Question

For all values of $\theta$, the value of $3–\cos \theta +\cos \left(\theta +\frac{\pi }{3}\right)$ lies in the interval

• A. $[-2,3]$
• B. $[-2,1]$
• C. $[2,4]$
• D. $[1,5]$

Answer 1

The correct option is C

$[2,4]$

Finding the interval of give expression:

Given, $3–\cos \theta +\cos \left(\theta +\frac{\pi }{3}\right)$

$$\begin{gathered} =3–\cos \theta +[\cos \theta \cos \frac{\pi }{3}–\sin \theta \sin \frac{\pi }{3}] \\ =3–\cos \theta +\left[\frac{(\cos \theta )}{2}–\frac{\sqrt{3}(\sin \theta )}{2}\right] \\ =3–\cos \theta +\frac{(\cos \theta )}{2}–\frac{\sqrt{3}\sin \theta }{2} \\ =3–\frac{\cos \theta }{2}–\frac{\sqrt{3}\sin \theta }{2} \end{gathered}$$

We know that $c-\sqrt{a^2+b^2}\le a\cos x+b\sin x+c\le c+\sqrt{a^2+b^2}$

Here, $a=-\frac{1}{2},b=\frac{3}{2},c=3$

Now, $\sqrt{a^2+b^2}=\sqrt{{\left(\frac{-1}{2}\right)}^2+{\left(\frac{-\sqrt{3}}{2}\right)}^2}$

$$\begin{gathered} =\sqrt{\frac{1}{4}+\frac{3}{4}} \\ =\sqrt{1} \\ =1 \end{gathered}$$

And $\begin{array}{rcl}c–\sqrt{a^2+b^2}& =& 3–1\\ & =& 2\end{array}$, $\begin{array}{rcl}c+\sqrt{a^2+b^2}& =& 3+1\\ & =& 4\end{array}$

$\therefore$the interval is $[2,4]$.

Hence, the correct option is C.