JEE PYQ: Complex Numbers Question 1
Question 1 - 2021 (16 Mar Shift 1)
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
(1) $8$
(2) $7$
(3) $6$
(4) $5$
Show Answer
Answer: (2)
Solution
$\frac{|z|+11}{(|z|-1)^2} \le \left(\frac{1}{\sqrt{2}}\right)^{-2} = 2$. So $|z| + 11 \le 2(|z|-1)^2 \Rightarrow |z|^2 - 4|z| - 21 \le 0 \Rightarrow (|z|-7)(|z|+3) \le 0 \Rightarrow |z| \le 7$.
BETA
