Complex Numbers and Quadratic Equation Test

Result

Complex Numbers and Quadratic Equation Test - 8

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Weekly Quiz Competition

Question 1
2 / -0.83

If the sum of the roots of the equation ax² + bx + c = 0 is equal to sum of the squares of their reciprocals, then bc² , ca² , ab² are in

A
A.P

B
G.P

C
H.P

D
A.G.P

Solution
Question 2
2 / -0.83

If k >0 and the product of the roots of the equation x² - 3kx + 2e^2logk -1 = 0 is 7 then the sum of the roots is

A
1

B
4

C
6

D
8

Solution

Product = 2e^2logk
Question 3
2 / -0.83

The number of real solution of the equation (9/10)^x = -3 + x - x² is

A
2

B
0

C
1

D
3

Solution

LHS always positive RHS always negative LHS ≠RHS
Question 4
2 / -0.83

If the roots of (x - a)(x - b) +(x - b)(x - c) + (x - c)(x - a) = 0 are equal then which of the following is not possible

A
a = b = c

B
a + b + c = 0

C
a + bw + cw²

D
a + bw² + cw = 0

Solution

°•°( x - a ) ( x - b ) + ( x - b ) ( x - c ) + ( x - c ) ( x - a ) = 0 .
==>x ²- bx - ax + ab + x ²- cx - bx + bc + x ²- ax - cx + ac = 0 .
==>3x ²- 2bx - 2ax - 2cx + ab + bc + ca = 0 .
==>3x ²- 2x( a + b + c ) + ( ab + bc + ca ) = 0 .
When equation is compared with Ax ²+ Bx + C = 0 .
Then , A = 3 .
B = 2( a + b + c ) .
And, C = ( ab + bc + ca ) .
•°•Discriminant ( D ) = b ²- 4ac .
= [ 2( a + b + c )]²- 4 ×3 ×( ab + bc + ca ) .
= 4( a + b + c )²- 12( ab + bc + ca ) .
= 4[ ( a + b + c )²- 3( ab + bc + ca ) ] .
= 4( a ²+ b ²+ c ²+ 2ab + 2bc + 2ca - 3ab - 3bc - 3ca ) .
= 4( a ²+ b ²+ c ²- ab - bc - ca ) .
= 2( 2a ²+ 2b ²+ 2c ²- 2ab - 2bc - 2ca ) .
= 2[ ( a - b )²+ ( b - c )²+ ( c - a )²] ≥0 .
[ °•°( a - b )²≥0, ( b - c )²≥0 and ( c - a )²≥0 ] .
This shows that both the roots of the given equation are real .
For equal roots, we must have : D = 0 .
Now, D = 0 .
==>( a - b )²+ ( b - c )²+ ( c - a )²= 0 .
==>( a - b ) = 0, ( b - c ) = 0 and ( c - a ) = 0 .
Question 5
2 / -0.83

The equation log₂ (3- x) + log₂ (1- x) = 3 has

A
One root

B
Two root

C
Infinite roots

D
No root

Solution
Question 6
2 / -0.83

If ax² + bx + c = 0 and bx² + cx + a = 0 have a common a ≠0

A
1

B
2

C
3

D
9

Solution

a α² + b α+ c = 0 and b α² + c α+ a = 0 ⇒ α= 1 ∴a + b + c = 0
Question 7
2 / -0.83

If a, b are the roots of x² + px + 1 = 0, and c, d are the roots of x² + qx + 1 = 0, then the value of E = (a - c)(b - c)(a+ d)(b+ d) is

A
p² - q²

B
q² - p²

C
q² + p²

D
None of these

Solution

We have x² + px + 1 = (x –a)(x –b)
Thus, E = (c –a)(c –b)(-d - a)(-d - b)
= (c² + pc + 1)[(-d² ) - pd + 1] = (c² + pc + 1)(d² - pd + 1) [∴a + b = -p]
But c² + qc + 1 = 0 and d² + qd + 1 = 0
∴E = (-qc + pc)(-qd - pd) = cd(q - p)(q + p) = cd(q² - p² ) = q² - p² [∴cd = 1]
Question 8
2 / -0.83

If x² + 2ax +10 - 3a >0 for each x ∈R, then

A
a <- 5

B
-5

C
a >5

D
2 < a < 5

Solution

For x² + 2ax + 10 - 3a >0
D <0 ⇒(a + 5) (a - 2) <0
Question 9
2 / -0.83

If a, b, c are real and a ≠b, then the roots of the equation 2(a - b)x² -11(a + b + c)x - 3(a - b) = 0 are

A
real and equal

B
real and unequal

C
purely imaginary

D
none of these

Solution

The discriminant D of the quadratic equation is given by
D = 121(a + b + c)² + 24(a –b)2
As a, b, c are real, 121(a + b + c)²> 0
Also, as a ≠b , (a - b)² >0 ; Thus, D >0
Therefore, the equation (1) has real and unequal roots.
Question 10
2 / -0.83

The minimum value of

A
2

B
4

C
6

D
None