{4}} , \sqrt{\dfrac{1}{4}+\dfrac{(2+\sqrt{3})^{2}}{4}}\:\Big]}$

$\sf{=\Big[-\sqrt{\dfrac{1}{4}+\dfrac{2^{2}+(\sqrt{3})^{2}+2(2)(\sqrt{3})}{4}} , \sqrt{\dfrac{1}{4}+\dfrac{2^{2}+(\sqrt{3})^{2}+2(2)(\sqrt{3})}{4}}\:\Big]}$

$\sf{=\Big[-\sqrt{\dfrac{1}{4}+\dfrac{4+3+4\sqrt{3}}{4}} , \sqrt{\dfrac{1}{4}+\dfrac{4+3+4\sqrt{3}}{4}}\:\Big]}$

$\sf{=\Big[-\sqrt{\dfrac{1+4+3+4\sqrt{3}}{4}} , \sqrt{\dfrac{1+4+3+4\sqrt{3}}{4}}\:\Big]}$

$\sf{=\Big[-\sqrt{\dfrac{8+4\sqrt{3}}{4}} , \sqrt{\dfrac{8+4\sqrt{3}}{4}}\:\Big]}$

$\sf{=\Big[-\sqrt{\dfrac{\cancel{4}(2+\sqrt{3})}{\cancel{4}}} , \sqrt{\dfrac{\cancel{4}(2+\sqrt{3})}{\cancel{4}}}\:\Big]}$

$\sf{=\Big[-\sqrt{2+\sqrt{3}} , \sqrt{2+\sqrt{3}}\:\Big]}$

Also, we know that here

f(x)= g(x)+3

Now,

$\longrightarrow\sf{-\sqrt{2+\sqrt{3}}\leq g(x)\leq \sqrt{2+\sqrt{3}}}$

On adding 3 to all sides, we get

$\longrightarrow\sf{-\sqrt{2+\sqrt{3}}+3\leq g(x)+3\leq \sqrt{2+\sqrt{3}}+3}$

$\longrightarrow\sf{-\sqrt{2+\sqrt{3}}+3\leq f(x)\leq \sqrt{2+\sqrt{3}}+3}$

So,

$\sf{Range\:of\:f(x)=\pink{\Big[-\sqrt{2+\sqrt{3}}+3, \sqrt{2+\sqrt{3}}+3\:\Big]}}$

Answer 2

Question :

$\bf{Range\:\:of\::\:\:f\left(x\right)\:=\:cosx\:+\:cos\left(x+\dfrac{\pi }{6}\right)\:+\:3}$

Answer :

$\bigstar\bf{\:Combine\:the\:ranges\:of\:all\:domain\:intervals\:to\:obtain\:the\:function\:range}\:$

${\frac{\sqrt{2+\sqrt{3}}}{2}+\frac{-\sqrt{3}\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{4}+3\le \:f\left(x\right)\le \frac{\sqrt{2+\sqrt{3}}}{2}+\cos \left(2\pi -\arccos \left(\dfrac{\sqrt{2+\sqrt{3}}}{2}\right)+\frac{\pi }{6}\right)+3}$

Additional Information :

$\bf{Function\:range} :$

The range of a function is defined as the set of resulting values of the dependent variable

$\mathrm{Domain\:of\:}\:\cos \left(x\right)+\cos \left(x+\frac{\pi }{6}\right)+3\::\quad -\infty \:<x<\infty \:$