## Complex Numbers Quiz-3

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 z is a complex number in the Argand plane such that arg⁡((z-3√3)/(z+3√3))=Ï€/3 then the lous of z is
•  |z-3i|=6
•  |z-3i|=6,Im (z)>0
•  |z-3i|=6,Im (z)<0
•  None of these
Solution

Q2.Let z1,z2,z3 be the affixes of the vertices of a triangle having the circumcentre at the origin. If z is the affix of it’s orthocentre, then z is equal to
•  (z1+z2+z3)/3
•  (z1+z2+z3)/2
•  z1+z2+z3
•  None of these
Solution

Q3.

•   |z|=5
•  |z|<5
•  |z|>5
•  None of these
Solution
(b) log√3⁡((|z|2-|z|+1)/(2+|z| ))<2
⇒ (|z|2-|z|+1)/(2+|z| )<(√3)2
⇒ | z|2-|z|+1<3(2+|z|)
⇒ |z|2-4|z|-5<0 ⇒ (|z|+1)(|z|-5)<0
⇒ -1<|z|<5⇒|z|<5 as |z|>0
∴ Locus of z is | z|<5

Q4.

•  1
•  6
•  2/3
•  3
Solution
(d) log3⁡5.log2527.log49⁡7/log81⁡3 =(log⁡5/log⁡3 .3/2.log⁡3/log⁡5 .1/2)/(1/4) =3

Q5.(a2-3a+2) x2+(a2-5a+6)x+a-2=r for three distinct values of x for some r∈R,if a+r is equal to
•  1
•  2
•  3
•  Does not exist
Solution
(b) Now, a2-3a+2=0 ⇒a=1,2 …(i)
and a2-5a+6=0 ⇒a=2 ,3 ...(ii)
⇒a-2-r=0 At a=2 [common value from Eqs. (i) and (ii)] r=0 So, a+r=2

Q6. The maximum value of |z| where z satisfies the condition |z+2/z|=2, is
•  √3-1
•  √3+1
• √3
•  √2+√3
Solution
(b) |z+2/z|=2
⇒|z|-2/|z| ≤2
⇒|z|2-2|z|-2≤0
This is a quadratic equation in |z|
∴|z|≤(2±√(4+8))/2≤1±√3
Hence, maximum value of | z| is 1+√3

Q7.If sec⁡Î± and tan⁡Î± are the roots of ax2+bx+c=0, then
•  a2-b2+2ac=0
•  a3+b3+c3-2abc=0
•  a4+4ab2 c=b4
•  None of these
Solution

Q8.The equation whose roots are reciprocal of the roots of the equation ax2+bx+c=0 is
•  bx2+cx+a=0
•  bx2+ax+c=0
•  cx2+ax+b=0
•  cx2+bx+a=0
Solution
(d) Let roots of the equation ax+bx+c=0 are Î± and Î²
∴ Î±+Î²=-b/a and Î±Î²=c/a Now, 1/Î±+1/Î²=(Î±+Î²)/Î±Î² =(-b/a)/(c/a)=(-b)/c And 1/Î±×1/Î²=1/(c/a)=a/c
∴ Required equation is x2-(-b/c)x+a/c=0 ⇒ cx2+bx+a=0
Alternate
To find the equation of reciprocal rots, interchange the coefficients of x2 and constant term in the given equation then required equation is cx2+bx+a=0

Q9.If z1=1+2i and z2=3+5i, then Re [(zc2 )z1/z2] is equal to
•  -31/17
•  17/22
•  -17/31
•  22/17
Solution
(d) (zc2 )z1=(3-5i)(1+2i)=13+i
∴ ((zc2 ) z1)/z2 =((13+i))/((3+5i))×((3-5i))/((3-5i))=(44-62i)/34
∴ Real part of (((zc2 ) z1)/z2 )=44/34=22/17

Q10. If a(p+q)2+2b pq+c=0 and a(p+r)2+2b pr+c=0, then qr=
•  p2+c/a
•  p2+a/c
•  p2+a/b
• p2+b/a
Solution
(a) We have, a(p+q)2+2 bpq+c=0 and, a(p+r)2+2 bpr+c=0
It is evident from these two equations, that q and r are roots of the equation a(p+x)2+2bpx+c=0 or,ax2+2x(a+b)p+ap2+c=0
∴Product of the roots=(ap2+c)/a
⇒qr=(ap2+c)/a=p2+c/a

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