Complex Number and Quadratic Equations Quiz-28
Assertion - Reasoning Type
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.
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: |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 |
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|
(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 Ï€/2
Statement 2: In equilateral triangle orthocentre coincides with centroid
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 Ï€/2
Statement 2: In equilateral triangle orthocentre coincides with centroid
Solution
(b)
(b)
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: a=b=c
Statement 2: Two quadratic equations with real coefficients cannot have only one imaginary root common
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
(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
Solution
(d)
(d)
Q5.
Statement 1: The product of all values of (cosα+i sinα )(3/5) is cos3α+i sin3α
Statement 2: The product of fifth roots of unity is 1
Statement 1: The product of all values of (cosα+i sinα )(3/5) is cos3α+i sin3α
Statement 2: The product of fifth roots of unity is 1
Solution
(b)
We have,
Let x=(cosθ+i sinθ )(3/5)
⇒x5=(cosθ+i sinθ )3
⇒x5-(cos3θ+i sin3θ )=0
⇒ Product of roots =cos3θ+i sin3θ 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
(b)
We have,
Let x=(cosθ+i sinθ )(3/5)
⇒x5=(cosθ+i sinθ )3
⇒x5-(cos3θ+i sin3θ )=0
⇒ Product of roots =cos3θ+i sin3θ 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)=2Ï€, then both z1 and z2 are purely real (z1 and z2 have principle arguments)
Statement 2: Principle argument of complex number lies in (-Ï€,Ï€)
Statement 1: If arg(z1 z2)=2Ï€, then both z1 and z2 are purely real (z1 and z2 have principle arguments)
Statement 2: Principle argument of complex number lies in (-Ï€,Ï€)
Solution
(a)
arg(z12 )=2Ï€⇒arg(z1 )+arg(z2 )=2Ï€⇒arg(z1 )=arg(z2 )=Ï€, as principal arguments are from –Ï€ to Ï€
Hence both the complex numbers are purely real. Hence both the statements are true and statement 2 is correct explanation of statement 1
(a)
arg(z12 )=2Ï€⇒arg(z1 )+arg(z2 )=2Ï€⇒arg(z1 )=arg(z2 )=Ï€, as principal arguments are from –Ï€ to Ï€
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: 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.
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.
(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: 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 |
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 >
(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 >
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: 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
Solution
(d)
f(x,y)=(2x-y)2+(x+y-3)2
Therefore, statement 1 is false as it represents a point (1, 2)
(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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