## Complex Number & Quadratic Equation Quiz-11

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)..

Q1. If z1,z2 and z3 are complex numbers such that |z1 |=|z2 |=|z3 |=|(1/z1)+(1/z2)+(1/z2)|=1, then |z1+z2+z3 | is
•  Equal to 1
•  Less than 1
•  Greater than 3
•  Equal to 3
Q2.If z1 z2 be complex numbers such that z1 ≠ z2 and |z1 |= |z2 |. If z1has positive real part and z2 has negative imaginary part, then[(z1+z2 )/(z1-z2 )] may be
•  Purely imaginary
•  Real and positive
•  Real and negative
•  None of these
Q3.  If the expression [mx-1+(1/x)] is non-negative for all positive real x, then the minimum value of m must be
•  -1/2
•  0
•  1/4
•   1/2
Solution
(c) We know that ax2 + bx + c ≥ 0 , ∀ x ∈ R,
If a > 0 and b2 - 4ac ≤ 0. So,
mx -1 + 1/x ≥ 0 ⇒ (mx2 - x + 1)/x ≥ 0
⇒mx2 - x + 1 ≥ 0 as x > 0
Now, mx2 - x + 1 ≥ 0 if m > 0 and 1-4m ≤ 0
⇒ m > 0 and m ≥ 1/4
Thus, the minimum value of m is ¼

Q4. If for complex numbers z1 z2, arg⁡(z1 )-arg⁡(z2 )=0, then |z1-z2 | is equal to
•  |z1 |+|z2 |
•  |z1 |-|z2 |
•  [|z1 |-|z2 |]
•  0
•   Straight lines passing through (2, 0)
•   Straight lines passing through (2, 0), (1, 1)
•   A line segment
•   A set of two rays
Solution

Q6. For positive integers n1,n2 the value of the expression (1+i)(n1 ) + (1 + i3 )(n1 ) + (1 + i5 )(n2 ) + (1 + i7 )(n2 ), where i=√(-1) is a real number if and only if
•  n1=n2+1
•  n1=n2-1
• n1=n2
•  n1>0,n2 > 0
Q7. If |z2+iz1 |=|z1 |+|z2 | and |z1 |=3 and |z2 |=4, then area of ∆ABC, if affixes of A,B and C are z1,z2 and [(z2-iz1)/(1-i)] respectively, is
•  5/2
•  0
•  25/2
•  25/4
Q8. x2-xy+y2-4x-4y+16=0 represents
•  A point
•  A Circle
•  A pair of straight lines
•   None of these
Solution
(a)
Given equation is
x2-(y+4)x+y2-4y+16=0
Since x is real, so,
D≥0
⇒(y+4)2-4(y2-4y+16)≥0
⇒ -3y2+24y-48≥0
^2-8y+16≤0
⇒(y-4)2≤0
⇒y-4=0
⇒y=4
Since the equation is symmetric in x and y, therefore x=4 only

Q9. Sum of common roots of the equations z3+2z2+2z+1=0 and z1985+z100+1=0 is
•  -1
•  1
•  0
•  1
Solution
(a)
We have,
z3+2z2+2z+1=0
⇒(z3+1)+2z(z+1)=0
⇒(z+1)(z2+z+1)=0
⇒z=-1,ω,ω2
Since z=-1 does not satisfy z1985+z100+1=0 while z=ω,ω2 satisfy it, hence sum is ω+ω2=-1

Q10. If the roots of the equation ax2-bx+c=0 are α,β then the roots of the equation b2 cx2-ab2 x+a3=0 are
•  1/(α3+αβ) , 1/(β3+αβ)
•  1/(α2+αβ) , 1/(β2+αβ)
•  1/(α4+αβ) , 1/(β4+αβ)
• None of these #### Written by: AUTHORNAME

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