NIMCET 2023 — Mathematics PYQ

Analysis of $x=|{\sin }^{-1}x|$  

The domain of ${\sin }^{-1}x$ is $[-1,1]$.  
Therefore, $x$ must be in the interval $[-1,1]$.  

Since the right-hand side, $|{\sin }^{-1}x|$, is always non-negative, the left-hand side $x$ must also be non-negative.  
Combining these conditions, $x$ must be in the interval $[0,1]$.  

For $x\in [0,1]$, ${\sin }^{-1}x\ge 0$, so $|{\sin }^{-1}x|={\sin }^{-1}x$.  
The equation becomes  
$$x={\sin }^{-1}x.$$

This equation can be rewritten as  
$$\sin x=x.$$

The only real solution to $\sin x=x$ is $x=0$.  

Therefore, $n_1=1$.  

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Analysis of $x=\sin x$  

The equation is $x=\sin x$.  

This is the same equation encountered in the previous section.  
The only real solution to $x=\sin x$ is $x=0$.  

Therefore, $n_2=1$.  

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Calculation of $n_2-n_1$}  

The value of $n_1$ is $1$.  
The value of $n_2$ is $1$.  

The difference is  
$$n_2-n_1=1-1=0.$$  

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\textbf{Final Answer:}  
$$n_2-n_1=0$$