Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

For the identity ${\sin }^{-1}A+{\cos }^{-1}A=\frac{\pi }{2}$ to hold, the argument $A$ must satisfy $-1\le A\le 1$. Here, $A=x^2-6x+8$. So, we require $-1\le x^2-6x+8\le 1$. The inequality $x^2-6x+8\ge -1$ simplifies to $(x-3{)}^2\ge 0$, which is always true for real $x$. The inequality $x^2-6x+8\le 1$ simplifies to $x^2-6x+7\le 0$. The roots of $x^2-6x+7=0$ are $x=\frac{6\pm \sqrt{36-28}}{2}=3\pm \sqrt{2}$. Thus, $x\in [3-\sqrt{2},3+\sqrt{2}]$. Since $3-\sqrt{2}\approx 1.586$ and $3+\sqrt{2}\approx 4.414$, the integer values of $x$ in this interval are $2,3,4$. Therefore, there are 3 possible integer values for $x$.