Engineering Mathematics Test - 1 - PDF Flipbook
Engineering Mathematics Test - 1
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Test-01Solutions
ENGINEERING MATHEMATICS
1. The recurrence relation
T(1) = 2
T(n) = 3T� 4 � + n
has the solution T(n) equal to
a) O(n)
b) O(log n)
c) O(n3/4)
d) None of the above
Answer: (a)
Solution:
Apply master’s therom,
For the recurrence relation,
T(n) = aT� 4 � + f(n)
Where F(n) = O(nk)
If a < bk then T(n) = O(nk)
for the given problem
a = 3, b = 4, k = 1,
∴ a < bk
Hence T(n) = O(n)
1
2. Two girls have picked 10 roses, 15 sunflowers and 14 daffodils.
What is the number of ways they can divide the flowers among
themselves?
a) 1638
b) 2100
c) 2640
d) None of the above
Answer: (c)
Solution:
The 10 roses can be distributed among the two girls in11 ways
(Use the formula, for finding number of ways to distribute r
similar balls in numbered boxes)
V (n, r) + C(n – 1 + r, r)
Hence, n = 2 and r = 10
Similarly, his 15 sunflowers can be distributed in 16 ways
The 14 daffodils can be distributed in 15 ways
Required number of ways = (11)(16)(15) = 2640.
3. Mala has a coloring book in which each English letter is drawn
two times. She wants to paint each of these 52 prints with one of
k colors, such that the colour-pairs used to colour any two letters
are different. Both prints of a letter can also be colored with the
same colour. What is the minimum value of k that satisfies this
requirement?
a) 9
b) 8
2
c) 7
d) 6
Answer: (d)
Solution:
With two colors say red and green, 4 color pairs possible
{RR, RG, GR, GG}(Cartesian product)
Number of colors pairs possible with 3 letters = 9
Number of colors pairs possible with 4 letters = 16
Number of colors pairs possible with 5 letters = 25
Since, we have 26 letters of alphabet, the required number of
colors = 6.
4. What is the minimum number of ordered pairs of non-negative
numbers that should be chosen to ensure that there are two pairs
(a, b) and (c, d) in the chosen set such that a ≡ c mod3 and b ≡
d mod5
a) 4
b) 6
c) 16
d) 24
Answer: (c)
Solution:
c mod 3 value is 0 or 1 or 2
d mod 5 value 0, 1, 2l 3 or 4
Number of combinations of pairs = 3.5 = 15(pigeonholes)
(1, 1) (1, 2), (1, 3) (1, 4) (1, 5)
3
(2, 1) (2, 2), (2, 3) (2, 4) (2, 5)
(3, 1) (3, 2), (3, 3) (3, 4) (3, 5)
By pigeonhole principle,
If we take any other pair of non-zone numbers, (c, d), then with
one of above pair, the given condition can be satisfied.
∴ Required number of pairs =16
5. What is the cardinality of the set of integers X defined below? X
= {n|1 ≤ n ≤ 123, z is not divisible by either 2, 3 or 5}
a) 28
b) 33
c) 37
d) 44
Answer: (b)
Solution:
Let n (2) = number of integers in the given are divisible by 2, 3
or 5
N(2∨3∨5) = n(2) + n(3) + n(5) – n(2∧3) – n(3∧5) – n(2∧5) +
n(2∧3∧5)
= 61 + 41 + 24 – 20 – 8 – 12 + 4 = 90
⟹ Cardinality of X = 123 – 90 = 33
6. Let an represent the number of bit strings of length n containing
two consecutive l's. What is the recurrence relation for an?
a) an-2 + an-1 + 2n-2
b) an-2 + 2an-1 + 2n-2
c) 2an-2 + an-1 + 2n-2
4
d) 2an-2 + 2an-1 + 2n-2
Answer: (a)
Solution:
Let an = number of bits’ strength of length n containing two
consecutives ones.
Case 1: if the first bit is zero, then remaining bits we can choose
in an-1 ways.
Case 2: if the first bit 1 and second bit is 0, then the remaining
bits we can choose in an-2 ways
Case 3: if the first bit 1 and second bit is also 1, then the
remaining bits we can choose in 2n-2 ways
∴ The recurrence relation from an is = an-1 + an-1 + 2n-2
7. A graph is planar if and only if,
a) it does not contain sub graphs homeomorphic to K5 and K3,3.
b) it does not contain sub graphs isomorphic to K5 or K3,3.
c) it does not contain sub graphs isomorphic to K5 and K3,3.
d) it does not contain sub graphs homeomorphic to K5 or K3,3.
Answer: (d)
Solution:
Statement of Kuratowski's theorem
8. Let G be an undirected connected graph with distinct edge
weights. Let emax be the edge with maximum weight and emax,
the edge with minimum weight. Which of the following
statements is false?
a) Every minimum spanning tree of G must contain emin,
5
b) If emax, is in a minimum spanning tree, then its removal must
disconnect G
c) No minimum spanning tree contains emax
d) G has a unique minimum spanning tree
Answer: (c)
Solution:
From the definition of minimal spanning tree, (a), (b) & (d) are
true, Option (c) need not be true
9. A graph G = (V, E) satisfies |E| ≤ 3|V| ‒ 6 The min-degree of G
is defined as mveiRn{deg ree(v)} Therefore, min-degree of G
cannot be
a) 3
b) 4
c) 5
d) 6
Answer: (d)
Solution:
Given that |E| ≤ (3|V| – 6)
⟹ 2|E| ≤ (6|V| – 12)
If min degree of G is k
Then k|v| ≤ 2|E|
∴ k|v| ≤ (6|v| – 12)
Which cannot be satisfied for k = 6.
6
10. How many graphs on n labelled vertices exist which have at
least (n2 – 3n)/2 edges?
a) (n^2 − n)/2 (n^2−3n)/2
b) ∑ ( n =^02−n)/2(n^2 − n)
c) (n^2 − n)/2 n
d) ∑( n =^02−n)/2
Answer: (d)
Solution:
Maximum number of edges possible on a simple graph with n
vertices = ( −1) = m(say)
2
Number of sample graph possible with at lest � 2−3 � = r(say)
2
edges
= ∑ = ( , )
= ∑ = 0 ( , )
Because, C(n, k) = C(n, n – k)
⟹ C(m, m) = C(m, 0)
(The least k binomial coefficient are same as first binomial
coefficient)
C0 = Cn, C1, = Cn-1,…
7
1 −2 2 −3
11. If the vector � 2 �is an eigen vector of = � 2 1 −6�
−1 −1 −2 0
then one of the eigen value of A is
a) 1
b) 2
c) 5
d) ‒1
Answer: (c)
Solution:
1 −2 2 −3
Given that X = � 2 �is an eigen vector of = � 2 1 −6�
−1 −1 −2 0
Consider AX = where is an eigen value of A
−2 2 −3 1
= � 2 1 −6� � 2 �
−1 −2 0 −1
−2 + 4 + 3 5 1
= � 2 + 2 + 6 � = �10� = 5 � 2 � =
−1 − 4 + 0 −5 −1
1
∴ � 2 � is an eigen vector of A corresponding to an eigen value
−1
= 5
12. If A, B, C are squares matrices of the same order then (ABC)-1
is equal be
a) C-1A-1B-1
b) C-1B-1A-1
c) A-1B-1C-1
8
d) A-1C-1B-1
Answer: (b)
Solution:
By the property of reversal law of inverse of product of three
matrices A, B, C we have (ABC)-1 = C-1B-1A-1
(∴(AB)-1 = B-1A-1)
Where A, B, C are square matrices of same order.
13. The Eigen values of the matrix �52 93� are
a) (5.13, 9.42)
b) (3.85, 2.93)
c) (9.00, 5.00)
d) (10.16, 3.84)
Answer: (d)
Solution:
Given A = �25 93�
Consider | − | = 0 where is an Eigen value of A
⇒ �5 − 3 � = 0
2 −
9
⇒ 2 − 14 + 39 = 0
⇒ = 7 + √10, 7 − √10
⇒ = 10.16, 3.84
9
14. A is a 3 × 4 matrix and AX = B is an inconsistent system of
equations. The highest possible rank of A is
a) 1
b) 2
c) 3
d) 4
Answer: (b)
Solution:
Given that A is 3 × 4 matrix and AX = B is inconsistent
AX = B
→ 3×4
→ � �3×5
If the system is inconsistent then ρ(A) < 3 and ρ(A) < ρ(A|B).
∴ The highest possible rank of A is 2.
15. Let A be a × real matrix such that A2 = I and Y be an n-
dimensional vector. Then the linear system of equation AX = Y
has
a) No solution
b) Unique solution
c) More than one but infinitely many dependent solutions.
d) Infinitely many dependent solutions.
Answer: (b)
Solution:
Given × × = ×1 and 2 =
⇒ =
10
⇒ is invertable i. e. A−1exists
⇒ A is non singular
∴ Unique solution exists.
16. If A is square symmetric real valued matrix of dimension 2n,
then the Eigen values of A are
a) 2n distinct real values
b) 2n real values not necessarily distinct
c) n distinct pairs of complex conjugate numbers
d) n pairs of complex conjugate numbers, not necessarily
distinct
Answer: (b)
Solution:
The number of Eigen values of × is n & eigen values of real
symmetric matrix are always real.
∴The number of eigen values of real symmetric matrix A of
order 2n (or dimension 2n) are 2n real values which may or may
not be repeated.
17. The function f(x) = ex is _______
a) Even
b) odd
c) Neither even nor odd
d) None
Answer: (c)
Solution:
(− ) = − ≠ ( ) (− ) ≠ − ( )
11
18. l i →m0 sin( /2) is
a) 0.5
b) 1
c) 2
d) Not defined
Answer: (a)
Solution:
xli→m0 sin mx = m [Standard limit]
x
∴ θli→m0 sin(θ/2) = 1
θ 2
19. In the Taylor series expansion of ex about x = 2, the coefficient
of (x – 2)4 is
a) 1
4!
b) 24
4!
c) 2
4!
d) 4
4!
Answer: (c)
Solution:
f (x) = ex about x = 2
Coefficient of (x – 2)4 = (2) = 2
4! 4!
12
20. What is the value of l i→m∞ �1 − 1 �2 ?
a) 0
b) e-2
c) e-1/2
d) 1
Answer: (b)
Solution:
l i→m∞ �1 − 1 �2 = l i→m∞ ��1 + − 1 � 2
�
= ( −1)2 = −2
21. The value of the integral ∫−
1+ 2
a) −
b) −
2
c)
2
d)
Answer: (d)
Solution:
∫− = 2 ∫0 1 = 2 −1 ] 0
1+ 2 1+ 2
= 2 � 2 − 0� =
22. The integral ∫− � − 6 � 6 sin( ) evaluates to
a) 6
b) 3
c) 1.5
13
d) 0
Answer: (b)
Solution:
∫− ( ) ( − ) = ( ) ℎ > 0
∴ ∫− 6 � − 6 � sin( ) = 6 = 3
6
23. The directional derivative of f(x, y) = 2x2 + 3y2 + z2 at point
P(2, 1, 3) in the direction of the vector = ⃗ − 2 � ⃗ is
a) 4/√5
b) −4/√5
c) √5/4
d) −√5/4
Answer: (b)
Solution:
F(x, y) = 2x2 + 3y2 + Z2, P(2, 1, 3) And � = ̅ ‒ 2 � )
Directional derivative = f(p).|�� �� �| = −4
√5
24. The vector field F = ̅ − ̅ (where i and j are unit vectors) is
a) divergence free, but not irrotational
b) irrotational, but not divergence free
c) divergence free and irrotational
d) neither divergence free nor irrotational
Answer: (c)
14
Solution:
� = xi ‒ yj
div � = 1 ‒ 1 = 0
⟹is solenoid (divergence free)
�
curl � = � �= 0� ⟹ � is irrotational
−
25. A fair coin is tossed 3 times in succession. If the first toss
produces a head, then the probability of getting exactly two
heads in three tosses is
a) 1
8
b) 1
2
c) 3
8
d) 3
4
Answer: (b)
Solution:
Sample space = {HTT, HTH, HHT, HHH}
Required probability = 2/4 =1/2
26. A random variable is uniformly distributed over the interval 2
to 10. Its variance will be
a) 16/3
b) 6
c) 256/9
d) 36
15
Answer: (a)
Solution:
Var(x) = ( − )2 = for a < x < b = 16/3
12
27. If the difference between the expectation of the square of a
random variable [E(X2)] and the square of the expectation of the
random variable [E(X)]2 is denoted by R, then,
a) R = 0
b) R < 0
c) R ≥ 0
d) R > 0
Answer: (d)
Solution:
R = var(X) = E(X2) – [E(X)]2 >0)
Variance never be negative
28. A fair coin is tossed till a head appears for the first time. The
probability that the number of required tosses is odd, is
a) 1/3
b) 1/2
c) 2/3
d) 3/4
Answer: (c)
Solution:
Req. probability = 1 + �21�3 + �12�3+…..∝ = 1
2 2
1−14
= 2
3
16
29. In an experiment, positive and negative values are equally
likely to occur. The probability of obtaining at most one
negative value in five trials is
a) 1
32
b) 2
32
c) 3
32
d) 6
32
Answer: (d)
Solution:
P(X ≤ 1) = P(X = 0) + P(X = 1)
= 5 0×�21�5+ 5 1× �12�5
= 1+5 = 6
32 32
30. Let U and V be two independent zero mean Gaussian random
variables of variances 1 and 1 respectively. The probability
4 9
P(3V≥2U) is
a) 4/9
b) 1/2
c) 2/3
d) 5/9
Answer: (b)
Solution:
Difference between the two normal random, variables is also
normal random variable.
17
P(3V ≥ 2U) = P(3V – 2U ≥ 0)
P(Z ≥ 0) =1/2
18
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