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## Discrete Mathematics MCQ Set 1

1. Leibniz rule gives the
a) Nth derivative of addition of two function
b) Nth derivative of division of two functions
c) Nth derivative of multiplication of two functions
d) Nth derivative of subtraction of two function

Answer: c [Reason:] Leibniz rule is Hence Leibniz theorem gives nth derivative of multiplication of two functions u and v.

2. Leibniz theorem is applicable if n is a
a) Rational Number
b) Negative Integer
c) Positive Integer
d) Decimal Number

Answer: c [Reason:] Leibniz rule is For all n > 0, i.e n should be positive Hence Leibniz theorem gives nth derivative of multiplication of two functions u and v if n is a positive integer.

3. If nth derivative of xy3 + x2y2 + x3y0 = 0 then order of its nth differential equation is,
a) n
b) n+1
c) n+2
d) n+3

Answer: d [Reason:] 1. If we differentiae this equation n times then terms comes in nth order differential equation is yn+3 , yn+2, yn+1, yn, yn-1, yn-2, yn-3. Hence order of differential equation becomes n+3. 2. By Leibniz rule differentiating it n times, we get Xyn+3 + nyn+2 + x2yn+2 + 2nxyn+1 + 2n(n-1)yn + x3yn + 3nx2yn-1 + 6n(n-1)xyn-2 + 6n(n-1)(n-2)yn-3 = 0 Hence order of differential equlation becomes n+3.

4. Find nth derivative of x2y2 + xy1 + y = 0
a) x2yn+2 + (2n+1)xyn+1 + (n2+1)yn = 0
b) x2yn+2 + nxyn+1 + (n2+1)yn = 0
c) x2yn+2 + (2n+1)xyn+1 + n2yn = 0
d) x2yn+2 + 2nxyn+1 + n2yn = 0

Answer: a [Reason:] x2y2 + xy1 + y = 0 By Leibniz theorem

x2yn+2 + n(2x)(yn+1) + n(n-1)(2)(yn) + xyn+1 + nyn + yn= 0

x2yn+2 + xyn+1(2n+1) + yn(n2+1) = 0.

5. The nth derivative of x2y2 + (1-x2)y1 + xy = 0 is,
a) x2yn+2 + yn(2n2-2n-2nx+x) – yn-1(2n2-3n)=0
b) x2yn+2 + yn+1(2nx-x2) + yn(2n2-2n-2nx+x) – yn-1(2n2-3n)=0
c) x2yn+2 + yn+1(2nx+1-x2) + yn(2n2-2n-2nx+x)-yn – 1(2n2-3n)=0
d) x2yn+2 + yn+1(2nx+1-x2) + xyn2n2-yn – 1(2n2-3n)=0

Explanation: x2y2 + (1-x2)y1 + xy = 0

Differenetiating n times by Leibniz Rule

x2yn+2 + 2nxyn+1 + 2n(n-1)yn + (1-x2)yn+1 – 2nxyn – 2n(n-1)yn-1 + xyn + nyn-1 = 0

x2yn+2 + yn+1(2nx+1-x2) + yn(2n2-2n-2nx+x) – yn-1(2n2-3n)=0.

6. Find nth derivative of xnSin(nx) Answer: a [Reason:] Y = xnSin(nx) By Leibniz Rule , put u = xn and v = Sin(nx), we get 7. If y(x) = tan-1x then
a) (yn+1)(0) = (n-1)(yn-1)0
b) (yn+1)(0) = n(n-1)(yn-1)0
c) (yn+1)(0) = -(n-1)(yn-1)0
d) (yn+1)(0) = -n(n-1)(yn-1)0

Answer: d [Reason:] Y = tan-1x Y1 = 1/(1+x2) (1+x2)y1 = 1 By Leibniz Rule, (1+x2)yn+1 + 2nxyn + n(n-1)yn-1 = 0 Put x=0, gives → (yn+1)(0) = -n(n-1)(yn-1)(0).

8. If y = sin-1x, then
a) (1-x2)yn+2 – xyn+1(2n-1) = nyn(2n-1)
b) x2yn+2 – xyn+1(2n-1) = nyn(2n-1)
c) (1-x2)yn+2 – 2nxyn+1 = nyn(2n-1)
d) (1-x2)yn+2 – xyn+1(2n-1) +nyn(2n-1)=0

Answer: a [Reason:] Y = sin-1x Differentiating it Y1 = 1/√(1-x2 )

(1-x2)(y1)2= 1

Again Differentiating we get (1-x2)2y1y2 – 2x(y1)2 = 0

(1-x2)y2 = xy1

By Leibniz Rule, Diff it n times, (1-x2)yn+2 – 2xnyn+1 – 2n(n-1)yn = xyn+1 + nyn

(1-x2) yn+2 – xyn+1(2n-1) = nyn(2(n-1)+1)

(1-x2) yn+2 – xyn+1(2n-1) = nyn(2n-1).

9. If y = cos-1x, then
a) (yn+2)(0) = -n(2n-1) yn(0)
b) (yn+2)(0) = n(2n-1) yn(0)
c) (yn+2)(0) = n(n-2) yn(0)
d) (yn+2)(0) = n(n-3) yn(0)

Answer: b [Reason:] y = cos-1x

Differentiating it Y1 = 1/√(1-x2 )

(1-x2)(y1)2= 1

Again Differentiating we get (1-x2)2y1y2 – 2x(y1)2 = 0

(1-x2)y2 = xy1

By Leibniz Rule, Diff it n times, (1-x2)yn+2 – 2xnyn+1 – 2n(n-1)yn = xyn+1 + nyn

(1-x2) yn+2 – xyn+1(2n-1) = nyn(2(n-1)+1)

(1-x2) yn+2 – xyn+1(2n-1) = nyn(2n-1) At x=0, we get (yn+2)(0) = n(2n-1) yn(0).

## Discrete Mathematics MCQ Set 2

1. If f(t) = sinhat, then its Laplace transform is
a) eat
b) s ⁄ s2-a2
c) a ⁄ s2-a2
d) Exists only if ‘t’ is complex

Answer: c [Reason:] The Laplace transform of a function is given by put f(t) = sinhat On solving, a ⁄ s2-a2 is obtained.

2. If f(t) = coshat, its Laplace transform is given by
a) s ⁄ s2-a2
b) s+a ⁄ s-a
c) Indeterminate
d) (sinh(at))2

Answer: a [Reason:] The Laplace transform of a function is given by put f(t) = coshat On solving, s ⁄ s2-a2 is obtained.

3. If f(t) = eat sin(bt), then its Laplace transform is given by
a) s2-a2 ⁄ (s – a)2
b) b ⁄ (s + a)2 + b2
c) b ⁄ (s – a)2 + b2
d) Indeterminate

Answer: c [Reason:] The Laplace transform of a function is given by put f(t) = eatsin(bt) On solving, we get the b ⁄ (s – a)2 + b2.

4. If f(t) = eat cos(bt), then its Laplace transform is
a) 2a3 ⁄ (s2 + a2)
b) s+a ⁄ s-a
c) Indeterminate
d) s-a ⁄ (s – a)2 + b2

Answer: d [Reason:] The Laplace transform of a function is given by put f(t) = eatcos(bt) Solve the above integral, to obtain s-a ⁄ (s – a)2 + b2.

5. If f(t) = eat sinh(bt) then its Laplace transform is
a) e-as ⁄ s
b) s+a ⁄ (s – a)2 + b2
c) b ⁄ (s – a)2 – b2
d) Does not exist

Answer: c [Reason:] The Laplace transform of a function is given by put f(t) = eatsinh(bt) On solving, we get the b ⁄ (s – a)2 – b2.

6. If f(t) = 1a sinh(at), then its Laplace transform is
a) 1⁄s2-a2
b) 2a ⁄ (s – b)2 + b2
c) n! ⁄ (s – a)n-1
d) Does not exist

Answer: a [Reason:] The Laplace transform of a function is given by put f(t) = f(t) = 1⁄a sinh(at) On solving the above integral, we get the 1⁄s2-a2.

7. If f(t) = tn ⁄ n, then its Laplace transform is Answer: d [Reason:] The Laplace transform of a function is given by put f(t) = tn ⁄ n On solving, we obtain the Laplace transform of the required function.

8. If f(t) = 1√Πt, then its Laplace transform is Answer: b [Reason:] The Laplace transform of a function is given by put f(t) = 1 ⁄ √Πt The solution for the above question is obtained by solving the above integral.

9. If f(t) = t2 a sinat, then its Laplace transform is
a) b ⁄ (s + a)2 + b2
b) 2a ⁄ (s – b)2 + b2
c) Indeterminate
d) s ⁄ (s2 + a2)2

Answer: d [Reason:] The Laplace transform of a function is given by put f(t) = t⁄2a sinat Integrate to obtain, the required transform s ⁄ (s2 + a2)2.

10. If f(t) = δ(t), then its Laplace transform is
a) s + a ⁄ (s – a)2 + b2
b) a3 ⁄ (s2 + a2)2
c) 1
d) Does not exist

Answer: c [Reason:] The Laplace transform of a function is given by put f(t) = δ(t) Solve the above integral to obtain 1 as RHS.

11. If f(t) = te-at, then its Laplace transform is Answer: a [Reason:] The Laplace transform of a function is given by put f(t) = te-at On solving, the required answer is obtained.

12. If f(t) = u(t), then its Laplace transform is Answer: b [Reason:] The Laplace transform of a function is given by put f(t) = u(t) to solve the problem.

13. f(t) = t, then its Laplace transform is Answer: d [Reason:] The Laplace transform of a function is given by put f(t) = t to solve the problem. Answer: d [Reason:]The Laplace transform of a function is given by put f(t) = 1⁄b eatsinh(bt) to solve the problem.

15. If L { f(t) } = F(s), then L { kf(t) } =
a) F(s)
b) k F(s)
c) Does not exist
d) F(sk)

Answer: b [Reason:] This is the Linearity property of Laplace transform.

## Discrete Mathematics MCQ Set 3

1. In a Binomial Distribution, if ‘n’ is the number of trials and ‘p’ is the probability of success, then the mean value is given by
a) np
b) n
c) p
d) np(1-p)

Answer: a [Reason:] For a discrete probability function, the mean value or the expected value is given by For Binomial Distribution P(x)= nCx px q(n-x), substitute in above equation and solve to get µ = np.

2. In a Binomial Distribution, if p, q and n are probability of success, failure and number of trials respectively then variance is given by
a) np
b) npq
c) np2q
d) npq2

Answer: b [Reason:] For a discrete probability function, the variance is given by Where µ is the mean, substitute P(x)=nCx px q(n-x) in the above equation and put µ = np to obtain V = npq.

3. If ‘X’ is a random variable, taking values ‘x’, probability of success and failure being ‘p’ and ‘q’ respectively and ‘n’ trials being conducted, then what is the probability that ‘X’ takes values ‘x’? Use Binomial Distribution
a) P(X = x) = nCx px qx
b) P(X = x) = nCx px q(n-x)
c) P(X = x) = xCn qx p(n-x)
d) P(x = x) = xCn pn qx

Answer: b [Reason:] It is the formula for Binomial Distribution that is asked here which is given by P(X = x) = nCx px q(n-x).

4. If ‘p’, ‘q’ and ‘n’ are probability pf success, failure and number of trials respectively in a Binomial Distribution, what is its Standard Deviation ?
a) (np)12
b) (pq)12
c) (np)2
d) (npq)12

Answer: d [Reason:] The variance (V) for a Binomial Distribution is given by V = npq Standard Deviation = (variance)12 = (npq)12.

5. In a Binomial Distribution, the mean and variance are equal
a) True
b) False

Answer: b [Reason:] Mean = np Variance = npq ∴ Mean and Variance are not equal.

6. It is suitable to use Binomial Distribution only for
a) Large values of ‘n’
b) Fractional values of ‘n’
c) Small values of ‘n’
d) Any value of ‘n’

Answer: c [Reason:] As the value of ‘n’ increases, it becomes difficult and tedious to calculate the value of nCx.

7. For larger values of ‘n’, Binomial Distribution
a) loses its discreteness
b) tends to Poisson Distribution
c) stays as it is
d) gives oscillatory values

Answer: b [Reason:] Where m = np is the mean of Poisson Distribution.

8. In a Binomial Distribution, if p = q, then P(X = x) is given by
a) nCx (0.5)n
b) nCn (0.5)n
c) nCx p(n-x)
d) nCn p(n-x)

Answer: a [Reason:] If p = q, then p = 0.5 Substituting in P(x)=nCx px q(n-x) we get nCn (0.5)n.

9. Binomial Distribution is a
a) Continuous distribution
b) Discrete distribution
c) Irregular distribution
d) Not a Probability distribution

Answer: b [Reason:] It is applied to a discrete Random variable, hence it is a discrete distribution.

## Discrete Mathematics MCQ Set 4

1. Error is the Uncertainty in measurement
a) True
b) False

Answer: a [Reason:] In the term of mathematics, “Error tells the person how much correct or certain its measurement is.”

2. Relative error in x is
a) δx
b) δxx
c) δxx * 100
d) 0

Answer: b [Reason:] Option ‘a’ is called absolute error. Option ‘b’ is called relative error. Option ‘c’ is called Percentage error.

3. Find the percentage change power in the circuit if error in value of resistor is 1% and that of voltage source is .99%
a) z should be homogeneous and of order n
b) z should not be homogeneous but of order n
c) z should be implicit
d) z should be the function of x and y only

Answer: a [Reason:] Power is given by P = V2R Taking log on both sides, log(P) = 2log(V) – log(r) Differentiating it , δpp = 2δVVδrr Multiplying by 100 we get, %P = 2%V – %r = (2*.99) – 1 = 0.98%.

4. Magnitude of error can be negative or positive
a) True
b) False

Answer: b [Reason:] Magnitude of error can not be negative.Negative or positive sign only shows the increase or decrease in the quatity.

5. Given the kinetic energy of body is T = 12 mv2 . If the mass of body changes from 100 kg to 100 kg and 500 gm and velocity of a body changes from 1600 mt/sec to 1590 mt/sec. Then find the approximate change in T.
a) 960000 J decrease in value
b) 960000 J increase in value
c) 450000 J decrease in value
d) 450000 J increase in value

Answer: a [Reason:] Given T = 12 mv2 Now taking log and differentiating, δT = 0.5[v2 δm + 2mvδv] Now, v = 1600 mt/sec m = 100kg δv = -10 δm = 0.5 Then, δT = -960000 J => decrese in value of T by 960000 J.

6. The speed of a boat is given by, v = k(1t – at), where k is the constant and l us the distance travel by boat in time t and a is the acceleration of water. If there is an change in ‘l’ from 2cm to 1cm in time 2sec to 1sec. If the acceleration of water changes from 0.95 mt/sec2 to 2 mt/sec2 find the motion of boat.
a) -2
b) 2
c) 0.5
d) -0.5

Answer: a [Reason:] Given, v = k(1t – at) Differentiating it we get Put, l = 2cm , t = 2sec , a = 0.95 mt/sec2 and δl = 1cm ,δt = 1 sec⁡and δa = -1.05 mt/sec2 we get, δvv = 2.

7. The relative error in the volume of figure having hemispherical ends and a body of right circular cylinder is, if error in radius(r) is 0 and in height(h) is 1.
a) 1/(h + 43 r)
b) 1/(h + 23 r)
c) h/(h + 43 r)
d) r/(h + 43 r)

Answer: a [Reason:] Given V = πr2 h + 43 πr3 Now since error in radius is zero , it should be treated as constant, Hence, 8. If n resistors of unequal resistances are connected in parallel,and the percenrage error in all
resistors are k then,total error in parallel combination is Answer: b [Reason:] Given 1r = 1a + 1b + 1c +⋯.. + 1n Differentiating all, – 1r2 dr = – 1a2 da – 1b2 db – ….- 1n2 dn Now, 1r2 dr = + 1a2 da + 1b2 db + ….+ 1n2 dn Multiplying by 100 and putting da = db = ……. = dn =k.

9. The approximate value of function f(x + δx, y + δy) is
a) f + ∂f∂x dx + ∂f∂y dy
b) ∂f∂x dx + ∂f∂y dy
c) f – ∂f∂x dx + ∂f∂y dy
d) ∂f∂x dx – ∂f∂y dy

Answer: a [Reason:] f(x + δx, y + δy) = f(x,y) + df = f + ∂f∂x dx + ∂f∂y dy.

10. At a distance of 120ft from the foot of a tower the elevation of its top is 60o. If the possible error in measuring the distance and elevation are 1 inch and 1 minute, find approximate error in calculated height.
a) 0.342
b) 0.284
c) 0.154
d) 0.986

Answer: b [Reason:] Tan(z) = hx h = x Tan(z) Taking log and then differentiate we get, ∂hh = ∂xx + 1Tan(z) Sec2 (z)δz Now h = 120 tan(60o) = 120√3 Putting, δx = 112 ft ,δz = π(60*180) Putting the values we get, δh = 0.284.

11. Find the approximate value of (1.04)3.01
a) 1.14
b) 1.13
c) 1.11
d) 1.12

Answer: d [Reason:] Let, f(x,y) = xy Now, ∂f∂x = yx(y-1) and ∂f∂y = xy log⁡(x) Putting, x = 1, y = 3 δx = 0.04, δy = 0.01 Now, df = δx ∂f∂x + δy∂f∂y =0.12 Hence, f(x + δx,y + δy) = (1.04)3.01 = 1.12.

12. Find the approximate value of [0.982+2.012+1.942 ](12)
a) 1.96
b) 2.96
c) 0.04
d)-0.04

Answer: b [Reason:] Let f(x,y,z) = (x2+y2+z2 )(12) ……………..(1) Hence, x = 1, y = 2, z = 2 so that, dx = -0.02, dy = 0.01, dz = -0.06 From (1), ∂f∂x = xf

∂f∂y = yf

∂f∂z = zf

And df= ∂f∂x dx + ∂f∂y dy + ∂f∂z dz = ((xdx + ydy + zdz))/f = (-0.02 + 0.02 – 0.12)/3 = -0.04

Hence,

[0.982+2.012+1.942 ](12) = f(1, 2, 2) + df = 3-0.04 = 2.96.

13. Find the approximate value of log⁡(11.01-log⁡(10.1)), Given log(10) = 2.30 and and log(8.69) = 2.16, all the log are in base ‘e’.
a) 2.1654
b) 2.1632
c)2.1645
d) 2.1623

Answer: d [Reason:] Let, f(x,y) = log(x-log(y)) Now by differentiating, Hence, df = 0.0023

Hence, f(x + δx, y + δy) = log⁡(11.01 – log⁡(10.1) )= 2.16 + df = 2.1623.

14. Find approximate value of e10.19.09 ,given e90 = 1.22 * 1039.
a) 2.41 * 1039
b) 2.42 * 1039
c) 2.43 * 1039
d) 2.44 * 1039

Answer: b [Reason:] Let, f(x,y) = exy = exy Now by differentiating, ∂f∂x = yexy and ∂f∂y = xexy

Now , putting, x = 10, y = 9, δx = .01 and δy = .09 We get, ∂f∂x = 1.09* 1040 and ∂f∂y = 1.22* 1040

Hence, df = 1.27* 1039

Hence, f(x + δx, y + δy) = e10.19.09 = 2.42 * 1039.

## Discrete Mathematics MCQ Set 5

1. If f(t) = 1, then its Laplace Transform is given by
a) s
b) 1s
c) 1
d) Does not exist

Answer: b [Reason:] The Laplace Transform of a functions is given by put f(t) = 1 On simplifying, we get 1s.

2. If f(t) = tn where, ‘n’ is an integer greater than zero, then its Laplace Transform is given by
a) n!
b) tn+1
c) n! ⁄ sn+1
d) Does not exist

Answer: c [Reason:]The Laplace Transform of a functions is given by f(t) = tn On simplifying, we get n! ⁄ sn+1.

3. If f(t)=√t, then its Laplace Transform is given by
a) 12
b) 1s
c) √π ⁄ 2√s
d) Does not exist

Answer: c [Reason:]The Laplace Transform of a functions is given by Put f(t)=√t On Solving, we get √π ⁄ 2√s.

4. If f(t) = sin(at), then its Laplace Transform is given by
a) cos(at)
b) 1 ⁄ asin(at)
c) Indeterminate
d) a ⁄ s2+a2

Answer: d [Reason:]The Laplace Transform of a functions is given by Put f(t) = sin(at) On solving, we get a ⁄ s2+a2.

5. If f(t) = tsin(at) then its Laplace Transform is given by
a) 2as ⁄ (s2+a2)2
b) a ⁄ s2+a2
c) Indeterminate
d) √π ⁄ 2√s

Answer: a [Reason:]The Laplace Transform of a functions is given by Put f(t) = tsin(at) On Solving, we get 2as ⁄ (s2+a2)2.

6. If f(t) = eat, its Laplace Transform is given by
a) a ⁄ s2+a2
b) √π ⁄ 2√s
c) 1 ⁄ s-a
d) Does not exist

Answer: c [Reason:] The Laplace Transform of a functions is given by Put f(t) = eat On solving the above integral, we obtain 1 ⁄ s-a.

7. If f(t) = tp where p > – 1, its Laplace Transform is given by
a) √π ⁄ 2√s
b) f(t) = tsin(at)
c) γ(p+1) ⁄ sp+1
d) Does not exist

Answer: d [Reason:]The Laplace Transform of a functions is given by Put f(t) = tp On Solving, we get γ(p+1) ⁄ sp+1.

8. If f(t) = cos(at), its Laplace transform is given by
a) s ⁄ s2+a2
b) a ⁄ s2+a2
c) √π ⁄ 2√s
d) Does not exist

Answer: a [Reason:]The Laplace Transform of a functions is given by Put f(t) = cos(at) On solving the above integral, we get s ⁄ s2+a2.

9. If f(t) = tcos(at), its Laplace transform is given by
a) 1 ⁄ s-a
b) s2 – a2 ⁄ (s2+a2)2
c) Indeterminate
d) s2at

Answer: b [Reason:]The Laplace Transform of a functions is given by Put f(t) = tcos(at) On solving the above integral, using suitable rules of integration we get the answer s2 – a2 ⁄ (s2+a2)2.

10. If f(t) = sin(at) – atcos(at), then its Laplace transform is given by
a) Indeterminate form is encountered
b) a3 ⁄ (s2 + a2)2
c) 2a3 ⁄ (s2 – a2)2
d) 2a3 ⁄ (s2 + a2)2

Answer: d [Reason:]The Laplace Transform of a functions is given by Put f(t) = sin(at) – atcos(at) On solving the above integral, we obtain the answer2 a3 ⁄ (s2 + a2)2.

11. If f(t) = sin(at) – atcos(at), then its Laplace transform is given by Answer: d [Reason:]The Laplace Transform of a functions is given by Put f(t) = sin(at) – atcos(at) On solving, we obtain 2as2 ⁄ (s2+a2)2

12. If f(t) = cos(at) – atsin(at), then its Laplace transform is given by Answer: b [Reason:]The Laplace Transform of a functions is given by Put f(t) = cos(at) – atsin(at) On solving, we obtain a3 ⁄ (s2 + a2)2.

13. If f(t) = cos(at) + atsin(at), its Laplace transform is given by Answer: c [Reason:]The Laplace Transform of a functions is given by Put f(t) = cos(at) + atsin(at) to solve the problem.

14. If f(t) = sin(at + b), its Laplace transform is given by Answer: b [Reason:]The Laplace Transform of a functions is given by Put f(t) = sin(at + b) to solve the problem. Answer: c [Reason:]The Laplace Transform of a functions is given by Put f(t) = cos(at + b) to solve the problem.