Question

If $({\sin }^{-1}x{)}^2+({\sin }^{-1}y{)}^2+({\sin }^{-1}z{)}^2=\frac{3{\pi }^2}{4},$ then

• A. maximum value of $3x-y+4z$ is $8$
• B. maximum value of $3x-y+4z$ is $6$
• C. minimum value of $3x-y+4z$ is $-8$
• D. minimum value of $3x-y+4z$ is $-6$

Answer 1

Answer: (A), (C)

minimum value of $3x-y+4z$ is $-8$

Given : $({\sin }^{-1}x{)}^2+({\sin }^{-1}y{)}^2+({\sin }^{-1}z{)}^2=\frac{3{\pi }^2}{4}$
Since ${\sin }^{-1}x\in [-\frac{\pi }{2},\frac{\pi }{2}]$
So, the given equation is true when
$\Rightarrow {\sin }^{-1}x=\pm \frac{\pi }{2},{\sin }^{-1}y=\pm \frac{\pi }{2},{\sin }^{-1}z=\pm \frac{\pi }{2}$
$\Rightarrow x=\pm 1,y=\pm 1,z=\pm 1$
$\therefore$ Maximum value of $3x-y+4z=8$ and
Minimum value of $3x-y+4z=-8$