Question

The value of $si{n}^{-1}(x^2-4x+6)+co{s}^{-1}(x^2-4x+6)$ for all $xϵR$ is

• A. $\frac{\pi }{2}$
• B. $\pi$
• C. $0$
• D. none of these

Answer 1

The correct option is D none of these

${\sin }^{-1}\left(x\right)ϵ[-1,1]$

Therefore

$-1\le x^2-4x+6\le 1$

$-1\le (x-2{)}^2-4+6\le 1$

$-1\le (x-2{)}^2+2\le 1$

$-3\le (x-2{)}^2\le -1$

Now, $(x-2{)}^2>0$ for all $xϵR$.

Hence the inequality

$-3\le (x-2{)}^2\le -1$ yields no real real values of x.

Therefore,

$|x^2-4x+6|$ cannot be less than or equal to 1.

Hence ${\sin }^{-1}(x^2-4x+6)$ is no possible.

The, same goes for $\cos x$.

Hence the answer is none of these.