Question

Let $f(x)=\cos \left(2x+\frac{\pi }{3}\right)+\sin \left(\frac{3x}{2}-\frac{\pi }{4}\right)$. Which of the following statement(s) is/are CORRECT?

• A. Fundamental period of the function $f(x)$ is $4\pi$.
• B. Number of values of $x$ in $[-\pi ,\pi ]$ for which $f(x)=0$, is 4
• C. Number of solutions of the equation $f(x)=2$ which lie in the interval $[0,2\pi ]$, is 1 .
• D. Number of solutions of the equation $f(x)=2$ which lie in the interval $[0,2\pi ]$, is 4 .

Answer 1

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

Obviously the period of $f(x)$ is L.C.M. of $\pi ,\frac{4\pi }{3}$ which is $4\pi$ Ans. $\Rightarrow$ (A)
$\text{ Now }f(x)=0\Rightarrow \cos \left(2x+\frac{\pi }{3}\right)=\sin \left(\frac{\pi }{4}-\frac{3x}{2}\right)=\cos \left(\frac{\pi }{4}+\frac{3x}{2}\right)\Rightarrow 2x+\frac{\pi }{3}=2n\pi \pm \left(\frac{\pi }{4}+\frac{3x}{2}\right)$
With positive sign, $2x+\frac{\pi }{3}=2n\pi +\frac{\pi }{4}+\frac{3x}{2}\Rightarrow \frac{x}{2}=2n\pi -\frac{\pi }{12}\Rightarrow x=4n\pi -\frac{\pi }{6},n=0,\pm 1,\pm 2,\dots$.
For $n=0,x=\frac{-\pi }{6}$ is the only value in $[-\pi ,\pi ]$
[Online test-2, P-2]
For $n=1$, no solution
Now with negative sign, $2x+\frac{\pi }{3}=2n\pi -\frac{\pi }{4}-\frac{3x}{2}\Rightarrow \frac{7x}{2}=2n\pi -\frac{\pi }{4}-\frac{\pi }{3}$
$\Rightarrow \frac{7x}{2}=2n\pi -\frac{7\pi }{12}\Rightarrow x=\frac{4n\pi }{7}-\frac{\pi }{6}$
For $n=0,x=\frac{-\pi }{6};$ For $n=1,x=\frac{4\pi }{7}-\frac{\pi }{6}=\frac{17\pi }{42}$
For $n=2,x=\frac{8\pi }{7}-\frac{\pi }{6}=\frac{41\pi }{42};$ For $n=-1,x=\frac{-4\pi }{7}-\frac{\pi }{6}=\frac{-31\pi }{42}$
Solution set is $\left\{\frac{-\pi }{6},\frac{17\pi }{12},\frac{41\pi }{42},\frac{-31\pi }{42}\right\}$ i.e. 4 Ans. $\Rightarrow$(B)
Now $f(x)=2\Rightarrow \cos \left(2x+\frac{\pi }{3}\right)+\sin \left(\frac{3x}{2}-\frac{\pi }{4}\right)=2$
This is possible only if $\cos \left(2x+\frac{\pi }{3}\right)=1$ and $\sin \left(\frac{3x}{2}-\frac{\pi }{4}\right)=1$
$\Rightarrow 2x+\frac{\pi }{3}=2n\pi ,n\in I\Rightarrow x=n\pi -\frac{\pi }{6}$....(1)
Now $\sin \left(\frac{3x}{2}-\frac{\pi }{4}\right)=1\Rightarrow \frac{3x}{2}-\frac{\pi }{4}=2n\pi +\frac{\pi }{2}\Rightarrow \frac{3x}{2}=2n\pi +\frac{3\pi }{4}\Rightarrow x=\frac{4n\pi }{3}+\frac{\pi }{2}$...(2)
Clearly $x=\frac{11\pi }{6}$ satisfies both the equations (1) and (2). $\Rightarrow$ (C)