Question

Let $f_1\left(x\right)={\sin }^{-1}\left(\cos \right({\sin }^{2}x\left)\right),f_2\left(x\right)={\cos }^{-1}\left(\sin \right({\cos }^{2}x\left)\right),f_3\left(x\right)={\sin }^{-1}\left(\cos \right({\cos }^{2}x\left)\right),f_4\left(x\right)={\cos }^{-1}\left(\sin \right({\sin }^{2}x\left)\right)$,
then which of the following is/are correct?
where [.] denotes Greatest Integer function.

• A. $\sum _4^{i=1}{f}_i\left(x\right)=2\pi -2$
• B. $\sum _4^{i=1}(-1{)}^i{f}_i(x)=0$
• C. $\sum _4^{i=1}(-1{)}^{\left[\frac{i}{3}\right]}{f}_i(x)=0$
• D. $f_1\left(x\right)>f_2\left(x\right)\Rightarrow \cos 2x<0$

Answer 1

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

The correct options are
A $\sum _4^{i=1}{f}_i\left(x\right)=2\pi -2$
B $\sum _4^{i=1}(-1{)}^i{f}_i(x)=0$
C $\sum _4^{i=1}(-1{)}^{\left[\frac{i}{3}\right]}{f}_i(x)=0$

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

$\sum _4^{i=1}{f}_i\left(x\right)=\frac{\pi }{2}-{\sin }^{2}x+\frac{\pi }{2}-{\cos }^{2}x+\frac{\pi }{2}-{\cos }^{2}x+\frac{\pi }{2}-{\sin }^{2}x=2\pi -2$

$\sum _4^{i=1}(-1{)}^i{f}_i(x)=-(\frac{\pi }{2}-{\sin }^{2}x)+(\frac{\pi }{2}-{\cos }^{2}x)+-(\frac{\pi }{2}-{\cos }^{2}x)+(\frac{\pi }{2}-{\sin }^{2}x)=0$

$\sum _4^{i=1}(-1{)}^{\left[\frac{i}{3}\right]}{f}_i(x)=(\frac{\pi }{2}-{\sin }^{2}x)+(\frac{\pi }{2}-{\cos }^{2}x)+-(\frac{\pi }{2}-{\cos }^{2}x)-(\frac{\pi }{2}-{\sin }^{2}x)=0$)

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