## Complex Number & Quadratic Equation Quiz-8

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 #### Written by: AUTHORNAME

AUTHORDESCRIPTION ## Want to know more

Please fill in the details below:

## Latest NEET Articles\$type=three\$c=3\$author=hide\$comment=hide\$rm=hide\$date=hide\$snippet=hide

Name

ltr
item
BEST NEET COACHING CENTER | BEST IIT JEE COACHING INSTITUTE | BEST NEET, IIT JEE COACHING INSTITUTE: Complex Numbers and Quadratic Equations-Quiz-8