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## Discrete Mathematics MCQ Set 1

1. is
a) 0
b) 1
c) 2
d) 3

Answer: a [Reason:] 2. Value of limx → 0⁡(1+Sin(x))Cosec(x)
a) e
b) 0
c) 1
d) ∞

Answer: a [Reason:] limx → 0⁡(1+Sin(x))Cosec(x)

Put sin(x) = t we get

limt → 0⁡(1+t)(1t) = e.

3. Value of limx → 0⁡(1 + cot(x))sin(x)
a) e
b) e2
c) 1e
d) Can not be solved

Answer: a [Reason:] 4. a) True
b) False

Answer: a [Reason:] It is a property of limits.

5. a) True
b) False

Answer: b [Reason:] 6. Evaluate limx → 1⁡[(xx – 1) / (xlog(x))]
a) ee
b) e
c) 1
d) e2

Answer: c [Reason:] limx → 1⁡ [(xx – 1) / (xlog(x))] = (00)

By L hospital rule,

limx → 1⁡ [xx (1+xlog(x))/ (1+xlog(x))] = limx → 1⁡ [xx] = 1.

7. Find n for which , has non zero value.
a) >=1
b) >=2
c) <=2
d) ~2

Answer:b [Reason:] Hence, limit have non zero limit, if n ≠ 0 and (n-1) ≠ 0 and (n-2) >= 0 means n >= 2.

8. Find the value of limx → 0⁡(Sin(2x))Tan2 (2x) ?
a) e0.5
b) e-0.5
c) e-1
d) e

Answer: b [Reason:] 9. Evaluate a) 14
b) 13
c) 12
d)1

Answer: c [Reason:] ## Discrete Mathematics MCQ Set 2

1. Distance travelled by any object is
a) Double integral of its accelecration
b) Double integral of its velocity
c) Double integral of its Force
d) Double integral of its Momentum

Answer: b [Reason:] We know that, x(t) = ∫∫a(t) dtdt.

2. Find the distance travelled by a car moving with acceleration given by a(t)=t2 + t, if it moves from t = 0 sec to t = 10 sec, if velocity of a car at t = 0sec is 40 km/hr.
a) 743.3km
b) 883.3km
c) 788.3km
d) 783.3km

Answer: d [Reason:] Add constant automatically We know that, 3. Find the distance travelled by a car moving with acceleration given by a(t)=Sin(t), if it moves from t = 0 sec to t = π/2 sec, if velocity of a car at t=0sec is 10 km/hr.
a) 10.19 km
b) 19.13 km
c) 15.13 km
d) 13.13 km

Answer: d [Reason:] Add constant automatically We know that, 4. Find the distance travelled by a car moving with acceleration given by a(t)=t2 – t, if it moves from t = 0 sec to t = 1 sec, if velocity of a car at t = 0sec is 10 km/hr.
a) 11922 km
b) 11915 km
c) 12912 km
d) 11912 km

Answer: b [Reason:] Add constant automatically We know that, 5. Find the value of ∫∫ xy dxdy over the area bpunded by parabola y=x2 and x = -y2,is
a) 167
b) 124
c) –16
d) –112

Answer: b [Reason:] 6. Find the value of ∫∫ xydxdy over the area b punded by parabola x = 2a and x2 = 4ay, is
a) a44
b) a43
c) a53
d) a23

Answer: b [Reason:] 7. Find the integration of ∫∫0x (x2 + y2) dxdy, where x varies from 0 to 1
a) 43
b) 53
c) 23
d) 1

Answer: c [Reason:] Add constant automatically 8. Evaluate the value of , where y varies from 0 to 1.
a) 1112
b) 146
c) 116
d) 117

Answer: c [Reason:] 9. Evaluate ∫∫[x2 + y2 – a2 ]dxdy where, x and y varies from –a to a.
a) –23 a4
b) –43 a4
c) –43 a5
d) –23 a5

Answer: b [Reason:] Equation of circle is given by x2 + b2 = a2 10. Find the area inside a ellipse of minor-radius ‘b’ and major-radius ‘a’.
a) –43 a2
b) –43 ab2
c) 43 ab
d) –43

Answer: c [Reason:] 11. Find the value of where, y varies from 0 to 1.
a) 16946
b) 8945
c) 1645
d) 16945

Answer: d [Reason:] 12. Find the value of integral a) 315
b) 215
c) 230
d) 115

Answer: b [Reason:] ## Discrete Mathematics MCQ Set 3

1. Three companies A, B and C supply 25%, 35% and 40% of the notebooks to a school. Past experience shows that 5%, 4% and 2% of the notebooks produced by these companies are defective. If a notebook was found to be defective, what is the probability that the notebook was supplied by A?
a) 4469
b) 2569
c) 1324
d) 1124

Answer: b [Reason:] Let A, B and C be the events that notebooks are provided by A, B and C respectively. Let D be the event that notebooks are defective Then, P(A) = 0.25, P(B) = 0.35, P(C) = 0.4 P(D|A) = 0.05, P(D|B) = 0.04, P(D|C) = 0.02 P(A│D) = (P(D│A)*P(A))/(P(D│A) * P(A) + P(D│B) * P(B) + P(D│C) * P(C) ) =(0.05*0.25)/((0.05*0.25)+(0.04*0.35)+(0.02*0.4)) = 2000/(80*69)

= 2569.

2. A box of cartridges contains 30 cartridges, of which 6 are defective. If 3 of the cartridges are removed from the box in succession without replacement, what is the probability that all the 3 cartridges are defective? Answer: b [Reason:] Let A be the event that the first cartridge is defective. Let B be the event that the second cartridge is defective. Let C be the event that the third cartridge is defective. Then probability that all 3 cartridges are defective is P(A ∩ B ∩ C) Hence, P(A ∩ B ∩ C) = P(A) * P(B|A) * P(C | A ∩ B) = (630) * (529) * (428)

= (6 * 5 * 4)(30 * 29 * 28).

3. Two boxes containing candies are placed on a table. The boxes are labelled B1 and B2. Box B1 contains 7 cinnamon candies and 4 ginger candies. Box B2 contains 3 cinnamon candies and 10 pepper candies. The boxes are arranged so that the probability of selecting box B1 is 13 and the probability of selecting box B2 is 23. Suresh is blindfolded and asked to select a candy. He will win a colour TV if he selects a cinnamon candy. What is the probability that Suresh will win the TV (that is, she will select a cinnamon candy)?
a) 733
b) 633
c) 1333
d) 2033

Answer: c [Reason:] Let A be the event of drawing a cinnamon candy. Let B1 be the event of selecting box B1. Let B2 be the event of selecting box B2.

Then, P(B1) =13 and P(B2) = 23 P(A) = P(A ∩ B1) + P(A ∩ B2) = P(A|B1) * P(B1) + P(A|B2)*P(B2) = (711) * (13) + (311) * (23)

= 1333.

4. Two boxes containing candies are placed on a table. The boxes are labelled B1 and B2. Box B1 contains 7 cinnamon candies and 4 ginger candies. Box B2 contains 3 cinnamon candies and 10 pepper candies. The boxes are arranged so that the probability of selecting box B1 is 13 and the probability of selecting box B2 is 23. Suresh is blindfolded and asked to select a candy. He will win a colour TV if he selects a cinnamon candy. If he wins a colour TV, what is the probability that the marble was from the first box?
a) 713
b) 137
c) 733
d) 633

Answer: a [Reason:] Let A be the event of drawing a cinnamon candy. Let B1 be the event of selecting box B1. Let B2 be the event of selecting box B2.

Then, P(B1) = 13 and P(B2) = 23 Given that Suresh won the TV, the probability that the cinnamon candy was selected from B1 is P(B1|A) = (P(A|B1) * P( B1) ) /( P(A│B1 ) * P( B1 ) + P(A│B1 ) * P(B2) ) = 713.

5. Suppose box A contains 4 red and 5 blue coins and box B contains 6 red and 3 blue coins. A coin is chosen at random from the box A and placed in box B. Finally, a coin is chosen at random from among those now in box B. What is the probability a blue coin was transferred from box A to box B given that the coin chosen from box B is red?
a) 1529
b) 1429
c) 12
d) 710

Answer: a [Reason:] Let E represent the event of moving a blue coin from box A to box B. We want to find the probability of a blue coin which was moved from box A to box B given that the coin chosen from B was red. The probability of choosing a red coin from box A is P(R) = 79 and the probability of choosing a blue coin from box A is P(B) = 59. If a red coin was moved from box A to box B, then box B has 7 red coins and 3 blue coins. Thus the probability of choosing a red coin from box B is 710 . Similarly, if a blue coin was moved from box A to box B, then the probability of choosing a red coin from box B is 610. Hence, the probability that a blue coin was transferred from box A to box B given that the coin chosen from box B is red is given by = 1529.

6. An urn B1 contains 2 white and 3 black chips and another urn B2 contains 3 white and 4 black chips. One urn is selected at random and a chip is drawn from it. If the chip drawn is found black, find the probability that the urn chosen was B1.
a) 47
b) 37
c) 2041
d) 2141

Answer: d [Reason:] Let E1, E2 denote the vents of selecting urns B1 and B2 respectively. Then P(E1) = P(E2) = 12 Let B denote the event that the chip chosen from the selected urn is black . Then we have to find P(E1 /B). By hypothesis P(B /E1) = 35 and P(B /E2) = 47 By Bayes theorem P(E1 /B) = (P(E1)*P(B│E1))/((P(E1) * P(B│E1)+P(E2) * P(B│E2)) ) = ((1/2) * (3/5))/((1/2) * (3/5)+(1/2)*(4/7) ) = 21/41.

7. At a certain university, 4% of men are over 6 feet tall and 1% of women are over 6 feet tall. The total student population is divided in the ratio 3:2 in favour of women. If a student is selected at random from among all those over six feet tall, what is the probability that the student is a woman?
a) 25
b) 35
c) 311
d) 1100

Answer: c [Reason:] Let M be the event that student is male and F be the event that the student is female. Let T be the event that student is taller than 6 ft. P(M) = 25 P(F) = 35 P(T|M) = 4100 P(T|F) = 1100

P(F│T) = (P(T│F) * P(F))/(P(T│F) * P(F) + P(T│M) * P(M))

= ((1/100) * (3/5))/((1/100) * (3/5) + (4/100) * (2/5) )

= 311.

8. Previous probabilities in Bayes Theorem that are changed with help of new available information are classified as
a) independent probabilities
b) posterior probabilities
c) interior probabilities
d) dependent probabilities

Answer: b [Reason:] None.

## Discrete Mathematics MCQ Set 4

1. Transfer function may be defined as
a) Ratio of out to input
b) Ratio of laplace transform of output to input
c) Ratio of laplace transform of output to input with zero initial conditions
d) None of the above

Answer: c [Reason:] Transfer function may be defined as the ratio of laplace transform of output to input with zero initial conditions.

2. Poles of any transfer function is define as the roots of equation of denominator of transfer function.
a) True
b) False

Answer: a [Reason:] Let transfer function be defined as G(s)/H(s), then poles of transfer function may be defined as H(s)=0.

3. Zeros of any transfer function is define as the roots of equation of numerator of transfer function.
a) True
b) False

Answer: a [Reason:] Let transfer function be defined as G(s)/H(s), then zeros of transfer function may be defined as G(s)=0.

4. Find the poles of transfer function which is defined by input x(t)=5Sin(t)-u(t) and output y(t)=Cos(t)-u(t).
a) 4.79, 0.208
b) 5.73, 0.31
c) 5.89, 0.208
d) 5.49, 0.308

Answer: a [Reason:] Given ,y(t) =Cos(t) – u(t) and x(t) = 5Sin(t) – u(t),

Hence, transfer function Roots of equation s2 – 5s + 1 = 0 is s = 4.79, 0.208.

5. Find the equation of transfer function which is defined by y(t)-∫0t y(t)dt + ddt x(t) – 5Sin(t) = 0 Answer: c [Reason:] 6. Find the poles of transfer function given by system d2dt2 y(t) – ddt y(t) + y(t) – ∫0t x(t)dt = x(t)
a) 0, 0.7 ± 0.466
b) 0, 2.5 ± 0.866
c) 0, 0 .5 ± 0.866
d) 0, 1.5 ± 0.876

Answer: c [Reason:] We know that, 7. Find the transfer function of a system given by equation d2dt2 y(t-a) + x(t) + 5 ddt y(t) = x(t-a).
a)(e-as-s)/(1+e-as s2)
b)(e-as-5s)/(e-as s2)
c) (e-as-s)/(2+e-as s2)
d) (e-as-5s)/(1+e-as s2)

Answer: d [Reason:] Given, d2dt2 y(t-a) + x(t) + 5 ddt y(t) = x(t-a).

Taking laplace transform, s2 Y(s) e-sa + X(s) + 5sY(s) = e-as X(s)

Hence, H(s) = Y(s)X(s) =(e-as-5s)/(1+e-as s2).

8. Any system is said to be stable if and only if
a) It poles lies at the left of imaginary axis
b) It zeros lies at the left of imaginary axis
c) It poles lies at the right of imaginary axis
d) It zeros lies at the right of imaginary axis

Answer: a [Reason:] Any system is said to be stable if and only if it poles lies at the left of imaginary axis.

9. The system given by equation 5 d3dt3 y(t) + 10 ddt y(t) – 5y(t) = x(t) + ∫0t x(t)dt, is
a) Stable
b) Unstable
c) Has poles 0, 0.455, -0.236±1.567
d) Has zeros 0, 0.455, -0.226±1.467

Answer: a [Reason:] Hence system is Unstabe.

10. Find the laplace transform of input x(t) if the system given by d3dt3 y(t) – 2 d2dt2 y(t) –ddt y(t) + 2y(t) = x(t), is stable.
a) s + 1
b) s – 1
c) s + 2
d) s – 2

Answer: b [Reason:] d3dt3 y(t) – 2 d2dt2 y(t) –ddt y(t) + 2y(t) = x(t),

Taking laplae transform,

(s3 – 2s2 – s + 2)Y(s) = X(s) H(s) = Y(s)X(s) = 1(s-1)(s+1)(s+2)

For the system to be stable, X(s) = s – 1.

11. The system given by equation y(t – 2a) – 3y(t – a) + 2y(t) = x(t – a),is
a)Stable
b)Unstable
c) Marginally stable
d) 0

Answer: a [Reason:] Given, y(t – 2a) – 3y(t – a) + 2y(t) = x(t – a), then ## Discrete Mathematics MCQ Set 5

1. The curvature of a function f(x) is zero. Which of the following functions could be f(x)
a) ax + b
b) ax2 + bx + c
c) sin(x)
d) cos(x)

Answer: a [Reason:] The expression for curvature is Given that k = 0 we have f (x) = 0 f(x) = a + b.

2. The curvature of the function f(x) = x2 + 2x + 1 at x = 0 is
a) 32
b) 2
c) ∣2 / 532
d) 0

Answer: c [Reason:] The expression as we know is 3.The curvature of a circle depends inversely upon its radius r
a) True
b) False

Answer: a [Reason:]Using parametric form of circle x = r.cos(t) : y = r.sin(t) 4.Find the curvature of the function f(x) = 3x3 + 4680x2 + 1789x + 181 at x = -520
a) 1
b) 0
c) ∞
d) -520

Answer: b [Reason:] For a Cubic polynomial the curvature at x = -b3a is zero because f is zero at that point. Looking at the form of the given point we can see that x = -46803*3 Thus, curvature is zero.

5.Let c(f(x)) denote the curvature function of given curve f(x). The value of c(c(f(x))) is observed
to be zero. Then which of the following functions could be f(x)
a) f(x) = x3 + x + 1
b) f(x)2 + y2 = 23400
c) f(x) = x19930 + x + 90903
d) No such function exist

Answer: b [Reason:] We know that the curvature of a given circle is a constant function. Further, the curvature of any constant function is zero. Thus, we have to choose the equation of circle from the options.

6. The curvature of the function f(x) = x3 – x + 1 at x = 1 is given by
a) ∣65
b) ∣35
c) ∣6⁄ 532
d) ∣3⁄ 532

Answer: c [Reason:] Using expression for curvature we have 7.The curvature of a function depends directly on leading coefficient when x=0 which of the following could be f(x)
a) y = 323x3 + 4334x + 10102
b) y = x5 + 232x4 + 232x2 + 12344
c) y = ax5 + c
d) f(x) = x3 – x + 1

Answer: d [Reason:] Using formula for curvature

y = 33x2 + 112345x + 8945

Observe numerator which is Now this second derivative must be non zero for the above condition asked in the question Looking at all the options we see that only quadratic polynomials can satisfy this.

8. Given x = k1ea1t : y = k2ea2t it is observed that the curvature function obtained is zero. What is the relation between a1 and a2
a) a1 ≠ a2
b) a1 = a2
c) a1 = (a2)2
d) a2 = (a1)2

Answer: b [Reason:] Using formula for Curvature in parametric form 9.The curvature function of some function is given to be k(x) = 1 [2 + 2x + x2]32 then which of the following functions
could be f(x)
a) x22 + x + 101
b) x24 + 2x + 100
c) x2 + 13x + 101
d) x3 + 4x2 + 1019

Answer: a [Reason:]The equation for curvature Is Answer: d [Reason:] If the scaling factor is then the function can be written as f(x) = eax Now using curvature formula we have 