## Complex Numbers Quiz-8

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 ecos⁡x -e-cos⁡x =4, then the value of cos⁡x is
•  loge⁡(2+√5)
•  -loge⁡(2+√5)
•  loge⁡(-2+√5)
•  None of these
Solution
(d) Let ecos⁡x =y.
Then, ecos⁡x -e(-cos⁡x )=4
⇒y-1/y=4
⇒y2-4y-1=0
⇒y=2± √5
⇒y=2+√5 as y>0
⇒ecosx =2+√5⇒cos⁡x=loge(2+√5)
Clearly, loge(2+√5)>1 and cos⁡x≤1
So, there is no value of cos⁡x satisfying the given equation

Q2.If the sum of the squares of the roots of the equation x2-(sin⁡Î±-2)x-(1+sin⁡Î± )=0 is least, then Î±=
•  Ï€/4
•  Ï€/3
•  Ï€/2
•  Ï€/6
Solution
(c) Let p,q be the roots of the given equation.
Then, p2+q2=(p+q)2-2pq
⇒p2+q2=(sin⁡Î±-2 )2+2(1+sin⁡Î±)
⇒p2+q2=sin2Î±-2 sin⁡Î±+6=(sin⁡Î±-1 )2 +5
Clearly, p2+q2 is last when sin⁡Î±-1=0
⇒sin⁡Î±=1
⇒Î±=Ï€/2
Q3.  Number of real roots of the equation (6-x)4+(8-x)4=16 is
•   4
•  2
•  0
•  None of these
Solution
(b) Put (6-x+8-x)/2=y
⇒x=7-y (y-1)4+(y+1)4=16
⇒y4+6y2+1=8 ⇒y4+6y2-7=0
⇒y2=1,-7 ⇒y2=1 (∵y2=-7 is not possible)
⇒y=±1 ⇒x=6,8
∴ Total number of real roots are 2.

Q4. If x2+ax+10=0 and x2+bx-10=0 have a common root, then a2-b2 is equal to
•  10
•  20
•  30
•  40
Solution
(d) Let Î± be a common root of the equations x2+ax+10=0 and x2+bx-10=0.
Then, Î±2+a Î±+10=10 and, Î±2+b Î±-10=0

Adding and subtracting these two equations, we get 2 Î±2+Î±(a+b)=0 and (a-b)Î±+20=0 ⇒Î±=-(a+b)/2 and Î±=-20/(a-b) ⇒-(a+b)/2=-20/(a-b)⇒a2-b2=40

Q5.If a and b are the non-zero distinct roots of x2+ax+b=0, then the least value of x2+ax+b is
•  2/3
•  9/4
•  -9/4
•  1
Solution
(c) Since a,b are roots of x2+ax+b=0.
Therefore, a2+a2+b=0 and, b2+ab+b=0
⇒b=-2 a2 and b+a+1=0
⇒-2 a2+a+1=0
⇒2 a2-a-1=0
⇒a=1 or, a=-1/2
Now, a=1,
⇒b=-2 [∵b+a+1=0] and,a=-1/2⇒b=-1/2
But, a≠b.
Therefore, a=1,b=-2
∴ Least value of x2+ax+b is -((a2-4b)/4)=-((1+8)/4)=-9/4

Q6. For a≠b, if the equation x2+ax+b=0 and x2+bx+a=0 have a common root, then the value of a+b equals
•  -1
•  0
• 1
•  2
Solution
(a) Let Î± be the common root for both the equations x2+ax+b=0 and x2+bx+a=0, then Î±2+aÎ±+b=0 And Î±2+bÎ±+a=0 ⇒ Î±2/((a2-b2))=Î±/(b-a)=1/(b-a) ∴ Î±2=-(a+b) and Î±=1 Hence, a+b=-1

Q7.The expression tan⁡{i log⁡((a-ib)/(a+ib)) } reduces to
•  ab/(a2+b2 )
•  2ab/(a2-b2 )
•  ab/(a2-b2 )
•  2ab/(a2+b2 )
Solution
(b) We have, log⁡((a-ib)/(a+ib)) =log⁡(a-ib)-log⁡(a+ib) =[log⁡√(a2+b2 )+i tan-1⁡((-b)/a) ]-[log⁡√(a2+b2 )+i tan-1⁡(b/a) ] =-2i tan-1⁡(b/a) ∴i log⁡((a-ib)/(a+ib))=2 tan-1⁡(b/a)=tan-1⁡((2 b/a)/(1-b2/a2 ))=tan-1⁡((2 ab)/(a2-b2 )) ⇒tan⁡{i log⁡((a-ib)/(a+ib)) }=tan⁡{tan-1⁡((2 ab)/(a2-b2 )) }=(2 ab)/(a2-b2

Q8.7(2 log7⁡5 ) is equal to
•  log7⁡35
•  5
•  25
•  log72⁡5
Solution
(c) 7(2 log7⁡5 )=7(log7⁡(52) ) =(5)2=25 [∵alogax =x;x>0,x≠0,1]

Q9.The value of expression 2(1+Ï‰)(1+Ï‰2 )+3(2+Ï‰)(2+Ï‰2 )+4(3+Ï‰)(3+Ï‰2 )+⋯+(n+1)(n+Ï‰)(n+Ï‰2 ), where Ï‰ is an imaginary cube root of unity is
•  {n(n+1)/2}2
•  {n(n+1)/2}2-n
•  {n(n+1)/2}2+n
•  None of these
Solution
(c) {n(n+1)/2}2+n

Q10. If cos⁡Î±+2 cos⁡Î²+3 cos⁡Î³=sin⁡Î±+ 2sin⁡Î²+ 3sin⁡Î³=0, then the value of sin⁡3 Î±+ 8sin 3 Î²+ 27sin 3 Î³ is
•  sin⁡(Î±+Î²+Î³)
•  3sin⁡(Î±+Î²+Î³)
•  18sin⁡(Î±+Î²+Î³)
• sin⁡(Î±+2Î²+3Î³)
Solution
(c) Let Î±=cos⁡Î±+i sin⁡Î±,b=cos⁡Î²+i sin⁡Î² and, c=cos⁡Î³+i sin⁡Î³
Then, a+2b+3c =(cosÎ±+2 cos⁡Î²+3 cos⁡Î³ )+i(sin⁡Î±+2 sinÎ²+3 sin⁡Î³ )=0
⇒a3+8 b3+27 c3=18 abc
⇒cos⁡3 Î±+8 cos⁡3 Î²+27 cos⁡3 Î³=18 cos⁡(Î±+Î²+Î³) and, sin⁡3 Î±+8 sin3 Î²+27 sin⁡3 Î³=18 sin⁡(Î±+Î²+Î³)

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