Complex Numbers and Quadratic Equation Test

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Complex Numbers and Quadratic Equation Test - 9

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Weekly Quiz Competition

Question 1
2 / -0.83

The set of values of x for which the inequality [x]² - 5[x] + 6 < 0 (where [.] denote the greatest integral function) hold good is

A
2 < [x] <3

B
2 < x <4

C
2 < x <3

D
2 < x < 4

Solution
Question 2
2 / -0.83

If the sum of n terms of an A.P. is 2n² + 5n, then find the 4th term.

A
4n –3

B
3n –4

C
4n + 3

D
3n + 4

Solution

Sum of first n terms of AP, S_n = 2n² + 5n

Now choose n =1 and put in the above formula,
First term = 2+5 = 7

Now put n=2 to get the sum of first two terms = 2x4 + 5x 2= 8 + 10 = 18

This means

first term + second term = 18

but first term =7 as calculated above

so, second term = 18-7 = 11

So common difference becomes, 11-7 = 4

So the AP becomes, 7, 11, 15, ....
nth term = a + (n - 1)d = 7 + (n-1)4 = 7 + 4n - 4 = 3 + 4n
Question 3
2 / -0.83

If the harmonic mean between the roots of (5 + √2)x² - bx + (8 + 2 √5) = 0 is 4, then the value of b is

A
2

B
3

C
4 - √5

D
4 + √5

Solution

given that harmonic mean between the roots of the
given equation is 4.
Question 4
2 / -0.83

If α, βare the roots of x² + ax - b = 0 and γ,δare the roots of x² + ax + b = 0 then (α - γ) (α - δ) (β - δ) (β - γ) =

A
4b²

B
b²

C
2b²

D
3b²

Solution

α + β = -a,αβ = -b;γ+δ = -a,γδ = b
(α² + a α + b) (β² + a β +b) = 4b²
Question 5
2 / -0.83

If the ratio of the roots of ax² + 2bx + c = 0 is same as the ratio of the roots of px² + 2qx + r = 0 then

A

B

C

D

Solution

Let α,βbe roots of ax³ +bx+c=0
γ,δbe roots of px² +2qx+r=0
α/β= γ/δ…………(1) and
β/α= δ/γ………….(2)
(1) + (2)
⇒α/β+ β/α= γ/δ+ δ/γ
= [(2² +β² )/αβ+ 2]= [γ² +δ² ]/γδ+ 2
⇒[(α)² +(β)² +2 αβ]/αβ= [γ² +δ² +2 γδ]/γδ
= [(α+β)² ]/αβ= [(γ+δ)² ]/γδ
⇒(4b² /a² )/(c/a) = (4q² /p² )/(r/p)
⇒b² /ac = q² /pr.
Question 6
2 / -0.83

The roots of the equation (b - c) x² + 2 (c - a)x +(a - b) = 0 are always

A
Real

B
real and equal

C
real and distinct

D
imaginary

Solution

Correct Answer :- A

Explanation : (b-c)x² + 2(c-a)x + (a-b) = 0

b² - 4ac

[2(c-a)² - 4(b-c)(a-b)]

4(c-a)² -4(b-c)(a-b)

4[(c-a)² - (b-c)(a-b)]

4[c² + a² - 2ac -(ab - b² - ca + bc)]

4[c2 + a2 - 2ac -ab + b² + ca - bc)]

4{a² + b² + c² - ab - bc - ca}

4(a² + b² + c² ) -2(2ab + 2bc + 2ca)

4(a² + b² + c² ) -2{(a+b+c)² - a² - b² - c² }

4(a² + b² + c² ) -2(a+b+c)² + 2a² + 2b² + 2c²

4(a² + b² + c² ) -2(a+b+c)² >0 (Real)
Question 7
2 / -0.83

If a ∈Z and the equation (x - a)(x -10) +1 = 0 has integral roots, then values of ‘a ’are

A
10, 8

B
12, 10

C
12, 8

D
10, 12

Solution

(x –a)(x –10) = -1
Since a is an integer and x is also integer,
So, (x –a) &(x –10) can be 1 or –1
For (x –10) = 1, x = 11
(x –a) = -1, 11 –a = -1, a = 12
For (x –10) = -1, x = 9
(x –a) = 1, 9 –a = 1, a = 8
Question 8
2 / -0.83

The value of λin order that the equations 2x² + 5 λx + 2 = 0 and 4x² + 8 λx + 3 = 0 have a common root is given by

A
1

B
-1

C
±1

D
3

Solution
Question 9
2 / -0.83

If both roots of the equation x² - 2ax + a² -1 = 0 lie in the interval (-3,4) then sum of the integral parts of ‘a ’is

A
0

B
2

C
4

D
–1

Solution

x² - 2ax + a² -1 = 0 ⇒ (x - a)² = 1
Roots are a - 1 and a + 1
Both roots lie in (-3, 4) ∴a - 1 >-3, a + 1 <4
Ie -2
Question 10
2 / -0.83

Number of rational roots of the equation |x² - 2x - 3| + 4x = 0 is

A
1

B
2

C
3

D
4

Solution

Use extreme value formula