Engineering Mathematics Test - 5 - PDF Flipbook
Engineering Mathematics Test - 5
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Test-05Solutions
ENGINEERING MATHEMATICS
1. The recurrence equation T(1) = 1 and T(n) = 2T(n – 1) + n (n
≥2) evaluates to
a) 2n+1 – n – 1
b) 2n – n
c) 2n+1 – 2n – 2
d) 2n + n
Answer: (a)
Solution:
Given that T(n) = 2T(n – 1) + n …... (1)
Complementary function = C1.2n
Substituting in (1), we have (An + B) = 2{(A(n – 1) + B} + n
Comparision coefficient of n and constants
We get
A = –1 and B = 2
The solution is
T(n) = C1.2n – n – 2
Applying initial condition, we get C1 = 2
T(n) = 2n+1 – n – 2
2. In how many ways can b blue balls and r red balls be distributed
in n distinct boxes?
a) ( + −1)!( + −1)
( −1)! !( −1)! !
b) ( +( + )−1)!
( −1)!( −1)!( + )!
c) !
! !
1
d) ( +( + )−1)!
!( + −1)!
Answer: (a)
Solution:
Number of ways to distribute b similar balls in n distinct boxes
= C (n – 1 + b, b)
Similar,
Number of ways to distribute b similar balls in n distic boxes =
C (n – 1 + r, r)
By product rule, required number of ways = C (n – 1 + b, b).C(n
– 1 + r, r)
3. Consider the recurrence relation a1 = 8, an = 6n2 + 2n + an-1. Let
a99 = K x 104. The value of K is_____.
Answer: 198
Solution:
The recurrence relation is an-an-1 = 6n2 + 2n
Complementary function = C1
Let particuler solution = (An2 + BN + C)n
Where, A, B & C are undetermined coefficents
Substituting in the recurrence relation, and solving we get
A = 2, B = 4, and C = 2
The solution is An = C1 + 2n3 + 4n2 + 2n
Using the initial condition, we have
a1= 8 ⟹ 8 = C1 + 8
⟹ C1 = 0
A99 = 2{(993) + 2(992) + 99}
2
= 2{(100 − 1)3 + 2(100 − 1)2 + (100 – 1)}
=104 (198)
K = 198
4. The number of integers between 1 and 500 (both inclusive) that
are divisible by 3 or 5 or 7 is _____.
Answer: 271
Solution:
Number of integers divisible by 3 or 5 or 7 = n(3Λ ∨ 7) = n(3)
+n(5) + n (7) – n (3 Λ 5) – n(5 Λ 7) – n(3 Λ 7) + n(3 Λ 5 Λ 7)
= �5300� + �5500� + �5700� − �51050� − �53050� − �52010� + �510005�
= 166 + 100 + 71 – 33 – 14 – 23 +4 = 271
5. How many perfect matching are there in a complete graph of 6
vertices?
a) 15
b) 24
c) 30
d) 60
Answer: (a)
Solution:
Number of perfect matching in a complete graph with 2n
vertices = (2n – 1) .(2n – 3). (2n – 5) ..5.3.1.
Number of perfect matching in a complete graph with 6 vertices
= (5).(3).(1) = 15
3
6. What is the number of vertices in an undirected connected graph
with 27 edges, 6 vertices of degree 2, 3 vertices of degree 4 and
remaining of degree 3?
a) 10
b) 11
c) 18
d) 19
Answer: (d)
Solution:
Sum of degrees of vertices = 2(number of edges)
⟹ 6(2) + 3(4) + (n – 9)3 = 2(27)
When n = number of vertices in G
7. Which of the following graphs has an Eulerian circuit?
a) Any k-regular graph where k is an even number.
b) A complete graph on 90 vertices.
c) The complement of a cycle on25 vertices.
d) None of the above.
Answer: (c)
Solution:
In connected graph G
Euler circuit exists if G has no vertices of odd degree
(a) K –regular graph may or may not be connected
Option (a) may not be true
(b) (b) in K90, we have 90 vertices with odd degree
Euler’s circuit does not exist
4
(c) in the complement of C25, degree of each vertex is 22
Euler’s circuit exist
8. G is a graph on z vertices and 2n-2 edges. The edges of G can be
partitioned into two edge disjoint spanning trees. Which of the
following in NOT true for G?
a) For every subset of /r vertices, the induced sub graph has at
most 2k-2 edges
b) The minimum cut in G has at least two edges
c) There are two edge-disjoint paths between every pair of
vertices
d) There are two vertex-disjoint paths between every pair of
vertices
Solution:
G is a wheel graph on n vertices (n ≥ 4). For a wheel graph, all
the given options are true.
9. Consider the set S: {a, b , c, d}. Consider the following 4
partitions 1, 2, 3, 4 on S : 1 = {� � � � � � � �}, 2 = � � � � , � � � �, 3 =
�� � � � � , ̅�, 4 = � � , � , ,̅ ̅�. Let ≺ be the partial order on the set
of partitions S’={ 1, 2, 3, 4} defined as follows: ≺ if and
only refines . The poset diagram for (S',≺) is
a)
5
b)
c)
d)
Answer: (c)
Solution:
If 1is a refinment of 2, then any two elements that are in the
same block (subset) odf 1, must also lie in the same block of
2
4is related to all other partions
Every partition is related 1
2and 3are not comparable
The poset diagram is shown in below
6
10. How many onto (or surjective) functions are there from an n-
element (n ≥ 2) set to a 2- element set?
a) 2n
b) 2n – 1
c) 2n – 2
d) 2(2n – 2)
Answer: (c)
Solution:
No of surjections from an ’m’ element set to an ‘n’ element set
= nm – C(n, 1) (n – 1)m + C(n, 2) (n – 2)m + (–1)n-1 C(n, n – 1) 1m
Put m = n & n = 2
= 2n – C(2, 1) 1n = 2n – 2
11. For matrices of same dimension M, N and scalar c, which one
of these properties DOES NOT ALWAYS hold?
a) ( ) =
b) ( ) = ( )
c) ( + ) = +
d) =
Answer: (d)
Solution:
Since matrix multiplication is not commutative MN ≠ NM
7
12. Consider a 3 3 real symmetric matrix S such that two of its
eigen values are ≠ 0, ≠ 0 with respective eigen vectors
1 1
� 2� � 2�. ≠ 1 1 2 2 3 3
3 3 If then + + equals
a) a
b) b
c) ab
d) 0
Answer: (d)
Solution:
Since the Eigen vectors of a real symmetric matrix are pair-wise
orthogonal. (i.e., dot product= 0) i.e., x1y1 + x2y2 + x3y3 = 0
13. The sum of Eigen values of the matrix, [M] is where
215 650 795
[M] = �655 150 835�
485 355 550
a) 915
b) 1355
c) 1640
d) 2180
Answer: (a)
Solution:
The sum of Eigen values of M = Trace
‘M’ = 215 + 150 + 550 = 915
8
6 044
14. The rank of the matrix �−2 14 8 18 � is ____
14 −14 0 −10
Answer: 2
Solution:
6 044
�−2 14 8 18 �
14 −14 0 −10
1 ↔ 2
−2 14 8 18
~� 6 0 4 4 �
14 −14 0 −10
2 → 2 + 3 1; 3 → 3 + 7 1
−2 14 8 18
~ � 0 42 28 58 �
0 84 56 116
3 → 3 − 2 2
−2 14 8 18
~ � 0 42 28 58�
0 000
∴Required rank = 2
478
15. If any two columns of a determinant = �3 1 5� are
962
interchanged, which one of the following statements regarding
the value of the determinant is CORRECT?
a) Absolute value remains unchanged but sign will change.
b) Both absolute valueand sign will change.
c) Absolute value will change but sign will not change.
d) Both absolute value and sign will remain unchanged.
9
Answer: (a)
Solution:
If we interchange any two rows (or columns), then the value of
the determinant will be multiplied by –1
321 1
16. Given the matrices J = �2 4 2� and = � 2 �, the product
−1
126
KTJK is _____
Answer: 23
Solution:
321 1
KTJK = (1 2 −1) �2 4 2� � 2 �
1 2 6 −1
1
= (6 8 −1) � 2 � = 23
−1
17. The expression = ∫0 2 �1 − ℎ �2 ℎ for the volume of a
cone is equal to______.
a) ∫0 2 �1 − ℎ �2
b) ∫0 2 �1 − ℎ �2 ℎ
c) ∫0 2 �1 − � ℎ
d) ∫0 2 �1 − �2
Answer: (d)
Solution:
Integrating option (d) we obtain the volume of the cone as
10
= 1 2
3
18. Given = √−1, the value of the definite integral, =
∫0 /2 + is:
−
a) 1
b) –1
c) i
d) –i
Answer: (c)
Solution:
= = = 2 = = 1 � 2 × 2 − 0�
− 2
∫02 ∫02 � 22 �02
= 1 { + − 1} = − 2 − 2 =
2 2 2 2
19. Choose the most appropriate equation for the function drawn
as a thick line, in the plot below.
a) = − | |
b) = −( − | |)
c) = + | |
d) = −( + | |)
Answer: (b)
11
Solution:
At (2, –1), only the equation given in (b) is satisfied.
20. The contour on the x-y plane, where the partial derivative of x2
+ y2 with respect to y is equal to the partial derivative of 6y + 4x
with respect to 'x', is
a) y = 2
b) x = 2
c) x + y = 4
d) x – y = 0
Answer: (a)
Solution:
Given ( 2 + 2) = (6 + 4 ) ⇒ 2 = 4
∴ = 2
21. The double integral ∫0 ∫0 ( , ) is equivalent to
a) ∫0 ∫0 ( , )
b) ∫0 ∫ ( , )
c) ∫0 ∫ ( , )
d) ∫0 ∫0 ( , )
Answer: (c)
Solution:
The limits of the double integral are given by y = 0 to y = a & x
= 0 to x = y.
12
After changing the order of integration, the limits of integral are
given by
x = 0 to x = a and y = x to y = a
∴ ∫ = 0 ∫ = 0 ( , ) = ∫ = 0 ∫ = ( , )
22. The value of l i →m0 �1− 2 4� 2�� is
a) 0
b) 1
2
c) 1
4
d) Undefined
Answer: (c)
Solution:
l i →m0 �1− 2 4� 2�� �00�
Applying L’Hospital rule
l i →m0 �1− 2 4� 2�� = l i →m0 �2 8 3� 2��
= 1 l i →m0 � � 2 2�� = 1
4 4
13
23. If a vector field � is related to another field ̅ through � = ∇ ×
̅, which of the following is true?
Note: C and SC refer to any closed contour and any surface
whose boundary is C.
a) ∬ � ⃗ . � � � ⃗ = ∬ ⃗ . � � � � ⃗
b) ∬ ⃗ . � � � ⃗ = ∬ � ⃗ . � � � � ⃗
c) ∬ ∇ × � ⃗ . � � � ⃗ = ∬ ∇ × ⃗ . � � � � ⃗
d) ∬ ∇ × ⃗ . � � � ⃗ = ∬ � ⃗ . � � � � ⃗
Answer: (b)
Solution:
Using stokes Theorm for ̅ over the curve
‘C’ bounding an open surface Sc
∮ ̅.d ̅ = ∫ ∫ ̅ . ̅ =∫ ∫ � . ̅
24. f(x, y) = (x2 + xy) � + (y2 + xy) � . Its line integral over the
straight line from (x, y) = (0, 2) to (x, y) = (2, 0) evaluates to
a) –8
b) 4
c) 8
d) 0
Answer: (d)
14
Solution:
F(x, y) = (x2 + xy) � x + (y2 + xy) � y
∫ � .d ̅ =∫ 2 ( 2 + ) + ( 2 + )
X + y = 2 ⟹ dx = –dy
=∫ 2 [x2 + x(2 ‒ x)]dx + [(2 ‒ x)2 + x (2 ‒ x)](‒dx) = 0
25. If E denotes expectation, the variance of a random variable X
is given by
a) E(X2) – E2(X)
b) E(X2) + E2(X)
c) E(X2)
d) E2(X)
Answer: (a)
Solution:
Var (X) = E(X2) – [E(X)]2
26. Assume for simplicity that N people, all born in April (a month
of 30 days) are collected in a room, consider the event of at least
two people in the room being born on the same date of the
month (even if in different years e.g. 1980 and 1985). What is
the smallest N so that the probability of this exceeds 0.5 is?
a) 20
b) 7
15
c) 15
d) 16
Answer: (b)
Solution:
Let N = 2,
Probability that none of them born on the same day = 1 × 29
30
Required probability = 1 − 29 = 1 is not Grater than 0.5
30 30
Let N = 3,
Required probability = 1 − 29 × 28 = 0.098
30 30
Let N = 4,
Required probability = 1 − 29 × 28 × 27 = 0.188
30 30 30
Similar N = 7,
Required probability = 1 − 29 × 28 × 27 × 26 × 25 × 24
30 30 30 30 30 30
= 0.53 > 0.5
27. Let X be a real - valued random variable with E[X] and E[X2]
denoting the mean values of X and X2, respectively. The relation
which always holds true is
a) (E[X])2 > E[x2]
b) E[x2 ] ≥ (E[x])2
c) E[x2] = (E[x])2
d) E[X2] > (E[X])2
Answer: (b)
16
Solution:
V(x) = E(X)2 - [E(X)]2 ≥ 0
⟹ E(X)2 ≥ [E(X)]2
28. An unbiased coin is tossed an infinite number of times. The
probability that the fourth head appears at the tenth toss is
a) 0.067
b) 0.073
c) 0.083
d) 0.091
Answer: (c)
Solution:
Let p = 1 probability of getting a head any time q = 12; n = 9
2
we have to get ‘3’ heads and ‘6’ tails in first ‘9’ tosses, before
getting a head in the 10th toss
The required probability = �9 3 �21�3 �21�6� × 1 = 9 3 = 64
2 210 1024
= 0.082
29. A box contains 25 parts of which 10 are defective. Two parts
are being drawn simultaneously in a random manner from the
box. The probability of both the parts being good is
a) 7/20
b) 42/125
c) 25/29
d) 5/9
Answer: (a)
17
Solution:
Required probability = 15 2
25 2
= 7
20
30. A machine produces 0, 1 or 2 defective pieces in a day with
associated probability of 1/6, 2/3 and 1/6 respectively. Then
mean value and the variance of the number of defective pieces
produced by
a) 1 and 1/3
b) 1/3 and 1
c) 1 and 4/3
d) 1/3 and 4/3
Answer: (a)
Solution:
Mean = ΣX P(X) = �1 × 16� + �1 × 46� + 2�16� = 6 =1
6
Variance 2 = Σ 2 P(X) ‒ 2
= Σ��0 × 16� + �1 × 64� + �4 × 61� � = 8 ‒ 1 = 2 = 1
6 6 3
18
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