Polynomials Test

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Polynomials Test - 6

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Weekly Quiz Competition

Question 1
1 / -0

R₁ and R₂ are the remainders when x³ + 2x² – 5ax – 7 and x³ + ax² – 12x + 6 are divided by x + 1 and x – 1, respectively. If 2R₁ + R₂ = 6, find the value of a.

A
23/12

B
23/11

C
23/13

D
None of these

Solution

(– 1)³ + 2 (– 1)² – 5a(– 1) –7 = R₁
R₁ = – 6 + 5a
(1)³ + a (1)² – 12 (1) + 6 = R₂
R₂ = a – 5
2R₁ + R₂ = 6
– 12 + 10a – 5 + a = 6
11a = 23
a =
Question 2
1 / -0

The inequality b² + 5 > 16b - 23 is satisfied if

A
b > 92 (or) b < 14

B
b > 14 (or) b < 2

C
b = 14 (or) b = 2

D
b > 8 (or) b < 0

Solution

b² + 5 > 16b - 23
Or b² - 16b + 28 > 0
Solving this, we get
b > 14 (or) b < 2
Question 3
1 / -0

The factors of a² + b² + 2(ab + bc + ca) are (a + b+ m) and (a + b + nc). Find the value of m + n out of the given options.

A
0

B
2

C
4

D
6

Solution

a² + b² + 2ab + 2bc + 2ca + c² – c²
= (a + b + c)² – c²
= (a + b) (a + b + 2c)
On comparision we get
m = 0, n = 2
m + n = 0 + 2 = 2
Question 4
1 / -0

Choose the factor of (p² – q²) (r² – s²) – 4pqrs out of the given options.

A
(pr – qs+ qr + ps)

B
pr – qs+ qr – ps

C
Cannot be determined

D
None of these

Solution

(p² – q²) (r² – s²) – 4pqrs
= (pr – qs)² – (qr + ps)²
= (pr - qs + qr + ps)
Question 5
1 / -0

If b is the mean of interval 0 - 1, then the value of b + will be:

A
greater than 2

B
less than 2

C
less than 1

D
less then 0

Solution

The mean is the average of the numbers and, If b is the mean of interval 0 - 1, then the value of b + will surely be greater than 2.
Question 6
1 / -0

Which of the following options is true, if x² – 1 is a factor of ax⁶ + bx⁵ + cx² + dx + e .

A
a + c + e = b + d

B
a – c – e = b – d

C
a + c – e = b – d

D
a + c + e = d – b

Solution

x² – 1 = (x + 1) (x – 1)
Since (x + 1), (x –1) are factors of the given polynomial, consider (x + 1) is a factor a + c + e = b + d
Question 7
1 / -0

f(x) = ax¹⁰¹ + bx⁹⁷ + cx – 5, where a, b and c are constants. Also f(– 7) = 7. Find the value of (7).

A
– 17

B
– 7

C
14

D
21

Solution

f(x) = ax¹⁰¹+ bx⁹⁷ + cx – 5

f(-x) = -ax¹⁰¹- bx⁹⁷ - cx – 5
Putting x = -7
7 = -a7¹⁰¹- b7⁹⁷-7x - 5
a(7)¹⁰¹ + b(7)⁹⁷ + 7x = -12 ...(i)

f(7) = a(7)¹⁰¹ + b(7)⁹⁷ + 7x - 5
= -12 - 5 = -17
Question 8
1 / -0

What is the zero of the binomial ax + b²?

A
0

B

C

D

Solution

To find the zero of the binomial ax + b², substitute ax + b² = 0.
ax = -b²
x = -b²/a
Question 9
1 / -0

Choose the factors of x² + 11x + 6 from the following options:

A
(x – 3) (x + 2)

B
(x – 3) (x – 2)

C
(x + 3) (x – 2)

D
(x + 3) (x + 2)

Solution

+ 11 x + 6
= + 9x + 2x + 6
= x (x + 3+ 2 (x + 3)
= (x + 3) (+ 2)
Question 10
1 / -0

What is the value of (a + b)³ + (a - b)³ + 6a(a² - b²)?

A
6a³

B
8a³

C
10a³

D
12a³

Solution

Let a + b = A, a - b = B
A + B = 2a
(a + b)³ + (a - b)³ + 6a(a² -b²)
A³ + B³ + 3(A + B) AB = (A + B)³
= (2a)³ = 8a³
Question 11
1 / -0

The polynomials 2x³ + ax² + 3x – 5 and x³ + x² – 4x – a leave the same remainder when divided by x + 1. Find the value of a out of the following options.

A
a = 7

B
a = 1

C
a = 2

D
a = – 2

Solution

2 (-1)³ +a( -1)²+3(-1) – 5 = (-1)³ + (-1)² –4(-1) – a
-2 + a - 3 – 5 = -1 + 1 + 4 – a
2a = 14
a = 7
Question 12
1 / -0

Factorise x¹² -y¹² and choose the right answer.

A
(x – y) (x² + xy + y²) (x + y)
(x² – xy + y²) (x² + y²) (x⁴ – x² y² + y⁴)

B
(x + y) (x² – xy + y²) (x + y)
(x² – xy + y²) (x² + y²) (x⁴ – x² y² + y⁴)

C
(x + y) (x² + xy – y²) (x + y)
(x² – xy + y²) (x² + y²) (x⁴ – x² y² + y⁴)

D
(x – y) (x² – xy + y²) (x + y)
(x² – xy + y²) (x² + y²) (x⁴ – x² y² + y⁴)

Solution

x¹² –y¹²
(x⁶ + y⁶) (x⁶ – y⁶)
(x⁴ – x²y² + y⁴) (x² + y²)
(x – y) (x² + xy + y²) (x + y) (x² – xy + y²)
Question 13
1 / -0

If p(x) = 4x³ - 3x² + 2x + 1, q(x) = x³ - x² + x + 1 and r(x) = x² - 2x + 1, then the value of p(x) + 2q(x) - r(x) is

A
6x³ - 6x² + 6x + 2

B
6x³ - 6x² - 11x + 11

C
6x³ - 15x² + 6x + 11

D
6x³ - 15x² - 11x - 11

Solution

p(x) + 2q(x) - r(x)
= 4x³ - 3x²+ 2x + 1 + 2x³- 2x²+ 2x + 2 - x² + 2x - 1
= 6x³ - 6x² + 6x + 2
Question 14
1 / -0

If 2y³ + my² + 11y + m + 3 is exactly divisible by 2y - 3, then the value of m is

A
-105/12

B
- 105/13

C
105/13

D
none

Solution

2+ m + 11 + m + 3 = 0
m = - 7
Question 15
1 / -0

If p + q + r = 0, then p³ + q³ + r³equals

A
pqr

B
2pqr

C
3pqr

D
zero

Solution

Given: p + q + r = 0
p + q = – r ....(1)
Cubing both sides, we get
(p + q)³ = (–r)³
p³ + q³ + 3pq(p + q) = – r³
Put (p + q) = -r:
From equation (1), we get
p³ + q³ + r³ = 3pqr

Correct option is 3.
Question 16
1 / -0

If y - 2 and y - 1 are the factors of py² + 5y + r, then

A
r = 2p

B
p = r

C
r = -2p

D
None of these

Solution

If (y - 2) is a factor of py² + 5y + r, then P(2)² + 5(2) + r = 0.
4p + r + 10 = 0 â¦ (1)
If (y - 1) is a factor of py² + 5y + r, then p(1)² + 5(1) + r = 0.
p + 5 + r = 0 â¦ (2)
By solving (1) and (2), we get

p = and r =

So, r = 2p
Question 17
1 / -0

If p + q + r = 0, evaluate and choose the answer from the following:

A
1

B
2

C
3

D
4

Solution

=
= = 3
Question 18
1 / -0

Factorise (3 - 4y - 7y²)² - (4y + 1)² and choose the correct option:

A
(4 - 7y²) (2 - 8y - 7y²)

B
(7y²- 4) (2 - 8y - 7y²)

C
(4 - 7y²) (7y² + 8y - 2)

D
(7y²- 4) (7y² - 8y - 2)

Solution

(3 - 4y -7y²)² - (4y + 1)²
(4 - 7y²) (2 - 8y - 7y²)
Question 19
1 / -0

If (x + 1) and (x –1) are the factors of px³ –2x + q, then the values of p and q are

A
p = – 1, q = 2

B
p = 2, q = – 1

C
p = 2, q = 0

D
p = –2, q = –2

Solution

Since (x + 1) and (x -1) are factors of px³ –2x + q
Then p ( –1)³ – 2(1) + q = 0
– p + q = – 2 ………(1)
p(1)³ – 2(1) + q = 0
p + q = 2……… (2)
By solving we get p = 2 , q = 0
Question 20
1 / -0

When f(x) is divided by x+7/9, the remainder is

A
f(2/9)

B
f(5/9)

C
f(7/9)

D
f(-7/9)

Solution

When f(x) is divided by 2x+ 3
reminder will be f