Question

The inequality ${\sin }^{-1}\left(\sin 5\right)>x^2-4x$ holds if

• A. $x=2–\sqrt{9–2\pi }$
• B. $x=2+\sqrt{9–2\pi }$
• C. $x\in \left(2–\sqrt{9–2\pi },2+\sqrt{9–2\pi }\right)$
• D. $x>2+\sqrt{9–2\pi }$

Answer 1

The correct option is C

$x\in \left(2–\sqrt{9–2\pi },2+\sqrt{9–2\pi }\right)$

Explanation for the correct option:

Compute the value of $x$.

Since, $\sin \left(5\right)=\sin (5-2\mathrm{\pi })$

Therefore,

$$\begin{gathered} {\sin }^{-1}(\sin 5)>x^2-4x \\ \Rightarrow 5-2\mathrm{\pi }>x^2-4x \\ \Rightarrow 0>x^2-4x+\left(2\mathrm{\pi }-5\right) \end{gathered}$$

Now, find the roots of the quadratic equation.

$$\begin{gathered} x=\frac{4\pm \sqrt{{\left(-4\right)}^2-4(2\mathrm{\pi }-5)}}{2} \\ \Rightarrow x=\frac{4\pm \sqrt{4\left(4-(2\mathrm{\pi }-5)\right)}}{2} \\ \Rightarrow x=\frac{4\pm 2\sqrt{4-(2\mathrm{\pi }-5)}}{2} \\ \Rightarrow x=2\pm \sqrt{9-2\mathrm{\pi }} \end{gathered}$$

Since, the quadratic equation is less than zero and the value of $a$ is greater than zero.

So, the value of $x$ must lies between the roots of the quadratic equation.

Therefore, $x\in \left(2–\sqrt{9–2\pi },2+\sqrt{9–2\pi }\right)$.

Hence, option $\left(C\right)$ is the correct option.