## Complex Number and Quadratic Equations Quiz-28 Assertion - Reasoning Type

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.
Statement 1: |z1-a| < a ,|z2-b| < b ,|z3-c| < c, where a,b,c are positive real numbers, then |z1+z2+z3 | is greater than 2|a+b+c|
Statement 2: |z1± z2 |≤|z1 |+|z2 |
•  Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is False
•  Statement 1 is False, Statement 2 is True
Solution
(d)
|z1+z2+z3 |=|z1-a+z2-b+z3-c+(a+b+c)|
≤|z1-a|+|z2-b|+|z3-c|+|a+b+c|
≤2|a+b+c|
Hence, |z1+z2+z3 | is less than 2|a+b+c|
Q2.
Statement 1: If |z1 |=|z2 |=|z3 |,z1+z2+z3=0 and (z1 ),B(z2 ),C(z3) are the vertices of ∆ABC, then one of the values of arg⁡(z2+z3-2z1 )/(z3-z2) is π/2
Statement 2: In equilateral triangle orthocentre coincides with centroid
•  Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is False
•  Statement 1 is False, Statement 2 is True
Q3.  Let a,b,c be real such that ax2+bx+c=0 and x2+x+1=0 have a common root
Statement 1: a=b=c
Statement 2: Two quadratic equations with real coefficients cannot have only one imaginary root common
•  Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is False
•  Statement 1 is False, Statement 2 is True
Solution
(a)
x2+x+1 = 0
D=-3< 0
Therefore, x2+x+1 = 0 and ax2+bx+c = 0 have both the roots common. Hence, a = b = c

Q4.
Statement 1: If the roots of x5-40x4+Px3+Qx2+Rx+S=0 are in G.P. and sum of their reciprocal is 10, then |S|=64
Statement 2: x1 x2 x3 x4 x5=-S, where x1,x2,x3,x4,x5 are the roots of given equation
•  Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is False
•  Statement 1 is False, Statement 2 is True
Q5.
Statement 1: The product of all values of (cos⁡α+i sin⁡α )(3/5) is cos⁡3α+i sin⁡3α
Statement 2: The product of fifth roots of unity is 1
•  Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is False
•  Statement 1 is False, Statement 2 is True
Solution
(b)
We have,
Let x=(cos⁡θ+i sin⁡θ )(3/5)
⇒x5=(cos⁡θ+i sin⁡θ )3
⇒x5-(cos⁡3θ+i sin⁡3θ )=0
⇒ Product of roots =cos⁡3θ+i sin⁡3θ Also product of roots of the equation x5-1=0 is 1. Hence statement 2 is true. But it is not correct explanation of statement 1
Q6.
Statement 1: If arg⁡(z1 z2)=2π, then both z1 and z2 are purely real (z1 and z2 have principle arguments)
Statement 2: Principle argument of complex number lies in (-π,π)
•  Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is False
•  Statement 1 is False, Statement 2 is True
Solution
(a)
arg⁡(z12 )=2π⇒arg⁡(z1 )+arg⁡(z2 )=2π⇒arg⁡(z1 )=arg⁡(z2 )=π, as principal arguments are from –π to π
Hence both the complex numbers are purely real. Hence both the statements are true and statement 2 is correct explanation of statement 1

Q7.
Statement 1: Let f(x) be quadratic expression such that f(0)+f(1)=0. If -2 is one of the root of f(x)=0, then other root is 3/5.
Statement 2: If α,β are the zero’s of f(x)=ax2+bx+c, then sum of zero’s =-b/a, product of zero’s =c/a.
•  Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is False
•  Statement 1 is False, Statement 2 is True
Solution
(a)
Since, x=-2 is a root of f(x).
∴f(x)=(x+2)(ax+b)
But f(0)+f(1)=0
∴ 2b+3a+3b=0
⇒ -b/a=3/5
Hence, option (a) is correct.

Q8.
Statement 1: Let z be a complex number, then the equation z4+z+2=0 cannot have a root, such that |z| < 1
Statement 2: |z1+z2 |≤|z1 |+|z2 |
•  Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is False
•  Statement 1 is False, Statement 2 is True
Solution
(a)
Suppose there exists a complex number z which satisfies the given equation and is such that |z|< 1. Then,
z4+z+2=0 ⇒-2=z4+z ⇒|-2|=|z4+z|
⇒2≤|z4 |+|z|⇒2< 2, because |z|< 1
But 2< 2 is not possible. Hence given equation cannot have a root z such that |z|< 1
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Q9.  Consider a general expression of degree 2 in two variables as f(x,y)=5x2+2y2-2xy-6x-6y+9
Statement 1: f(x,y) can be resolved into two linear factors over real coefficients
Statement 2: If we compare f(x,y) with ax2+by2+2hxy+2gx+2fy+c=0, we have abc+2fgh-af2-bg2-ch2=0
•  Statement 1 is True, Statement 2 is True; Statement 2 is correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is True; Statement 2 is not correct explanation for Statement 1
•  Statement 1 is True, Statement 2 is False
•  Statement 1 is False, Statement 2 is True
Solution
(d)
f(x,y)=(2x-y)2+(x+y-3)2
Therefore, statement 1 is false as it represents a point (1, 2) #### Written by: AUTHORNAME

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