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Answer $B$

Given:

$$\sin (\sin^{-1}\frac{2}{5} + \cos^{-1}x) = 1\qquad \to(1)$$

But we know that:

$$\sin \frac{\pi}{2} = 1$$

Substituting this value in in $(1)$, we get:

$$\sin (\sin^{-1}\frac{2}{5} + \cos^{-1}x) = \sin\frac{\pi}{2}$$

On comparing $\sin$ with $\sin$, we get:

$$\sin^{-1}\frac{2}{5} + \cos^{-1}x = \frac{\pi}{2}\qquad \to (2)$$

Also, we know that:

$$\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}\qquad \to (3)$$

So, on comparing $(2)$ and $(3)$, we get:

$$x = \frac{2}{5}$$

$\therefore \;B$ is the correct option.

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