Mathematics Practice Questions Set for NDA/CDS & Defence Exams | Set-2|

C. x² – 7x – 44

Putting x = -4 in every option:

1) x² – 7x + 44 = 16 + 28 + 44 = 88

2) x² + 7x – 44 = 16 – 28 – 44 = -56

3) x² – 7x – 44 = 16 + 28 – 44 = 0

4) x² + 7x + 44 = 16 – 28 + 44 = 32

Since x = -4 satisfies the 3^rd equation;

∴ (x + 4) is a factor of (x² – 7x – 44)

B. -2

2x² + 6x + k = 0

Since α and β are the roots of this equation;

∴ Sum of roots = α + β = -6/2 = -3

And Multiplication of roots = αβ = k/2

Now,

Since k < 1;

∴ Maximum value of this expression will be -2. (As the first term will be negative always)

A. 1 only

Since HCF (b, c) = 1;

Suppose b = 2, c = 3 and d = 8 then a = b × c = 2 × 3 = 6;

Statement 1:

HCF(c, bd) = HCF (3, 16) = 1

And HCF(c, d) = HCF (3, 8) = 1

Hence HCF(c, bd) = HCF(c, d) is true.

Statement 1:

LCM(a, d) = LCM(6, 8) = 24

And LCM(c, bd) = LCM(3, 16) = 48

Hence LCM(a, d) = LCM(c, bd) is not true.

∴ Statement 1 only is correct.

B. 13

Let N = 2⁴⁰

Taking ‘log’ both the sides;

⇒ log₁₀N = 40(log₁₀2)

⇒ log₁₀N = 40 × 0.301 = 12.04

Taking ‘anti-log’ both the sides;

⇒ N = (10)^12.04

∴ Number of digits in 2⁴⁰ = 13

A. 2/3

(a² – 5a + 3)x² + (3a – 1)x + 2 = 0;

Let the roots of this equation are ‘y’ and ‘2y’;

∴ Sum of roots = 3y = [-(3a – 1)/(a² – 5a + 3)]      —- (1)

And Product of roots = 2y² = 2/(a² – 5a + 3)        —- (2)

∴ (Product) / (Sum) = 2y²/3y = 2/-(3a – 1)

⇒ y = [-3/(3a – 1)]

From equation 1;

3y = [-(3a – 1)/(a² – 5a + 3)]

Putting the value of y;

⇒ 3 × [-3/(3a – 1)] = [-(3a – 1)/(a² – 5a + 3)]

⇒ 9a² – 45a + 27 = 9a² – 6a + 1

⇒ 45a – 6a = 26

⇒ 39a = 26

⇒ a = 2/3

A. 4

We know that;

4¹/9 = Remainder will be 4;

4²/9 = Remainder will be 7;

4³/9 = Remainder will be 1;

Now (4444)⁴⁴⁴⁴ = [(4444)³]¹⁴⁸¹ × (4444)¹

When dividing this from 9 it can be written as: [(4)³]¹⁴⁸¹ × (4)¹]

Since Remainder when 4³ is divided by 9 is 1 hence dividing [(4)³]¹⁴⁸¹ × (4)¹] by 9, Remainder will be 4 only.

D. (A ∩ B) – C 

A = {x : x is a multiple of 7} = {7, 14, 21, 28, 35………..}

B = {x : x is a multiple of 5} = {5, 10, 15, 20, 25………..}

C = {x : x is a multiple of 35} = {35, 70, 105, 140……..}

Checking all the options:

(A – B) ∪ C = {7, 14, 21, 28, 42 …..} ∪ {35, 70, 105, 140……..} = {7, 14, 21, 28, 35, 42, 49…}

Hence it is not a null set.

(A – B) – C = {7, 14, 21, 28, 42 …..} – {35, 70, 105, 140……..} = {7, 14, 21, 28, 42, 49….}

Hence it is not a null set.

(A ∩ B) ∩ C = {35, 70, 105, 140……} ∩ {35, 70, 105, 140……..} = {35, 70, 105, 140……..}

Hence it is not a null set.

(A ∩ B) – C = {35, 70, 105, 140……} – {35, 70, 105, 140……..} = ∅

Hence it is a null set.

B. 2

Given

x = 2 + 2^2/3 + 2^1/3

⇒ (x – 2) = (2^2/3 + 2^1/3)

Taking cube both the sides;

⇒ x³ – 8 – 6x² + 12x = 4 + 2 + 3 × 2^2/3 × 2^1/3 (2^2/3 + 2^1/3)

⇒ x³ – 14 – 6x² + 12x = 6 (x – 2)

⇒ x³ – 14 – 6x² + 12x – 6x + 12 = 0

⇒ x³ – 6x² + 6x = 2

D. 0

x log₁₀(10/3) + log₁₀3 = log₁₀(2 + 3^x) + x

⇒ x log₁₀10 – x log₁₀3 + log₁₀3 = log₁₀(2 + 3^x) + x

⇒ x – x log₁₀3 + log₁₀3 = log₁₀(2 + 3^x) + x

⇒ log₁₀3^-x + log₁₀3 = log₁₀(2 + 3^x)

⇒ log₁₀(3^-x) × 3 = log₁₀(2 + 3^x)

⇒ (3^-x) × 3 = (2 + 3^x)

⇒ 3/3^x = (2 + 3^x)

Suppose 3^x = p;

⇒ 3/p = (2 + p)

⇒ p² + 2p – 3 = 0

⇒ (p + 3) (p – 1) = 0

⇒ p = -3, 1

∴ 3^x = 1

⇒ 3^x = 3⁰

∴ x = 0

B. 9

a³ = 335 + b³ and a = 5 + b;

∴ a³ – b³ = 335 and a – b = 5

We know that;

(a³ – b³) = (a – b) (a² + b² + ab)

⇒ (a³ – b³) = (a – b) ((a – b)² + 3ab)

Putting the values;

⇒ 335 = 5 × (25 + 3ab)

⇒ 67 = (25 + 3ab)

⇒ 3ab = 42

⇒ ab = 14

And we know that (a – b) = 5

∴ a = 7 and b = 2

∴ a + b = 9