Question

$\sin \theta +\sqrt{3}\cos \theta =6x-x^2-11$, $0\le \theta \le 4\pi ,x\in R$, holds for

• A. no values of $x$ and two values of $\theta$
• B. one value of $x$ and two value of $\theta$
• C. two values of $x$ and two values of $\theta$
• D. two point of values of $(x,\theta )$

Answer 1

Answer: (B), (D)

The correct options are
B one value of $x$ and two value of $\theta$
D two point of values of $(x,\theta )$

$\sin \theta +\sqrt{3}\cos \theta =6x-x^2-11$
$\Rightarrow \sin \theta +\sqrt{3}\cos \theta =-2-(x-3{)}^2$
But maximum value of L.H.S is 2 and maximum value of R.H.S is $-2$,
thus solution is only possible if,
$x=3$ and $\sin \theta +\sqrt{3}\cos \theta =-2$
$\Rightarrow \frac{1}{2}\sin \theta +\frac{\sqrt{3}}{2}\cos \theta =-1$
$\Rightarrow \sin \frac{\pi }{6}\sin \theta +\cos \frac{\pi }{6}\cos \theta =-1$
$\Rightarrow \cos (\theta -\frac{\pi }{6})=-1$
therefore solution in the given interval is,
$\theta =\pi +\frac{\pi }{6},3\pi +\frac{\pi }{6}$
Hence, options 'B' and 'D' are correct.