Question

Let $f ( x )=\cos \left(2 x +\frac{\pi}{3}\right)+\sin \left(\frac{3 x }{2}-\frac{\pi}{4}\right)$. 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)
$\text { Now } f(x)=0 \Rightarrow \cos \left(2 x+\frac{\pi}{3}\right)=\sin \left(\frac{\pi}{4}-\frac{3 x}{2}\right)=\cos \left(\frac{\pi}{4}+\frac{3 x}{2}\right) \Rightarrow 2 x+\frac{\pi}{3}=2 n \pi \pm\left(\frac{\pi}{4}+\frac{3 x}{2}\right)$
With positive sign, $2 x +\frac{\pi}{3}=2 n \pi+\frac{\pi}{4}+\frac{3 x }{2} \Rightarrow \frac{ x }{2}=2 n \pi-\frac{\pi}{12} \Rightarrow x =4 n \pi-\frac{\pi}{6}, n =0, \pm 1, \pm 2, \ldots$.
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, $2 x +\frac{\pi}{3}=2 n \pi-\frac{\pi}{4}-\frac{3 x }{2} \Rightarrow \frac{7 x }{2}=2 n \pi-\frac{\pi}{4}-\frac{\pi}{3}$
$\Rightarrow \frac{7 x }{2}=2 n \pi-\frac{7 \pi}{12} \Rightarrow x =\frac{4 n \pi}{7}-\frac{\pi}{6}$
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 $f(x)=2 \Rightarrow \cos \left(2 x+\frac{\pi}{3}\right)+\sin \left(\frac{3 x}{2}-\frac{\pi}{4}\right)=2$
This is possible only if $\cos \left(2 x+\frac{\pi}{3}\right)=1$ and $\sin \left(\frac{3 x}{2}-\frac{\pi}{4}\right)=1$
$\Rightarrow 2 x +\frac{\pi}{3}=2 n \pi, n \in I \Rightarrow x = n \pi-\frac{\pi}{6}$....(1)
Now $\sin \left(\frac{3 x }{2}-\frac{\pi}{4}\right)=1 \Rightarrow \frac{3 x }{2}-\frac{\pi}{4}=2 n \pi+\frac{\pi}{2} \Rightarrow \frac{3 x }{2}=2 n \pi+\frac{3 \pi}{4} \Rightarrow x =\frac{4 n \pi}{3}+\frac{\pi}{2}$...(2)
Clearly $x=\frac{11 \pi}{6}$ satisfies both the equations (1) and (2). $\Rightarrow$ (C)