Complex Numbers Quiz-9

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. Consider the following statements: 1. The equation x2-cx+d=0 and x2-ax+b=0 have common root and second equation has equal roots if ac=2(b+d). 2. If Î± is a root of the equation 4x2+2x-1=0, then the other root is 4Î±3-3Î±. 3. The expression (x-1)(x-3)(x-4)(x-6)+10 is positive for all real values of x. Which of these is/are correct?
•  Only (3)
•  Only (2)
•  All of the above
•  None of these
Solution
(c) All of the above

Q2.If x2-3x+2 be a factor of x4-px2+q,then (p,q) is equal to
•  (3, 4)
•  (4, 5)
•  (4, 3)
•  (5, 4)
Solution
(d) Given that x2-3x+2 be a factor of x4-px2+q=0 ...(i)
⇒(x2-3x+2)=0
⇒(x-2)(x-1)=0
⇒x=2,1
On putting these values in Eq. (i), we get 4p-q-16=0 ...(ii)
and p-q-1=0 ...(iii)
On solving Eqs.(ii) and (iii), we get p=5 and q=4
⇒(p,q)=(5,4)
Q3.  If Î±,Î² and Î³ are the roots of equation x3-3x2+x+5=0, then y=∑Î±2+Î±Î²Î³ satisfies the equation
•   y3+y+2=0
•  y3-y2-y-2=0
•  y3+3y2-y-3=0
•  y3+4y2+5y+20=0
Solution
(b) Here, Î±+Î²+Î³=3,Î±Î²+Î²Î³+Î³Î±=1 and Î±Î²Î³=-5
Now, y=Î±2+Î²2+Î³2+Î±Î²Î³ =(Î±+Î²+Î³)2-2(Î±Î²+Î²Î³+Î³Î±)+Î±Î²Î³ =(3)2-2(1)-5 ⇒y=2
So, y=2 satisfies the equation y3-y2-y-2=0

Q4. The sum of non-real roots of the equation (x2+x-2)(x2+x-3)=12, is
•  1
•  -1
•  -6
•  6
Solution
(b) The given equation is (x2+x-2)(x2+x-3)=12
⇒(y-2)(y-3)=12, where y=x2+x
⇒y2-5y-6=0
⇒y=6,-1
⇒x2+x=6 or x2+x=1
⇒x2+x-6=0 or, x2+x-1=0
⇒x=-3,2,Ï‰,Ï‰2
∴Sum of real roots =-3+2=-1

Q5.If y=2(1/logx(8) ), then x is equal to
•  y
•  y2
•  y3
•  None of these
Solution
(c) Given, y=2(1/logx⁡(8) )=2log8⁡(x) ⇒y=2log2⁡∛x =∛x ⇒x=y3

Q6. If the imaginary part of the expression (z-1)/eiÎ¸ +eiÎ¸/(z-1) be zero, then the locus of z is
•  A straight line parallel to x-axis
•  A parabola
• A circle of radius 1 and centre (1, 0)
•  None of the above
Solution
(c) Let z-1=reiÎ±
∴ (x-1)+iy=r(cos⁡Î± + isin⁡Î±)
∴ r2=(x-1)2+y2 and tan⁡Î±=y/(x-1)
The expression (z-1)/eiÎ¸ +eiÎ¸/(z-1)=rei(Î±-Î¸) +1/r e(-i(Î±-Î¸) )
Which is given as real
∴ r sin⁡(Î±- Î¸)-1/r sin⁡(Î±-Î¸)=0
⇒ r-1/r=0
⇒ r2=1
⇒ (x-1)2+y2=1
Which is a circle with centre (1, 0) and radius 1

Q7.If Î±,Î² be the roots of the quadratic equation ax^2+bx+c=0 and k be a real number, then the condition so that Î±<k<Î² is given by
•  ac>0
•  ak2+bk+c=0
•  ac<
•  a2 k2+abk+ac<0
Solution
(d) Since, Î±,Î² are the roots of the equation ax2+bx+c=0.
∴ ax2+bx+c=a(x-Î±)(x-Î²)
⇒ Î±,Î² be the roots of ax2+bx+c=0.
Also Î±<k<Î² So, a(k-Î±)(k-Î²)<0
Also, a2 k2+abk+ac=a(ak2+bk+c)=a2 (k-Î±)(k-Î²)<0 ⇒ a2 k2+abk+ac<0

Q8.
•  cos⁡Î± + isin⁡Î±
•  cos(Î±/2) - isin⁡(Î±/2)
•  e(iÎ±/2)
•  ∛(eiÎ± )
Solution

Q9.If Î±,Î²,Î³ are the roots of the equation x3-7x+7=0, then 1/Î±4 +1/Î²4 +1/Î³4 is
•  7/3
•  3/7
•  4/7
•  7/4
Solution
(b) Here, ∑Î±=0, ∑Î±Î²=-7, Î±Î²Î³=-7
∴ 1/Î±4 +1/Î²4 +1/Î³4 =(Î±4 Î²4 +Î²4 Î³4 + Î³4 Î²4)/(Î±4 Î²4 Î³4 ) =(∑Î±4 Î²4 )/(Î±4 Î²4 Î³4) ...(i)
Now, ∑Î±Î² ∑Î±Î² ∑Î±Î² ∑Î±Î²=(∑Î±Î²)2 (∑Î±Î²)2
⇒(-7)2=[Î±2 Î²2+Î²2 Î³2+Î³2 Î±2+2Î±Î²Î³(Î±+Î²+Î³)][Î±2 Î²2+Î²2 Î³2+Î³2 Î±2+2Î±Î²Î³(Î±+Î²+Î³)] =(Î±4 Î²4 + Î²4 Î³4+Î³4 Î±4 )(Î±4 Î²4 +Î²4 Î³4+Î³4 Î±4 ) [∑Î± = Î±+Î²+Î³ =0] =Î±4 Î²4+Î²4 Î³4+Î³4 Î±4+2Î±4 Î²2 Î³2+2Î±2 Î²4 Î³2+2Î±2 Î²2 Î³4 =∑Î±4 Î²4+2Î±2 Î²2 Î³2 [(∑Î±)2-2∑Î±Î²] =∑Î±4 Î²4+2Î±2 Î²2 Î³2 [0-2×(-7)] =∑Î±4 Î²4+2(-7)2 (2×7)
⇒∑Î±4 Î²4=(-7)4+4(-7)3
⇒ ∑Î±4 Î²4=(-7)3 (-7+4)=-3(-7)3
On putting this value in Eq. (i), we get
1/Î±2 +1/Î²4 +1/Î³4 =(-3(-7)3)/(-7)4 =(-3)/(-7)=3/7

Q10. The number of real roots of f(x)=0, where f(x)=(x-1)(x-2)(x-3)(x-4) lying in the interval (1, 3) is
•  1
•  2
•  3
• 4
Solution
(a) Given, f(x)=(x-1)(x-2)(x-3)(x-4)
The real roots are 1, 2, 3, 4
Hence, only 2 lies in the interval (1, 3)

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