Complex Numbers and Quadratic Equations

More Topic from Mathematics

Question 1

Let α and  β be the roots of equations x2-6x-2 = 0. If ann- βn, for n≥1, the value of a10-2a8/2a9 is equal to 
  • 6

  • -6

  • 3

  • -3

Solution

C.

3

α and β are the roots of the equation
x2-6x-2 =0
or
α=6x+2 
α2 = 6α +2
α10= 6 α9+2α8 ... (i)
β10= 6 β9+2β8 ... (ii)
On subtracting eq (ii) from eq(i), we get
α10- β10= 6 ( α99) + 2 (α88)
a10 = 6a9 + 2a8 (∴ an = αn- βn)
⇒ a10 -2a8 = 6a9   
⇒ a10-2a8/2a9  = 3

Question 2

A body weighing 13 kg is suspended by two strings 5 m and 12 m long, their other ends being fastened to the extremities of a rod 13 m long. If the rod be so held that the body hangs immediately below the middle point. The tensions in the strings are 

  • 12 kg and 13 kg

  • 5 kg and 5 kg

  • 5 kg and 12 kg

  • 5 kg and 13 kg

Solution

C.

5 kg and 12 kg


straight T subscript 2 space cos space open parentheses straight pi over 2 minus straight theta close parentheses space equals space straight T subscript 1 space cosθ
rightwards double arrow space straight T subscript 1 space cos space straight theta space equals space straight T subscript 2 space sin space straight theta
straight T subscript 1 space sin space straight theta space plus space straight T subscript 2 space cos space straight theta space equals space 13
because space OC space equals space CA space space equals CB
rightwards double arrow space angle space AOC space equals space angle OAC space and space angle COB space equals space angle OBC
therefore space sin space straight theta space equals space sin space straight A space equals space 5 over 13 space and space cos space straight theta space equals 12 over 13
rightwards double arrow space straight T subscript 1 over straight T subscript 2 space equals space 5 over 12 space rightwards double arrow space straight T subscript 1 space equals space 5 over 12 space and space cos space straight theta space equals space 12 over 13
rightwards double arrow space straight T subscript 1 over straight T subscript 2 space equals space 5 over 12 space rightwards double arrow space straight T subscript 1 space equals space 5 over 12 straight T subscript 2
straight T subscript 2 space open parentheses 5 over 12.5 over 13.12 over 13 close parentheses space equals space 13
straight T subscript 2 space open parentheses fraction numerator 169 over denominator 12.13 end fraction close parentheses space equals space 13
straight T subscript 2 space equals space 12 space kgs
rightwards double arrow space straight T subscript 1 space equals space 5 space kgs
Question 3

A complex number z is said to be unimodular, if |z|= 1. suppose z1 and z2 are complex numbers such that fraction numerator straight z subscript 1 minus 2 straight z subscript 2 over denominator 2 minus straight z subscript 1 begin display style stack straight z subscript 2 with minus on top end style end fraction is unimodular and z2 is not unimodular. Then, the point z1 lies on a

  • straight line parallel to X -axis

  • straight line parallel to Y -axis

  • circle of radius 2

  • circle of radius square root of 2

Solution

C.

circle of radius 2

If z unimodular, then |z| = 1, also, use property of modulus i.e. straight z top enclose straight z space equals space vertical line straight z vertical line squared
Given, z2 is not unimodular i.e |z2|≠1 and fraction numerator straight z subscript 1 minus 2 straight z subscript 2 over denominator 2 minus straight z subscript 1 begin display style stack straight z subscript 2 with minus on top end style end fraction is unimodularfraction numerator straight z subscript 1 minus 2 straight z subscript 2 over denominator 2 minus straight z subscript 1 begin display style stack straight z subscript 2 with minus on top end style end fraction space equals space 1
rightwards double arrow vertical line z subscript 1 minus 2 z subscript 2 vertical line squared space equals space vertical line 2 minus z subscript 1 top enclose z subscript 2 end enclose vertical line squared
rightwards double arrow left parenthesis z subscript 1 minus 2 z subscript 2 right parenthesis left parenthesis stack z subscript 1 with bar on top space minus 2 stack z subscript 2 with bar on top right parenthesis space equals space left parenthesis 2 minus z subscript 1 stack z subscript 2 with bar on top right parenthesis left parenthesis 2 minus top enclose z subscript 1 z subscript 2 right parenthesis
left parenthesis because space z top enclose z space equals space vertical line z squared vertical line right parenthesis
rightwards double arrow space vertical line z subscript 2 vertical line squared space plus space 4 vertical line z subscript 2 vertical line squared minus space 2 top enclose z subscript 1 end enclose z subscript 2 space minus space 2 z subscript 1 top enclose z subscript 2 end enclose
rightwards double arrow space left parenthesis vertical line z subscript 1 vertical line squared minus 1 right parenthesis left parenthesis vertical line z subscript 1 vertical line minus 4 right parenthesis space equals space 0
vertical line z subscript 2 vertical line space not equal to 1
vertical line z subscript 1 vertical line space equals space 2
z subscript 1 space equals space x plus i y
x squared plus y squared space equals space left parenthesis 2 right parenthesis squared
therefore space point space straight z subscript 1 space lies space on space straight a space circle space of space radius space 2.

Question 4

All the values of m for which both roots of the equations x2 − 2mx + m2 − 1 = 0 are greater than −2 but less than 4, lie in the interval

  • −2 < m < 0

  • m > 3

  • −1 < m < 3 

  • 1 < m < 4 

Solution

C.

−1 < m < 3 

Equation x2 − 2mx + m2 − 1 = 0
(x − m)2 − 1 = 0
(x − m + 1) (x − m − 1) = 0
x = m − 1, m + 1 − 2 < m − 1 and m + 1 < 4
m > − 1 and m < 3 − 1 < m < 3.

Question 5

If (1 – p) is a root of quadratic equation x2 +px + (1-p)=0 , then its roots are

  • 0, 1

  • -1, 2

  • 0, -1

  • -1, 1

Solution

C.

0, -1

Since (1 - p) is the root of quadratic equation
x2 + px + (1 - p) = 0 ........ (i)
So, (1 - p) satisfied the above equation
∴ (1 - p)2 + p(1 - p) + (1 - p) = 0
(1 - p)[1 - p + p + 1] = 0 (1 - p)(2) = 0
⇒ p = 1 On putting this value of p in equation (i)
x2 + x = 0
⇒ x(x + 1) = 0 ⇒ x = 0, -1