### ${\log _{\frac{1}{{\sqrt 2 }}}}\left( {\frac{{|z| + 11}}{{{{(|z| - 1)}^2}}}} \right) \le 2$, $|z{|_{\max }} = ?$

Let a complex number z, $|z|\neq 1$, satisfy ${\log _{\frac{1}{{\sqrt 2 }}}}\left( {\frac{{|z| + 11}}{{{{(|z| - 1)}^2}}}} \right) \le 2$. Then, the largest value of |z| is equal to ........

(A) 6                                                (B) 7
(C) 5                                                (D) 8

[Based on JEE Main 2021]

Solution

We have, ${\log _{\frac{1}{{\sqrt 2 }}}}\left( {\frac{{|z| + 11}}{{{{(|z| - 1)}^2}}}} \right) \le 2$

Since, the base of the log < 1 when antilog is taken the inequality sign reverses.

So, $\frac{{|z| + 11}}{{{{(|z| - 1)}^2}}} \ge {\left( {\frac{1}{{\sqrt 2 }}} \right)^2}$

Or, $\frac{{|z| + 11}}{{{{(|z| - 1)}^2}}} \ge \frac{1}{2}$

Or, $2|z| + 22 \ge |z{|^2} - 2|z| + 1$

$\Rightarrow |z{|^2} - 4|z| - 21 \le 0$

$\Rightarrow (|z| - 7)(|z| + 3) \le 0$

$\Rightarrow |z| - 7 \le 0$, since |z| + 3 > 0

$\Rightarrow |z{|_{\max }} = 7$