Question

The number of solutions of the equation $2{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)-\pi x=0$ is equal to
A. 0
B. 1
C. 2
D. 3

Answer 1

In this problem, we have to find the number of solutions to the given equation. We can first rearrange the given equation for our convenience to find the answer. We can then divide 2 on both sides. We can take sine on both sides. We can then substitute the values for x to find the solutions for the given equation.

Complete step by step answer:
Here we have to find the number of solutions for the given equation.
$2{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)-\pi x=0$
We can now rearrange the given equation and write it as,
$\Rightarrow 2{{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\pi x$
We can now divide 2 on both sides, we get
$\Rightarrow {{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\dfrac{\pi x}{2}$
We can now take sine on both sides, were we can cancel the sine and its inverse in the left-hand sides, we get
$\Rightarrow \left( \dfrac{2x}{1+{{x}^{2}}} \right)=\sin \dfrac{\pi x}{2}$
We can now take x = 0 and substitute in the above step, we get

& \Rightarrow \left( \dfrac{2\left( 0 \right)}{1+0} \right)=\sin 0 \\\ & \Rightarrow 0=0 \\\ \end{aligned}$$ We can see that 0 satisfies both sides. We can now take x = 1 and substitute in the above step, we get $$\begin{aligned} & \Rightarrow \left( \dfrac{2}{1+{{1}^{2}}} \right)=\sin \dfrac{\pi }{2} \\\ & \Rightarrow 1=1 \\\ \end{aligned}$$ We can now take x = -1 and substitute, we get $$\begin{aligned} & \Rightarrow \left( \dfrac{2\left( -1 \right)}{1+{{\left( -1 \right)}^{2}}} \right)=\sin \dfrac{\pi \left( -1 \right)}{2} \\\ & \Rightarrow -1=-1 \\\ \end{aligned}$$ Therefore, the given equation satisfies the values of 1, 0, -1. **So, the correct answer is “Option D”.** **Note:** We should always remember some of the degree values such as $$\sin 0=0,\sin \dfrac{\pi }{2}=1$$. We should also know that the number of solutions is nothing but the values of x. When the values of x satisfies the given equation, then the solutions are correct.