Question

Let $f\left(x\right)={\sin }^{2}x+{\cos }^{4}x+2$ and $g\left(x\right)=\cos \left(\cos x\right)+\cos \left(\sin x\right)$. Also let period of $f\left(x\right)$ and $g\left(x\right)$ be $T_1$ and $T_2$ respectively then

• A. $T_1=2T_2$
• B. $2T_1=T_2$
• C. $T_1=T_2$
• D. $T_1=4T_2$

Answer 1

The correct option is A $T_1=2T_2$

Solution- $f\left(x\right)=f(x+T)$ (: they are periodic)

$\begin{array}{cc}f(x+\pi )& ={\sin }^2(x+\pi )+{\cos }^4(x+7)+2\\ & =[-\sin x{]}^2+[-\cos x{]}^4+2\\ & ={\sin }^{2}x+{\cos }^{4}x+2=f\left(x\right)\\ T_1=& \pi \\ g\left(x\right)=\cos \left(\cos x\right)+\cos \left(\sin x\right)& \\ g(x+\frac{\pi }{2})& =\cos \left(\cos (x+\frac{\pi }{2})\right)+\cos \left(\sin (x+\frac{\pi }{2})\right)\\ & =\cos (-\sin x)+\cos \left(\cos x\right)\\ & =\mathrm{ses}\left(\sin x\right)+\cos \left(\cos x\right)=g\left(x\right)\\ T_2& =\frac{\pi }{2}\\ \Rightarrow & T_2=\frac{T_1}{2}\text{or}{T}_1=2T_2\end{array}$

Hence, (A) is the correct option.