Complex Number and Quadratic Equation Quiz-6
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. Number of positive integers n for which n^2 + 96 is a perfect square is
Solution
(c)
Let m be a positive integer for which
n^2 + 96 = m^2
⇒m^2 - n^2 = 96
⇒(m + n)(m - n) = 96
⇒(m + n){(m + n) - 2n} = 96
⇒m+n and m-n must be both even
As 96 = 2×48 or 4×24 or 6×16 or 8×12, hence, number of solutions is 4
(c)
Let m be a positive integer for which
n^2 + 96 = m^2
⇒m^2 - n^2 = 96
⇒(m + n)(m - n) = 96
⇒(m + n){(m + n) - 2n} = 96
⇒m+n and m-n must be both even
As 96 = 2×48 or 4×24 or 6×16 or 8×12, hence, number of solutions is 4
Q2.The greatest positive argument of complex number satisfying |z-4| = Re(z) is
Solution
(d)
|z - 4| = Re(z)
⇒ √((x - 4)^2 + y^2 ) = x
⇒x^2 - 8x + 16 + y^2 = x^2
⇒y^2 = 8(x - 2)
The given relation represents the part of the parabola with focus (4, 0) lying above x-axis and the imaginary axis as the directrix. The two tangents from directrix are at right angle. Hence greatest positive argument of z is π/4
(d)
⇒ √((x - 4)^2 + y^2 ) = x
⇒x^2 - 8x + 16 + y^2 = x^2
⇒y^2 = 8(x - 2)
The given relation represents the part of the parabola with focus (4, 0) lying above x-axis and the imaginary axis as the directrix. The two tangents from directrix are at right angle. Hence greatest positive argument of z is π/4
Q3. If x and y are complex numbers, then the system of equations (1 + i)x + (1 - i)y = 1 ,2ix + 2y = 1 + i has
Solution
(c)
Observing carefully the system of equations, we find
(1 + i)/2i = (1 - i)/2 = 1/(1 + i)
Hence, there are infinite number of solutions
(c)
Observing carefully the system of equations, we find
(1 + i)/2i = (1 - i)/2 = 1/(1 + i)
Hence, there are infinite number of solutions
Q4. For the equation 3x^2 + px + 3 = 0 , p > 0, if one of the root is square of the other, then p is equal to
Solution
(c)
Let α , α^2 be the roots of 3x^2 + px + 3 = 0. Now, S = α + α^2 = -p/3 , p = α^3 = 1
⇒ α = 1 , ω , ω^2 (where ω = (-1 + √3 i)/2)
α + α^2 = -p/3
⇒ ω + ω^2 = -p/3
⇒-1=-p/3⇒p=3
(c)
Let α , α^2 be the roots of 3x^2 + px + 3 = 0. Now, S = α + α^2 = -p/3 , p = α^3 = 1
⇒ α = 1 , ω , ω^2 (where ω = (-1 + √3 i)/2)
α + α^2 = -p/3
⇒ ω + ω^2 = -p/3
⇒-1=-p/3⇒p=3
Q5.
If α,β are the roots of the equation x^2 - 2x + 3 = 0. Then the equation whose roots are P = α^3 - 3α^2 + 5α - 2 and Q = β^3 - β^2 + β + 5 is
Solution
(c)
Given, α,β are roots of equation
x^2 - 2x + 3 = 0
⇒ α^2 - 2α + 3 = 0 (1)
And β^2 - 2β + 3 = 0 (2)
⇒ α^2 = 2α - 3 ⇒α^3 =2α^2 - 3α
⇒ P = (2α^2 - 3α) - 3α^2 + 5α - 2
⇒ -α^2 + 2α - 2 = 3 - 2 = 1, [Using (1)]
Similarly, we have Q = 2
Now, sum of root is 3 and product of roots is 2. Hence, the required equation is x^2 - 3x + 2 = 0
(c)
Given, α,β are roots of equation
x^2 - 2x + 3 = 0
⇒ α^2 - 2α + 3 = 0 (1)
And β^2 - 2β + 3 = 0 (2)
⇒ α^2 = 2α - 3 ⇒α^3 =2α^2 - 3α
⇒ P = (2α^2 - 3α) - 3α^2 + 5α - 2
⇒ -α^2 + 2α - 2 = 3 - 2 = 1, [Using (1)]
Similarly, we have Q = 2
Now, sum of root is 3 and product of roots is 2. Hence, the required equation is x^2 - 3x + 2 = 0
Q6. If centre of a regular hexagon is at origin and one of the vertices on Argand diagram is 1 + 2i, then its perimeter is
Solution
(d)
Let the vertices be z_0,z_1,…,z_5 w.r.t. centre O at origin and |z_0 |=√5
Now ∆OA_2 A_3 is equilateral ⇒ OA_2 = OA_3 = A_2 A_3 = √5
=|z_0 ||cosθ+i sinθ-1| Perimeter =6√5
(d)
Now ∆OA_2 A_3 is equilateral ⇒ OA_2 = OA_3 = A_2 A_3 = √5
=|z_0 ||cosθ+i sinθ-1| Perimeter =6√5
Q7.
Solution
(c)
(c)
Q8.
P(x) is a polynomial with integral coefficients such that for four distinct integers a , b , c , d ; P(a) = P(b) = P(c) = P(d) = 3. If P(e) = 5 (e is an integer), then
Solution
(d)
P(a) = P(b) = P(c) = P(d) = 3
⇒P(x) = 3 has a,b,c,d as its roots
⇒P(x) - 3 = (x - a)(x - b)(x - c)(x - d)Q(x) [∵Q(x) has integral coefficient]
Given P(e) = 5, then
(e - a)(e - b)(e - c)(e - d) Q(e)=5
This is possible only when at least three of the five integers (e - a),(e - b)(e - c)(e - d)Q(e) are equal to 1 or -1. Hence, two of them will be equal, which is not possible. Since a,b,c,d are distinct integers, therefore P(e) = 5 is not possible
(d)
P(a) = P(b) = P(c) = P(d) = 3
⇒P(x) = 3 has a,b,c,d as its roots
⇒P(x) - 3 = (x - a)(x - b)(x - c)(x - d)Q(x) [∵Q(x) has integral coefficient]
Given P(e) = 5, then
(e - a)(e - b)(e - c)(e - d) Q(e)=5
This is possible only when at least three of the five integers (e - a),(e - b)(e - c)(e - d)Q(e) are equal to 1 or -1. Hence, two of them will be equal, which is not possible. Since a,b,c,d are distinct integers, therefore P(e) = 5 is not possible
Q9.If α,β are the roots of ax^2 + bx + c = 0 and a + b , β + h are the roots of px^2 + qx + r = 0 , then h =
Solution
(c)
We have
α + β = -b/a , αβ = c/a
a + h + β + h = -q/p ,(α + h)(β + h) = r/p
⇒ α + β + 2h = -q/p
⇒ -b/a + 2h = (-q)/p [∵a + b = -b/a]
⇒ h = 1/2 (b/a - q/p)
(c)
We have
α + β = -b/a , αβ = c/a
a + h + β + h = -q/p ,(α + h)(β + h) = r/p
⇒ α + β + 2h = -q/p
⇒ -b/a + 2h = (-q)/p [∵a + b = -b/a]
⇒ h = 1/2 (b/a - q/p)
Q10. If t and c are two complex numbers such that |t| ≠ |c|, |t| = 1 and z = (at + b)/(t - c), z = x + iy. Locus of z is (where a,b are complex numbers)
Solution
(c)
z = (at + b)/(t - c)
⇒ t = (b + cz)/(z - a)
Now, |t|= 1
⇒|( b + cz )/( z -a )|=1
⇒|( z + b/c )/( z - a )| = 1/|c| =(≠ 1 as |c| ≠ |t|)
⇒ locus of z is a circle
(c)
z = (at + b)/(t - c)
⇒ t = (b + cz)/(z - a)
Now, |t|= 1
⇒|( b + cz )/( z -a )|=1
⇒|( z + b/c )/( z - a )| = 1/|c| =(≠ 1 as |c| ≠ |t|)
⇒ locus of z is a circle
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