Engineering Mathematics MCQ Quiz - Objective Question with Answer for Engineering Mathematics - Download Free PDF

Explanation:

Let the 7 consecutive even numbers be:

(x - 6), (x - 4), (x - 2), x, (x + 2), (x + 4), (x + 6)

Here, x is the middle number, and the other numbers are spaced evenly around x, increasing or decreasing by 2. Consecutive even numbers are separated by a difference of 2.

Step 2: Calculating the average

The average of these numbers is given as 58. Using the formula for the average, we can calculate x:

Average = Sum of all numbers ÷ Total number of numbers

Sum = (x - 6) + (x - 4) + (x - 2) + x + (x + 2) + (x + 4) + (x + 6)

Sum = 7x (since all other terms cancel out when added)

Average = 58 = 7x ÷ 7

x = 58

So, the middle number of the sequence is 58. The numbers are:

(58 - 6), (58 - 4), (58 - 2), 58, (58 + 2), (58 + 4), (58 + 6)

Which simplifies to:

52, 54, 56, 58, 60, 62, 64

Step 3: Finding the difference between the highest and lowest numbers

The highest number in the sequence is 64, and the lowest number is 52. The difference between them is:

Difference = Highest number - Lowest number

Difference = 64 - 52

Difference = 12

Correct Answer: The difference between the highest even number and the lowest even number is 12

Analysis of the Problem:

We are given two matrices, 'A' and 'B', where the order of matrix 'A' is 3 × 4 (3 rows and 4 columns). We are tasked with determining the order of matrix 'B' such that both the matrix products AB and BA are defined. To solve this, we need to recall the rules of matrix multiplication:

• For the product AB to be defined, the number of columns in A must equal the number of rows in B.
• For the product BA to be defined, the number of columns in B must equal the number of rows in A.

Step-by-Step Solution:

1. Matrix A has dimensions 3 × 4:

This means matrix A has 3 rows and 4 columns.

2. For AB to be defined:

The number of columns in A (which is 4) must equal the number of rows in B. Therefore, the number of rows in B is 4.

3. For BA to be defined:

The number of columns in B must equal the number of rows in A (which is 3). Therefore, the number of columns in B is 3.

4. Order of matrix B:

From the above two conditions, the dimensions of matrix B must be 4 × 3 (4 rows and 3 columns).

Correct Option:

The correct order of matrix B is 4 × 3, which corresponds to Option 4.

The correct answer is Average (x1, x2 ... x9, m) = m and Average (x2, x3 ... x9) > m

Explanation:

Let's take the example of it

The nine numbers x1, x2, x3 ... x9, are in ascending order
Let  the nine number is 1,1,1,1,1,1,1,1,10
Their Avg will be = 18/9 = 2

Also, this satisfies that their average m is strictly greater than all the first eight numbers. 

now verifing

Average (x1, x2 ... x9, m) = m
Avg[1,1,1,1,1,1,1,1,10,2] will be 20/10 = 2
Which satisfies the given statement

Now checking Average (x2, x3 ... x9) > m

Average (1,1,1,1,1,1,1,10) will 17/8 = 2.12
Which is greater then 2

So it is also satisfied  Average (x2, x3 ... x9) > m

Conclusion:-

So, Average (x1, x2 ... x9, m) = m and Average (x2, x3 ... x9) > m is the true statement.

The correct answer is Option 4) [T, T].

Key Points

• We are given two sets of values for (p, q, r):
  • Case 1: (T, F, T)
  • Case 2: (F, T, T)
• The expression is: p ∨ (¬q → (p ∧ r))
• We'll evaluate it step-by-step for both cases.

Case 1: p = T, q = F, r = T

• ¬q = T (since q is F)
• p ∧ r = T ∧ T = T
• ¬q → (p ∧ r) = T → T = T
• p ∨ T = T ∨ T = T

Case 2: p = F, q = T, r = T

• ¬q = F
• p ∧ r = F ∧ T = F
• ¬q → (p ∧ r) = F → F = T (implication is true when antecedent is false)
• p ∨ T = F ∨ T = T

Final result: [T, T]

Additional Information

• ¬q means logical NOT of q.
• → (implication) is only false when LHS is true and RHS is false.
• ∨ (OR) is true if at least one operand is true.
• ∧ (AND) is true only if both operands are true.

The correct answer is Option 2) The maximum degree of a vertex in a graph can be d, where d is the number of vertices in the graph.

Key Points

• This statement is false because in a simple undirected graph (no self-loops), the maximum degree of a vertex is n - 1, where n is the number of vertices.
• A vertex cannot be connected to itself, so it can be connected to at most all other n - 1 vertices.

 Additional Information

• Option 1 - True: Not all graphs with an Eulerian cycle (visiting each edge exactly once) necessarily have a Hamiltonian cycle (visiting each vertex exactly once). These are two distinct properties.
• Option 3 - True: A graph with n vertices and no cycles is a tree or forest, and such a graph has at most n - 1 edges.
• Option 4 - True: If the sum of degrees is at least 21 in a 7-vertex graph, then by the Pigeonhole Principle, at least one vertex must have degree ≥ 4 (since 21/7 = 3 and degrees are integers).

Concept:

Relation between mode, median and mean is given by:

Mode = 3 × median – 2 × mean

Calculation:

Given:

Mode – median = 2

As we know

Mode = 3 × median – 2 × mean

Now, Mode = median + 2

⇒ (2 + median) = 3median – 2mean   

⇒ 2Median - 2Mean = 2

⇒ Median - Mean = 1

Concept:

Let B = |A¹²¹ - A¹²⁰|

B = |A¹²⁰ × (A – I)| 

B = |A¹²⁰| × |A – I|

Calculation:

A = \(\left[ {\begin{array}{*{20}{c}} 8&5\\ 7&6 \end{array}} \right]\)

Now, calculating matrix [A – I]

[A – I] = \(\left[ {\begin{array}{*{20}{c}} 8&5\\ 7&6 \end{array}} \right] - {\rm{\;}}\left[ {\begin{array}{*{20}{c}} 1&0\\ 0&1 \end{array}} \right]\)

[A – I] = \(\left[ {\begin{array}{*{20}{c}} 7&5\\ 7&5 \end{array}} \right]\)

Now determinant of |A – I|,

|A – I| = \(\left| {\begin{array}{*{20}{c}} 7&5\\ 7&5 \end{array}} \right|\)

|A – I| = 0   (Since two rows are repeated, therefore determinant = 0)

Hence, |A¹²¹ - A¹²⁰| =  A120|A – I|  = 0

Calculation:

A bag contains 3 white, 2 blue and 5 red balls.

Total number of balls = 3 + 2 + 5 = 10

Number of balls that are not red = 10 - 5 = 5

Concept:

Rank:

The rank of a matrix is a number equal to the order of the highest order non-vanishing minor, that can be formed from the matrix.

For matrix A, it is denoted by ρ(A).

The rank of a matrix is said to be r if,

• There is at least one non-zero minor of order r.
• Every minor of matrix A having order higher than r is zero. 

Property of Rank of Matrix:

​ρ(AB) ≤ min [ρ(A), ρ(B)]

Calculation:

Given:

ρ(A) = 2, ρ(B) = 3

Using properties

ρ(AB) ≤ min [ρ(A), ρ(B)]

ρ(AB) ≤ min (2,3)

⇒ ρ(AB) ≤  2

Alternate Method

Let order of matrix A be 2 × m and order of matrix B be m × 3 (∵ for multiplication we need the column of A and row of B to be same)

∴ Order of matrix AB will be 2 × 3

Using properties

ρ(AB) ≤ min (Row, Column)

⇒ ρ(AB) ≤ min (2, 3)  [only when the column of A and row of B is the same]

⇒ ρ(AB) ≤ 2.

∵ we don't know the dimension of A and B, we cannot predict the exact rank of AB but its maximum rank will be 2.

Important Points

Other properties of rank of a matrix are:

• The rank of a matrix does not change by elementary transformation, we can calculate the rank by changing the matrix into Echelon form. In the Echelon form, the rank of a matrix is the number of non-zero rows of the matrix.
• The rank of a matrix is zero if the matrix is null.
• ρ(A) ≤ min (Row, Column)
• ρ(AB) ≤ min [ρ(A), ρ(B)]
• ρ(AT A) = ρ(A AT) = ρ(A) = ρ(AT)
• If A and B are matrices of the same order, then ρ(A + B) ≤ ρ(A) + ρ(B) and ρ(A - B) ≥ ρ(A) - ρ(B).
• If Aθ is the conjugate transpose of A, then ρ(Aθ) = ρ(A) and ρ(A Aθ) = ρ(A).
• The rank of a skew-symmetric matrix cannot be one.

EXPLANATION:

Let A, B and C be the matrices a, b, and c be scalars and the sizes of matrices are such that the operations can be performed then

The multiplication of real-valued square matrices of the same dimension is:

Associative, which means that for any three matrices A, B, and C of the same dimension, the following holds true: (AB)C = A(BC)

Not Commutative, which means that the order of the matrices matters. In other words, AB is not necessarily equal to BA, unless one or both of the matrices are diagonal or scalar matrices.

Not always positive definite, since the determinant of the product of two matrices is equal to the product of the determinants of the matrices, and the determinant of a matrix is positive if and only if the matrix is invertible. Therefore, the product of two invertible matrices is invertible, and hence positive definite. However, the product of two non-invertible matrices may not be invertible, and hence not positive definite.

The order of the multiplication matters, so it is not always possible to compute the matrices. That is, AB is not necessarily equal to BA, unless one or both of the matrices are diagonal or scalar matrices.

Additional Information 

Properties of Matrix addition and scalar multiplication:

Commutative Property of addition

• A + B = B + A

Associative Property of addition

• A + (B + C) = (A + B) + C
• A + O = O + A Where O is the appropriate zero matrix

Distributive Property of addition

• c(A + B) = cA + cB
• (a + b)C = aC + bC
• (ab)C = a(bc)

Concept:

Trapezoidal rule states that for a function y = f(x)

xn = x0 + nh, where n = Number of sub-intervals

h = step-size

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{2}\left[ {\left( {{y_0} + {y_n}} \right) + 2\left( {{y_1} + {y_2} + {y_3} + \ldots + {y_{n - 1}}} \right)} \right]\)   ---(1)

For a trapezoidal rule, a number of sub-intervals must be a multiple of 1.

Calculation:

\(\begin{array}{*{20}{c}} x&:&0&1&2\\ {f\left( x \right)}&:&4&3&{12} \end{array}\)

Here: x₀ = 4, x₁ = 3, x₂ = 12, h = 1

From equation (1);

\(\mathop \smallint \limits_0^2 f\left( x \right)dx = \frac{h}{2}\left[ {\left( {{x_0} + {x_2}} \right) + 2\left( {{x_1}} \right)} \right]\)

\( = \frac{1}{2}\left[ {\left( {{4} + {12}} \right) + 2\left( {{3}} \right)} \right]={22\over2}=11\)

Key Points:

Apart from the trapezoidal rule, other numerical integration methods are:

Simpson’s one-third rule:

For applying this rule, the number of subintervals must be a multiple of 2.

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{3}\left[ {\left( {{y_0} + {y_n}} \right) + 4\left( {{y_1} + {y_3} + {y_5} + \ldots + {y_{n - 1}}} \right) + 2\left( {{y_2} + {y_4} + {y_6} + \ldots + {y_{n - 2}}} \right)} \right]\)     ..2)

Simpson’s three-eighths rule:

For applying this rule, the number of subintervals must be a multiple of 3.

\(\mathop \smallint \nolimits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{{3h}}{8}\left[ {\left( {{y_0} + {y_n}} \right) + 3\left( {{y_1} + {y_2} + {y_4} + {y_5} + \ldots } \right) + 2\left( {{y_3} + {y_6} + \ldots } \right)} \right]\)

Concept:

If rank of 3 × 3 homogeneous matrix is less than 3 then the corresponding equations will have non-trivial solutions.

Explanation:

For non-trivial solution

\(\left| {\begin{array}{*{20}{c}} p&q&r\\ q&r&p\\ r&p&q \end{array}} \right| = 0\)

R₁ = R₁ + R₂ + R₃

\(\left| {\begin{array}{*{20}{c}} {p + q + r}&{\;q + r + p}&{r + p + q}\\ q&r&p\\ r&p&q \end{array}} \right| = 0\)

\(\left( {p + q + r} \right)\left| {\begin{array}{*{20}{c}} 1&{\;1}&1\\ q&r&p\\ r&p&q \end{array}} \right| = 0\)

∴ p + q + r = 0

(OR)

\(\left| {\begin{array}{*{20}{c}} 1&{\;1}&1\\ q&r&p\\ r&p&q \end{array}} \right| = 0\)

Concept:

From the properties of a matrix,

The rank of m × n matrix is always ≤ min {m, n}

If the rank of matrix A is ρ(A) and rank of matrix B is ρ(B), then the rank of matrix AB is given by

ρ(AB) ≤ min {ρ(A), ρ(B)}

If n × n matrix is singular, the rank will be less than ≤ n

Calculation:

Given:

AX = B

Where A is 2 × 4 matrices and b ≠ 0

The order of A^T is 4 × 2

The order of A^TA is 4 × 4

Rank of (A) ≤ min (2, 4) = 2

Rank of (A^T) ≤ min (2, 4) = 2

Rank (A^TA) ≤ min (2, 2) = 2

As the matrix A^TA is of order 4 × 4, to have a unique solution the rank of A^TA should be 4.

Therefore, the unique solution of this equation is not possible.

Trace of a matrix

Let A be k × k matrix. Then its trace is denoted by trace(A) or tr(A), is  the sum of its diagonal elements

\(tr\left( A \right) = \mathop \sum \limits_{k = 1}^k {A_{kk}}\)

Calculation:

The given matrix is

\(\left[ {\begin{array}{*{20}{c}} 1&2&3\\ 4&5&6\\ 7&8&9 \end{array}} \right]\)

1, 5, 9 are diagonal elements.

tr(A) = A₁₁ + A₂₂ + A₃₃

tr(A) = 1 + 5 + 9

tr(A) = 15

Concept:

Properties of Determinant of a Matrix:

• If each entry in any row or column of a determinant is 0, then the value of the determinant is zero.
• For any square matrix say A, |A| = |AT|.
• If we interchange any two rows (columns) of a matrix, the determinant is multiplied by -1.
• If any two rows (columns) of a matrix are same then the value of the determinant is zero.

Calculation:

\(\left| {\begin{array}{*{20}{c}} 1&a&{b + c}\\ 1&b&{c + a}\\ 1&c&{a + b} \end{array}} \right|\)

Apply C2 → C2 + C3 

\( = \left| {\begin{array}{*{20}{c}} 1&{a + b + c}&{b + c}\\ 1&{a + b + c}&{c + a}\\ 1&{a + b + c}&{a + b} \end{array}} \right|\)

Taking common (a + b + c) from column 2, we get

\(= \left( {a + b + c} \right)\left| {\begin{array}{*{20}{c}} 1&1&{b + c}\\ 1&1&{c + a}\\ 1&1&{a + b} \end{array}} \right|\)

As we can see that the first and the second column of the given matrix are equal. 

We know that, if any two rows (columns) of a matrix are same then the value of the determinant is zero.

∴ \(\left| {\begin{array}{*{20}{c}} 1&a&{b + c}\\ 1&b&{c + a}\\ 1&c&{a + b} \end{array}} \right|\)= 0