Question

If ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$, then $(x^{500}+y^{500}+z^{500})-(x^{501}+y^{501}+z^{501})$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (A)

$0$

We know that the range of ${\sin }^{-1}x$ is $[-\frac{\pi }{2},\frac{\pi }{2}]$

Given:${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$

$\Rightarrow {\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{\pi }{2}+\frac{\pi }{2}+\frac{\pi }{2}$

$\Rightarrow {\sin }^{-1}x=\frac{\pi }{2}{\sin }^{-1}y=\frac{\pi }{2}{\sin }^{-1}z=\frac{\pi }{2}$

$\Rightarrow x=\sin \frac{\pi }{2},y=\sin \frac{\pi }{2},z=\sin \frac{\pi }{2}$

$\Rightarrow x=1,y=1,z=1$

$(x^{500}+y^{500}+z^{500})-(x^{501}+y^{501}+z^{501})$

$=(1^{500}+1^{500}+1^{500})-(1^{501}+1^{501}+1^{501})$

$=(1+1+1)-(1+1+1)=3-3=0$