Question

Let $f_1\left(x\right)={\sin }^{-1}\left(\cos \right({\sin }^{2}x\left)\right),$ $f_2\left(x\right)={\cos }^{-1}\left(\sin \right({\cos }^{2}x\left)\right),$ $f_3\left(x\right)={\sin }^{-1}\left(\cos \right({\cos }^{2}x\left)\right)$ and $f_4\left(x\right)={\cos }^{-1}\left(\sin \right({\sin }^{2}x\left)\right).$ Then which of the following is (are) CORRECT?

• A. $\sum _{i=1}^4{f}_i\left(x\right)=2\pi -2$
• B. $\sum _{i=1}^4(-1{)}^i{f}_i(x)=0$
• C. $\sum _{i=1}^4(-1{)}^{\left[i/3\right]}{f}_i(x)=0,$ where $[.]$ denotes the greatest integer function
• D. $f_1\left(x\right)>f_2\left(x\right)\Rightarrow \cos 2x>0$

Answer 1

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

$f_1\left(x\right)>f_2\left(x\right)\Rightarrow \cos 2x>0$

$f_1\left(x\right)=\frac{\pi }{2}-{\sin }^{2}x$
$f_2\left(x\right)=\frac{\pi }{2}-{\cos }^{2}x$
$f_3\left(x\right)=\frac{\pi }{2}-{\cos }^{2}x$
$f_4\left(x\right)=\frac{\pi }{2}-{\sin }^{2}x$

$f_1\left(x\right)=f_4\left(x\right)$ and $f_3\left(x\right)=f_2\left(x\right)$
$\sum _{i=1}^4{f}_i\left(x\right)=f_1\left(x\right)+f_2\left(x\right)+f_3\left(x\right)+f_4\left(x\right)$
$=2(\frac{\pi }{2}-{\sin }^{2}x+\frac{\pi }{2}-{\cos }^{2}x)$
$=2(\pi -1)$

$\sum _{i=1}^4(-1{)}^i{f}_i(x)=-f_1(x)+f_2(x)-f_3(x)+f_4(x)=0$

$\sum _{i=1}^4(-1{)}^{\left[i/3\right]}{f}_i(x)=f_1(x)+f_2(x)-f_3(x)-f_4(x)=0$

$f_1\left(x\right)>f_2\left(x\right)$
$\Rightarrow -{\sin }^{2}x>-{\cos }^{2}x$
$\Rightarrow {\cos }^{2}x-{\sin }^{2}x>0$
$\Rightarrow \cos 2x>0$