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The trigonometric equation ${{\sin }^{-1}}x=2{{\sin }^{-1}}2\alpha $ has a real solution if
A. $\left| \alpha \right|>\dfrac{1}{\sqrt{2}}$
B. $\dfrac{1}{2\sqrt{2}}<\left| \alpha \right|<\dfrac{1}{\sqrt{2}}$
C. $\left| \alpha \right|>\dfrac{1}{2\sqrt{2}}$
D. $\left| \alpha \right|\le \dfrac{1}{2\sqrt{2}}$

Answer
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Hint: To solve this question, we should know the values of the function ${{\sin }^{-1}}x$ for which we can get a real solution of x. The range of values of ${{\sin }^{-1}}x$ is given by $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$. As the L.H.S varies in the range $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$, R.H.S should also vary in the range $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$, to get the real solutions of x. By writing the relation $-\dfrac{\pi }{2}\le 2{{\sin }^{-1}}2\alpha \le \dfrac{\pi }{2}$. Dividing by 2 and applying Sin on all the terms, we get the range of $\alpha $ for which we get a real solution in x.

Complete step-by-step solution:
The given equation in the question is ${{\sin }^{-1}}x=2{{\sin }^{-1}}2\alpha $ and we are asked to find the range of $\alpha $for which we get a real solution in x. To get the real solution in x, we know that the R.H.S should be in the range of the function ${{\sin }^{-1}}x$ , then only we get the real solution of x.
We know that the range of ${{\sin }^{-1}}x$ is given by $\left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]$.
So, we can write that $-\dfrac{\pi }{2}\le 2{{\sin }^{-1}}2\alpha \le \dfrac{\pi }{2}$
Dividing by 2 in all the terms, the inequality becomes,
$-\dfrac{\pi }{4}\le {{\sin }^{-1}}2\alpha \le \dfrac{\pi }{4}$
We can apply Sin to the whole inequality as Sin is an increasing function in the interval $\left[ -\dfrac{\pi }{4},\dfrac{\pi }{4} \right]$, the inequality doesn’t change, the inequality becomes
$\begin{align}
  & \sin \left( -\dfrac{\pi }{4} \right)\le 2\alpha \le \sin \left( \dfrac{\pi }{4} \right) \\
 & \dfrac{-1}{\sqrt{2}}\le 2\alpha \le \dfrac{1}{\sqrt{2}} \\
\end{align}$
Dividing by 2, we get
$\dfrac{-1}{2\sqrt{2}}\le \alpha \le \dfrac{1}{2\sqrt{2}}\to \left( 1 \right)$
If we get an inequality as $-a\le x\le a$, we can write the inequality as $\left| x \right|\le a$,
Similarly equation-1 becomes,
$\left| \alpha \right|\le \dfrac{1}{2\sqrt{2}}$
$\therefore $The range of $\alpha $ for real solutions of x is given by $\left| \alpha \right|\le \dfrac{1}{2\sqrt{2}}$. The answer is option (D).

Note: Students should be careful while applying the sine or cosine function to the inequality. A sine function is increasing in the interval $\left[ -\dfrac{\pi }{4},\dfrac{\pi }{4} \right]$, Hence the inequality did not change in the process. If the question is asked in ${{\cos }^{-1}}x$ terms, depending on the range, the inequality changes accordingly.