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Complex Numbers and Quadratic Equations

JEE Main 2024 (Online) 27th January Morning ShiftNumericalMedium📈 Topic in 16% papers+4
Ques. 1JEE Main 2024, 27 January (Morning Shift)

If satisfies the equation x2+x+1=0 and (1+ )7=A+B +C 2, A, B, C 0, then 5(3A-2B-C) is equal to .

Answer

5

Solution

= or 2 ⇒ 3=1, 1+ =- 2

(1+ )7=(- 2)7=- 14

=- 12 2=- 2=1+

⇒ A=1, B=1, C=0

(Adding k(1+ + 2)=0 keeps 3A-2B-C=1.)

∴ 5(3A-2B-C)=5(3-2-0)=5

Answer: 5

Keep practising — Complex Numbers and Quadratic Equations

Ques. 2JEE Main 2024, 27 January (Evening Shift)

If , are the roots of the equation, x2-x-1=0 and Sn=2023 n+2024 n, then :

  1. 2S12=S11+S10
  2. S12=S11+S10
  3. S11=S10+S12
  4. 2S11=S12+S10
Ques. 3JEE Main 2024, 27 January (Evening Shift)

Let the complex numbers and 1 lie on the circles |z-z0 |2=4 and |z-z0 |2=16 respectively, where z0=1+i. Then, the value of 100| |2 is .

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