Question

If $\sin \theta +\sqrt{3}\cos \theta =6x-x^2-11$, where $0\le \theta \le 4\pi ,x\in R$, then which of the following options is/are correct ?

• A. ${\theta }_{min}=\frac{7\pi }{6}$
• B. ${\theta }_{max}=\frac{19\pi }{6}$
• C. There are two pairs of values for $(x,\theta )$
• D. There are four pairs of values for $(x,\theta )$

Answer 1

Answer: (A), (B), (C)

There are two pairs of values for $(x,\theta )$

$\sin \theta +\sqrt{3}\cos \theta =-(x^2-6x+11)$
$\Rightarrow \sin \theta +\sqrt{3}\cos \theta =-\left[\right(x-3{)}^2+2]$
Range of L.H.S. is $[-2,2]$
Range of R.H.S. is $(-\infty ,-2]$
So the equality will hold only when
L.H.S. $=$ R.H.S. $=-2$
For L.H.S.$=-2$
$\Rightarrow \sin \theta +\sqrt{3}\cos \theta =-2$
$\Rightarrow \frac{1}{2}\sin \theta +\frac{\sqrt{3}}{2}\cos \theta =-1$
$\Rightarrow \cos (\theta -\frac{\pi }{6})=-1$
$\Rightarrow \theta -\frac{\pi }{6}=(2n+1)\pi$
$\Rightarrow \theta =\frac{\pi }{6}+(2n+1)\pi$

Now, for $0\le \theta \le 4\pi$
$\theta =\frac{7\pi }{6},\frac{19\pi }{6}$
For R.H.S. $=-2$
$-\left[\right(x-3{)}^2+2]=-2$
$\Rightarrow x=3$
$\therefore (x,\theta )=(3,\frac{7\pi }{6}),(3,\frac{19\pi }{6})$