Maximum value of the expression $|\sqrt{1+\sin2x}-\sqrt{1-\sin2x}|$ in $[0,\pi]$

Question

Maximum value of the expression $|\sqrt{1+\sin2x}-\sqrt{1-\sin2x}|$ in $[0,\pi]$ is
$(A)\sqrt2\hspace{1cm}(B)\frac{1}{\sqrt2}\hspace{1cm}(C)1\hspace{1cm}(D)2$

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Maximum value of the expression $|\sqrt{1+\sin2x}-\sqrt{1-\sin2x}|$ is attained when its square attains maximum value.
$|\sqrt{1+\sin2x}-\sqrt{1-\sin2x}|^2=4\sin^2x$
Its maximum value is $4$
So the maximum value of $|\sqrt{1+\sin2x}-\sqrt{1-\sin2x}|$ is $2$.
In my textbook,answer is $\sqrt2$.Am i wrong here.

Answer 1

The square is $$2-2\sqrt{1-\sin^2(2x)}=2-2|\cos(2x)|\leq2$$

which is not $4\sin^2(x)$

Answer 2

$\sqrt{1+x}$ is increasing and $\sqrt{1-x}$ decreasing in the range $[-1,1]$ so that the maximum absolute difference occurs at one end of the domain, and

$$f(-1)=f(1)=\sqrt2.$$

Answer 3

Observe that $1+\sin(2x)=(\cos(x)+\sin(x))^2 \Rightarrow \sqrt {1+\sin(2x)}=\cos(x)+\sin(x)$

Similarly $1-\sin(2x)=(\cos(x)-\sin(x))^2 \Rightarrow \sqrt {1-\sin(2x)}=\cos(x)-\sin(x)$

So $|\sqrt {1+\sin(2x)}-\sqrt {1-\sin(2x)}|=|\cos(x)+\sin(x)-\cos(x)+\sin(x)|=|2\sin(x)|$

For $x \in [0,\pi] $ maximum value of $\sin(x)=1$

Therefore $|\sqrt {1+\sin(2x)}-\sqrt {1-\sin(2x)}|=|2\sin(x)| \leq2$