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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

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

 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$