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##### COMPLEX NUMBERS AND QUADRATIC EQUATIONS Quiz-1

Complex numbers and quadratic equations is a segment of maths that deals with crucial theorems and concepts along with various formulae. It comprises of linear and quadratic equations along with roots related to the complex number's set (known as complex roots)..

•  2
•  1
•  0
•  3
Q2.If z=x+iy(x,y∈ R,x≠-1/2), the number of value of z safisfying

(n ∈ N, n>1) is
•  0
•  1
•  2
•  3
Q3.  If α,β,γ are the roots of x^3 -x^2 -1=0 then the value of (1+α)/(1-α)+(1+β)/(1-β)+(1+γ)/(1-γ) is equal to
•   -5
•  -6
•  -7
•  -2
Solution
(a) ∑ α=1,∑ αβ=0, αβγ = 1
∑(1+α) / (1-α) = -∑(-α+1-2)/(1-α) = ∑(2/(1-α)-1)
= 2∑〖1/(1-α)-3〗
Now,
1/((x-α)) +1/((x-β)) +1/((1-γ)) = ( 3x^2- 2x)/( x^3 -x^2 -1)
⇒ 1/(1-α) +1/(1-β) +1/(1-γ) =(3-2)/(1-1-1)=-1
⇒ (1+α)/(1-α)=-5

Q4. If the equation |x^2+bx+c|=k has four roots, then
Solution
(a) For the equation to have four real roots, the line y=k must intersect y=|x^2+bx+c| at four points
∴D>0 and k∈(0,-D/4)

Q5. The value of z satisfying the equation log⁡z+log⁡〖z^2 〗+⋯+log⁡〖z^n =0 is
Q6. If a(p+q)^2+2bpq+c=0 and a(p+r)^2+2bpr+c=0 (a≠0), then
•  qr = p^2
•  qr = p^2 + c/a
• qr = -p^2
•  None of these
Solution
(b) Given,
a(p+q)^2+2bpq+c=0 and a(p+r)^2+2bpr+c=0
⇒q and r satisfy the equation a(p+x)^2+2bpx+c=0
⇒q and r are the roots of ax^2 + 2(ap +bp )x + c+ ap^2 = 0
⇒qr= product of roots =(c+ ap^2)/a = p^2 + c/a

Q7. The value of m for which one of the roots of x^2 -3x+ 2m = 0 is double of one of the roots of x^2-x+m=0 is
•  -2
•  1
•  2
•  None of these
Solution
(a) Let α be the root of x^2-x+m=0 and 2α be the root of x^2-3x+2m=0.
Then, α^2-α+m=0 and 4α^2-6α+2m=0
Eliminating α,m^2=-2m
⇒m=0,m=-2

Q8. Roots of the equations are (z+1)^5=(z-1)^5 are
Q9.Total number of integral values of ‘a’ so that x^2-(a+1)x+a-1=0 has integral roots is equal to
•  1
•  2
•  4
•  None Of These
Solution
(a) x^2-(a+1)x+a-1=0
⇒(x-a)(x-1)=1
Now, a∈I and we want x to be an integer.
Hence, x-a=1,x-1=1 or x-a=-1,x-1=-1
⇒a=1 in both cases

Q10. z_1,z_2,z_3,z_4 are distinct complex numbers representing the vertices of a quadrilateral ABCD taken in order. If z_1-z_4=z_2-z_3 and arg⁡〖[(z_4-z_1)/(z_2-z_1)]〗=π/2, then the quadrilateral is:
•  Rectangle
•  Rhombus
•  Square
• Trapezium
Solution
The first condition implies that (z_1+z_3)/2=(z_2+z_4)/2, i.e., diagonals AC and BD bisect each other. Hence, quadrilateral is a parallelogram. The second condition implies that the angle between AD and AB is 90°. Hence the parallelogram is a rectangle #### Written by: AUTHORNAME

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