Complex Numbers and Quadratic Equations-Quiz-8

Complex Number & Quadratic Equation Quiz-8

Dear Readers,

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 | z - 2 - i | = |z||sin⁡(π/4 - arg⁡z ) |, then locus of z is

•  A pair of straight lines
•  Circle
•  Parabola
•  Ellipse

Solution
(c) br We have,
|(x-2) + i(y-1)| = |z| | 1/√2 cos⁡θ - 1/√2 sin⁡θ |
Where θ = arg⁡z
√( ( x - 2 )^2 + ( y - 1 )^2 ) = 1/√2 |x - y|
Which is a parabola

Q2.If z = i log⁡[( 2 - √(-3 ) ) ] then cos⁡z =

•  -1
•  -1/2
•  1
•  1/2

Solution
(d)
z = ilog⁡[(2 - √3)]
⇒ e^iz = e^(i^2 log⁡[( 2 - √3 )] ) = e^( -log ⁡[(2 - √3)] )
⇒ e^iz = e^log⁡[( 2 - √3 )^(-1) ] = e^log⁡[ (2 - √3) ] = (2 + √3)
⇒ cos⁡z = (e^iz + e^(-iz) )/2 = ( (2 + √3) +( 2 - √3 ) )/2 = 2

Q3.  If x , y ∈ R satisfy theEquation x^2 + y^2 - 4x - 2y + 5 = 0, then the value of the expression [ ( √x - √y )^2 + 4√xy ) ]/( x + √xy ) is

•  √2 + 1
•  ( √2 + 1 )/2
•  ( √2 - 1 )/2
•  ( √2 + 1 )/√2

Solution
(d)
f(x , y) = ( x - 2 )^2 + (y - 1 )^2 = 0
⇒ x = 2 and y = 1
∴ E =( (√2 - 1 )^2 + 4√2 )/( 2 + √2 ) = ( √2 + 1 )^2 /( √2( √2 + 1) ) = ( √2 + 1 )/√2

Q4. The least value of the expression x^2 + 4y^2 + 3z^2 - 2x - 12y - 6z + 14 is

•  1
•  No least value
•  0
•  None of these

Solution
(a)
Let, f(x , y , z) = x^2 + 4y^2 + 3z^2 - 2x - 12y - 6z + 14
= ( x - 1 )^2 + ( 2y - 3 )^2 + 3(z - 1)^2 + 1
For the least value of f(x,y,z),
x - 1 = 0 , 2y - 3 = 0 and z - 1 = 0
∴ x = 1 , y = 3/2 , z = 1
Hence the least value of f(x,y,z) is f(1 , 3/2 , 1 ) = 1

Q5. If A(z_1 ) , B(z_2 ) , C(z_3) are the vertices of the triangle ABC such that (z_1 - z_2 )/( z_3 - z_2 ) = ( 1/√2 ) + (i/√2), the triangle ABC is

•  Equilateral
•  Right angled
•  Isosceles
•  Obtuse angled

Solution
 (c)

( z_1 - z_2 )/( z_3 - z_2 ) = 1/√2 + i/√2 = e^(iπ/4) br ∠CBA = π/4
Also,
|z_1 - z_2 | = | z_3 - z_2 |
Hence, ∆ABC is isosceles

Q6. 76. If the roots of the equation, x^2 + 2ax + b = 0, are real and distinct and they differ by at most 2m, then b lies in the interval

•  (a^2 , a^2 + m^2 )
•  (a^2 - m^2 , a^2 )
• [ a^2 - m^2 , a^2 )
•  None of these

Solution
(c)
Let the roots be α , β
∴ α + β = -2a and αβ = b
Given,
|α - β| ≤ 2m
⇒ |α - β|^2 ≤ (2m)^2
⇒ (α + β )^2 - 4ab ≤ 4m^2
⇒ 4a^2 - 4b ≤ 4m^2
⇒ a^2 - m^2 ≤ b and discriminant D > 0 or 4a^2 - 4b > 0
⇒ a^2 - m^2 ≤ b and b < a^2
Hence, b ∈ [a^2 - m^2 , a^2]

Q7.  If x is real, then the maximum value of ( 3x^2 + 9x + 17 )/( 3x^2 + 9x + 7 ) is

•  1/4
•  41
•  1
•  17/7

Solution


(b)
Let,
y = ( 3x^2 + 9x + 17 )/( 3x^2 + 9x + 7 )
⇒ 3x^2 (y - 1) + 9x(y - 1) + 7y - 17 = 0
Since x is real, so,
D ≥ 0
⇒ 81(y - 1)^2 - 4×3(y - 1)(7y - 17) ≥ 0
⇒ (y - 1)(y - 41) ≤ 0 ⇒ 1 ≤ y ≤ 41
Therefore, the maximum value of y is 41

Q8. If a < 0 , b > 0 then √a √b is equal to

•  -√(|a| b)
•  √(|a| b) i
•  √(|a| b)
•  None of these

Solution
(b)
Verify by selecting particular values of a and b
Let a = -9 and b = 4. Then,
√a √b = √( -9 ) √4 =( 3i )( 2 ) = 6i
From option (a), we have
-√( |a|b ) = -√(|-9|×4) = -√36 = -6
From option (b), we have
√( |a|bi ) = √( |-9|×4 ) i = 6i

Q9. The inequality |z - 4| < |z - 2| represents the region given by

•  Re(z) ≥ 0
•  Re(z) < 0
•  Re(z) > 0
•  None of these

Solution
(d)
|z - 4| < |z - 2|
⇒ |(x - 4) + iy| < |(x - 2) + iy|
⇒ ( x - 4 )^2 + y^2 < ( x - 2 )^2 + y^2
⇒ -8x+16< -4x+4
⇒ 4x - 12 > 0
⇒ x > 3
⇒ Re(z) > 3

Q10. 

•  0
•  
  
•  
  
• 

Solution
(c)