Question

If $\sin \theta =a$ for exactly one value of $\theta \in [0,\frac{7\pi }{3}]$ then the value of a is

• A. $\frac{\sqrt{3}}{2}$
• B. 1
• C. 0
• D. -1

Answer 1

Answer: (B), (D)

The correct options are
B 1
D -1

We have, $\theta \in [0,\frac{7\pi }{3}]$

$=[0,2\pi +\frac{\pi }{3}]$

$=[0,2\pi ]+[2\pi ,\frac{7\pi }{3}]$

$=T+[2\pi ,\frac{7\pi }{3}]$ Where T stands for time period of the sinx.

Now $sin\theta$ achieves a minimum value of -1 once in its time period, and maximum value of 1, only once in its time period.

Now the given interval is

$T+[2\pi ,\frac{7\pi }{3}]$

In the interval of $[2\pi ,\frac{7\pi }{3}]$, sinx neither achieves 1 nor achieves -1.

Hence $a=1$ or $a=-1$ for the equation $sin\theta =a$ to have only one solution in the given interval of $\theta$.