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

Given:

x + 1/x = 1

Formula used:

x² + 1/x² = (x + 1/x)² - 2

x³ + 1/x³ = (x + 1/x) × (x² + 1/x²) - (x + 1/x)

Calculation:

1. Find x² + 1/x²:

⇒ x² + 1/x² = (x + 1/x)² - 2

⇒ x² + 1/x² = 1² - 2

⇒ x² + 1/x² = 1 - 2

⇒ x² + 1/x² = -1

2. Find x³ + 1/x³:

⇒ x³ + 1/x³ = (x + 1/x) × (x² + 1/x²) - (x + 1/x)

⇒ x³ + 1/x³ = 1 × (-1) - 1

⇒ x³ + 1/x³ = -1 - 1

⇒ x³ + 1/x³ = -2

∴ The value of x³ + 1/x³ is -2.

Given:

p(x) = 2x³ – 3x² – 2x + 3 and q(x) = 3x² + 8x + 5

Concept:

H.C.F. of two or more equations is the greatest factor that divides each of them exactly.

Calculation:

The factors of p(x) = 2x³ – 3x² – 2x + 3

⇒ x² × (2x – 3) – 1 × (2x – 3)

⇒ (x² – 1) × (2x – 3)

⇒ (x – 1) × (x + 1) × (2x – 3)

And, the factors of q(x) = 3x² + 8x + 5

⇒ 3x² + 5x + 3x + 5

⇒ x × (3x + 5) + 1 × (3x + 5)

⇒ (3x + 5) × (x + 1)

Given:

The product of {(341)491 × (625)317 × (6374)1793} and we are to find its unit digit.

Formula used:

To find the unit digit of a product, we multiply the unit digits of the individual numbers raised to their respective powers.

Calculation:

For (341)⁴⁹¹, the unit digit of 341 is 1. Any number ending in 1 raised to any power will have a unit digit of 1.

For (625)³¹⁷, the unit digit of 625 is 5. Any number ending in 5 raised to any power will have a unit digit of 5.

For (6374)¹⁷⁹³, the unit digit of 6374 is 4. The unit digit of numbers ending in 4 alternates in a cycle of 2 when raised to powers: 4¹ has a unit digit of 4, and 4² has a unit digit of 6, then it repeats.

Since 1793 is odd, 4 raised to an odd power will have a unit digit of 4.

Now, 1 (from 341⁴⁹¹) × 5 (from 625³¹⁷) × 4 (from 6374¹⁷⁹³) = 20

∴ The unit digit in the given product is 0.

Given:

Expression = (a + b)³

Formula Used:

Binomial expansion formula for (x + y)ⁿ

Calculations:

(a + b)³ = (a + b) × (a + b) × (a + b)

⇒ (a + b)³ = (a² + 2ab + b²) × (a + b)

⇒ (a + b)³ = a × (a² + 2ab + b²) + b × (a² + 2ab + b²)

⇒ (a + b)³ = a³ + 2a²b + ab² + a²b + 2ab² + b³

⇒ (a + b)³ = a³ + (2a²b + a²b) + (ab² + 2ab²) + b³

⇒ (a + b)³ = a³ + 3a²b + 3ab² + b³

⇒ (a + b)3 = a3 + b3 + 3ab(a + b)

∴ (a + b)3 = a3 + b3 + 3ab(a + b)

Given:

7x + 3y = 4 and 2x + y = 2

Calculation:

7x + 3y = 4      ----(1)

2x + y = 2     ----(2)

By solving,

equation (1) - 3 × equation(2)

⇒ 7x + 3y - 3(2x + y) = 4 - (3 × 2)

⇒ x = -2

Putting the value of x in equation (1)

7x + 3y = 4

⇒ 7 × (-2) + 3y = 4

⇒ y = 6

∴ The value of x and y is -2 and 6.

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:

\(\begin{array}{l} \frac{1}{x} = \frac{1}{{\sqrt{10}{\rm{\;}} + {\rm{\;}}3}}\\ = {\rm{\;}}\frac{{\sqrt{10} {\rm{\;}} - {\rm{\;}}3}}{{\left( {\sqrt{10} + {\rm{\;}}3} \right)\left( {\sqrt{10} {\rm{\;}} - {\rm{\;}}3} \right)}}\\ = {\rm{\;}}\frac{{\sqrt{10} {\rm{\;}} - {\rm{\;}}3 }}{{{{\left( {\sqrt{10} } \right)}^2} - {{\left( {3} \right)}^2}}} \end{array}\)

⇒ 1/x = √10 - 3

\( \Rightarrow x - \;\frac{1}{x} = \;\sqrt 10 + 3\; -\sqrt10 + 3 = 6\)     ----(1)

Squaring both side of (1),

\( \Rightarrow (x - \;\frac{1}{x})^2 = \;(6\;)^2\)

\( \Rightarrow {x^2} - 2x\frac{1}{x} + \;\frac{1}{{{x^2}}} = 36\)

\( \Rightarrow {x^2} - 2 + \;\frac{1}{{{x^2}}} = 36\)

\( \Rightarrow {x^2} + \;\frac{1}{{{x^2}}} = 38\)    -----(2)

\( ∴ \;{x^3} - \;\frac{1}{{{x^3}}}\; = \left( {\;x - \;\frac{1}{x}\;} \right)\left( {\;{x^2} + x\frac{1}{x} + \;\frac{1}{{{x^2}}}\;} \right)\)

\(\Rightarrow \;{x^3} - \;\frac{1}{{{x^3}}}\; = \left( {\;x - \;\frac{1}{x}\;} \right)\left( {\;{x^2} + \;\frac{1}{{{x^2}}} + 1} \right)\)

\(\Rightarrow \;{x^3} - \;\frac{1}{{{x^3}}}\; = 6 \times (38 + 1)\)

\(x^3 - \frac{1}{x^3} = 234\)

∴ The required value is 234.

 Shortcut TrickGiven:

x = √10 + 3

Formula used: 

\(\rm If ~x -\frac{1}{x} = a \)

⇒ \(x^3 - \frac{1}{x^3} = a^3 + 3a\)

Calculation:

x = √10 + 3

⇒ 1/x = √10 - 3

⇒ \(x -\frac{1}{x} = 6\) 

⇒ \(x^3 - \frac{1}{x^3} = 6^3 + 3\times 6\)

⇒ \(x^3 - \frac{1}{x^3} = 234\)

∴ 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

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:

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

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

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