## Complex Numbers Quiz-1

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. For real values of x, the expression ((x-b)(x-c))/((x-a)) will assume all real values provided
•  a ≤c ≤b
•  b ≥ a ≥c
•  b ≤ c ≤a
•  a ≥ b ≥c
Solution
(b) Let m=((x-b)(x-c))/(x-a)
⇒x2-(b+c+m)x+(bc+am)=0
Since x is real, we must have (b+c+m)2-4(bc+am)≥0
⇒m2+2(b+c-2a)m+(b-c)2≥0 for all m
⇒4(b+c-2a)2-4(b-c)2≤0
⇒(b+c-2a)2-(b-c)2≤0
⇒(b+c-2a+b-c)(b+c-2a-b+c)≤0
⇒2(b-a)2(c-a)≤0
⇒(a-b)(a-c)≤0
⇒b≤a≤c or, c≤a≤b

Q2.

•  {4,1/4}
•  {2,1/2}
•  {1,2}
•  {8,1/8}
Solution

Q3.  If x-c is a factor of order m of the polynomial f(x) of degree n(1&#60m&#60n), then x=c is a root of the polynomial
•   fm (x)
•  fm-1 (x)
•  fn (x)
•  None of these
Solution
(b) Since x-c is a factor of order m of the polynomial f(x)
∴f(x)=(x-c)m Ï•(x), where Ï•(x) is a polynomial of degree n-m
⇒f(x),f' (x)…fm-1 (x) are all zero for x=c but fm (x)≠0 at x=c
⇒x=c is root of f(x),f' (x),…,fm-1 (x)

Q4. Let
x1, x2 be the roots of the equation x2-3x+p=0 and let  x3, x4 be the roots of the equation x2-12x+q=0. If the numbers  x1, x2, x3, x4 (in order) form an increasing G.P., then
•  p=2,q=16
•  p=2,q=32
•  p=4,q=16
•  p=2,q=32
Solution
(b) It is given that x1,x2 are roots of x2-3x+p=0
⇒x1+x2=3,x1 x2=p x3,x4 are roots of x2-12x+q=0 ⇒x3+x4=12 and x3 x4=q
It is given that x1,x2,x3,x4 form an increasing G.P.
Therefore, x1=a,x2=ar,x3=ar2,x4=ar3, where r>1
Now, {(x1+x2=3
⇒a(1+r)=3 x3+x4=12
⇒ar2 (1+r)=12)}
⇒r=2 and a=1
∴x1=1,x2=2,x3=4,x4=8
Thus, p=x1 x2=2 and q=x3 x4=32

Q5.If (√3+i)10=a+ib, then a and b are respectively
•  128 & 128√3
•  64 &-64√3
•  512 &-512√3
•  None of these
Solution
(c) We have, (√3+i)10=a+ib ⇒i10 (1-i√3)10=a+i b
⇒-(-2Ï‰)10=a+i b [∵Ï‰=-1/2+i √3/2]
⇒-210 Ï‰10=a+i b
⇒-210 Ï‰=a+i b
⇒-210 ((-1)/2+i √3/2)=a+i b
⇒29-29 √3 i=a+i b
⇒a=29 and b=-29 √3

Q6. If Ï‰ is an imaginary cube root of unity and x=a+b, y=aÏ‰+bÏ‰2,z=aÏ‰2+bÏ‰, then x2+y2+z2 is equal to
•  6ab
•  3ab
•  6a2b2
•  3a2b2
Solution
(a) x2+y2+z2=(a+b)2+Ï‰2 (a+bÏ‰)2+(aÏ‰2+bÏ‰)2
=a2+b2+2ab+a2 Ï‰2+b2 Ï‰4+2abÏ‰3+a2 Ï‰4+b2 Ï‰2+2abÏ‰3 =a2 (1+Ï‰+Ï‰2 )+b2 (1+Ï‰+Ï‰2 )+6ab [∵Ï‰4=Ï‰] =6ab [∵1+Ï‰+Ï‰2=0]

Q7.The conjugate of the complex number (1+i)2/(1-i) is
•  1-i
•  1+i
•  -1+i
•  -1-i
Solution
(d) Given complex number is (1+i)2/(1-i)=((1+i2+2i))/(1-i)×(1+i)/(1+i)=(2i+2i2)/(1+1)=i-1
∴ Required conjugate is -i-1

Q8.Number of non-zero integral solutions of the equation (1-i)n=2n is
•  1
•  2
•  Infinite
•  None of these
Solution
(d) We have, (1-i)n=2n
⇒|1-i|n=|2|n
⇒(√2)n=2n
⇒2n/2=2n
⇒2n/2=1
⇒n=0
So, there is no non-zero integral solution of the given equation

Q9.If the roots of the equation qx2+px+q=0 are complex, where p,q are real, then the roots of the equation x2-4qx+p2=0 are
•  Real and unequal
•  Real and equal
•  Imaginary
•  None of these
Solution
(a) The given equations are qx2+px+q=0 …(i)
and x2-4qx+p2=0 …(ii)
Since, root of the Eq. (i) are complex, therefore p2-4q2<0
Now, discriminant of Eq. (ii) is 16q2-4p2=-4(p2-4q2 )>0
Hence, roots are real and unequal.

Q10. If z1,z2,z3,z4 represent the vertices of a rhombus taken in the anticlockwise order, then
•  z1+z2+z3+z4=0
•  z1+z2=z3+z4
•  amp (z2-z4)/(z1-z3 )=Ï€/2
• amp (z1-z2)/(z3-z4 )=Ï€/2
Solution
(c) Since the diagonals of a rhombus bisect each other at right-angle
∴(z1+z3)/2=(z2+z4)/2
⇒z1+z3=z2+z4
Also, ∠AOB=Ï€/2
⇒arg⁡ (z2-z4)/(z1-z3 )=Ï€/2

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