## 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

### View Answer

2. Leibniz theorem is applicable if n is a

a) Rational Number

b) Negative Integer

c) Positive Integer

d) Decimal Number

### View Answer

3. If nth derivative of xy_{3} + x^{2}y_{2} + x^{3}y_{0} = 0 then order of its nth differential equation is,

a) n

b) n+1

c) n+2

d) n+3

### View Answer

_{n+2}, y

_{n+1}, y

_{n}, y

_{n-1}, y

_{n-2}, y

_{n-3}. Hence order of differential equation becomes n+3. 2. By Leibniz rule differentiating it n times, we get Xy

_{n+3}+ ny

_{n+2}+ x

^{2}y

_{n+2}+ 2nxy

_{n+1}+ 2n(n-1)y

_{n}+ x

^{3}y

_{n}+ 3nx

^{2}y

_{n-1}+ 6n(n-1)xy

_{n-2}+ 6n(n-1)(n-2)y

_{n-3}= 0 Hence order of differential equlation becomes n+3.

4. Find nth derivative of x^{2}y_{2} + xy_{1} + y = 0

a) x^{2}y_{n+2} + (2n+1)xy_{n+1} + (n^{2}+1)y_{n} = 0

b) x^{2}y_{n+2} + nxy_{n+1} + (n^{2}+1)y_{n} = 0

c) x^{2}y_{n+2} + (2n+1)xy_{n+1} + n^{2}y_{n} = 0

d) x^{2}y_{n+2} + 2nxy_{n+1} + n^{2}y_{n} = 0

### View Answer

^{2}y

_{2}+ xy

_{1}+ y = 0 By Leibniz theorem

x^{2}y_{n+2} + n(2x)(y_{n+1}) + n(n-1)(2)(y_{n}) + xy_{n+1} + ny_{n} + y_{n}= 0

x^{2}y_{n+2} + xy_{n+1}(2n+1) + y_{n}(n^{2}+1) = 0.

5. The nth derivative of x^{2}y_{2} + (1-x^{2})y_{1} + xy = 0 is,

a) x^{2}y_{n+2} + y_{n}(2n^{2}-2n-2nx+x) – y_{n-1}(2n^{2}-3n)=0

b) x^{2}y_{n+2} + y_{n+1}(2nx-x^{2}) + y_{n}(2n^{2}-2n-2nx+x) – y_{n-1}(2n^{2}-3n)=0

c) x^{2}y_{n+2} + y_{n+1}(2nx+1-x^{2}) + y_{n}(2n^{2}-2n-2nx+x)-y_{n – 1}(2n^{2}-3n)=0

d) x^{2}y_{n+2} + y_{n+1}(2nx+1-x^{2}) + xy_{n}2n^{2}-y_{n – 1}(2n^{2}-3n)=0

### View Answer

Explanation: x^{2}y_{2} + (1-x^{2})y_{1} + xy = 0

Differenetiating n times by Leibniz Rule

x^{2}y_{n+2} + 2nxy_{n+1} + 2n(n-1)y_{n} + (1-x^{2})y_{n+1} – 2nxy_{n} – 2n(n-1)y_{n-1 }+ xy_{n} + ny_{n-1} = 0

x^{2}y_{n+2} + y_{n+1}(2nx+1-x^{2}) + y_{n}(2n^{2}-2n-2nx+x) – y_{n-1}(2n^{2}-3n)=0.

6. Find nth derivative of x^{n}Sin(nx)

### View Answer

^{n}Sin(nx) By Leibniz Rule , put u = x

^{n}and v = Sin(nx), we get

7. If y(x) = tan^{-1}x then

a) (y_{n+1})(0) = (n-1)(y_{n-1})0

b) (y_{n+1})(0) = n(n-1)(y_{n-1})0

c) (y_{n+1})(0) = -(n-1)(y_{n-1})0

d) (y_{n+1})(0) = -n(n-1)(y_{n-1})0

### View Answer

^{-1}x Y

_{1}= 1/(1+x

^{2}) (1+x

^{2})y

_{1}= 1 By Leibniz Rule, (1+x

^{2})y

_{n+1}+ 2nxy

_{n}+ n(n-1)y

_{n-1}= 0 Put x=0, gives → (y

_{n+1})(0) = -n(n-1)(y

_{n-1})(0).

8. If y = sin^{-1}x, then

a) (1-x^{2})y_{n+2} – xy_{n+1}(2n-1) = ny_{n}(2n-1)

b) x^{2}y_{n+2} – xy_{n+1}(2n-1) = ny_{n}(2n-1)

c) (1-x^{2})y_{n+2} – 2nxy_{n+1} = ny_{n}(2n-1)

d) (1-x^{2})y_{n+2} – xy_{n+1}(2n-1) +ny_{n}(2n-1)=0

### View Answer

^{-1}x Differentiating it Y1 = 1/√(1-x

^{2})

(1-x^{2})(y1)^{2}= 1

Again Differentiating we get
(1-x^{2})2y_{1}y_{2} – 2x(y1)^{2} = 0

(1-x^{2})y_{2} = xy_{1}

By Leibniz Rule, Diff it n times,
(1-x^{2})y_{n+2} – 2xny_{n+1} – 2n(n-1)y_{n} = xy_{n+1} + ny_{n}

(1-x^{2}) y_{n+2} – xy_{n+1}(2n-1) = ny_{n}(2(n-1)+1)

(1-x^{2}) y_{n+2} – xy_{n+1}(2n-1) = ny_{n}(2n-1).

9. If y = cos^{-1}x, then

a) (y_{n+2})(0) = -n(2n-1) y_{n}(0)

b) (y_{n+2})(0) = n(2n-1) y_{n}(0)

c) (y_{n+2})(0) = n(n-2) y_{n}(0)

d) (y_{n+2})(0) = n(n-3) y_{n}(0)

### View Answer

^{-1}x

Differentiating it
Y1 = 1/√(1-x^{2} )

(1-x^{2})(y1)^{2}= 1

Again Differentiating we get
(1-x^{2})2y_{1}y_{2} – 2x(y1)^{2} = 0

(1-x^{2})y_{2} = xy_{1}

By Leibniz Rule, Diff it n times,
(1-x^{2})y_{n+2} – 2xny_{n+1} – 2n(n-1)y_{n} = xy_{n+1} + ny_{n}

(1-x^{2}) y_{n+2} – xy_{n+1}(2n-1) = ny_{n}(2(n-1)+1)

(1-x^{2}) y_{n+2} – xy_{n+1}(2n-1) = ny_{n}(2n-1)
At x=0, we get
(y_{n+2})(0) = n(2n-1) y_{n}(0).

## Discrete Mathematics MCQ Set 2

1. If f(t) = sinhat, then its Laplace transform is

a) e^{at}

b) s ⁄ s^{2}-a^{2}

c) a ⁄ s^{2}-a^{2}

d) Exists only if ‘t’ is complex

### View Answer

^{2}-a

^{2}is obtained.

2. If f(t) = coshat, its Laplace transform is given by

a) s ⁄ s^{2}-a^{2}

b) s+a ⁄ s-a

c) Indeterminate

d) (sinh(at))^{2}

### View Answer

^{2}-a

^{2}is obtained.

3. If f(t) = e^{at} sin(bt), then its Laplace transform is given by

a) s^{2}-a^{2} ⁄ (s – a)^{2}

b) b ⁄ (s + a)^{2} + b^{2}

c) b ⁄ (s – a)^{2} + b^{2}

d) Indeterminate

### View Answer

^{at}sin(bt) On solving, we get the b ⁄ (s – a)

^{2}+ b

^{2}.

4. If f(t) = e^{at} cos(bt), then its Laplace transform is

a) 2a^{3} ⁄ (s^{2} + a^{2})

b) s+a ⁄ s-a

c) Indeterminate

d) s-a ⁄ (s – a)^{2} + b^{2}

### View Answer

^{at}cos(bt) Solve the above integral, to obtain s-a ⁄ (s – a)

^{2}+ b

^{2}.

5. If f(t) = e^{at} sinh(bt) then its Laplace transform is

a) e^{-as} ⁄ s

b) s+a ⁄ (s – a)^{2} + b^{2}

c) b ⁄ (s – a)^{2} – b^{2}

d) Does not exist

### View Answer

^{at}sinh(bt) On solving, we get the b ⁄ (s – a)

^{2}– b

^{2}.

6. If f(t) = ^{1}⁄_{a} sinh(at), then its Laplace transform is

a) 1⁄s^{2}-a^{2}

b) 2a ⁄ (s – b)^{2} + b^{2}

c) n! ⁄ (s – a)^{n-1}

d) Does not exist

### View Answer

^{2}-a

^{2}.

7. If f(t) = t^{n} ⁄ n, then its Laplace transform is

### View Answer

^{n}⁄ n On solving, we obtain the Laplace transform of the required function.

8. If f(t) = ^{1} ⁄ _{√Πt}, then its Laplace transform is

### View Answer

9. If f(t) = ^{t}⁄_{2} a sinat, then its Laplace transform is

a) b ⁄ (s + a)^{2} + b^{2}

b) 2a ⁄ (s – b)^{2} + b^{2}

c) Indeterminate

d) s ⁄ (s^{2} + a^{2})^{2}

### View Answer

^{2}+ a

^{2})

^{2}.

10. If f(t) = δ(t), then its Laplace transform is

a) s + a ⁄ (s – a)^{2} + b^{2}

b) a^{3} ⁄ (s^{2} + a^{2})^{2}

c) 1

d) Does not exist

### View Answer

11. If f(t) = te^{-at}, then its Laplace transform is

### View Answer

^{-at}On solving, the required answer is obtained.

12. If f(t) = u(t), then its Laplace transform is

### View Answer

13. f(t) = t, then its Laplace transform is

### View Answer

### View Answer

^{at}sinh(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(^{s}⁄_{k})

### View Answer

## 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)

### View Answer

^{n}C

_{x}p

^{x}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) np^{2}q

d) npq^{2}

### View Answer

^{n}C

_{x}p

^{x}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) = ^{n}C_{x} p^{x} q^{x}

b) P(X = x) = ^{n}C_{x} p^{x} q^{(n-x)}

c) P(X = x) = ^{x}C_{n} q^{x} p^{(n-x)}

d) P(x = x) = ^{x}C_{n} p^{n} q^{x}

### View Answer

^{n}C

_{x}p

^{x}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)^{1⁄2}

b) (pq)^{1⁄2}

c) (np)^{2}

d) (npq)^{1⁄2}

### View Answer

^{1⁄2}= (npq)

^{1⁄2}.

5. In a Binomial Distribution, the mean and variance are equal

a) True

b) False

### View Answer

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’

### View Answer

^{n}C

_{x}.

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

### View Answer

8. In a Binomial Distribution, if p = q, then P(X = x) is given by

a) ^{n}C_{x} (0.5)^{n}

b) ^{n}C_{n} (0.5)^{n}

c) ^{n}C_{x} p^{(n-x)}

d) ^{n}C_{n} p^{(n-x)}

### View Answer

^{n}C

_{x}p

^{x}q

^{(n-x)}we get

^{n}C

_{n}(0.5)

^{n}.

9. Binomial Distribution is a

a) Continuous distribution

b) Discrete distribution

c) Irregular distribution

d) Not a Probability distribution

### View Answer

## Discrete Mathematics MCQ Set 4

1. Error is the Uncertainty in measurement

a) True

b) False

### View Answer

2. Relative error in x is

a) δx

b) ^{δx}⁄_{x}

c) ^{δx}⁄_{x} * 100

d) 0

### View Answer

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

### View Answer

^{V2}⁄

_{R}Taking log on both sides, log(P) = 2log(V) – log(r) Differentiating it ,

^{δp}⁄

_{p}= 2

^{δV}⁄

_{V}–

^{δr}⁄

_{r}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

### View Answer

5. Given the kinetic energy of body is T = ^{1}⁄_{2} mv^{2} . 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

### View Answer

^{1}⁄

_{2}mv

^{2}Now taking log and differentiating, δT = 0.5[v

^{2}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(^{1}⁄_{t} – 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/sec^{2} to 2 mt/sec^{2} find the motion of boat.

a) -2

b) 2

c) 0.5

d) -0.5

### View Answer

^{1}⁄

_{t}– at) Differentiating it we get Put, l = 2cm , t = 2sec , a = 0.95 mt/sec

^{2}and δl = 1cm ,δt = 1 secand δa = -1.05 mt/sec

^{2}we get,

^{δv}⁄

_{v}= 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 + ^{4}⁄_{3} r)

b) 1/(h + ^{2}⁄_{3} r)

c) h/(h + ^{4}⁄_{3} r)

d) r/(h + ^{4}⁄_{3} r)

### View Answer

^{2}h +

^{4}⁄

_{3}πr

^{3}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

### View Answer

^{1}⁄

_{r}=

^{1}⁄

_{a}+

^{1}⁄

_{b}+

^{1}⁄

_{c}+⋯.. +

^{1}⁄

_{n}Differentiating all, –

^{1}⁄

_{r2}dr = –

^{1}⁄

_{a2}da –

^{1}⁄

_{b2}db – ….-

^{1}⁄

_{n2}dn Now,

^{1}⁄

_{r2}dr = +

^{1}⁄

_{a2}da +

^{1}⁄

_{b2}db + ….+

^{1}⁄

_{n2}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

### View Answer

^{∂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

### View Answer

^{h}⁄

_{x}h = x Tan(z) Taking log and then differentiate we get,

^{∂h}⁄

_{h}=

^{∂x}⁄

_{x}+

^{1}⁄

_{Tan(z)}Sec

^{2}(z)δz Now h = 120 tan(60

^{o}) = 120√3 Putting, δx =

^{1}⁄

_{12}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

### View Answer

^{y}Now,

^{∂f}⁄

_{∂x}= yx

^{(y-1)}and

^{∂f}⁄

_{∂y}= x

^{y}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 [0.98^{2}+2.01^{2}+1.94^{2} ]^{(1⁄2)}

a) 1.96

b) 2.96

c) 0.04

d)-0.04

### View Answer

^{2}+y

^{2}+z

^{2})

^{(1⁄2)}……………..(1) Hence, x = 1, y = 2, z = 2 so that, dx = -0.02, dy = 0.01, dz = -0.06 From (1),

^{∂f}⁄

_{∂x}=

^{x}⁄

_{f}

^{∂f}⁄_{∂y} = ^{y}⁄_{f}

^{∂f}⁄_{∂z} = ^{z}⁄_{f}

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,

[0.98^{2}+2.01^{2}+1.94^{2} ]^{(1⁄2)} = 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

### View Answer

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

14. Find approximate value of e^{10.19.09} ,given e^{90} = 1.22 * 10^{39}.

a) 2.41 * 10^{39}

b) 2.42 * 10^{39}

c) 2.43 * 10^{39}

d) 2.44 * 10^{39}

### View Answer

^{xy}= e

^{xy}Now by differentiating,

^{∂f}⁄

_{∂x}= ye

^{xy}and

^{∂f}⁄

_{∂y}= xe

^{xy}

Now , putting, x = 10, y = 9, δx = .01 and δy = .09
We get, ^{∂f}⁄_{∂x} = 1.09* 10^{40} and ^{∂f}⁄_{∂y} = 1.22* 10^{40}

Hence, df = 1.27* 10^{39}

Hence, f(x + δx, y + δy) = e^{10.19.09} = 2.42 * 10^{39}.

## Discrete Mathematics MCQ Set 5

1. If f(t) = 1, then its Laplace Transform is given by

a) s

b) ^{1}⁄_{s}

c) 1

d) Does not exist

### View Answer

^{1}⁄

_{s}.

2. If f(t) = t^{n} where, ‘n’ is an integer greater than zero, then its Laplace Transform is given by

a) n!

b) t^{n+1}

c) n! ⁄ s^{n+1}

d) Does not exist

### View Answer

^{n}On simplifying, we get n! ⁄ s

^{n+1}.

3. If f(t)=√t, then its Laplace Transform is given by

a) ^{1}⁄_{2}

b) ^{1}⁄_{s}

c) √π ⁄ 2√s

d) Does not exist

### View Answer

4. If f(t) = sin(at), then its Laplace Transform is given by

a) cos(at)

b) 1 ⁄ a^{sin(at)}

c) Indeterminate

d) a ⁄ s^{2}+a^{2}

### View Answer

^{2}+a

^{2}.

5. If f(t) = tsin(at) then its Laplace Transform is given by

a) 2as ⁄ (s^{2}+a^{2})^{2}

b) a ⁄ s^{2}+a^{2}

c) Indeterminate

d) √π ⁄ 2√s

### View Answer

^{2}+a

^{2})

^{2}.

6. If f(t) = e^{at}, its Laplace Transform is given by

a) a ⁄ s^{2}+a^{2}

b) √π ⁄ 2√s

c) 1 ⁄ s-a

d) Does not exist

### View Answer

^{at}On solving the above integral, we obtain 1 ⁄ s-a.

7. If f(t) = t^{p} where p > – 1, its Laplace Transform is given by

a) √π ⁄ 2√s

b) f(t) = tsin(at)

c) γ(p+1) ⁄ s^{p+1}

d) Does not exist

### View Answer

^{p}On Solving, we get γ(p+1) ⁄ s

^{p+1}.

8. If f(t) = cos(at), its Laplace transform is given by

a) s ⁄ s^{2}+a^{2}

b) a ⁄ s^{2}+a^{2}

c) √π ⁄ 2√s

d) Does not exist

### View Answer

^{2}+a

^{2}.

9. If f(t) = tcos(at), its Laplace transform is given by

a) 1 ⁄ s-a

b) s^{2} – a^{2} ⁄ (s^{2}+a^{2})^{2}

c) Indeterminate

d) s^{2}at

### View Answer

^{2}– a

^{2}⁄ (s

^{2}+a

^{2})

^{2}.

10. If f(t) = sin(at) – atcos(at), then its Laplace transform is given by

a) Indeterminate form is encountered

b) a^{3} ⁄ (s^{2} + a^{2})^{2}

c) 2a^{3} ⁄ (s^{2} – a^{2})^{2}

d) 2a^{3} ⁄ (s^{2} + a^{2})^{2}

### View Answer

^{3}⁄ (s

^{2}+ a

^{2})

^{2}.

11. If f(t) = sin(at) – atcos(at), then its Laplace transform is given by

### View Answer

^{2}⁄ (s

^{2}+a

^{2})

^{2}

12. If f(t) = cos(at) – atsin(at), then its Laplace transform is given by

### View Answer

^{3}⁄ (s

^{2}+ a

^{2})

^{2}.

13. If f(t) = cos(at) + atsin(at), its Laplace transform is given by

### View Answer

14. If f(t) = sin(at + b), its Laplace transform is given by

### View Answer

15. If f(t) = cos(at + b) , its Laplace transform is given by