Verified Solution • 2025–26 Session moderate

QJEEM09Ma97b3: JEE Main 2009 Mathematics - Class 11 JEEM09Ma97b3 Solution: If then the maximum value of is equal to :

If  ∣z−4z∣=2,\,\left| {z - {4 \over z}} \right| = 2, then the maximum value of  ∣z∣\,\left| z \right| is equal to :

Multiple Choice Options

  • A 5+1\sqrt 5 + 1
  • B 2
  • C 2+22 + \sqrt 2
  • D 3+1\sqrt 3 + 1

Step-by-Step Verified Solution

Given that ∣z−4z∣=2\left| {z - {4 \over z}} \right| = 2

Now ∣z∣=∣z−4z+4−z∣≤∣z−4z∣+4∣z∣\left| z \right| = \left| {z - {4 \over z} + {4 \over { - z}}} \right| \le \left| {z - {4 \over z}} \right| + {4 \over {\left| z \right|}}

⇒∣z∣≤2+4∣z∣\Rightarrow \left| z \right| \le 2 + {4 \over {\left| z \right|}}

⇒∣z∣2−2∣z∣−4≤0\Rightarrow {\left| z \right|^2} - 2\left| z \right| - 4 \le 0

⇒(∣z∣−2+202)(∣z∣−2−202)≤0\Rightarrow \left( {\left| z \right| - {{2 + \sqrt {20} } \over 2}} \right)\left( {\left| z \right| - {{2 - \sqrt {20} } \over 2}} \right) \le 0

(∣z∣−(1+5))(∣z∣−(1−5))≤0\left( {\left| z \right| - \left( {1 + \sqrt 5 } \right)} \right)\left( {\left| z \right| - \left( {1 - \sqrt 5 } \right)} \right) \le 0

⇒(−5+1)≤∣z∣≤(5+1)\Rightarrow \left( { - \sqrt 5 + 1} \right) \le \left| z \right| \le \left( {\sqrt 5 + 1} \right)

⇒∣z∣max⁡=5+1\Rightarrow {\left| z \right|_{\max }} = \sqrt 5 + 1

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