# Describe $f(S)$ where S is the solution set of equation $\lfloor5\sin x\rfloor+\lfloor\cos x\rfloor+6=0$

If $$\lfloor 5\sin x\rfloor+\lfloor \cos x\rfloor+6=0$$ then the solution set of range of $$f(x)=\sqrt{3}\cos x+\sin x$$

I tried solving this question by doing the following $$5\sin x$$ ranges from $$[-5,5]$$ So $$\lfloor 5 \sin x\rfloor$$ has the range $$-5,-4,-3,-2,-1,1,2,3,4,5$$.

Similar logic for $$\lfloor\cos x\rfloor$$; $$-5$$ and $$-1$$ satisfies $$\lfloor 5\sin x \rfloor+ \lfloor\cos x\rfloor+6=0$$

But substituting these values in the second equation spent yield me the answer

But seems like I'm totally wrong as the correct answer would be $$\left(\frac{3×\sqrt3+4}{5},-1\right)$$

I think Im missing something conceptual here.

Could anyone help me out?

Note :

$$\lfloor \cdot \rfloor$$ indicates floor function

• It is possible only when $\lfloor 5 \sin x\rfloor =-5$ and $\lfloor \cos x\rfloor =-1$ Apr 7 at 13:43
• @mathophile Ya sorry i wrote mistakenly incomplete. Infact i did find that Apr 7 at 13:44
• We assume $[y]$ is meant to be the floor function? Apr 7 at 13:47
• Hi Elizabeth - you can use \lfloor and \rfloor to represent the floor function - usually "[$\cdot$]" means the integer function. Apr 7 at 14:14
• @student91 'JEE Advanced' Exam. Heard of it? Apr 11 at 10:01

I think you are misinterpreting the question (Its wording is a bit weird imo). I would word it like this

When $$x$$ satisfies $$\lfloor 5\sin x\rfloor+\lfloor \cos x\rfloor+6=0,$$ then what values can the function $$f(x)=\sqrt 3\cos x+\sin x$$ obtain for these $$x$$.

Here is an image showing the situation for the interval $$[0,2\pi]$$: In this picture, we have:

• Orange region is where $$\lfloor\cos x\rfloor = -1$$.
• Red region is where $$\lfloor 5\sin x\rfloor = -5$$.
• Green region is where $$\lfloor 5\sin x\rfloor+\lfloor\cos x\rfloor+6=0$$.

And you can solve it as follows:

Step 1: Determine the possible values of $$x$$: First we assume that $$x\in[0,2\pi]$$ because everything is periodic.

We need that $$\sin x< -\frac45$$ and $$\cos x< 0$$, because then $$\lfloor 5\sin x\rfloor=-5$$ and $$\lfloor \cos x\rfloor=-1$$ so the equation is satisfied. Clearly, $$\cos x< 0$$ gives $$\frac\pi2< x<\frac{3\pi}2$$ (The orange region). You can also solve $$\sin x<-\frac45$$ to obtain $$2\pi-\sin^{-1}\frac45< x<\pi+\sin^{-1}\frac45$$ (the red region).

So we find that $$x$$ is in the interval $$\left(2\pi-\sin^{-1}\frac45,\frac{3\pi}2\right)$$ (The green region).

Step 2: Find the values that $$f$$ takes in this interval. First we look at the value of $$f$$ at the boundary points $$2\pi-\sin^{-1}\frac45$$ and $$\frac{3\pi}2$$. We have that $$f\left(2\pi-\sin^{-1}\frac45\right) = -\sqrt 3\sqrt{1-\left(-\frac45\right)^2}-\frac45=-\frac{3\sqrt 3+4}5$$ and $$f(\frac{3\pi}2)=-1$$, so we at least know that the range contains these values.

Now we have to see if the function $$f$$ has a minimum or maximum in this interval. It does not (we see this in the image), so we are done.

Since $$-1\leqslant \lfloor\cos x\rfloor$$ always holds, we have to have $$-1\leqslant -6-\lfloor 5\sin x\rfloor$$, i.e. $$\lfloor 5\sin x\rfloor \leqslant -5$$.

Since $$-5\leqslant \lfloor 5\sin x\rfloor$$ always holds, we have to have $$\lfloor 5\sin x\rfloor =-5\iff -1\leqslant \sin x\lt -\frac 45\tag1$$ and $$\lfloor\cos x\rfloor=-1\iff -1\leqslant \cos x\lt 0\tag2$$ These are sufficient.

Solving $$(1)(2)$$ for $$0\leqslant x\lt 2\pi$$, we get $$\alpha\lt x\lt\dfrac{3\pi}{2}$$ where $$\sin\alpha=-\dfrac 45$$ with $$\dfrac{7}{6}\pi\lt \alpha\lt\dfrac{3}{2}\pi$$ since $$\sin\alpha=-\dfrac 45\lt -\dfrac 12=\sin\dfrac{7\pi}{6}$$.

We have $$f(x)=\sqrt 3\cos x+\sin x=2\sin\bigg(x+\frac{\pi}{3}\bigg)$$ and $$\bigg(\frac{3\pi}{2}\lt\bigg)\alpha+\frac{\pi}{3}\lt x+\frac{\pi}{3}\lt\frac{11\pi}{6}\bigg(\lt 2\pi\bigg)$$ Since $$y=\sin x$$ is strictly increasing for $$\dfrac{3\pi}{2}\lt x\lt 2\pi$$, the range is $$2\sin\bigg(\alpha+\frac{\pi}{3}\bigg)\lt f(x)\lt 2\sin\bigg(\frac{11\pi}{6}\bigg)$$ i.e. $$2\bigg(\sin\alpha\cos\frac{\pi}{3}+\cos\alpha\sin\frac{\pi}{3}\bigg)\lt f(x)\lt -1$$ i.e. $$2\bigg(-\frac 45\cdot\frac 12-\sqrt{1-\bigg(-\frac 45\bigg)^2}\cdot\frac{\sqrt 3}{2}\bigg)\lt f(x)\lt -1$$ i.e. $$\color{red}{\frac{-4-3\sqrt 3}{5}\lt f(x)\lt -1}$$