Quantitative Aptitude PYQs

Given:

Cartesian product A × A has 16 elements among which are

(0, 2) and (1, 3).

Concept:

The cardinality of a set is a measure of the number of elements of the set. For example, the set contains 3 elements and therefore has a cardinality of 3.

Calculation:

The cardinality of A × A,

|A × A| = n² = 16 = 4²

A = {0, 1, 2, 3}

Hence, yes it is possible to determine set A.

Set A = {0, 1, 2, 3}

Hence, A × A also contains the element (3, 2)

Given: A = {1, 2, 3, ..., 20}. Define a relation R from A to A by R = {(x, y) : 4x - 3y = 1}, where x, y ∈ A.

Concept:

Calculation:

A = {1, 2, 3, ..., 20}

R = {(x, y) : 4x - 3y = 1}

\(y = {{4x-1} \over 3}\)

a relation R from A to A, so finding the value of y which is present in set A.

R = {(1,1); (4,5); (7,9); (10,13); (13,17)}

If we put x = 16 then y = 21 which is not present in set A.

Hence, the domain of R is {1, 4, 7, 10, 13). So, statement 1 is not correct.

Hence, the range of R is {1, 5, 9, 13, 17). So, statement 2 is correct.

The co-domain of R is set A = {1, 2, 3, ..., 20}, which is different from the range of R.

So, statement 3 is not correct.

∴ The correct answer is option (2).

Concept:

A function from a set X to a set Y assigns to each element of X exactly one element of Y. The set X is called the domain of the function and the set Y is called the codomain of the function.

Explanation:

Statement 1: The relation f defined by

 \(f(x)= \begin{cases}x^3, & 0 \leq x \leq 2 \ 4 x, & 2 \leq x \leq 8\end{cases}\) is a function.

For 0, 1, 2 for the function x³ the values are 0, 1, 8

For 2, 3, 4, 5, 6, 7, 8 for the function 4x the values are

8, 12, 16, 20, 24, 28, 32

So, for x = 2, the function has the same value i.e., 8 (for x3 and 4x).

Hence f(x) is a function.

Statement 2: The relation g defined by 

\(g(x)= \begin{cases}x^2, & 0 \leq x \leq 4 \ 3 x, & 4 \leq x \leq 8\end{cases}\) is a function.

For 0, 1, 2, 3, 4 for the function x2 the values are 0, 1, 4, 9, 16.

For 4, 5, 6, 7, 8 for the function 3x the values are 12, 15, 18, 21, 24.

So, for x = 4 the function has different values i.e., 16 (for x2) and

12 (for 3x).

Hence g(x) is a not function.

∴ Correct answer is option (1)

Explanation:

Let A = {1, 2, 3, 4}, B = {3, 4, 5, 6}

Now, evaluate each of the statements,

Statement 1. A = (A ∪ B) ∪ (A - B),

A ∪ B = {1, 2, 3, 4, 5, 6}; A - B = {1, 2}

RHS,

(A ∪ B) ∪ (A - B) = {1, 2, 3, 4, 5, 6} = A ∪ B \(\neq\) A

Hence, statement 1 is not correct.

Statement 2. A ∪ (B - A) = (A ∪ B)

B - A = {5, 6}; A ∪ B = {1, 2, 3, 4, 5, 6}

LHS,

A ∪ (B - A) = {1, 2, 3, 4, 5, 6} = A ∪ B

Hence, statement 2 is correct.

Statement 3. B = (A ∪ B) - (A - B)

RHS,

A ∪ B = {1, 2, 3, 4, 5, 6}; A - B = {1, 2}

(A ∪ B) - (A - B) = {3, 4, 5, 6} = B

Hence, statement 3 is correct.

∴ Correct answer is option (2).

Calculation:

Given that 

f(y) ≠ 0, f(1) = 0.5

\(f(x-y)=\frac{f(x)}{f(y)}\)

\(⇒ f(x)={f(y)}.{f(x-y)}\)

 

Now putting the values,

f(2) + f(3) + f(4) + f(5) + f(6) = \({{1} \over{4}} +{{1} \over {8}} + {{1} \over{16}} +{{1} \over {32}}+{{1} \over{64}} \)

⇒ f(2) + f(3) + f(4) + f(5) + f(6) = \( = {{16 + 8 + 4 +2+ 1} \over{64}} = {{31} \over{64}} \)

Hence, f(2) + f(3) + f(4) + f(5) + f(6) is \(\frac{31}{64}\).

Concept:

Inverse Trigonometric and Trigonometric Identities:

• The expression involves a composition of trigonometric and inverse trigonometric functions.
• Let θ = cos⁻¹(√5/3). Then, cos(θ) = √5/3 and the angle lies in [0, π].
• If we need to find cot(θ/2), we can use the identity:
• cot(θ/2) = (1 + cosθ)/sinθ
• This is a standard identity used to evaluate half-angle expressions in trigonometry.

 

Calculation:

Let θ = cos⁻¹(√5/3)

⇒ cosθ = √5/3

⇒ sinθ = √(1 − cos²θ) = √(1 − 5/9) = √(4/9) = 2/3

⇒ cot(θ/2) = (1 + cosθ)/sinθ = (1 + √5/3)/(2/3)

⇒ cot(θ/2) = ( (3 + √5)/3 ) × (3/2 ) = (3 + √5)/2 × 3/1 = (3 + √5) × (1/2)

⇒ 2 × cot(θ/2) = 2 × (3 + √5)/2 = 3 + √5

∴ Hence, the value of 2 cot(1/2 cos⁻¹(√5/3)) is 3 + √5.

Given: 

sec-1 p - cosec-1q = 0, where p > 0, q > 0

Calculation:

sec-1 p - cosec-1q = 0

sec-1 p = cosec-1q

As, sec θ and cosec θ are equal at θ = 45º 

Hence, p = q = \( √{2}\)

p-2 + q-2 = 2p-2 ​

p-2 + q-2 = 2 × (1/√2)²

p-2 + q-2  = 1

Formula Used:

cos-1x = θ ⇒ x = cos θ 

In a Right angled triangle,

\(cos:\theta = {B\over H}\), \(sin:\theta= {P\over H}\)

 

Explanation:

Let, \(\theta=\cos ^{-1}\left(\frac{3}{\sqrt{17}}\right)\)

⇒ \(cos:\theta=\frac{3}{\sqrt{17}} = {B\over H}\)

By using the Pythagoras theorem,

P ² = H² - B²

⇒ P2 = 17 - 9 =  8

⇒ P = \(2 \sqrt{2} \)

⇒ \(sin:\theta= {P\over H}=\frac{2{\sqrt2}}{\sqrt{17}} \)

⇒ \(sin^2:\theta= \frac{8}{17} \)

Therefore

\(1+\sin ^2\theta = 1+\frac{8}{17} \)

∴ \(1+\sin ^2\theta = \frac{25}{17} \)

Given:

tan (π cos θ) = cot (π sin θ), \(0<\theta<\frac{\pi}{2}\)

Concept:

tan (90 - θ) = cot θ

sin (A + B) = sin A cos B + cos A sin B

\(sin:{45} = cos:{45} = {1 \over {\sqrt2}}\)

Calculation:

tan (π cos θ) = tan (90 - π sin θ)

π cos θ = 90 - π sin θ

180 (cos θ + sin θ) = 90

cos θ + sin θ = 1/2

Multiply \(1 \over {\sqrt{2}}\) on both sides,

\({1 \over {\sqrt2}}cos θ + {1 \over {\sqrt2}}sin θ = {1 \over {\sqrt2}}{1 \over2}\)

\(sin: 45 :cos θ +cos: 45 :sinθ = {1 \over {2\sqrt2}}\)

\(sin({\theta + {\pi \over4}}) = {1 \over {2\sqrt2}}\)

Squaring both sides,

\(sin^2({\theta + {\pi \over4}}) = {1 \over 8}\)

\(8 \sin ^2\left(\theta+\frac{\pi}{4}\right) = 1\)

Given:

 \(\tan \alpha=\frac{1}{7}\), \(\sin \beta=\frac{1}{\sqrt{10}}\); \(0<\alpha, \beta<\frac{\pi}{2}\)

Formula Used:

• cos (A + B) = cos A cos B - sin A sin B
• sin 2A = 2 sin A sin B
• cos 2A = 2cos2A - 1

In a right-angled triangle, P = Perpendicular, B = Base, H = Hypotenuse

P² + B² = H²

Calculation:

In a right-angled triangle, P = Perpendicular, B = Base, H = Hypotenuse

\(\tan \alpha=\frac{1}{7} = \frac{P}{B}\)

\(\cos \alpha = \frac{B}{H} =\frac{7}{\sqrt{50}}\), \(\sin \alpha = \frac{P}{H} =\frac{1}{\sqrt{50}}\)

In a right-angled triangle, P = Perpendicular, B = Base, H = Hypotenuse

\(\sin \beta=\frac{1}{\sqrt{10}} = \frac{P}{H}\)

\(\cos \beta = \frac{B}{H} =\frac{3}{\sqrt{10}}\)

cos (α + 2β) = cos α cos 2β - sin α sin 2β

cos (α + 2β) = cos α (2cos² β - 1) - sin α (2sin β cos β)

Now, Put the values,

\(cos (α + 2β) = \frac{7}{\sqrt{50}} [2 (\frac{3}{\sqrt{10}})^2- 1] - \frac{1}{\sqrt{50}} (2\frac{1}{\sqrt{10}}\frac{3}{\sqrt{10}})\)

\(cos (α + 2β) = \frac{7}{\sqrt{50}} [ \frac{9}{{5}}- 1] - \frac{1}{\sqrt{50}} (\frac{3}{5})\)

\(cos (α + 2β) = \frac{7}{\sqrt{50}} (\frac{4}{{5}}) - \frac{1}{\sqrt{50}} (\frac{3}{5})\)

\(cos (α + 2β) = \frac{28}{5\sqrt{50}} - \frac{3}{5\sqrt{50}}\)

\(cos (α + 2β) = \frac{25}{5\sqrt{50}}\)

\(cos (α + 2β) = \frac{5}{\sqrt{50}}\)

\(cos (α + 2β) = \sqrt \frac{25}{50}\)

\(cos (α + 2β) = \frac{1}{\sqrt 2}\)

Given: 

(1 - x)4 + (5 - x)4 = 82

Formula Used:

(a - b)² = a² - 2ab + b²

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca

Calculation:

 

Hence, (x - 2) and (x - 4) are the two factors of the equation

(1 - x)4 + (5 - x)4 = 82

Expanding it by using the above formula 

⇒ (1 + x² - 2x)² + (25 + x² - 10x)² = 82

Again, using the above formula

⇒ x⁴ − 4x³ + 6x² − 4x + 1 + x⁴ − 20x³ + 150x² − 500x + 625 = 82

⇒ x⁴ − 4x³ + 6x² − 4x + 626 + x⁴ − 20 x³ + 150x² − 500x = 82

⇒ 2x⁴ − 24x³ + 156x² − 4x + 626 − 500x = 82

⇒ 2x⁴ − 24x³ + 156x² − 504x + 544 = 0

⇒ x⁴ - 12x³ + 78x² - 252x + 272 = 0

As, (x - 2)(x - 4) = x² - 6x - 8,

So by dividing x4 - 12x3 + 78x2 - 252x + 272 by (x2 - 6x - 8) we get x2 - 6x + 34,

⇒ (x2 - 6x - 8)(x2 - 6x + 34) = x4 - 12x3 + 78x2 - 252x + 272

Now calculating Discriminant for (x2 - 6x + 34), a = 1, b = -6, c = 34

\(D = b^2-4ac\)

\(D = {-6^2-4(1)(34)}\)

\(D = {-172}\)

D < 0, therefore this equation has imaginary roots.

Hence, The equation (1 - x)4 + (5 - x)4 = 82 has two real roots and two imaginary roots.

Given: (1 - x)4 + (5 - x)4 = 82

Explanation:

Expanding: (1 - x)4 + (5 - x)4 = 82

⇒ (1 + x2 - 2x)2 + (25 + x2 - 10x)2 = 82

⇒ x4 − 4x3 + 6x2 − 4x + 1 + x4 − 20x3 + 150x2 − 500x + 625 = 82

⇒ x4 − 4x3 + 6x2 − 4x + 626 + x4 − 20 x3 + 150x2 − 500x = 82

⇒ 2x4 − 24x3 + 156x2 − 4x + 626 − 500x = 82

⇒ 2x4 − 24x3 + 156x2 − 504x + 544 = 0

⇒ x4 - 12x3 + 78x2 - 252x + 272 = 0

Now, Sum of the roots of a power 4 equation is \(-{Coefficient: of :x^3\over Coefficient: of :x^4}\)

\(Sum: of : Roots = {12 \over 1} = 12\)

Concept:

1. 1, ω and ω² are called cube roots of unity where\(ω = {-1 +i \sqrt{3} \over 2}:and: ω^2 = {-1 -i \sqrt{3} \over 2}\)
2. Properties of the cube roots of unity: 

• 1 + ω + ω2 = 0
• ω³ = 1

3. Roots of x² + x + 1 = 0 are ω & ω²

 

Calculation:

Given:

 z³ + 2z² + 2z + 1 = 0

⇒ z³ + 2z² + 2z + 1 = 0

⇒ (z + 1) × (z² + z + 1) = 0

⇒ z + 1 = 0, (z2 + z + 1) = 0

⇒ z = -1, ω and ω²

∴ Roots of the equation z3 + 2z2 + 2z + 1 = 0 are – 1, ω and ω2.

Concept:

1. 1, ω and ω2 are called cube roots of unity where\(ω = {-1 +i \sqrt{3} \over 2}:and: ω^2 = {-1 -i \sqrt{3} \over 2}\)
2. Properties of the cube roots of unity:

• 1 + ω + ω2 = 0
• ω3 = 1
• ω³ⁿ = 1

3. Roots of x2 + x + 1 = 0 are ω & ω2

 

Calculation:

Given: z1985 + z100 + 1 = 0

Put z = -1

⇒ (-1)1985 + (-1)100 + 1 = 0

⇒ \(-1 + 1 + 1 \neq 0\)

z = -1 does not satisfy the equation

Put z = ω

⇒ (ω)1985 + (ω)100 + 1 = 0

As, ω³ⁿ = 1

⇒ ω2 + ω + 1 = 0

z = ω satisfy the equation

Put z = ω²

⇒ (ω2)1985 + (ω2)100 + 1 = 0

⇒ ω2870 + ω200 + 1 = 0

As, ω3n = 1

⇒ ω2 + ω + 1 = 0

z = ω² satisfy the equation

∴ Roots of the equation z1985 + z100 + 1 = 0 are ω and ω².

Concept:

1. 1, ω and ω2 are called cube roots of unity where\(ω = {-1 +i \sqrt{3} \over 2}:and: ω^2 = {-1 -i \sqrt{3} \over 2}\)
2. Properties of the cube roots of unity: 

• 1 + ω + ω2 = 0
• ω3 = 1

3. Roots of x2 + x + 1 = 0 are ω & ω2

 

Calculation:

Equation I:

 z3 + 2z2 + 2z + 1 = 0

⇒ z3 + 2z2 + 2z + 1 = 0

⇒ (z + 1) × (z2 + z + 1) = 0

⇒ z + 1 = 0, (z2 + z + 1) = 0

⇒ z = -1, ω and ω2

Equation II: 

Put z = -1

⇒ (-1)1985 + (-1)100 + 1 = 0

⇒ \(-1 + 1 + 1 \neq 0\)

z = -1 does not satisfy the equation

Put z = ω

⇒ (ω)1985 + (ω)100 + 1 = 0

As, ω3n = 1

⇒ ω2 + ω + 1 = 0

z = ω satisfy the equation

Put z = ω2

⇒ (ω2)1985 + (ω2)100 + 1 = 0

⇒ ω2870 + ω200 + 1 = 0

As, ω3n = 1

⇒ ω2 + ω + 1 = 0

z = ω2 satisfy the equation

The roots of the equation z1985 + z100 + 1 = 0 are ω and ω2.

Therefore, there are two common roots which are ω and ω2.

Concept:

For a quadratic equation, ax2 +bx + k = 0, the roots are given by, 

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Comparison of Discriminant,

• For real roots, \(D = {b^2-4ac} >0\)
• For real and equal roots, \(D = {b^2-4ac} = 0\)
• For imaginary roots, \(D = {b^2-4ac} < 0\)

 

Explanation:

(a + b) x2 - (a + b + c) x + k = 0,

⇒ \(D = {[-(a+b+c)]^2-4(a+b)(k)}\)

⇒ \(D = {a^2+b^2+c^2 +2ab+2bc+2ca-4ak-4bk}\)

As k = c/2,

⇒ \(D = {a^2+b^2+c^2 +2ab+2bc+2ca-2ac-2bc}\)

⇒ \(D = {a^2+b^2+c^2 +2ab}\)

⇒ \(D = {(a+b)^2+c^2 }\) which is +ve & ≠ 0

Hence, the roots are Real and unequal.

Concept:

For a quadratic equation, ax2 +bx + k = 0, the roots are given by, 

\(x = {-b \pm \sqrt{D} \over 2a}\)

Comparison of Discriminant,

• For real roots, \(D = {b^2-4ac} >0\)
• For real and equal roots, \(D = {b^2-4ac} = 0\)
• For imaginary roots, \(D = {b^2-4ac} < 0\)

 

Calculation:

Now, for (a + b) x2 - (a + b + c) x + k = 0,

As, k = c

⇒ (a + b) x2 - (a + b + c) x + c = 0,

⇒ \(D = {[-(a+b+c)]^2-4(a+b)(c)}\)

Since,

(A + B + C) = A² + B² + C² + 2AB + 2BC + 2CA

Therefore,

⇒ \(D = {a^2+b^2+c^2 +2ab+2bc+2ca-4ac-4bc}\)

⇒ \(D = {a^2+b^2+c^2 +2ab-2ac-2bc}\)

⇒ \(D = {a^2+b^2+(-c)^2 +2ab+2a(-c)+2b(-c)}\)

⇒ \(D = {(a+b-c)^2 }\)      -----(1)

Therefore, root of given equations 

⇒ \(x = {(a+b+c) \pm \sqrt{{(a+b-c)^2}} \over 2(a+b)}\)

⇒ \(x = {{a+b+c+a+b-c} \over 2(a+b)}, {{a+b+c -a-b+c} \over 2(a+b)}\)

⇒ \(x = {2{(a+b)} \over 2(a+b)}, {{2c} \over 2(a+b)}\)

∴  \(x = {1}, {{c} \over {a+b}}\)

Calculation:

(1 + x)n = 1 + T1x + T2x2 + T3x3 + ... + Tnxn    ----(1)

At n = 3 ⇒ T_nxⁿ = T₃x³

From equation (1)

(1 + x)³ = 1 + T₁x + T2x2 + T3x3 

1 + 3x + 3x² + x³ = 1 + T1x + T2x2 + T3x3 

On comparing but side,

T₁ = 3, T₂ = 3, T₃ = 1          

Now, required sum 

S = T1 + 2T2 + 3T3 + ... + nTn 

At n = 3, it become

S = T1 + 2T2 + 3T3  

S = 3 + 2 × 3 + 3 × 1 

S = 12

From the above option (4), at n = 3

n2^n-1 = 3 × 2^3-1 =

n2n-1  = 3 × 4 = 12 

∴ Correct answer is option (D).

Calculation:

(1 + x)n = 1 + T1x + T2x2 + T3x3 + ... + Tnxn    ----(1)

At n = 3 ⇒ Tnxn = T₃x³

From equation (1)

(1 + x)3 = 1 + T1x + T2x2 + T3x3 

1 + 3x + 3x2 + x3 = 1 + T1x + T2x2 + T3x3 

On comparing but side,

T1 = 3, T2 = 3, T3 = 1          

Now, required sum 

S = 1 - T1 + 2T2 - 3T3 + ... + (-1)nnTn 

At n = 3, it become

S = 1 - T1 + 2T2 - 3T3 

S = 1 - 3 + 2 × 3 - 3 × 1 

S = 1 

∴ Correct answer is option (D).

Calculation:

(1 + x)n = 1 + T1x + T2x2 + T3x3 + ... + Tnxn    ----(1)

At n = 3 ⇒ Tnxn = T3x3

From equation (1)

(1 + x)3 = 1 + T1x + T2x2 + T3x3 

1 + 3x + 3x2 + x3 = 1 + T1x + T2x2 + T3x3 

On comparing but side,

T1 = 3, T2 = 3, T3 = 1          

Now, required sum 

S = T1 + T2 + T3 + ... + Tn 

At n = 3, it become

S = T1 + T2 + T3 

S = 3 + 3 + 1 = 7 

From option (B), 

2ⁿ-1 = 2³-1 = 7

∴ Correct answer is option (B).