Question

Match the following :
List - I List - II
1. Period of $a.2\pi$
$\cos x\cos ({60}^0-x)\cos ({60}^0+x)$
2. Period of $3\sin 2x+100\cot 4x$ $b.\frac{\pi }{2}$
3. Period of $\sin x+\cos x$ $c.\pi$
4. Period of ${\sin }^{4}x+{\cos }^{4}x$ $d.\frac{2\pi }{3}$

• A. $1-d,2-c,3-a,4-b$
• B. $1-d,2-c,3-b,4-a$
• C. $1-a,2-b,3-c,4-d$
• D. $1-b,2-c,3-a,4-d$

Answer 1

The correct option is A $1-d,2-c,3-a,4-b$

1. $\cos x\cos (60-x)\left(\cos \right(60+x\left)\right)$
$=\frac{\cos x}{2}\left[\cos \right(120)+\cos 2x]\left[2\cos \right(A)\cos B=\cos (A+B)+\cos (A-B\left)\right]$
$=\frac{\cos x}{2}[-\frac{1}{2}+2{\cos }^{2}x-1]$
$=\frac{2{\cos }^{3}x-\frac{3}{2}\cos x}{2}$
$=\frac{4{\cos }^{3}x-3\cos x}{4}$
$f\left(x\right)=\frac{\cos 3x}{4}[\because \cos 3x=4{\cos }^{3}x-3\cos x]$
$\therefore$ Period of $f\left(x\right)=\frac{2\pi }{3}$
2. Period of $3\sin 2x+100cot4x$
$\Rightarrow LCMof(\pi ,\frac{\pi }{4})=\pi$ [concept - 1]
3. Period of $\sin x+\cos x$
$\sin x+\cos x=\sqrt{2}\cos (x-\frac{\pi }{4})$
Period of $\cos (x-\frac{\pi }{4})=2\pi$
4. Period of ${\sin }^{4}x+{\cos }^{4}x$
${\sin }^{4}x+{\cos }^{4}x=1-\frac{(\sin 2x{)}^2}{2}=\frac{}{2}-\left(\frac{1-\cos 4x}{2}\right)2=\frac{1-\cos 4x}{4}[\cos 2x=1-2{\sin }^{2}x]$
Period of $\left(\frac{1-\cos 4x}{4}\right)=\frac{2\pi }{4}=\frac{\pi }{2}$