## COMPLEX NUMBERS AND QUADRATIC EQUATIONS Quiz-3

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 Î±,Î² be the roots of the equation u^2-2u+2=0 and if cot⁡Î¸= x + 1, then [( x + Î± )^n - (x + Î²)^n ]/[Î± - Î²] is equal to
•  sin⁡nÎ¸ / sin^n⁡Î¸
•  cos⁡nÎ¸ / cos^n⁡Î¸
•  sin⁡nÎ¸ / cos^n⁡Î¸
•  cos⁡nÎ¸ / sin^n⁡Î¸
Solution
(a)
u^2 - 2u + 2 = 0 ⇒ u = 1 ± i
⇒( (x + Î±)^n - ( x + Î²)^n ) /( Î± - Î²)
=( [ (cot⁡Î¸ - 1) + (1 + i) ]^n - [ ( cot⁡Î¸ - 1 ) + (1 - i) ]^n )/2i (∵cot⁡Î¸ - 1 = x)
=( (cos⁡Î¸ + isin⁡Î¸ )^n - (cos⁡Î¸ - isin⁡Î¸ )^n )/( sin^n⁡Î¸ 2i)
=( 2i sin⁡nÎ¸ )/( sin^n⁡Î¸ 2i )
=sin⁡nÎ¸ / sin^n⁡Î¸

Q2.If the cube roots of unity are 1 , Ï‰ , Ï‰^2, then the roots of the equation (x- 1)^3 + 8 = 0 are
•  -1 , 1 + 2Ï‰ , 1 + 2Ï‰^2
•  -1 , 1 - 2Ï‰ , 1 - 2Ï‰^2
•  -1 , -1 , -1
•  None of these
Solution
(b)
( (x - 1)/( -2 ) )^3 = 1
⇒(x - 1)/ (-2) = 1 , Ï‰ , Ï‰^2
⇒x = -1 , 1 - 2Ï‰ , 1 , -2Ï‰^2

Q3.  Suppose A is a complex number and n∈N, such that A^n = [(A + 1)]^n = 1, then the least value of n is
•   3
•  6
•  9
•  12
Solution
(b)
Let A = x + iy . Given,
| A | = 1 ⇒ x^2 + y^2 = 1 and
| A + 1 | = 1
⇒( x + 1 )^2 + y^2 = 1
⇒x = -1/2 and y = ±√3/2
⇒A = Ï‰ or Ï‰^2
⇒(Ï‰)^n = (1 + Ï‰)^n = (-Ï‰^2 )^n
Therefore, n must be even and divisible by 3

Q4. If a,b,c,d are four consecutive terms of an increasing A.P. then the roots of the equation (x - a) (x - c) + 2(x - b) (x - d) = 0 are
•  Non-real complex
•  Real and equal
•  Integers
•  Real and distinct
Solution
(d)
Given that a < b < c < d. let
f(x) = (x - a)( x - c) + 2(x - b)(x - d)
⇒f(b) = (b - a)(b - c) < 0
and f(d) = (d - a)(d - c) > 0
Hence, f(x) = 0 has one root in (b , d). Also, f(a)f(c) < 0. So the other root lies in (a , c). Hence, roots of the equation are real and distinct

Q5. If the equations ax^2 + bx + c = 0 and x^3 + 3x^2 + 3x + 2 = 0 have two common roots, then
•  a=b=c
•  a=b≠c
•  a=-b=c
•  None of these
Solution
(a) By observation x=-2 satisfies equation x^3 + 3x^2 + 3x + 2 = 0 then we have
(x + 2)( x^2 + x + 1) = 0
x^2 + x + 1 = 0 has non-real roots
Since non-real roots occur in conjugate pair,
x^2 + x + 1 = 0 and ax^2 + bx + c = 0 are identical
⇒a=b=c

Q6. If Î± and Î² , Î± and Î³ , Î± and Î´ are the roots of the equations ax^2 + 2bx + c = 0, 2bx^2 + cx + a = 0 and cx^2 + ax + 2b = 0, respectively, where a,b and c are positive real numbers, then Î± + Î±^2 =
•  abc
•  a + 2b + c
• -1
•  0
Solution
(c)
Since Î± is root of all equations
aÎ±^2 + 2bÎ± + c = 0
2bÎ±^2 + cÎ± + Î± = 0
cÎ±^2 + aÎ± + 2b = 0
Adding we get ( a + 2b + c )( Î±^2 + Î± + 1 ) = 0
a + 2b + c ≠ 0 as a , b , c > 0
⇒ Î±^2 + Î± + 1 = 0 or Î±^2 + Î± = -1

Q7. If Î±,Î² are the roots of x^2 + px + q = 0 and x^2n + p^n x^n + q^n = 0 and if ( Î±/Î² ) , ( Î²/Î± ) are the roots of x^n + 1 + (x + 1)^n = 0, then n(∈N)
•  Must be an odd integer
•  May be any integer
•  Must be an even integer
•  Cannot say anything
Solution
(c)
We have,
Î± + Î² =-p and Î±Î² = q (1)
Also, since Î± , Î² are the root of x^2n + p^n x^n + q^n = 0,
we have ,
Î±^2n + p^n Î±^n + q^n = 0 and Î²^2n + p^n Î²^n + q^n = 0
Subtracting the above relations, we get
(Î±^2n - Î²^2n ) + p^n (Î±^n - Î²^n ) = 0
∴ Î±^n + Î²^n = -p^n (2)
Given, Î±/Î² or Î²/Î± is a root of x^n + 1 + (x + 1)^n = 0. So,
( Î±/Î² )^n + 1 + [ (Î±/Î²) + 1 ]^n = 0
⇒( Î±^n + Î²^n ) + ( Î± + Î² )^n = 0
⇒ -p^n + (-p)^n = 0 [Using (1) and (2)]
It is possible only when n is even

Q8. If z^2 + z |z| + |z^2 | = 0, then the locus of z is
•  A circle
•  TA straight line
•  A pair of straight lines
•  None of these
Solution
(c)
z^2 + z|z| + |z|^2 = 0
⇒( z/|z| )^2 + z/|z| + 1 = 0
⇒z/|z| = Ï‰ , Ï‰^2 ⇒ z = Ï‰ |z| or z = Ï‰^2 |z|
⇒x + iy = |z| (( -1 )/2 + ( i√3 ) / 2) or x + iy =| z |((-1)/2 - (i√3)/2)
⇒x = -1/2 |z|, y = |z|√3/2 or x = -|z|/2, y =- (|z| √3)/2
⇒y + √3 x = 0 or y - √3 x = 0 ⇒ y^2 - 3x^2 = 0

Q9. If |z|=1 and w=(z-1)/(z+1) (where z≠-1), then Re(w) is
•  0
•
•
•
Solution
(a)
Since, | z | = 1 and w = (z - 1)/(z + 1)
⇒ z = (1 + w)/(1 - w)
⇒ | z |=( | 1 + w |)/(| 1 - w |)
⇒ | 1 - w | = | 1 + w | [ ∵ | z | = 1 ]
⇒ 1 + | w |^2 - 2 Re (w) = 1 + | w |^2 + 2 Re (w)
⇒ Re (w) = 0

Q10. If the equation x^2 + ax + b = 0 has distinct real roots and x^2 + a| x | + b = 0 has only one real root, then which of the following is true
•  b = 0 , a > 0
•  b = 0 , a < 0
•  b > 0 , a < 0
• b<0,a>0
Solution
(a) Since the equation x^2+ax+b=0 has distinct real roots and x^2+a|x|+b=0 has only one real root, so one root of the equation x^2+ax+b=0 will be zero and other root will be negative. Hence, b=0 and a>0

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