Polynomials Test

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Polynomials Test - 3

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Weekly Quiz Competition

Question 1
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 2
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 3
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 4
1 / -0

If b is the mean of interval 0 - 1, then the value of b + 1/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 + 1/b will surely be greater than 2.
Question 5
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 6
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 7
1 / -0

What is the zero of the binomial ax + b²?

A
0

B
b/a

C
-a/b

D
-b²/a

Solution

To find the zero of the binomial ax + b², substitute ax + b² = 0.
ax = -b²
x = -b²/a
Question 8
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

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Polynomials Test - 3

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Polynomials Test - 3

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SHARING IS CARING

Question 1

b > 92 (or) b < 14

b > 14 (or) b < 2

b = 14 (or) b = 2

b > 8 (or) b < 0

Solution

Question 2

0

2

4

6

Solution

Question 3

(pr – qs+ qr + ps)

pr – qs+ qr – ps

Cannot be determined

None of these

Solution

Question 4

greater than 2

less than 2

less than 1

less then 0

Solution

Question 5

a + c + e = b + d

a – c – e = b – d

a + c – e = b – d

a + c + e = d – b

Solution

Question 6

– 17

– 7

14

21

Solution

Question 7

0

b/a

-a/b

-b2/a

Solution

Question 8

6a3

8a3

10a3

12a3

Solution

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-b²/a

6a³

8a³

10a³

12a³