Complex Numbers and Quadratic Equation Test

Result

Complex Numbers and Quadratic Equation Test - 10

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
2 / -0.83

The set of real values of x satisfying |x-1| < 3 and |x - 1| > 1

A
[2, 4]

B
(-∞,2)∪(4,∞)

C
[-2,0]∪[2,4]

D
[0, 2]

Solution
Question 2
2 / -0.83

Let f(x) be a polynomial for which the remainders when divided by x –1, x –2, x –3 respectively 3, 7, 13. Then the remainder of f(x) when divided by (x –1) (x –2) (x –3) is

A
f(x)

B
x² + x +1

C
x² +1

D
x + 2

Solution

P(x) when divided by x −1, x-2 and x −3 leaves remainder 3,7 and 13 respectively.
From polynomial-remainder theorem, P(1) = 3 and P(2) = 7 P(3) = 13
If the polynomial is divided by (x −1)(x −2)(x-3) then remainder must be of the form ax+b (degree of remainder is less than that of divisor)

⇒P(x) = Q(x)(x −1)(x −2)(x-3)+(ax² + bx + c), where Q(x) is some polynomial.
Substituting for x=1, x=2 and x=3:
P(1) = 3 = a+b+c
P(2) = 7 = 4a+2b+c
P(3) = 13 = 9a+3b+c
=>Subtract P(2) from P(1)
-3a - b = -4.............(1)
Subtract P(3) from P(2)
5a + b = 6..............(2)
From (1) and (2) Solving for a and b, we get
a = 1 and b = 1, c = 1
⇒Remainder = x² + x + 1
Question 3
2 / -0.83

The range of values of x which satisfy 5x + 2 <3x + 8 and

A
(-∞,1)∪(2,3)

B
(2, 3)

C
(2,∞)

D
R

Solution
Question 4
2 / -0.83

For x ∈R, the least value of

A
-1

B

C

D

Solution
Question 5
2 / -0.83

Suppose a² = 5a –8 and b² = 5b –8, then equation whose roots are a/b and b/a is

A
6x² –5x + 6 = 0

B
8x² –9x + 8 = 0

C
9x² –8x + 9 = 0

D
8x² + 9x + 8 = 0

Solution

a, b are roots of x² –5x + 8 = 0
a + b = 5, ab = 8:

∴required equation is
x² –(9/8)x + 1 = 0
⇒ 8x² –9x + 8 = 0
Question 6
2 / -0.83

If α,βare roots of ax² + bx + c = 0, then roots of a³ x² + abcx + c³ = 0 are

A
αβ, α+ β

B
α² β, αβ²

C
a β, a² β²

D
a³ , β³

Solution

Write a³ x² + abcx + c³ = 0 as
Question 7
2 / -0.83

If P(x) = ax² + bx + c and Q(x) = -ax² + dx + c, where ac ≠0, then P(x)Q(x) = 0 has

A
no real root

B
exactly two real roots

C
at least two distinct real roots

D
none of these

Solution

Let D₁ = b² –4ac and D₂ = d² + 4ac
As ac ≠0, either ac <0 or ac >0
If ac <0, then D₁ >0
In this case P(x) = 0 has distinct two real roots
If ac >0, the D₂ >0
In this case Q(x) = 0 has two distinct real roots.
Thus, in either case P(x)Q(x) = 0 has at least two distinct real roots.
Question 8
2 / -0.83

If the product of the roots of the equation x² –5kx + 2e^{4 ln k} - 1 = 0 is 31, then sum of the root is

A
-10

B
5

C
-8

D
10

Solution

We have product of the roots = 2e^{4 ln k} - 1 = 31 (given)

⇒ k⁴ = 16 ⇒k⁴ - 16 = 0
⇒(k –2)(k + 2)(k² + 4) = 0 ⇒k = 2, -2
as k >0, we get k = 2.
Sum of the roots = 5k = 10
Question 9
2 / -0.83

Let α, βbe the roots of the equation x² - px + r = 0 and (α/2), 2 βbe the roots of the equation x² - qx + r = 0. Then the value of r is

A

B

C

D

Solution

But α+ β= p and α+ 4 β= 2q ⇒β= (1/3)(2q –p) and α= (2/3)(2p –q)
Question 10
2 / -0.83

The sum of all the real roots of the equation |x –2|² + |x –2| –2 = 0 is

A
7

B
4

C
1

D
none of these

Solution

Putting y = |x –2|,we can rewrite the given equation as
y² + y –2 = 0 or y² + 2y –y –2 = 0
⇒y(y + 2) –(y + 2) = 0 ⇒(y –1)(y + 2) = 0
as y = |x –2| > 0, y + 2 > 2
∴y –1 = 0 ⇒y = 1 ⇒|x –2| = 1
⇒x –2 = ±1 ⇒x = 2 ±1 ⇒x = 3 or 1