Question

Consider the equation ${\sin }^{-1}(x^2-6x+\frac{17}{2})+{\cos }^{-1}k=\frac{\pi }{2}.$ Then

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

Answer 1

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

the equation has a unique solution if $k=-\frac{1}{2}$

For the equation to have a solution,
we must have $(x^2-6x+\frac{17}{2})=k$
$\Rightarrow (x-3{)}^2-\frac{1}{2}=k$
$\Rightarrow (x-3{)}^2=k+\frac{1}{2}$ where $k\in [-1,1]$

Now, for real roots,
$k+\frac{1}{2}\ge 0$
$\Rightarrow k\ge -\frac{1}{2}$
But $k\in [-1,1]$
$\therefore k\in [-\frac{1}{2},1]$

For unique solution,
$k+\frac{1}{2}=0$
$\Rightarrow k=\frac{-1}{2}$