## Complex Numbers Quiz-6

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 (cos⁡Î¸+ i sin⁡Î¸) (cos⁡2Î¸+ i sin⁡2Î¸)… (cos⁡nÎ¸+i sin nÎ¸)=1, then the value of Î¸ is
•  4mÏ€
•  2mÏ€/n(n+1)
•  4mÏ€/n(n+1)
•  mÏ€/n(n+1)
Solution

Q2.If x2+2ax+10-3a>0 for all x∈R, then
•  -5<a<2
•  a<-5
•  a>5
•  2<a<5
Solution
(a) We have, x2+2ax+10-3a>0 for all x∈R
⇒4a2-40+12a60;0 [Using: discriminant <0] ⇒a2+3a-10<0 ⇒(a+5)(a-2)<0⇒-5<a<2
Q3.  If Ï‰(≠1) be a cube root of unity and (1+Ï‰2 )n=(1+Ï‰4 )n, then the least positive value of n is
•   2
•  3
•  5
•  6
Solution
(b) We have, (1+Ï‰2 )n=(1+Ï‰4 )n
⇒ (-Ï‰)n=(-Ï‰2 )n ⇒ Ï‰n=1 ⇒ n=3 is least positive value of n

Q4. If Î±,Î² are the roots of the equation x^2-(1+n2 )x+1/2 (1+n2+n4 )=0, then Î±2+Î²2 is
•  n2
•  -n2
•  n4
•  -n4
Solution
(a) Here, Î±+Î²=1+n2 and Î±Î²=(1+n2+n4)/2 Now, Î±2+Î²2=(Î±+Î²)2-2Î±Î² =(1+n2 )2-(1+n2+n4 )=n2

Q5.For positive integers n1,n2 the value of the expression (1+i)(n1 )+(1+i3 )(n1 )+(1+i5 )(n2 )+(1+i7 )(n2 ),i=√(-1) is a real number if and only if
•  n1=n2+1
•  n1=n2-1
•  n1=n2
•  n1>0, n2>0
Solution

Q6. If the equation 2x2>-7x+1=0 and ax2+bx+2=0 have a common root, then
•  a=2,b=-7
•  a=-7/2,b=1
• a=4,b=-14
•  None of these
Solution
(c) Let Î± be a common root of the two equations.
Then, 2 Î±2-7 Î±+1=0 a Î±2+b Î±+2=0 ⇒Î±2/(-14-b)=Î±/(a-4)=1/(2b+7a) ⇒(a-4)/(2b+7a)=(b+14)/(4-a) ⇒(7a+2b)(b+14)+(a-4)2=0 Clearly, a=4,b=-14 satisfy this equation

Q7.If n is a positive integer greater than unity and z is a complex number satisfying the equation zn=(z+1)n, then
•  Im (z)<0
•  Im (z)>0
•  Im (z)=0
•  None of these
Solution
(d) Explaination not available

Q8.The Î±,Î² are the roots of the equation x2+ax+b=0, then 1/Î±2 +1/Î²2 is equal to
•  (a2-2b)/b2
•  (b2-2a)/b2
•  (a2+2b)/b2
•  (b2+2a)/b2
Solution
(a) Here, Î±+Î²=-a, Î±Î²=b ∴ 1/Î±2 +1/Î²2 =(Î±2+Î²2)/(Î±Î²)2 =((Î±+Î²)2-2Î±Î²)/(Î±Î²)2 =(a2-2b)/b2

Q9.Area of the triangle formed by 3 complex numbers 1+i,i-1,2i in the Argand plane is
•  1/2
•  1
•  √2
•  2
Solution

Q10. For two complex numbers z1,z2 the relation |z1+z2 |=|z1 |+|z2 | holds, if
•  arg⁡(z1 )+arg⁡(z2 )=Ï€/2
•  arg⁡(z1 )+arg⁡(z2 )=Ï€/2
•  z1z2=1
• |z1 |=|z2
Solution
(a) Since, |z1+z2 |=|z1 |+|z2 | ⇒ |z1 |2+|z2 |2+2|z1 ||z2 | cos⁡(Î¸1-Î¸2) =|z1 |2+|z2 |2+2|z1 ||z2 | ⇒ cos⁡(Î¸1-Î¸2)=1=cos⁡0° ⇒ Î¸1-Î¸2=0 ⇒ Î¸1=Î¸2 ⇒ arg⁡(z1 )=arg⁡(z2)

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