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

Hint: Consider the argument of the sine functions as 5 radians. Find the range in which 5 radians lie in terms of the integral multiple of $\dfrac{\pi }{2}$, to do this divide 5 by 1.57. Now, use the information that ${{\sin }^{-1}}\left( \sin x \right)=x$ for $-\dfrac{\pi }{2}\le x\le \dfrac{\pi }{2}$ to simplify the inverse sine function. Form a quadratic inequality and factorize it into two linear factors, use the formula that if $\left( x-a \right)\left( x-b \right)<0$ then $x\in \left( a,b \right)$, where $a< b$, to get the answer.

Complete step-by-step solution:
Here we have been provided with the inequality ${{\sin }^{-1}}\left( \sin 5 \right)>{{x}^{2}}-4x$ and we are asked to find the solution set for this. But first we need to simplify the inverse sine function in the L.H.S.
Now, we can see that in the argument of the sine functions we have the angle 5, as nothing is provided so we will consider it as angle in radian and not degrees. We know that ${{\sin }^{-1}}\left( \sin x \right)=x$ for $x\in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. So we need to check the range of 5 radians in terms of $\dfrac{\pi }{2}$.
Considering the function ${{\sin }^{-1}}\left( \sin 5 \right)$ we have, on dividing 5 with $\dfrac{\pi }{2}$, i.e. nearly 1.57 we get nearly 3.184 that means we can say that: -
$\begin{align}
  & \Rightarrow \dfrac{3\pi }{2}\le 5\le \dfrac{5\pi }{2} \\
 & \Rightarrow \dfrac{3\pi }{2}-2\pi \le 5-2\pi \le \dfrac{5\pi }{2}-2\pi \\
 & \Rightarrow \dfrac{-\pi }{2}\le 5-2\pi \le \dfrac{\pi }{2} \\
\end{align}$
So we can write the given expression as:
\[\Rightarrow {{\sin }^{-1}}\left( \sin 5 \right)={{\sin }^{-1}}\left( \sin \left( \left( 5-2\pi \right)+2\pi \right) \right)\]
Using the relation $\sin \left( 2n\pi +\theta \right)=\sin \theta $ we get,
\[\begin{align}
  & \Rightarrow {{\sin }^{-1}}\left( \sin 5 \right)={{\sin }^{-1}}\left( \sin \left( 5-2\pi \right) \right) \\
 & \Rightarrow {{\sin }^{-1}}\left( \sin 5 \right)=5-2\pi \\
\end{align}\]
Therefore, the inequality can be simplified as,
$\begin{align}
  & \Rightarrow 5-2\pi >{{x}^{2}}-4x \\
 & \Rightarrow {{x}^{2}}-4x+\left( 2\pi -5 \right)<0 \\
\end{align}$
Now, we need to factorize the above quadratic expression into the product of two linear factors, so substituting it equal to 0 and using the discriminant formula given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$, where a is the coefficient of ${{x}^{2}}$, b is the coefficient of $x$ and c is the constant term we get,
$\begin{align}
  & \Rightarrow {{x}^{2}}-4x+\left( 2\pi -5 \right)=0 \\
 & \Rightarrow x=\dfrac{4\pm \sqrt{16-4\left( 2\pi -5 \right)}}{2} \\
 & \Rightarrow x=\dfrac{4\pm \sqrt{36-8\pi }}{2} \\
 & \Rightarrow x=2\pm \sqrt{9-2\pi } \\
\end{align}$
Therefore, we can say that $\left( x-\left( 2+\sqrt{9-2\pi } \right) \right)$ and $\left( x-\left( 2-\sqrt{9-2\pi } \right) \right)$ are the two factors of the quadratic polynomial. So we can write the inequality as,
$\Rightarrow \left( x-\left( 2+\sqrt{9-2\pi } \right) \right)\left( x-\left( 2-\sqrt{9-2\pi } \right) \right)<0$
Clearly we can see that $\left( 2-\sqrt{9-2\pi } \right)<\left( 2+\sqrt{9-2\pi } \right)$, so using the formula that if $\left( x-a \right)\left( x-b \right)<0$ then $x\in \left( a,b \right)$, where $a$\therefore x\in \left( 2-\sqrt{9-2\pi },2+\sqrt{9-2\pi } \right)$
Hence, option (c) is the correct answer.

Note: Note that you cannot assume 5 as 5 degrees because there is no symbol regarding that which only means it is a real number must be considered as a radian. Do not directly remove the inverse trigonometric and trigonometric functions if the angle doesn’t lie in the defined range. You may get the answer wrong while doing so.