## Complex Number & Quadratic Equation Quiz-4

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. Let f(x) = ax^2 - bx+ c^2 , b ≠ 0 and f(x) ≠ 0 for all x ∈ R. Then:
•  a + c^2 < b
•  4a + c^2 > 2b
•  9a - 3b + c^2 < 0
•  None of these
Solution
(b)
Here, ax^2 - bx + c^2 = 0 does not have real roots. So,
D < 0 ⇒ b^2 - 4ac^2 < 0 ⇒ a > 0
Therefore, f(x) is always positive. So,
f(2) > 0 ⇒ 4a - 2b + c^2 > 0

Q2.The number of real roots of the equation x^2 - 3| x | + 2 = 0 is
•  2
•  1
•  4
•  3
Solution
(c)
We have,
|x|^2 - 3|x| + 2 = 0
⇒(|x| - 1)(|x| - 2) = 0
⇒|x| = 1 , 2 ⇒ x = ±1 , ±2

Q3.  If z_1 and z_2 are the complex roots of the equation (x - 3)^3 + 1 = 0, then z_1 + z_2 equals to
•   1
•  3
•  5
•  7
Solution
(d)
(x - 3)^3 + 1 = 0
⇒((x - 3)/ -1))^3 = 1
⇒(x - 3)/( -1 ) = 1 , ω , ω^2
⇒x=2 , 3 - ω , 3 - ω^2
Hence, the run of complex root is 6 - (ω + ω)^2 = 6 + 1 = 7

Q4. If x^2 + px + 1 is factor of the expression ax^3 + bx + c , then
•  a^2 - c^2 = ab
•  a^2 + c^2 = -ab
•  a^2 - c^2 = -ab
•  None of these
Solution
(a)
Given that x^2 + px + 1 is a factor of ax^3 + bx + c. Then let ax^3 + bx + c = (x^2 + px + 1)(ax + λ), where λ is a constant. Then equation the coefficients of like powers of x on both sides, we get
0 = ap + λ , b = pλ + a ,c = λ
⇒p = -λ/a = -c/a
Hence,
b = (-c/a)c + a
or ab = a^2 - c^2

•  1
•  e^( -π/2 )
•  e^( -π )
•  None of these
Solution
(a)
i = cos⁡[π/2] + i sin⁡[π/2] = e^(iπ/2)
⇒i^i = ( e^(iπ/2) )^i = e^(-π/2)
⇒z= (i)^((i)^i ) = [i^e]^(-π/2)
⇒|z| = 1

Q6. The number of roots of the equation √(x - 2) (x^2 - 4x + 3 ) = 0 is
•  Three
•  Four
• One
•  Two
Solution
(d) Given equation is satisfied by x = 1 , 2 , 3. But for x = 1 , √(x - 2) is not defined. Hence, number of roots is 2 and the roots are x=2 and 3

Q7.  Total number of values of a so that x^2 - x - a = 0 has integral roots, where a ∈ N and 6 ≤ a ≤ 100, is equal to
•  2
•  4
•  6
•  8
Solution

(d)
x^2 - x - a = 0 ,D = 1 + 4a = odd
D must be perfect square of some odd integer. Let
D = (2λ + 1)^2
⇒ 1 + 4a = 1 + 4λ^2 + 4λ
⇒ a = λ(λ + 1)
Now,
a ∈ [6 , 100]
⇒ a = 6 , 12 , 20 , 30 , 42 , 56 , 72 , 90
Thus a can attain eight different values
Q8. If a,b,care the sides of the triangle ABC such that a ≠ b ≠ c and x^2 - 2(a + b + c)x + 3λ(ab + bc+ ca ) = 0 has real roots, then
•  λ < 4/3
•  λ > 5/3
•  λ ∈ (4/3 , 5/3)
•  λ ∈ (1/3 , 5/3)
Solution
(a)
Since, a,b and c are the sides of a ∆ ABC, then
|a - b| < |c| ⇒ a^2 + b^2 - 2ab < c^2
Similarly, b^2 + c^2 - 2bc < a^2, c^2 + a^2 - 2ca < b^2
(a^2 + b^2 + c^2 ) < 2(ab + bc + ca)
⇒ (a^2 + b^2 + c^2) / (ab + bc + ca )< 2 …..(i)
Also, D ≥ 0 ,(a + b + c)^2 - 3λ(ab + bc + ca) ≥ 0
⇒ (a^2 + b^2 + c^2) / ( ab + bc + ca) > 3λ-2 ………(ii)
From Eqs. (i) and (ii),
3λ - 2 < 2 ⇒ λ < 4/3

Q9.Suppose A,B,C are defined as A=a^2 b+ab^2-a^2 c-ac^2,B=b^2 c+bc^2-a^2 b-ab^2 and C=a^2 c+ac^2-b^2 c-bc^2, where a>b>c>0 and the equation Ax^2+Bx+C=0 has equal roots, then a,b,c are in
•  A.P.
•  G.P.
•  H.P.
•  A.G.P.
Solution
(c)
A = a(b - c)(a + b + c)
B = b(c - a)(a + b + c)
C = c(a - b)(a + b + c)
Now,
Ax^2 + Bx + C = 0
⇒(a + b + c) {a(b - c) x^2 + b(c - a)x + c(a - b)} = 0
Given that roots are equal. Hence,
D = 0
⇒b^2 (c - a^2 ) - 4ac(b - c)(a - b) = 0
⇒b^2 c^2 - 2ab^2 c + b^2 a^2 - 4a^2 bc + 4acb^2 + 4a^2 c^2- 4abc^2 = 0
⇒b^2 c^2 + b^2 a^2 + 4a^2 c^2 + 2ab^2 c - 4a^2 bc - 4abc^2 = 0
⇒(bc + ab - 2ac)^2 = 0
⇒bc + ab = 2ac
⇒1/a + 1/c = 2/b
⇒a , b , c are in H.P.

Q10. Consider the equation x^2 + 2x - n = 0, where n ∈ N and n ∈ [15 , 100]. Total number of different values of ‘n’ so that the given equation has integral roots is
•  8
•  3
•  6
• 4
Solution
(a)
x^2 + 2x - n = 0
⇒(x + 1)^2 = n + 1
⇒x = -1 ± √(n + 1)
Thus, n+1 should be a perfect square. Now,
n ∈ [5 , 100]
⇒ n + 1 ∈[6 , 101]
Perfect square values of n+1 are 9 ,16 ,25 ,36 ,49 ,64 ,81 ,100. Hence, number of values is 8 #### Written by: AUTHORNAME

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