Question

If $f:R\to Y$ defined by $f\left(x\right)=\sqrt{3}\sin x+\cos x+4$ is onto function then Y is .

Answer 1

$f:\to y$ as $f\left(x\right)=\sqrt{3}\sin x+\cos x+4$ is onto.

So, we need to find the range of f.

Now; $\sqrt{3}\sin x+\cos x+4=2\left(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x\right)+4$

Since $\sin \frac{\pi }{3}=\frac{\sqrt{3}}{2}$ and $\cos \frac{\pi }{3}=\frac{1}{2}$ , thus:

$2\left(\frac{\sqrt{3}}{2}\sin x+\frac{1}{2}\cos x\right)=2\cos \left(x-\frac{\pi }{3}\right)$

$f\left(x\right)=2\cos \left(x-\frac{\pi }{3}\right)\le 1$

Then, we have $-1\le \cos \left(x-\frac{\pi }{3}\right)\le 1$

Or ,$-2\le 2\cos \left(x-\frac{\pi }{3}\right)\le 2$

Or, $2\le 2\cos \left(x-\frac{\pi }{3}\right)+4\le 6$

Or, $2\le f\left(x\right)\le 6$

$y=\left[2,6\right]$