Question

Let $A=\{x:[5\sin x]+[\cos x]+6=0,x\in R\}$, where $[.]$ represents the greatest integer function. If $f\left(x\right)=\sqrt{3}\sin x+\cos x\forall x\in A$, then

• A. value of $f\left(x\right)$ is less than $\tan \frac{2\pi }{3}$
• B. value of $f\left(x\right)$ is less than $2\cos \left(\pi \right)$
• C. value of $f\left(x\right)$ is more than $\frac{4-3\sqrt{3}}{5}$
• D. value of $f\left(x\right)$ is more than $\frac{-3-4\sqrt{3}}{5}$

Answer 1

The correct option is A value of $f\left(x\right)$ is less than $\tan \frac{2\pi }{3}$

$\left[5\sin x\right]+\left[\cos x\right]+6=0$

We have,

$-5\le \left[5\sin x\right]\le 5$ and $-1\le \left[\cos x\right]\le 1$.

Hence,

$\left[5\sin x\right]=-5$ and $\left[\cos x\right]=-1$

$\Rightarrow -5<5\sin x<-4$ and $-1<\cos x<0$

So $x$ must lie in the third quadrant. Hence,

$\pi +{\sin }^{-1}\left(\frac{4}{5}\right)<x<\frac{3\pi }{2}$

Now,

$f\left(x\right)=\sqrt{3}\sin x+\cos x$

$\Rightarrow f\left(x\right)=2\sin (x+\frac{\pi }{6})$

$f^{\prime }\left(x\right)=2\cos (x+\frac{\pi }{6})$

$f^{\prime }\left(x\right)=0$ $\Rightarrow 2\cos (x+\frac{\pi }{6})=0$

$\Rightarrow x=\frac{4\pi }{3}$

$f^{\prime \prime }\left(\frac{4\pi }{3}\right)>0$

$f\left(\frac{4\pi }{3}\right)=-2$

This is a point of local minima.

Let us consider the endpoints.

$f\left(\frac{3\pi }{2}\right)=-\sqrt{3}=\tan \frac{2\pi }{3}$

$f(\pi +{\sin }^{-1}\frac{4}{5})=\frac{-3-4\sqrt{3}}{5}$

$-2<f(\pi +{\sin }^{-1}\frac{4}{5})<-\sqrt{3}$

Hence, for $x\forall A$, $-2<f\left(x\right)<-\sqrt{3}$