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

Given:

If x + 1/x = √7, find the value of x³ + 1/x³.

Formula used:

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

Calculation:

Here, a = √7

⇒ x³ + 1/x³ = (√7)³ - 3(√7)

⇒ x³ + 1/x³ = (7√7) - 3√7

⇒ x³ + 1/x³ = 4√7

∴ The correct answer is option (1).

Given:

\(\rm 8\left(\frac{0.2\times 0.2\times 0.2+0.04\times 0.04\times 0.04}{0.4\times 0.4\times 0.4+0.08\times 0.08\times 0.08}\right)+9\)

Formula Used:

\(\rm a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)

Calculation:

Let a = 0.2 and b = 0.04

Numerator = \(\rm (0.2)^3 + (0.04)^3 = a^3 + b^3\)

Denominator terms: 0.4 = 2a, 0.08 = 2b

Denominator = \(\rm (0.4)^3 + (0.08)^3 = (2a)^3 + (2b)^3 = 8a^3 + 8b^3 = 8(a^3 + b^3)\)

Fraction = \(\rm \frac{a^3 + b^3}{8(a^3 + b^3)} = \frac{1}{8}\)

Expression = \(\rm 8 \times \frac{1}{8} + 9\)

Expression = 1 + 9

Expression = 10

∴ The simplified value of the expression is 10.

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)

If x + 1 / x = 15, then find the value of x² + 1 / x².

Solution:

Square both sides:

(x + 1 / x)² = 15²

x² + 2 × (1 / x) × x + 1 / x² = 225

x² + 2 + 1 / x² = 225

x² + 1 / x² = 225 - 2

x² + 1 / x² = 223

∴ The value of x² + 1 / x² is 223.

Given:

\(\frac{(375 \times 375)+(125 \times 125) - (375 \times 125)}{(375 \times 375 \times 375)+(125 \times 125 \times 125)}\)

Formula Used:

a³ + b³ = (a + b)(a² + b² - ab)

Solution:

1/(a + b) = (a2 + b2 - ab)/(a3 + b3)

According to the question, 

⇒ 1/(375 + 125)

⇒ 1/500

∴ The correct answer is 1/500.

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:

x - (1/x) = (- 6)

Formula used:

If x - (1/x) = P, then

x + (1/x) = √(P² + 4) 

If x + (1/x) = P, then

x3 + (1/x3) = (P³ - 3P)

x⁵ - (1/x⁵) = {x³ + (1/x³)} × {x² - 1/x²} + {x - (1/x)}

Calculation:

x - (1/x) = (- 6)

x + (1/x) = √{(- 6)2 + 4} = √40 = 2√10

So, x² - 1/x² = (x + 1/x) (x - 1/x) = 2√10 × (-6) = -12√10

and x3 + (1/x3) =  (√40)³ - 3√40

⇒ 40√40 - 3√40 = 37 × 2√10 = 74√10

Now,

x5 - (1/x5) = {x3 + (1/x3)} × {x2 - 1/x2} + {x - (1/x)}

⇒ {74√10 × (-12√10)} + (- 6)

⇒ - 74 × 12 × (√10 × √10) - 6

⇒ (- 8880) - 6 = - 8886

∴ The correct answer is  - 8886.

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.

Given:

\(a + \frac{1}{a} = 7\)

Formula used:

(a + 1/a) = P ; then

(a² + 1/a²) = P² - 2

(a3 + 1/a3) = P³ - 3P

\(a^5 + \frac{1}{a^5} \) = (a² + 1/a²) × (a³ + 1/a³) - (a + 1/a)

Calculation:

a + (1/a) = 7

⇒ (a2 + 1/a2) = (7)2 - 2 = 49 - 2 = 47

⇒ (a3 + 1/a3) = (7)³ - (3 × 7) = 343 - 21 = 322

a⁵ + (1/a⁵) = (a2 + 1/a2) × (a3 + 1/a3) - (a + 1/a)

⇒ 47 × 322 - 7

⇒ 15134 - 7 = 15127

 ∴ The correct answer is 15127.

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:

(a + b + c) = 19

(a2 + b2 + c2) = 155

Formula used:

a² + b² + c² - (ab + bc + ca) = (1/2) × [(a - b)2 + (b - c)2 + (c - a)2]

Calculation:

a + b + c = 19

Squaring both sides

⇒ (a + b + c)² = (19)²

⇒ a2 + b2 + c2 + 2 × (ab + bc + ca) = 361

⇒ 155 + 2 × (ab + bc + ca) = 361

⇒ 2 × (ab + bc + ca) = (361 - 155)

⇒ (ab + bc + ca) = 206/2 = 103

Now,

a2 + b2 + c2 - (ab + bc + ca) = (1/2) × [(a - b)2 + (b - c)2 + (c - a)2]

⇒ 2 × (155 - 103) = (a - b)2 + (b - c)2 + (c - a)2

⇒ (a - b)2 + (b - c)2 + (c - a)2 = 104

∴ The correct answer is 104.

Given:

x2 + (1/x2) = 7

Formula used:

x² + (1/x²) = P

then x + (1/x) = √(P + 2)

and x - (1/x) = √(P - 2)

⇒ x2 - (1/x2) = {x + (1/x)} × {x - (1/x)}

Calculation:

x2 + (1/x2) = 7

⇒ x + (1/x) = √(7 + 2) = √9

⇒ x + (1/x) = 3

⇒ x - (1/x) = -√(7 - 2)

⇒ x - (1/x) = - √5 {0 < x < 1}

x2 - (1/x2) = {x + (1/x)} × {x - (1/x)}

⇒ 3 × (- √5)

∴ The correct answer is - 3√5.

Mistake Points

Please note that 

0 < x < 1

so

1/x > 1

so

x + 1/x > 1

and

x - 1/x < 0 (because 0 < x < 1 and 1/x > 1 so x - 1/x < 0)

so

(x - 1/x)(x + 1/x) < 0