complex-numbers-quiz-10

Complex Numbers Quiz-10

Dear Readers,

As per analysis for previous years, it has been observed that students preparing for JEE MAINS find Mathematics out of all the sections to be complex to handle and the majority of them are not able to comprehend the reason behind it. This problem arises especially because these aspirants appearing for the examination are more inclined to have a keen interest in Mathematics due to their ENGINEERING background. 

Furthermore, sections such as Mathematics are dominantly based on theories, laws, numerical in comparison to a section of Engineering which is more of fact-based, Physics, and includes substantial explanations. By using the table given below, you easily and directly access to the topics and respective links of MCQs. Moreover, to make learning smooth and efficient, all the questions come with their supportive solutions to make utilization of time even more productive. Students will be covered for all their studies as the topics are available from basics to even the most advanced. 

Q1. If 2+i√3 is a root of the equation x²+px+q=0, then the value of (p,q) is

•  (-7, 4)
•  (-4, 7)
•  (4, -7)
•  (7, -4)

Solution
(b) Since, 2+i√3 is a root of the equation x²+px+q=0, then the other root will be 2-i√3 

 ∴ 2+i√3+2-i√3=-p 

 ⇒ p=-4 And (2+i√3)(2-i√3=q) 

 ⇒ q=7

 ∴ The value of (p,q) is (-4,7)

Q2.Product of the real roots of the equation t² x²+|x|+9=0,(t≠0)

•  Is always positive
•  Is always negative
•  Does not exist
•  None of these

Solution
(d) We have, t² x²+|x|+9>0 for all x∈R So, given equation has no real root

Q3.  If |z₁ |=|z₂ |=|z₃ |=1 and z₁,z₂,z₃ represent the vertices of an equilateral triangle, then

•   z₁+z₂+z₃=0 and z₁ z₂ z₃=1
•  z₁+z₂+z₃=1 and z₁ z₂ z₃=1
•  z₁ z₂+z₂ z₃+z₃ z₁=0 and z₁+z₂+z₃=0
•  z₁ z₂+z₂ z₃+z₃ z₁=0 and z₁ z₂ z₃=1

Solution

Q4. If x=√(1+√(1+√(1+⋯∞)) ) , then x is equal to

•  (1+√5)/2
•  (1-√5)/2
•  (1±√5)/2
•  None of these

Solution
(a) We have, x=√(1+√(1+√(1+⋯∞)) ) 

 ⇒x=√(1+x) 

 ⇒x²=1+x ⇒x²-x-1=0 

 ⇒x=(1±√(1+4))/2=(1±√5)/2 

 As x>0, we take only x=(1+√5)/2.

Q5.If x²+ax+1 is a factor of ax³+bx+c, then

•  b+a+a²=0,a=c
•  b-a+a²=0,a=c
•  b+a-a²=0,a=0
•  None of these

Solution
 (d) x²+ax+1 must divide ax³+bx+c. 

 Now, (ax³+bx+c)/(x²+ax+1)=(a(x-a)+(b-a+a³ )x+c+a²)/(x²+ax+1) 

 Reminder must be zero 

 ⇒b-a+a³=0,a²+c=0

Q6. If x²-2r a_r x+r=0;r=1,2,3 are three quadratic equations of which each pair has exactly one root common, then the number of solutions of the triplet (a₁ ,a₂ ,a₃ ) is

•  1
•  2
• 9
•  27

Solution
(b) We have, x²-2 a₁ x+1=0 …(i)

 x²-4a₂ x+2=0 …(ii) 

 x²-6a₃ x+3=0 …(iii) 

 Let α,β;β,γ and γ,α be the pairs of roots of equations (i),(ii) and (iii) respectively. 

Then, α+β=2a₁,α β=1 …(iv)

 β+γ=4a_2,β γ=2 …(v) 

 γ+α=6a_3,γ α=3 …(vi)

 Now, α β=1,β γ=2 and γ α=3 

 ⇒(α β)(β γ)(γ α)=1×2×3

⇒α,β γ=± √6 ∴α=± √(3/2) ,β=± √(2/3) ,γ=± √6 and, α+β-2a₁,β+γ=4a₂,γ+α=6a₃ 

⇒α+β+γ=a₁+2a₂+3a₃ 

∴α=a₁-2a₂+3a₃,β=a₁+2a₂-3a_3,γ=-a₁+2a₂+3a₃ 

 Thus, we have the following sets of simultaneous linear equations:

 a₁-2a₂+3a₃=√(3/2) a₁-2a₂+3a₃=-√(3/2) a₁+2a₂-3a₃=√(2/3) and,a₁+2a₂-3a₃=-√(2/3) -a₁+2a₂+3a₃=√6 -a₁+2a₂+3a₃=-√6 

 Hence, there are two triplets (a₁,a₂,a₃)

Q7.If ω is an imaginary cube root of unity, n is a positive integer but not a multiple of 3, then the value of 1+ωⁿ+ω²ⁿ is

•  3
•  ω+2
•  0
•  ω²+1

Solution
(c) Since, n is not a multiple of 3, therefore=3m+1,n=3m+2, where m is a positive integer.

 For n=3m+1, 1+ωⁿ+ω²ⁿ=1+ω^(3m+1)+ω^(2(3m+1)) =1+ω^3m ω+(ω³ )^2m ω²=1+ω+ω²=0 

 Similarly, for n=3m+2

 ∴ 1+ωⁿ+ω²ⁿ=1+ω^(3m+2)+ω^2(3m+2) =1+ω^3m.ω²+(ω³ )^2m.ω³.ω=0 [∵ ω³=1] 

Q8. Let a∈R. If the origin and the non-real roots of 2z²+2z+a=0 form the three vertices of an equilateral triangle in the argand plane, then a=

•  1
•  2
•  -1
•  None of these

Solution
(d) We have, 2z²+2z+a=0⇒z=(-2±√(4-8a))/4=(-1±√(1-2a))/2

 For z to be non-real, we must have 4-8a<0⇒a>1/2 Let z₁=(-1+√(1-2a))/2 and z₂=(-1-√(1-2a))/2 

 Now, origin and points representing z₁ and z₂ will form an equilateral triangle in the argand plane, 

if z₁²+z₂²=z₁ z₂ [∵z₁₂+z₂²+z₃²=z₁ z₂+z₂ z₃+z₃ z₁] ⇒(z₁+z₂ )²=3 z₁ z₂ ⇒1=3a/2⇒a=2/3

 Clearly, a=2/3 satisfies the condition a>1/2 Hence, a=2/3 

Q9.If a=cos⁡θ+i sin⁡θ,then (1+a)/(1-a) is equal to

•  cot ⁡θ/2
•  cot⁡ θ
•  i cot⁡ θ/2
•  i tan⁡ θ/2

Solution
(c) (1+a)/(1-a)=(e^((-iθ)/2) (1+e^iθ))/(e^((-iθ)/2) (1-e^iθ))=(e^(-i(θ/2) )+e^{(i θ/2)})/(e^(-i(θ/2) )-e^{(-i θ/2)} ) = (2 cos⁡θ/z)/(-2i sin⁡θ/2 )=i cot⁡θ/2

Q10. The value of k for which the equation (k-2) x²+8x+k+4=0 has both roots real, distinct and negative, is

•  0
•  2
•  3
• -4

Solution
(c) Let f(x)=(k-2) +8x+k+4 If f(x)=0 has both negative roots, then we must have 

 (i) Discriminant >0 

 (ii) Vertex of y=f(x) is on left side of y-axis

 (iii) (k-2)f(0)>0

 Now, (i) Discriminant >0 

 ⇒64-4(k-2)(k+4)>0 

 ⇒k²+2k-24<0⇒-6<k<4 …(i) (ii) Vertex is on left side of y-axis 

 ⇒-8/2(k-2) <0⇒k-2>0⇒k>2 …(ii) (iii) (k-2)f(0)>0 ⇒(k-2)(k+4)>0⇒k<-4 or k>2 …(iii) 

 From (i),(ii) and (iii), we obtain k∈(2,4) Hence, k=3