JEE MAIN - Mathematics (2021 - 16th March Morning Shift - No. 5)
Let a complex number z, |z| $$\ne$$ 1,
satisfy $${\log _{{1 \over {\sqrt 2 }}}}\left( {{{|z| + 11} \over {{{(|z| - 1)}^2}}}} \right) \le 2$$. Then, the largest value of |z| is equal to ____________.
satisfy $${\log _{{1 \over {\sqrt 2 }}}}\left( {{{|z| + 11} \over {{{(|z| - 1)}^2}}}} \right) \le 2$$. Then, the largest value of |z| is equal to ____________.
5
8
6
7
Explanation
$${{|z| + 11} \over {{{(|z| - 1)}^2}}} \ge {1 \over 2}$$
$$2|z| + 22 \ge {(|z| - 1)^2}$$
$$2|z| + 22 \ge \,|z{|^2} - 2|z| + 1$$
$$|z{|^2} - 4|z| - 21 \le 0$$
$$(|z| - 7)(|z| + 3) \le 0$$
$$ \Rightarrow \,|z| \le 7$$
$$ \therefore $$ $$|z{|_{\max }} = 7$$
$$2|z| + 22 \ge {(|z| - 1)^2}$$
$$2|z| + 22 \ge \,|z{|^2} - 2|z| + 1$$
$$|z{|^2} - 4|z| - 21 \le 0$$
$$(|z| - 7)(|z| + 3) \le 0$$
$$ \Rightarrow \,|z| \le 7$$
$$ \therefore $$ $$|z{|_{\max }} = 7$$
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