Complex Numbers Quiz-5

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 one root of the equation x2+(l-3i)x-2(l+i)=0 is -l+i, then the other root is
•  -1-i
•  (-l-i)/2
•  i
•  2i
Solution
(d) Let another root of equation x+(1-3i)x-2(1+i)=0 is Î± ∴ Î±+(-1+i)=-(1-3i) ⇒ Î±=2i

Q2.In the argand plane, if O,P and Q represent respectively the origin O and the complex numbers z and z+iz respectively, then ∠OPQ is
•  Ï€/4
•  Ï€/3
•  Ï€/2
•  2Ï€/3
Solution
(c) Clearly, angle between z and iz is a right angle ∴∠OPQ=Ï€/2
Q3.  If z1,z2,z3,…,n are n nth roots of unity, then for k=1,2,…,n
•   |zk |=k|z(n+1) |
•  |z(k+1) |= k|zk |
•  |z(k+1) |=|zk |+|z(k+1) |
•  |z(k) |= |zk+1 |
Solution
(d) We have, zk=e((i 2 Ï€ k)/n), k=0,1,2,…,n-1
∴|zk |=|e((i 2 Ï€ k)/n) |=1 for all=0,1,2,…n-1 ⇒|zk |=|zk+1 | for all k=0,1,2,…,n-1

Q4. If both the roots of the equation ax2+bx+c=0 are zero, then
•  b=c=0
•  b=0,c≠0
•  b≠0,c=0
•  b≠0,c≠0
Solution
(a) Let Î±,Î² be the two roots of the equation ax2+bx+c=0. Then, Î±+Î²=-b/a and Î± Î²=c/a ⇒-b/a=0 and c/a=0 [∵Î±=Î²=0] ⇒b=0,c=0

Q5.If (x2-bx)/(ax-c)=(Î»-1)/(Î»+1) has roots equal in magnitude and opposite in sign then the value of Î» is
•  (a-b)/(a+b)
•  (a+b)/(a-b)
•  c
•  1/c
Solution
(a) The given equation is x2 (Î»+1)-x{b(Î»+1)+a(Î»-1)}+c(Î»-1)=0
This equation has roots equal in magnitude but opposite in sign
∴ Sum of the roots =0
⇒(b(Î»+1)+a(Î»-1))/(Î»+1)=0
⇒Î»=(a-b)/(a+b)

Q6. Let n=2006!. Then 1/log2⁡n +1/log3⁡n +...+1/log2006⁡n is equal to
•  2006
•  2005
• 2005!
•  1
Solution
(d) Given, n=2006!
∴ 1/log2⁡n +1/log3⁡n +...+1/log2006⁡n =logn2+logn3+...+logn2006 =logn(2.3.4.….2006) =logn(2006!)=lognn = 1

Q7.All the values of m for which both roots of the equation x2-2mx+m2-1=0 are greater than -2 but less than 4 lie in the interval
•  m>3
•  -1<m<3
•  1<m<4
•  -2<m<0
Solution
(b) According to the equation, D≥0,-2<-b/2a<4,f(4)>0 and f(-2)>0
Now, D≥0; 4m2-4m2+4≥0 ⇒ 4>0 ∀ m∈R ...(i)
-2<-b/2a<4; -2<(2m/2.1)<4 ⇒ -2<m<4 ...(ii)
f(4)>0 ⇒16-8m+m2-1>0 ⇒(m-3)(m-5)>0 ⇒ -∞<m<3 and 5<m<∞ ...(iii)
And f(-2)>0 ⇒ 4+4m+m2-1>0 ⇒ (m+3)(m+1)>0 ⇒ -∞<m<-3 and -1<m<∞ ...(iv)
∴ From Eqs. (i), (ii), (iii) and (iv), we get m lie between -1 and 3

Q8. If (5z2)/(11z1 ) is purely imaginary, then the value of |(2z1+3z2)/(2z1-3z2 )| is
•  37/33
•  2
•  1
•  3
Solution
(c) Let (5z2)/(11z1 )=iy ⇒ z2/z1 =11/5 iy Now, |(2z1+3z2)/(2z1-3z2 )|=|(2+3 z2/z1 )/(2-3 z2/z1 )|=|(2+33/5 iy)/(2-33/5 iy)|=1

Q9.

•  -1
•  0
•  1
•  None of these
Solution

Q10. If Î±,Î² and Î³ are the roots of the equation x3-8x+8=0, then ∑Î±2 and ∑1/Î±Î² are respectively
•  0 and -16
•  16 and 18
•  -16 and 0
• 16 and 0
Solution
(d) Here, Î±+Î²+Î³=0,Î±Î²+Î²Î³+Î³Î±=-8,Î±Î²Î³=-8 ...(i)
∴(Î±+Î²+Î³)2=0 ⇒ Î±2+Î²2+Î³2+2(Î±Î²+Î²Î³+Î³Î±)=0
⇒ ∑Î±2=-2(-8)=16 [from Eq. (i)] And 1/Î±Î²+1/Î²Î³+1/Î³Î±=(Î±+Î²+Î³)/Î±Î²Î³
⇒ 1/(∑Î±Î²)=0/(-8)=0 [from Eq. (i)]

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