Question

If the equation ${\sin }^{-1}\left(\sin k\right)=k-2\pi$, is possesing a real solution, then

• A. sum of all possible integral values of $k=18$
• B. Number of real roots for $k=3$
• C. Number of real roots for $k=4$
• D. $k_{max}-k_{min}=2\forall k\in Z$

Answer 1

Answer: (A), (D)

$k_{max}-k_{min}=2\forall k\in Z$

Given : ${\sin }^{-1}\left(\sin x\right)=k-2\pi$,
we know that, ${\sin }^{-1}\left(\sin x\right)=x-2\pi$ if
$\Rightarrow \frac{\pi }{2}\le x-2\pi \le \frac{\pi }{2}$
$\Rightarrow \frac{3\pi }{2}\le x\le \frac{5\pi }{2}$
$\Rightarrow k\in [\frac{3\pi }{2},\frac{5\pi }{2}]$
It can be observed that there are infinitely many values for $k$ satisfying the given equation.
So, possible integral values for $k=5,6,7$
$\Rightarrow$sum $=18,k_{max}-k_{min}=2$