Identities MCQ Quiz - Objective Question with Answer for Identities - Download Free PDF
Last updated on Jun 3, 2025
Latest Identities MCQ Objective Questions
Identities Question 1:
If
Answer (Detailed Solution Below)
Identities Question 1 Detailed Solution
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).
Identities Question 2:
Simplify the following expression.
Answer (Detailed Solution Below)
Identities Question 2 Detailed Solution
Given:
Formula Used:
Calculation:
Let a = 0.2 and b = 0.04
Numerator =
Denominator terms: 0.4 = 2a, 0.08 = 2b
Denominator =
Fraction =
Expression =
Expression = 1 + 9
Expression = 10
∴ The simplified value of the expression is 10.
Identities Question 3:
(a + b)3 = ?
Answer (Detailed Solution Below)
Identities Question 3 Detailed Solution
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)
Identities Question 4:
If
Answer (Detailed Solution Below)
Identities Question 4 Detailed Solution
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.
Identities Question 5:
Simplify:
Answer (Detailed Solution Below)
Identities Question 5 Detailed Solution
Given:
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.
Top Identities MCQ Objective Questions
If x −
Answer (Detailed Solution Below)
Identities Question 6 Detailed Solution
Download Solution PDFGiven:
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
If x = √10 + 3 then find the value of
Answer (Detailed Solution Below)
Identities Question 7 Detailed Solution
Download Solution PDFGiven:
x = √10 + 3
Formula used:
a2 - b2 = (a + b)(a - b)
a3 - b3 = (a - b)(a2 + ab + b2)
Calculation:
⇒ 1/x = √10 - 3
Squaring both side of (1),
∴ 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.
If
Answer (Detailed Solution Below)
Identities Question 8 Detailed Solution
Download Solution PDFGiven:
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.
If p – 1/p = √7, then find the value of p3 – 1/p3.
Answer (Detailed Solution Below)
Identities Question 9 Detailed Solution
Download Solution PDFGiven:
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.
If a + b + c = 14, ab + bc + ca = 47 and abc = 15 then find the value of a3 + b3 +c3.
Answer (Detailed Solution Below)
Identities Question 10 Detailed Solution
Download Solution PDFGiven:
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.
If
Answer (Detailed Solution Below)
Identities Question 11 Detailed Solution
Download Solution PDFGiven:
Formula used:
(a + 1/a) = P ; then
(a2 + 1/a2) = P2 - 2
(a3 + 1/a3) = P3 - 3P
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.
The sum of values of x satisfying x2/3 + x1/3 = 2 is:
Answer (Detailed Solution Below)
Identities Question 12 Detailed Solution
Download Solution PDFFormula used:
(a + b)3 = a3 + b3 + 3ab(a + b)
Calculation:
⇒ x2/3 + x1/3 = 2
⇒ (x2/3 + x1/3)3 = 23
⇒ x2 + x + 3x(x2/3 + x1/3) = 8
⇒ x2 + 7x - 8 = 0
⇒ x2 + 8x - x - 8 = 0
⇒ x (x + 8) - 1 (x + 8) = 0
⇒ x = - 8 or x = 1
∴ Sum of values of x = -8 + 1 = - 7.If a + b + c = 0, then (a3 + b3 + c3)2 = ?
Answer (Detailed Solution Below)
Identities Question 13 Detailed Solution
Download Solution PDFFormula 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
If (a + b + c) = 19 and (a2 + b2 + c2) = 155, find the value of (a - b)2 + (b - c)2 + (c - a)2.
Answer (Detailed Solution Below)
Identities Question 14 Detailed Solution
Download Solution PDFGiven:
(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.
If
Answer (Detailed Solution Below)
Identities Question 15 Detailed Solution
Download Solution PDFGiven:
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
x2 - (1/x2) = {x + (1/x)} × {x - (1/x)}
⇒ 3 × (- √5)
∴ The correct answer is - 3√5.
Mistake Points
Please note that
0
so
1/x > 1
so
x + 1/x > 1
and
x - 1/x 1 so x - 1/x
so
(x - 1/x)(x + 1/x)