Question

If we have an expression as $\left[ 2\cos x \right]+\left[ \sin x \right]=-3$ , then the range of the function, $f\left( x \right)=\sin x+\sqrt{3}\cos x$ in $\left[ 0,2\pi \right]$ where [] denotes the greatest integer function.
A. $\left[ -\sqrt{3},\sqrt{3} \right]$
B. $\left[ -2,\sqrt{3} \right]$
C. $\left[ -3,-1 \right]$
D. $\left[ -2,-\sqrt{3} \right]$

Answer 1

We have been given the equation $\left[ 2\cos x \right]+\left[ \sin x \right]=-3$. We will use this equation and find the domain in which x lies in $\left[ 0,2\pi \right]$. Then that domain will be the domain for that given function f(x). Then, as f(x) is an equation in both sin and cos, we will change it into an only cos or only sin equation and put the values of the endpoints of our domain in the then obtained function. Using that, we will find the required range.

Complete step by step solution:
Now, we have been given that $\left[ 2\cos x \right]+\left[ \sin x \right]=-3$. Since, both the cos and sin functions have their values between $\left[ -1,1 \right]$ , the value of $2\cos x$ will lie between $\left[ -2,2 \right]$ .
So, the given condition is only possible if $\left[ 2\cos x \right]=-2$ and $\left[ \sin x \right]=-1$ .
Now, we know that if $\left[ x \right]=t$ where x is any real number and t is an integer, $x\in [t,t+1)$ . From this we can find the domain for x.
Thus, we have:
$\begin{aligned} & \left[ \sin x \right]=-1 \\\ & \Rightarrow -1\le \sin x < 0 \\\ \end{aligned}$
Thus, in $\left[ 0,2\pi \right]$, we have the value of x in the domain:
$x\in \left( \pi ,2\pi \right)$ ......(i)
Now, we also have:
$\begin{aligned} & \left[ 2\cos x \right]=-2 \\\ & \Rightarrow -2\le 2\cos x < -1 \\\ & \Rightarrow -1\le \cos x < \dfrac{-1}{2} \\\ \end{aligned}$
Thus, in $\left[ 0,2\pi \right]$, we have the value of x in the domain:
$x\in \left( \dfrac{2\pi }{3},\dfrac{4\pi }{3} \right)$ .....(ii)
Now, the required domain for ‘x’ will be the intersection of the domain we found out in the two cases.
Thus, we have:
$\begin{aligned} & x\in \left( \pi ,2\pi \right)\cap \left( \dfrac{2\pi }{3},\dfrac{4\pi }{3} \right) \\\ & \Rightarrow x\in \left( \pi ,\dfrac{4\pi }{3} \right) \\\ \end{aligned}$
Thus the range of x is given as:
$x\in \left( \pi ,\dfrac{4\pi }{3} \right)$
Now, we have $f\left( x \right)=\sin x+\sqrt{3}\cos x$
We need to singly change our given function the form of either only a sine function or only a cosine function. Here, we will change it in the sine function.
Now, we can do that transformation by multiplying the numerator and the denominator of the function by the square root of the sum of the square of the coefficients of $\sin x$ and $\cos x$ in the function.
Doing the above mentioned, we get:
$\begin{aligned} & f\left( x \right)=\sin x+\sqrt{3}\cos x \\\ & \Rightarrow f\left( x \right)=\sqrt{{{1}^{2}}+{{\left( \sqrt{3} \right)}^{2}}}\dfrac{\left( \sin x+\sqrt{3}\cos x \right)}{\sqrt{{{1}^{2}}+{{\left( \sqrt{3} \right)}^{2}}}} \\\ & \Rightarrow f\left( x \right)=\sqrt{1+3}\dfrac{\left( \sin x+\sqrt{3}\cos x \right)}{\sqrt{1+3}} \\\ \end{aligned}$
$\begin{aligned} & \Rightarrow f\left( x \right)=\sqrt{4}\dfrac{\left( \sin x+\sqrt{3}\cos x \right)}{\sqrt{4}} \\\ & \Rightarrow f\left( x \right)=2\left( \dfrac{\sin x}{2}+\dfrac{\sqrt{3}\cos x}{2} \right) \\\ & \Rightarrow f\left( x \right)=2\left( \dfrac{1}{2}\sin x+\dfrac{\sqrt{3}}{2}\cos x \right) \\\ \end{aligned}$
Now, we know that $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$. Thus, we can change our given function in the form of $\sin \left( A+B \right)$ .
Now, we know that $\cos \dfrac{\pi }{3}=\dfrac{1}{2}$ and $\sin \dfrac{\pi }{6}=\dfrac{\sqrt{3}}{2}$ . Thus, we can put these values our function and make the required transformations.
As a result, we will get:
$\begin{aligned} & \Rightarrow f\left( x \right)=2\left( \dfrac{1}{2}\sin x+\dfrac{\sqrt{3}}{2}\cos x \right) \\\ & \Rightarrow f\left( x \right)=2\left( \cos \dfrac{\pi }{3}\sin x+\sin \dfrac{\pi }{3}\cos x \right) \\\ & \Rightarrow f\left( x \right)=2\left( \sin \left( x+\dfrac{\pi }{3} \right) \right) \\\ & \Rightarrow f\left( x \right)=2\sin \left( x+\dfrac{\pi }{3} \right) \\\ \end{aligned}$
Now, we have:
$\begin{aligned} & \pi < x < \dfrac{4\pi }{3} \\\ & \Rightarrow \pi +\dfrac{\pi }{3} < x+\dfrac{\pi }{3} < \dfrac{4\pi }{3}+\dfrac{\pi }{3} \\\ & \Rightarrow \dfrac{4\pi }{3} < x+\dfrac{\pi }{3} < \dfrac{5\pi }{3} \\\ \end{aligned}$
Thus, we can find the value of the function as:
$\dfrac{4\pi }{3} < x+\dfrac{\pi }{3} < \dfrac{5\pi }{3}$

& \Rightarrow \sin \dfrac{4\pi }{3} < \sin \left( x+\dfrac{\pi }{3} \right) < \sin \dfrac{5\pi }{3} \\\ & \Rightarrow 2\sin \dfrac{4\pi }{3} < 2\sin \left( x+\dfrac{\pi }{3} \right) < 2\sin \dfrac{5\pi }{3} \\\ & \Rightarrow 2\sin \dfrac{4\pi }{3} < f\left( x \right) < 2\sin \dfrac{5\pi }{3} \\\ \end{aligned}$$ Now, we know that the values of angles (in radians) between $$\dfrac{4\pi }{3}$$ and $$\dfrac{5\pi }{3}$$ lie between the third and fourth quadrants. Thus, we will have: $$\begin{aligned} & \Rightarrow 2\sin \dfrac{4\pi }{3} < f\left( x \right) < 2\sin \dfrac{5\pi }{3} \\\ & \Rightarrow 2\left( -1 \right) < f\left( x \right) < 2\left( -\dfrac{\sqrt{3}}{2} \right) \\\ & \Rightarrow -2 < f\left( x \right) < -\sqrt{3} \\\ \end{aligned}$$ Thus, the range of the function is $\left[ -2,-\sqrt{3} \right]$ **Hence, option (D) is the correct option.** **Note:** Here, we transformed our function into an only sine function. But it’s not necessary that we change that into sine only. We also could have changed the function into an only cosine function but we chose to change in into sine because this will change the function into a sum of angles in sine but to a difference of angles in cosine (as $\cos \left( A-B \right)=\cos A\cos B+ \sin A\sin B$ ) and it sum of angles is comparatively easy but there’s not any significant difference so you can choose either.