Complex Numbers and Quadratic Equations

Sponsor Area

Question
CBSEENMA11015553

If the roots of the equation bx2+ cx + a = 0 be imaginary, then for all real values of x, the expression 3b2x2 + 6bcx + 2c2 is

  • greater than 4ab

  • less than 4ab

  • greater than -4ab

  • less than 4ab

Solution

C.

greater than -4ab

As, bx2 + cx + a = 0 has imaginary roots
So, c2< 4ab
Now, 3b2x2 + 6bcx + 2c2
= 3(bx + c)2– c2≥ – c2≥ – 4ab

Sponsor Area

Question
CBSEENMA11015570

The conjugate of a complex number is fraction numerator 1 over denominator straight i minus 1 end fraction. Then the complex number is

  • fraction numerator negative 1 over denominator straight i minus 1 end fraction
  • fraction numerator 1 over denominator straight i plus 1 end fraction
  • fraction numerator negative 1 over denominator 1 plus straight i end fraction
  • fraction numerator 1 over denominator 1 plus straight i end fraction

Solution

C.

fraction numerator negative 1 over denominator 1 plus straight i end fraction

Question
CBSEENMA11015579

The quadratic equations x2 – 6x + a = 0 and x2 – cx + 6 = 0 have one root in common. The other roots of the first and second equations are integers in the ratio 4 : 3. Then the common root is

  • 1

  • 4

  • 3

  • 2

Solution

D.

2

Let α and 4β be roots of x2– 6x + a = 0 and
α, 3β be the roots of x2– cx + 6 = 0, then
α + 4β = 6 and 4αβ = a
α + 3β = c and 3αβ = 6.
We get αβ = 2 ⇒ a = 8
So the first equation is x2 – 6x + 8 = 0 ⇒ x = 2, 4
If α = 2 and 4β = 4 then 3β = 3
If α = 4 and 4β = 2, then 3β = 3/2 (non-integer)
∴ common root is x = 2.

Question
CBSEENMA11015585

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


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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
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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