Question

Ordered pair $(x,y)$ satisfying the system of simultaneous equation ${\sin }^{-1}x+{\sin }^{-1}2y=\pi$ and ${\cos }^{-1}x+{\cos }^{-1}{x}^2=0$ is

• A. $(-1,-\frac{1}{2})$
• B. $(1,-\frac{1}{2})$
• C. $(-1,\frac{1}{2})$
• D. $(1,\frac{1}{2})$

Answer 1

The correct option is D $(1,\frac{1}{2})$

${\sin }^{-1}x\in [\frac{-\pi }{2},\frac{\pi }{2}]$ and ${\sin }^{-1}2y\in [\frac{-\pi }{2},\frac{\pi }{2}]$

$\Rightarrow {\sin }^{-1}x+{\sin }^{-1}2y\in [-\pi ,\pi ]$

$\Rightarrow {\sin }^{-1}x+{\sin }^{-1}2y=\pi$ occus at extreme values of ${\sin }^{-1}x$ and ${\sin }^{-1}2y$

i.e., ${\sin }^{-1}x=\frac{\pi }{2}$ and ${\sin }^{-1}2y=\frac{\pi }{2}$

$\Rightarrow x=1,2y=1$$\Rightarrow x=1,y=1/2$ ...(1)

Now ${\cos }^{-1}x+{\cos }^{-1}{x}^2=0$

$\Rightarrow {\cos }^{-1}x=0;{\cos }^{-1}{x}^2=0$ and ${\cos }^{-1}x$ and ${\cos }^{-1}{x}^2$ both are non-negative terms

$\Rightarrow x=1$ and $x^2=1\Rightarrow x=1$ and $x=\pm 1$

$\Rightarrow x=1$ ...(2)

$\therefore$ from (1) and (2) solution of given equations would be $x=1,y=\frac{1}{2}$

$\Rightarrow$ (D) is correct option.