Complex Number & Quadratic Equation Quiz-11
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. If z1,z2 and z3 are complex numbers such that |z1 |=|z2 |=|z3 |=|(1/z1)+(1/z2)+(1/z2)|=1, then |z1+z2+z3 | is
Solution
(a)
(a)
Q2.If z1 z2 be complex numbers such that z1 ≠ z2 and |z1 |= |z2 |. If z1has positive real part and z2 has negative imaginary part, then[(z1+z2 )/(z1-z2 )] may be
Solution
(a)
(a)
Q3. If the expression [mx-1+(1/x)] is non-negative for all positive real x, then the minimum value of m must be
Solution
(c) We know that ax2 + bx + c ≥ 0 , ∀ x ∈ R,
If a > 0 and b2 - 4ac ≤ 0. So,
mx -1 + 1/x ≥ 0 ⇒ (mx2 - x + 1)/x ≥ 0
⇒mx2 - x + 1 ≥ 0 as x > 0
Now, mx2 - x + 1 ≥ 0 if m > 0 and 1-4m ≤ 0
⇒ m > 0 and m ≥ 1/4
Thus, the minimum value of m is ¼
(c) We know that ax2 + bx + c ≥ 0 , ∀ x ∈ R,
If a > 0 and b2 - 4ac ≤ 0. So,
mx -1 + 1/x ≥ 0 ⇒ (mx2 - x + 1)/x ≥ 0
⇒mx2 - x + 1 ≥ 0 as x > 0
Now, mx2 - x + 1 ≥ 0 if m > 0 and 1-4m ≤ 0
⇒ m > 0 and m ≥ 1/4
Thus, the minimum value of m is ¼
Q4. If for complex numbers z1 z2, arg(z1 )-arg(z2 )=0, then |z1-z2 | is equal to
Solution
(c)
(c)
Q5.
Solution
(d)
(d)
Q6. For positive integers n1,n2 the value of the expression (1+i)(n1 ) + (1 + i3 )(n1 ) + (1 + i5 )(n2 ) + (1 + i7 )(n2 ), where i=√(-1) is a real number if and only if
Solution
(d)
(d)
Q7. If |z2+iz1 |=|z1 |+|z2 | and |z1 |=3 and |z2 |=4, then area of ∆ABC, if affixes of A,B and C are z1,z2 and [(z2-iz1)/(1-i)] respectively, is
Solution
(d)
(d)
Q8.
x2-xy+y2-4x-4y+16=0 represents
Solution
(a)
Given equation is
x2-(y+4)x+y2-4y+16=0
Since x is real, so,
D≥0
⇒(y+4)2-4(y2-4y+16)≥0
⇒ -3y2+24y-48≥0
⇒^2-8y+16≤0
⇒(y-4)2≤0
⇒y-4=0
⇒y=4
Since the equation is symmetric in x and y, therefore x=4 only
(a)
Given equation is
x2-(y+4)x+y2-4y+16=0
Since x is real, so,
D≥0
⇒(y+4)2-4(y2-4y+16)≥0
⇒ -3y2+24y-48≥0
⇒^2-8y+16≤0
⇒(y-4)2≤0
⇒y-4=0
⇒y=4
Since the equation is symmetric in x and y, therefore x=4 only
Q9.
Sum of common roots of the equations z3+2z2+2z+1=0 and z1985+z100+1=0 is
Solution
(a)
We have,
z3+2z2+2z+1=0
⇒(z3+1)+2z(z+1)=0
⇒(z+1)(z2+z+1)=0
⇒z=-1,ω,ω2
Since z=-1 does not satisfy z1985+z100+1=0 while z=ω,ω2 satisfy it, hence sum is ω+ω2=-1
(a)
We have,
z3+2z2+2z+1=0
⇒(z3+1)+2z(z+1)=0
⇒(z+1)(z2+z+1)=0
⇒z=-1,ω,ω2
Since z=-1 does not satisfy z1985+z100+1=0 while z=ω,ω2 satisfy it, hence sum is ω+ω2=-1
Q10. If the roots of the equation ax2-bx+c=0 are α,β then the roots of the equation b2 cx2-ab2 x+a3=0 are
Solution
(b)
(b)
