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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$

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The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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