Complex Numbers and Quadratic Equation Test

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Complex Numbers and Quadratic Equation Test - 2

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Question 1
2 / -0.83

A
2 √5

B
4

C
3

D
2

Solution
Question 2
2 / -0.83

A
0

B
1

C
-1

D
2

Solution
Question 3
2 / -0.83

A
-1

B
-1/2

C
1

D
none of the above

Solution
Question 4
2 / -0.83

A
Straight line

B
Circle

C
Parabola

D
hyperbola

Solution
Question 5
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A
11

B
13

C
14

D
15

Solution
Question 6
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A
-2+2i

B
-2-2i

C
1-i

D
1-3i

Solution
Question 7
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A

B

C

D
None of the above

Solution
Question 8
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Find the value of

A

B

C

D

Solution
Question 9
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If the roots of the equation x² - 5x + 16 = 0 are a, b and the roots of the equation x² + px + q = 0 are (a² + b² ) and , then-

A
p = 1 and q = 56

B
p = 1 and q = – 6

C
p = – 1 and q = 56

D
p = – 1 and q = – 56

Solution
Question 10
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The number of the integer solutions of x² + 9 <(x + 3)² <8x + 25 is

A
1

B
2

C
3

D
None of these

Solution
Question 11
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A
-1

B
-1/2

C
1

D
2

Solution
Question 12
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The equatiobn π^x = –2x² + 6x – 9 has

A
No solution

B
One solution

C
Two solutions

D
Infinite solutions

Solution

The RHS of the expression has a <0 which means the graph will lie below the x-axis and π^x lies above the x-axis.Therefore,no solution.
Question 13
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Let a >0, b >0 and c >0. Then both the roots of the equation ax² + bx + c = 0

A
Are real and negative

B
Have negative real parts

C
Are rational numbers

D
Have positive real parts

Solution
Question 14
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The set of all solutions of the inequality <1/4 contains the set

A
(–∞, 0)

B
(–∞, 1)

C
(1, ∞)

D
(3, ∞)

Solution

(1/2)^{(x2 - 2x)} <(1/4)
(1/2)^{(x2 - 2x)} <(1/2)²
x² −2x >2......(as after multiplicative inverse sign of inequality changes)
x² −2x −2 >0
x² - 2x + 1 - 3 >
(x-1)² - 3 >0
(x-1)² >3
So for the above to hold good both the expression must be positive or both must be negative. After finding the solution the range of the solution will be
either x >3
(3,¥)
Question 15
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Consider y = , where x is real, then the range of expression y² + y – 2 is

A
[–1, 1]

B
[0, 1]

C
[–9/4, 0]

D
[–9/4, 1]

Solution
Question 16
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The values of k, for which the equation x² + 2(k – 1) x + k + 5 = 0 possess atleast one positive root, are

A
[4, ∞)

B
(∞, – 1] ∪[4, ∞)

C
[–1, 4]

D
(–∞, – 1]

Solution
Question 17
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If <= 4, then least and the highest values of 4x² are

A
0 &81

B
9 &81

C
36 &81

D
None of these

Solution
Question 18
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If coefficients of the equation ax² + bx + c = 0, a < 0 are real and roots of the equation are non –real complex and a + c + b <0, then

A
4a + c <2b

B
4a + c >2b

C
4a + c = 2b

D
none of these

Solution

Puting x=-1
a-b+c
but
a+c hence f(x)<0 for all real values of x
therefore
putting x=-2
we get f(X)=4a+c-2b <0
or 4a+c <2b

Question 19
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If the roots of the equation x² + 2ax + b = 0 are real and distinct and they differ by at most 2m, then b lies in the interval

A
(a² – m² , a² )

B
[a² – m² , a² )

C
(a² , a² + m² )

D
None of these

Solution

Question 20
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If the roots of the quadratic equation x² + px + q = 0 are tan 30 ºand tan 15 ºrespectively, then the value of 2 + q – p is

A
3

B
0

C
1

D
2

Question 21
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If (1 – p) is root of quadratic equation x² + px + (1 – p) = 0, then its roots are

A
0, 1

B
–1, 1

C
0, –1

D
–1, 2

Solution

As (1 –p) is root of the equation: x² + px + (1 –p) = 0
(1 –p)² + p(1 –p) + (1 –p) = 0
(1 –p)[1 –p + p + 1] = 0
(1 –p) = 0
p = 1
Therefore, given equation now becomes
x² + x = 0
x(x + 1) = 0
x = 0, -1

Question 22
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The values of x and y besides y can satisfy the equation (x, y ∈real numbers) x² – xy + y² – 4x – 4y + 16 = 0

A
2, 2

B
4, 4

C
3, 3

D
None of these

Question 23
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For what value of a and b the equation x⁴ – 4x³ + ax² + bx + 1 = 0 has four real positive roots ?

A
(–6, –4)

B
(–6, 5)

C
(–6, 4)

D
(6, –4)

Solution

Since all roots are positive,
coeff. of x shld be negatuve i.e., b is negative
let roots be m(>0), n(>0), p(>0), q(>0)
m+n+p+q = 4
mn+np+pq+mp+mq+nq = a
mnp+mnq+npq+mpq = -b
mnpq = 1
For the objective case,
put m=n=p=q =1
so a = 1+1+1+1+1+1 = 6
-b = 1+1+1+1 =4
so, a = 6 and b = -4

Question 24
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If a, b are roots of the equation ax² + bx + c = 0, then the value of a³ + b³ is

A

B

C

D

Question 25
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If z₁ = 2 + i, z₂ = 1 + 3i, then Re ( z₁ - z₂ ) =

A
ι

B
1

C
2 ι

D
2

Solution

The numbers in the questions are not very clear.
z₁ = 2 + i
z₂ = 1+3i
z₁ –z₂ = (2 –1) + i (1 –3)
= 1 –2i

Question 26
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If a, b are roots of the equation ax² + bx + c = 0 then the equation whose roots are 2a + 3b and 3a + 2b is

A
ab x² – (a + b) cx + (a + b)² = 0

B
ac x² – (a + c) bx + (a + c)² = 0

C
ac x² (a + c) bx – (a + c)² = 0

D
None of these

Question 27
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If S is the set of all real x such that is positive, then S contains

A
(– ∞, –3/2)

B
(–3/2, 1/4)

C
(–1/4, 1/2)

D
(–1/2, 3)

Solution

Question 28
2 / -0.83

A
-2-2i

B
-2+2i

C
i-1

D
2+2i

Solution

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Complex Numbers and Quadratic Equation Test - 2

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Complex Numbers and Quadratic Equation Test - 2

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SHARING IS CARING

Question 1

2 √5

4

3

2

Solution

Question 2

0

1

-1

2

Solution

Question 3

-1

-1/2

1

none of the above

Solution

Question 4

Straight line

Circle

Parabola

hyperbola

Solution

Question 5

11

13

14

15

Solution

Question 6

-2+2i

-2-2i

1-i

1-3i

Solution

Question 7

None of the above

Solution

Question 8

Solution

Question 9

p = 1 and q = 56

p = 1 and q = – 6

p = – 1 and q = 56

p = – 1 and q = – 56

Solution

Question 10

1

2

3

None of these

Solution

Question 11

-1

-1/2

1

2

Solution

Question 12

No solution

One solution

Two solutions

Infinite solutions

Solution

Question 13

Are real and negative

Have negative real parts

Are rational numbers

Have positive real parts

Solution

Question 14

(a² – m² , a² )

[a² – m² , a² )

(a² , a² + m² )