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The mid- point of the domain of the function $\mathrm{f}(\mathrm{x})=\sqrt{4-\sqrt{2 \mathrm{x}+5}}$ for real $\mathrm{x}$ is $-$
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The correct answer is:
$3 / 2$
$\begin{array}{l}
f(x)=\sqrt{4-\sqrt{2 x+5}} \\
4-\sqrt{2 x+5} \geq 0 \quad 2 x+5 \geq 0 \\
\sqrt{2 x+5} \leq 4 \quad x \geq-5 / 2 \\
x \leq \frac{11}{2} \\
x \in\left[-\frac{5}{2}, \frac{11}{2}\right] \\
\text { mid point }=\frac{-5 / 2+11 / 2}{2}=\frac{3}{2}
\end{array}$
f(x)=\sqrt{4-\sqrt{2 x+5}} \\
4-\sqrt{2 x+5} \geq 0 \quad 2 x+5 \geq 0 \\
\sqrt{2 x+5} \leq 4 \quad x \geq-5 / 2 \\
x \leq \frac{11}{2} \\
x \in\left[-\frac{5}{2}, \frac{11}{2}\right] \\
\text { mid point }=\frac{-5 / 2+11 / 2}{2}=\frac{3}{2}
\end{array}$
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