Question

Consider the equation ${\sin }^{-1}\left(x^2-6x+\frac{17}{2}\right)+{\cos }^{-1}k=\frac{\pi }{2}$, then

• A. the largest value of $k$ for which the equation has 2 distinct solution is 1 .
• B. the equation must have real root if $k\in \left(\frac{-1}{2},1\right)$.
• C. the equation must have real root if $k\in \left(-1,\frac{-1}{2}\right)$.
• D. the equation has unique solution if $k=\frac{-1}{2}$.

Answer 1

Answer: (A), (B), (D)

For the solution of equation, we must have $\left(x^2-6x+\frac{17}{2}\right)=k$, where $k\in [-1,1]$ and $\left(x^2-6x+\frac{17}{2}\right)\in [-1,1]$