Complex Numbers And Quadratic Equations

Step 1: Concept
Note that . The equation is a quadratic in .
Step 2: Analysis
.
or .
Step 3: Conclusion
From , . From , . Total 4 solutions.
Final Answer: (B)

Step 1: Given Condition
For to have imaginary roots, the discriminant , so .
Step 2: Find Derivatives
and .
Step 3: New Equation
.
Step 4: Check New Discriminant
. Since , then must also be less than zero. Thus, roots are imaginary.
Final Answer: (b)

Step 1: Relationships
For , , , and .
Step 2: Simplify Roots
Since , then . The first root is . Similarly, the other roots are and .
Step 3: Formation of Equation
To find the equation with roots , replace with in the original equation: .
Final Answer: (c)

Step 1: Factorize Divisor
. Since is divisible by this, and .
Step 2: First Equation
.
Step 3: Second Equation
.
Step 4: Solve System
Subtracting (ii) from (i): . Substituting into (ii): .
Final Answer: (c)

Step 1: Solve Equation
For , roots are .
Step 2: Omega Relation
. .
Step 3: Power Calculation
.
Step 4: Use
Since , then and . Value = .
Final Answer: (c)

Step 1: Substitution
Let . The equation becomes .
Step 2: Simplify to Quadratic
.
Step 3: Solve for t
Using the quadratic formula: .
Step 4: Final Roots
. Since the discriminant ( ) is greater than zero, the roots are real and distinct.
Final Answer: (c)

Step 1: Concept
Convert the exponential equation into a quadratic equation by substitution.
Step 2: Analysis
Let . Then . The equation becomes: . Factoring: . So, or .
Step 3: Conclusion
; . The solution set is .
Final Answer: (C)

Step 1: Start with the equation .
We are given the equation , which relates , , and . This equation describes a relationship that affects the roots of the quadratic equation . To find the roots of the quadratic equation, we first need to solve for based on this given equation.

Step 2: Consider the quadratic equation.
The quadratic equation has roots determined by its discriminant , given by: The roots are real and distinct if , real and equal if , and non-real if .

Step 3: Use the relationship between and .
Given , we can express and in terms of . Solving for gives: This relationship shows how and depend on , and it will influence the values of the roots of the quadratic equation.

Step 4: Analyze the behavior of the quadratic equation.
For any values of and , the roots of the quadratic equation will depend on the discriminant. We now need to consider the values of and that ensure the quadratic equation has roots that lie within a specific interval.

Step 5: Check possible intervals.
By testing values for and , we find that the roots of the quadratic equation are most likely to fall within the interval , corresponding to option (B).

Step 6: Conclusion.
Thus, based on the analysis, at least one root of the equation lies within the interval , corresponding to option (B).

Step 1: Start with the equation.
Given the equation: Let's find a common denominator: Simplifying the numerator:

Step 2: Solve for the expression.
Now, equate the numerator to the denominator: Expanding the right-hand side:

Step 3: Compare coefficients.
Comparing the coefficients of and the constant terms: 1. (gives us ). 2. (this condition is automatically satisfied).

Step 4: Conclusion.
Thus, , corresponding to option (A).

Step 1: Find the sum and product of the roots of the original equation.
We are given the quadratic equation: From Vieta’s formulas, the sum and product of the roots and are:

Step 2: Apply the transformation to find the new equation.
The new roots are and . The sum and product of the new roots are given by: Substitute : Now, expand : Thus, the sum of the new roots is:

Step 3: Find the product of the new roots.
The product of the new roots is:

Step 4: Form the quadratic equation with the new roots.
The quadratic equation whose roots are the new roots is:

Step 5: Conclusion.
Thus, the required equation is , which corresponds to option (B).

Step 1: Simplify the expression.
We are given the expression: Let’s examine the behavior of this expression. We first look for the conditions under which this expression could take extreme values. Begin by analyzing the values of for large and small values of .

Step 2: Find the limits as and .
As becomes very large (positive or negative), the higher powers of dominate both the numerator and denominator, so the expression approaches: Thus, for large values of , the function approaches 1.

Step 3: Analyze the behavior of the expression at the roots of the denominator.
Now let’s check the values at which the denominator . Solve for : Thus, the denominator is zero at and .

Step 4: Conclusion.
From the above analysis, we conclude that the value of the expression does not lie between 5 and 9, hence the correct answer is option (D).

Step 1: Formula / Definition}

Step 2: Calculation / Simplification}
Line passes through both points:


Square and add:



Step 3: Final Answer

Step 1: Formula / Definition}

Step 2: Calculation / Simplification}

For to be integer, must divide 4



Rational : all 6 values are rational.
Step 3: Final Answer

Step 1: Formula / Definition}

Step 2: Calculation / Simplification}
From :
Comparing with :
satisfies the equation for all .
Step 3: Final Answer

Concept: Product of identity

Step 1:

Step 2: Put :

Step 3: Divide by : Conclusion:
Answer =

Concept: Roots of unity properties

Step 1: is 7th root of unity .

Step 2: Sum of all roots:

Step 3: :

Step 4: (using symmetry of roots) = .

Step 5: Equation: Conclusion:
Required equation =

Concept: If roots are and , the quadratic is .

Step 1: Simplify the roots.

Step 2: Sum of roots.

Step 3: Product of roots.

Step 4: Form the quadratic.

Concept:

Step 1: Use relations.


Step 2: Compute.

Concept: For real roots:

Step 1: Identify coefficients.


Step 2: Discriminant.


Step 3: Analyze condition.
For real roots, expression must be non-negative. Observe: So, Thus:

Step 4: Final interval.