JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 11)
The least value of |z| where z is complex number which satisfies the inequality $$\exp \left( {{{(|z| + 3)(|z| - 1)} \over {||z| + 1|}}{{\log }_e}2} \right) \ge {\log _{\sqrt 2 }}|5\sqrt 7 + 9i|,i = \sqrt { - 1} $$, is equal to :
8
3
2
$$\sqrt 5 $$
Explanation
Let | z | = t, t $$ \ge $$ 0
$${e^{{{(t + 3)(t - 1)} \over {t + 1}}{{\log }_e}2}} \ge {\log _{\sqrt 2 }}16 = 8$$ ($$ \because $$ t + 1 > 0)
$${2^{{{(t + 3)(t - 1)} \over {t + 1}}}} \ge {2^3}$$
$${{(t + 3)(t - 1)} \over {t + 1}} \ge 3$$
$${t^2} + 2t - 3 \ge 3t + 3$$
$${t^2} - t - 6 \ge 0$$
$$t \in ( - \infty , - 2) \cup [3,\infty )$$ But t $$ \ge $$ 0
$$ \therefore $$ $$t \in [3,\infty )$$
$${e^{{{(t + 3)(t - 1)} \over {t + 1}}{{\log }_e}2}} \ge {\log _{\sqrt 2 }}16 = 8$$ ($$ \because $$ t + 1 > 0)
$${2^{{{(t + 3)(t - 1)} \over {t + 1}}}} \ge {2^3}$$
$${{(t + 3)(t - 1)} \over {t + 1}} \ge 3$$
$${t^2} + 2t - 3 \ge 3t + 3$$
$${t^2} - t - 6 \ge 0$$
$$t \in ( - \infty , - 2) \cup [3,\infty )$$ But t $$ \ge $$ 0
$$ \therefore $$ $$t \in [3,\infty )$$
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