Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Complex Numbers question

2021 · 16 Mar · Shift 1 · Q27
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Complex Numbers
  5. /2021 · 16 Mar · Shift 1 · Q27

Complex Numbers question

2021 · 16 Mar · Shift 1 · Q27

JEE MainMathematicsComplex NumbersMCQ+4 / −1
Let a complex number z, |z| eee 1, satisfy log⁡12(∣z∣+11(∣z∣−1)2)≤2{\log _{{1 \over {\sqrt 2 }}}}\left( {{{|z| + 11} \over {{{(|z| - 1)}^2}}}} \right) \le 2log2​1​​((∣z∣−1)2∣z∣+11​)≤2. Then, the largest value of |z| is equal to ‾\underline{\hspace{2cm}}​.
  1. A
    5
  2. B
    8
  3. C
    6
  4. D
    7
View written solutionFree

Correct answer: D

  1. We are given the inequality
log⁡12(∣z∣+11(∣z∣−1)2)≤2\log_{\frac{1}{\sqrt2}}\left(\frac{|z|+11}{(|z|-1)^2}\right)\le 2log2​1​​((∣z∣−1)2∣z∣+11​)≤2

with the condition that ∣z∣≠1|z|\ne 1∣z∣=1 (since the denominator would become 000).

Let

x=∣z∣(x≥0, x≠1).x=|z| \quad (x\ge 0,\ x\ne 1).x=∣z∣(x≥0, x=1).

Then the inequality becomes

log⁡12(x+11(x−1)2)≤2.\log_{\frac{1}{\sqrt2}}\left(\frac{x+11}{(x-1)^2}\right)\le 2.log2​1​​((x−1)2x+11​)≤2.
  1. Since the base of the logarithm is
12<1,\frac{1}{\sqrt2}<1,2​1​<1,

the logarithmic function is decreasing. Therefore,

log⁡12(A)≤2⟹A≥(12)2.\log_{\frac{1}{\sqrt2}}(A)\le 2 \quad \Longrightarrow \quad A\ge \left(\frac{1}{\sqrt2}\right)^2.log2​1​​(A)≤2⟹A≥(2​1​)2.

So we get

x+11(x−1)2≥12.\frac{x+11}{(x-1)^2} \ge \frac12.(x−1)2x+11​≥21​.
  1. Multiply both sides by 2(x−1)22(x-1)^22(x−1)2. Since (x−1)2>0(x-1)^2>0(x−1)2>0 for x≠1x\ne 1x=1, the inequality sign does not change:
2(x+11)≥(x−1)2.2(x+11) \ge (x-1)^2.2(x+11)≥(x−1)2.

Expand:

2x+22≥x2−2x+1.2x+22 \ge x^2-2x+1.2x+22≥x2−2x+1.

Bring all terms to one side:

0≥x2−4x−21.0 \ge x^2-4x-21.0≥x2−4x−21.

That is,

x2−4x−21≤0.x^2-4x-21 \le 0.x2−4x−21≤0.
  1. Factor the quadratic:
x2−4x−21=(x−7)(x+3).x^2-4x-21=(x-7)(x+3).x2−4x−21=(x−7)(x+3).

Hence,

(x−7)(x+3)≤0.(x-7)(x+3)\le 0.(x−7)(x+3)≤0.

So,

−3≤x≤7.-3\le x\le 7.−3≤x≤7.

But x=∣z∣≥0x=|z|\ge 0x=∣z∣≥0, therefore

0≤x≤7.0\le x\le 7.0≤x≤7.

Also x≠1x\ne 1x=1, but this does not affect the largest possible value.

  1. Therefore, the largest value of ∣z∣|z|∣z∣ is
7.\boxed{7}.7​.
  1. Checking options:
  • A: 555 — not largest
  • B: 888 — not allowed
  • C: 666 — not largest
  • D: 777 — correct

So the correct option is D.

PreviousNext

More from Complex Numbers

  • Let z and ω be two complex numbers such that ω=zz−2z+2,​z−3iz+i​​=1 and Re(ω) has minimum value. Then, the minimum value of n ∈ N for which ω n is real, is…2021 · Numerical
  • The least value of |z| where z is complex number which satisfies the inequality exp(∣∣z∣+1∣(∣z∣+3)(∣z∣−1)​loge​2)≥log2​​∣57​+9i∣,i=−1​, is equal to :2021 · MCQ
  • The area of the triangle with vertices A(z), B(iz) and C(z + iz) is :2021 · MCQ
  • Let S1, S2 and S3 be three sets defined as S1 = {z ∈ C : |z − 1|≤2​} S2 = {z ∈ C : Re((1 − i)z) ≥ 1} S3 = {z ∈ C : Im(z) ≤ 1} Then the set S1 ∩ S2 ∩ S3 :2021 · MCQ
  • If the equation a∣z∣2+αz+αz​+d=0 represents a circle where a, d are real constants then which of the following condition is correct?2021 · MCQ
  • Let z1, z2 be the roots of the equation z2 + az + 12 = 0 and z1, z2 form an equilateral triangle with origin. Then, the value of |a| is :2021 · Numerical
  • Let a complex number be w = 1 −3​ i. Let another complex number z be such that |zw| = 1 and arg(z) − arg(w) =2π​. Then the area of the triangle with vertices origin, z and w is equal to :2021 · MCQ
  • If z and ω are two complex numbers such that ∣zω∣=1 and arg(z)−arg(ω)=23π​, then arg(1+3zω1−2zω​) is :…2021 · MCQ