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For \mathrm{z} \in \mathbb{C} if the minimum value of \mathrm{(|z-3 \sqrt{2}|+|z-p \sqrt{2} i|)} is 5 \sqrt{2}, then a value of \mathrm{p} is____________.

Option: 1

3


Option: 2

\frac{7}{2}


Option: 3

4


Option: 4

\frac{9}{2}


Answers (1)

best_answer

Using triangle law

\mathrm{|z-3 \sqrt{2}|+|z-p \sqrt{2} i| \geqslant \mid(z-3 \sqrt{2})-(z-p \sqrt{2} i) }\\

                                           \mathrm{\geqslant|-3 \sqrt{2}+p \sqrt{2} i|} \\

                                          \mathrm{\geqslant \sqrt{18+2 p^{2}}}

Given that maximum value is \mathrm{5\sqrt{z}}

\mathrm{\therefore \quad \sqrt{18+2 p^{2}} =5 \sqrt{2}} \\

           \mathrm{18+2 p^{2} =50} \\

           \mathrm{p =\pm 4}

Hence correct option is 3

Posted by

Divya Prakash Singh

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