Algebra MCQ Quiz - Objective Question with Answer for Algebra - Download Free PDF

Expanding and combining like terms gives: x⁴ - 2x³ + 3x² + 2x² - 4x + 6 + 5 - 2x³

= x⁴ - 4x³ + 5x² - 4x + 11.

Therefore, A = 1, B = -4, C = 5, D = -4, E = 11.

A - B + C - D - E = 1 - (-4) + 5 - (-4) - 11 = 3

Given:

Formula used:

x - 1/x = √[(x+1/x)2 - 4]

= (x - 1/x)3 + 3 (x - 1/x)

Calculation:

⇒ 

⇒ x - 1/x = 4

⇒  = (x - 1/x)³ + 3 (x - 1/x)

⇒ 4³ + 3 × 4 =  64 + 12 = 76

∴ The correct option is 2

Given:

Concept:

System of equations

a1x + b1y = c1

a2x + b2y = c2

For Infinite solution

Calculation:

We have 

ax + by + c = 0       ----(1)

lx + my + n = 0       ----(2)

For infinite solutions, 

Important Points

For unique solution

For inconsistent solution

Given: 

(m + 1 / m) =√ 5

Concept used :

(x²-1/x²) = (x+1/x) × (x-1/x)

Formula used :

(p-1/p) = (p+1/p)² - 4

Calculation :

(m-1/m) = (m+1/m)² - 4

⇒ (m-1/m) = (√5)2 - 4

⇒ (m-1/m) = 5 - 4 =1

then.

⇒ (m²-1/m²) = (m+1/m) × (m-1/m) 

Put the value

⇒ (m2-1/m2) = (√5) × (1) 

⇒ (m2-1/m2) = √5

 Hence, the correct option is 3).

Given:

x - 1/x = 3

Concept used:

a³ - b³ = (a - b)³ + 3ab(a - b)

Calculation:

x³ - 1/x³ = (x - 1/x)³ + 3 × x × 1/x × (x - 1/x)

⇒ (x - 1/x)3 + 3(x - 1/x)

⇒ (3)³ + 3 × (3)

⇒ 27 + 9 = 36

∴ The value of x³ - 1/x³ is 36.

Alternate Method If x - 1/x = a, then x³ - 1/x³ = a³ + 3a

Here a = 3

x - 1/x³ = 3³ + 3 × 3

= 27 + 9

= 36

Given:

x = √10 + 3

Formula used: 

a² - b² = (a + b)(a - b)

a³ - b³ = (a - b)(a² + ab + b²)

Calculation:

⇒ 1/x = √10 - 3

     ----(1)

Squaring both side of (1),

    -----(2)

∴ The required value is 234.

 Shortcut TrickGiven:

x = √10 + 3

Formula used: 

⇒ 

Calculation:

x = √10 + 3

⇒ 1/x = √10 - 3

⇒  

⇒ 

⇒ 

∴ The required value is 234.

Given:

p – 1/p = √7

Formula:

P³ – 1/p³ = (p – 1/p)³ + 3(p – 1/p)

Calculation:

P³ – 1/p³ = (p – 1/p)³ + 3 (p – 1/p)

⇒ p³ – 1/p³ = (√7)³ + 3√7

⇒ p³ – 1/p³ = 7√7 + 3√7

⇒ p³ – 1/p³ = 10√7

Shortcut Trick x - 1/x = a, then x³ - 1/x³ = a³ + 3a

Here, a = √7                                                          ( put the value in required eqⁿ )

⇒p³ – 1/p³ = (√7)³ + 3 × √7 = 7√7 + 3√7

 ⇒p3 – 1/p3  = 10√7.

Hence; option 4) is correct.

Given:

a + b + c = 14, ab + bc + ca = 47 and abc = 15

Concept used:

a³ + b³ + c³ - 3abc = (a + b + c) × [(a + b + c)² - 3(ab + bc + ca)]

Calculations:

a³ + b³ + c³ - 3abc = 14 × [(14)² - 3 × 47]

⇒ a³ + b³ + c³ – 3 × 15 = 14(196 – 141)

⇒ a³ + b³ + c³ = 14(55) + 45

⇒ 770 + 45

⇒ 815

∴ The correct choice is option 1.

Formula used:

(a + b)³ = a³ + b³ + 3ab(a + b)

Calculation:

⇒ x^2/3 + x^1/3 = 2

⇒ (x^2/3 + x^1/3)³ = 2³

⇒ x² + x + 3x(x^2/3 + x^1/3) = 8

⇒ x² + 7x - 8 = 0

⇒ x² + 8x - x - 8 = 0

⇒ x (x + 8) - 1 (x + 8) = 0

⇒ x = - 8 or x = 1

Given:

3x² – ax + 6 = ax² + 2x + 2

⇒ 3x² – ax² – ax – 2x + 6 – 2 = 0

⇒ (3 – a)x² – (a + 2)x + 4 = 0

Concept Used:

If a quadratic equation (ax2 + bx + c=0) has equal roots, then discriminant should be zero i.e. b2 – 4ac = 0

Calculation:

⇒ D = B² – 4AC = 0

⇒ (a + 2)² – 4(3 – a)4 = 0

⇒ a² + 4a + 4 – 48 + 16a = 0

⇒ a² + 20a – 44 = 0

⇒ a² + 22a – 2a – 44 = 0

⇒ a(a + 22) – 2(a + 22) = 0

⇒ a = 2, -22

Formula used:

a3 + b3 + c3 - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

Calculation:

a + b + c = 0

a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

⇒ a3 + b3 + c3 - 3abc = 0 × (a2 + b2 + c2 - ab - bc - ca) = 0

⇒ a3 + b3 + c3 - 3abc = 0

⇒ a3 + b3 + c3 = 3abc 

Now, (a3 + b3 + c3)2 = (3abc)² = 9a2b2c2 

Given

2x⁵ + 2x³y³ + 4y⁴ + 5.

Concept

The degree of a polynomial is the highest of the degrees of its individual terms with non-zero coefficients.

Solution

Degree of the polynomial in 2x⁵ = 5

Degree of the polynomial in 2x³y³ = 6

Degree of the polynomial in 4y⁴ = 4

Degree of the polynomial in 5 = 0

Hence, the highest degree is 6

∴ Degree of polynomial = 6

Mistake Points  

One may choose 5 as the correct option due to x⁵ but the correct answer will be 6 as 2x³y³ has the highest power of 6.

Important Points

 The degree of a polynomial is the highest of the degrees of its individual terms with non-zero coefficients. Here for a specific value when x will be equal to y then the equation will be:

2x5 + 2x3y3 + 4y4 + 5

= 2x⁵ + 2x⁶ + 4x⁴ + 5

∴ The degree of the polynomial will be 6

Let my current age = x years and my cousin’s age = y years.

Three-fifths of my current age is the same as five-sixths of that of one of my cousins’,

⇒ 3x/5 = 5y/6

⇒ 18x = 25y

My age ten years ago will be his age four years hence,

⇒ x – 10 = y + 4

⇒ y = x – 14,

⇒ 18x = 25(x – 14)

⇒ 18x = 25x – 350

⇒ 7x = 350

∴ x = 50 years

Given:

x2 – x – 1 = 0

Formula used:

If the given equation is ax² + bx + c = 0

Then Sum of roots = -b/a

And Product of roots = c/a

Calculation:

As α and β are roots of x² – x – 1 = 0, then

⇒ α + β = -(-1) = 1

⇒ αβ = -1

Now, if (α/β) and (β/α) are roots then,

⇒ Sum of roots = (α/β) + (β/α)

⇒ Sum of roots = (α² + β²)/αβ

⇒ Sum of roots = [(α + β)² – 2αβ]/αβ

⇒ Sum of roots = (1)² – 2(-1)]/(-1) = -3

⇒ Product of roots = (α/β) × (β/α) = 1

Now, then the equation is,

⇒ x² – (Sum of roots)x + Product of roots = 0

⇒ x² – (-3)x + (1) = 0

⇒ x² + 3x + 1 = 0