## Complex Numbers Quiz-2

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.

•  1
•  2
•  3
•  4
Solution
(b) √(2+√5-√(6-3√5+√(14-6√5) ) )
=√(2+√5-√(6-3√5+√((9+5-6√5) )) )
=√(2+√5-√(6-3√5+√((3-√5)2 )) )
=√(2+√5-√(9-4√5) )
=√(2+√5-√((-2+√5)2 ))
=√(2+√5+2-√5) =2

Q2.If Î±≠Î² and Î±2=5Î±-3,Î²2=5Î²-3, then the equation having Î±/Î² and Î²/Î± as its roots is
•  3x2+19x+3=0
•  3x2-19x+3=0
•  3x2-19x-3=0
•  x2-16x+1=0
Solution
(b) Given, Î±2-5Î±+3=0 and Î²2-5Î²+3=0
⇒ Î±=(5±√13)/2 and Î²=(5±√13)/2
Since, Î±≠Î²
∴Î±=(5+√13)/2 and Î²=(5-√13)/2
Î±=(5-√13)/2 and Î²=(5+√13)/2
Now, Î±2+Î²2=(50+26)/4=19
And Î±Î²=1/4 (25-13)=3
∴ Required equation is x2-x(Î±/Î²+Î²/Î±)+Î±Î²/Î±Î²=0
⇒ x2-x((Î±2+Î²2)/Î±Î²)+1=0
⇒ 3x2-19x+3=0
Q3. For the equation 1/(x+a)-1/(x+b)=1/(x+c), if the product of the roots is zero, then the sum of the roots is
•   0
•  2ab/(b+c)
•  2bc/(b+c)
•  -2bc/(b+c)
Solution
(d) Since, (b-a)/(x2+(a+b)x+ab)=1/(x+c)
⇒x2+2ax+ab+ca-bc=0
Since, the product of roots is zero
Then, ab+ca-bc=0
⇒a=bc/(b+c)
∴ Sum of roots=-2a=(-2bc)/(b+c)

Q4. If the equation x3-3x+a=0 has distinct roots between 0 and 1, then the value of a is
•  2
•  1/2
•  3
•  None of these
Solution
(d) Let f(x)=x3-3x+a
If f(x) has distinct roots between 0 and 1.
Then, f' (x)=0 has a root between 0 and 1
But, f' (x)=0
⇒3x2-3=0
⇒x=± 1
Clearly, f' (x)=0 does not have any root between 0 and 1.
So, f(x) does not have distinct roots between 0 and 1 for any value of a

Q5.Let [x] denote the greatest integer less than or equal to x. Then, in [0,3] the number of solutions of the equation x2-3x+[x]=0, is
•  6
•  4
•  2
•  0
Solution
(c) We have the following cases:
CASE I
When x∈[0,1)
In this cases, we have [x]=0
∴x2-3x+[x]=0
⇒x2-3x=0
⇒x=0,3
⇒x=0
CASE II
When x∈[1,2)
In this case, we have [x]=1
∴x3-3x+[x]=0
⇒x2-3x+1=0
⇒x=(3±√5)/2
Clearly, these values of x do not belong to [1, 2].
So, the equation has no solution in [1, 2)
CASE III
When x∈[2,3)
∴x2-3x+[x]=0
⇒x2-3x+2=0
⇒x=1,2⇒x=2
Hence, the given equation has two solutions only

Q6. If the equations ax2+bc+c=0 and 2x2+3x+4=0 have a common root, then a∶b∶c
•  2:3:4
•  1:2:3
• 4:3:2
•  None of these
Solution
(a) The equation 2x2+3x+4=0 has complex roots which always occur in pairs.
So, the two equations have both roots common
∴a/2=b/3=c/4
⇒a∶b∶c=2∶3∶4

Q7.If Î± and Î² are the solutions of the quadratic equation ax2+bx+c=0 such that Î²=Î±(1/3), then
•  (ac)(1/3)+(ab)(1/3)+c=0
•  (a3 b)(1/4)+(ab3 )(1/4)+c=0
•  (a3 c)(1/4)+(ac3 )(1/4)+b=0
•  (a4 c)(1/3)+(ac4 )(1/3)+b=0
Solution

Q8.If Î± and Î² are roots of the quadratic equation x2+4x+3=0, then the equation whose roots are 2Î±+Î² and Î±+2Î² is
•  x2-12x+35=0
•  x2+12x-33=0
•  x2-12x-33=0
•  x2+12x+35=0
Solution
(d) Given, Î±,Î² are the roots of equation x2+4x+3=0
∴ Î±+Î²=-4 and Î±Î²=3
Now, 2Î±+Î²+Î±+2Î²=3(Î±+Î²)=-12
And (2Î±+Î²)(Î±+2Î²)=2Î±2+4Î±Î²+Î±Î²+2Î²2 =2(Î±+Î²)2+Î±Î² =2(-4)2+3=35
Hence, required equation is x2-(sum of roots)x+(product of roots)=0
⇒ x2+12x+35=0

Q9.The number of real solution of the equation (9/10)=-3+x-x2 is
•  0
•  1
•  2
•  None of these
Solution
(a) Let f(x)=-3+x-x2
Now, D=12-4(3)=-11<0
Here, coefficient of x2<0
∴ f(x)<0 Thus, LHS of the given equation is always positive whereas the RHS is always less than zero Hence, the given equation has no solution

Q10. If the roots of the equation 8x3-14x2+7x-1=0 are in GP, then the roots are
•  1,1/2,1/4
•  2, 4, 8
•  3, 6, 12
• None of these
Solution
(a) Since, the roots of the equation 8x3-14x2+7x-1=0 are in GP.
Let the roots be Î±/Î², Î±,Î±Î²,Î²≠0.
Then, the product of roots is Î±3=1/8⇒Î±=1/2 and hence,Î²=1/2. So, roots are 1, 1/2,1/4.

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