Q62.For a ∈C, let A = {z ∈C :Re (a + z) >Im (a + z)} and B = {z ∈C :Re (a + z) <Im (a + z)} . Then among the two statements: (S1) : If Re (a), Im (a) > 0, then the set A contains all the real numbers (S2) : If Re (a), Im (a) < 0, then the set B contains all the real numbers, (1) Only (S2) is true (2) only (S1) is true (3) Both are true (4) Both are false z2+8iz−15 : α −1311 i ∈S, α ∈R −{0}, then 242α2 is equal to
What This Question Tests
This question tests the understanding of real and imaginary parts of complex numbers and their behavior under addition, particularly when evaluating inequalities for specific conditions on the real and imaginary parts of a given complex number 'a'.
Concepts Tested
Formulas Used
Re(z1+z2) = Re(z1) + Re(z2)
Im(z1+z2) = Im(z1) + Im(z2)
📚 NCERT Sections This Tests
14.2 — Which Of The Statements Given In Exercise 14.1 Is True For P-Type
Physics Class 12 · Chapter 14
14.2 Which of the statements given in Exercise 14.1 is true for p-type semiconductos.
14.1 — In An N-Type Silicon, Which Of The Following Statement Is True:
Physics Class 12 · Chapter 14
14.1 In an n-type silicon, which of the following statement is true: (a) Electrons are majority carriers and trivalent atoms are the dopants. (b) Electrons are minority carriers and pentavalent atoms are the dopants. (c) Holes are minority carriers and pentavalent atoms are the dopants. (d) Holes are majority carriers and trivalent atoms are the dopants.
9.15 — Apply Mirror Equation And The Condition:
Physics Class 12 · Chapter 9
9.15 Apply mirror equation and the condition: (a) f < 0 (concave mirror); u < 0 (object on left) (b) f > 0; u < 0 (c) f > 0 (convex mirror) and u < 0 (d) f < 0 (concave mirror); f < u < 0 to deduce the desired result.
📋 Question Details
- Chapter
- Complex Numbers
- Topic
- Properties of complex numbers
- Year
- 2023
- Shift
- 11 Apr Shift 2
- Q Number
- Q62
- Type
- MCQ
- NCERT Ref
- Class 11 Mathematics Ch 5: Complex Numbers and Quadratic Equations
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