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## Discrete Mathematics MCQ Set 1

1. Time domain function of s a2 + s2 is given by
a) Cos(at)
b) Sin(at)
c) Cos(at)Sin(at)
d) None of the above

Answer: a [Reason:] L[Cos(at)] = s a2 + s2

L-1 [s a2 + s2 ] = Cos(at).

2. Inverse Laplace transform of 1/(s+1)(s-1)(s+2) is
a) –12 et + 16 e-t + 13 e2t
b) –12 e-t + 16 et + 13 e-2t
c) 12 e-t16 et13 e-2
d) –12 e-t + 16 e-t + 13 e-2

Answer: b [Reason:] 3. Inverse laplace transform of 1/(s-1)2 (s+5) is
a) 16 e – t136 et + 136 e-5t
b) 16 ett – 136 et + 136 e-5t
c) 16 e-tt2136 e-t + 136 e5t
d) 16 e-t t-136 e-t + 136 e5t

Answer: a [Reason:] 4. Find the inverse lapalce transform of a) 112 et113 Cos(-t) – 112 Sin(-t) – 1156 e-5t
b) 112 e-t113 Cos(t) – 112 Sin(t) – 1156 e5t
c) 112 et113 Cos(t) – 112 Sin(t) – 1156 e-5t
d) 112 et + 113 Cos(t) + 112 Sin(t) + 1156 e-5t

Answer: c [Reason:] 5. Find the inverse laplace transform of s(s2 + 4)2
a) 14 sin(2t)
b) t24 sin(2t)
c) t4 sin(2t)
d) t4 sin(2t2)

Answer: c [Reason:] 6. Final value theorem states that Answer: d [Reason:] Final value theorem states that 7. Initial value theorem states that Answer: a [Reason:] Initial value theorem states that 8. Find the value of x(∞) if a) 5
b) 4
c) 1220
d) 2

Answer: c [Reason:] Hence, by final value theorem, = 1220

9. Find the value of x(0) if a) 5
b) 4
c) 1220
d) 2

Answer: b [Reason:] Hence, by initial value theorem, = 2.

10. Find the inverse lapace of a) 13 et [Cos(t) – Cos(2t)].
b) 13 e-t [Cos(t) + Cos(2t)].
c) 13 et [Cos(t) + Cos(2t)].
d) 13 e-t [Cos(t) – Cos(2t)].

Answer: d [Reason:] 11. Find the inverse laplace transform of a) -e-t + 1 + et – 1
b)-e-t + 1 – et + 1
c) -e-t + 1 + et + 1
d)-e-t + 1 – et – 1

Answer: d [Reason:] Given, 12. Find the inverse lapalce transform of 1s(s-1)(s2+1)
a) 12 e-t + 12 Sin(-t) – 12 Cos(-t)
b) 12 et + 12 Sin(t) – 12 Cos(t)
c) 12 et + 12 Sin(t) + 12 Cos(t)
d) 12 et12 Sin(t) – 12 Cos(t)

Answer: b [Reason:] We know that, ## Discrete Mathematics MCQ Set 2

1. Consider the f(x, y) = x2 + y2 – a. For what values of a do we have critical points for the function
a) independent of a
b) for any real number except zero
c) a ∊ (0, +∞)
d) a ∊ (-1, 1)

Answer: a [Reason:] Consider fx = 2x and fy = 2y There is no a here. Thus, independent of a.

2. The critical point exist for the function f(x, y) = xn + xn-1 y +……+yn at (0,0)
a) True
b) False

Answer: b [Reason:] Counter example is with n=1 f(x, y) = x + y.

3. f(x, y) = sin(x).cos(y) Which of the following is a critical point
a) (Π4, Π4)
b) (- Π4, Π4)
c) (0, Π2)
d) (0, 0)

Answer: c [Reason:] fx = cos(x).cos(y) = 0 fy = – sin(x).sin(y) →(x, y) = (0, Π2).

4. The point (0,0) in the domain of f(x, y) = sin(xy) is a point of
b) Minima
c) Maxima
d) Constant

Answer: d [Reason:] Differentiating fxx = -y2.sin(xy) fyy = -x2.sin(xy) fxy = -yx.sin(xy) Observe that fxx. fyy – (fxy)2 Hence, it is a saddle point.

5. A man travelling onf(x, y) = sin(xy). His shadow passing through the origin in a straight line (sun travels with him overhead).
What is the slope of the line travelling on which would lead him to the lowest elevation.
a) There isn’t such a line
b) 1
c)-1
d) 0

Answer: a [Reason:] Differentiating yields fxx = -y2.sin(xy) fyy = -x2.sin(xy) fxy = -yx.sin(xy) Observe that (0,0) is an inconclusive point Hence, he will never reach the lowest elevation(because there isn’t such point.

6. let s(1) be the set of all critical points of f1(x, y) = g1(x).g2(y) and s(2) be the set of critical points of f2(g1(x), g2(y)) Which of the following is the right relation between s(1) and s(2), given that minimum number of elements in s(1) is 2.
a) s(1) = s(2)
b) s(1) ≠ s(2)
c) s(1) ∩ s(2) ≠ 0
d) depends on the functions

Answer: b [Reason:] Differentiating f1(g1(x), g2(y)) with respect to x and y seperately we get dx = f1x g1x (x) dy = f1y g1y (y) This implies g1x = 0 g1y = 0 Which are also the set of critical points of f1(x, y) Thus we have the relation as s(1) ∩ s(2) ≠ 0.

7. . Find the critical points
a) (0,0)
b) (1,1)
c) (2,22)
d) None exist

Answer: d [Reason:] Rewriting the function Differentiating with respect to y we get fy = -10 -10 ≠ 0 There exist no critical point.

8. Consider the vertical cone. The minimum value of the function in the region f(x,y) = c is
a) constant
b) 1
c) 0
d) -1

Answer: a [Reason:] f(x,y) = c is a level curve over which the function has constant value Hence, we have the answer as a constant.

## Discrete Mathematics MCQ Set 3

1. What is the saddle point
a)Point where function has maximujm value
b) Point where function has minimujm value
c) Point where function has zero value
d) Point where function neither have maximujm value nor minimum value

Answer: d [Reason:] Saddle point is a point where function have neither maximum nor minimum value.

2. Stationary point is a point where, function f(x,y) have,
a) ∂f∂x = 0
b) ∂f∂y =0
c) ∂f∂x = 0 & ∂f∂y = 0
d) ∂f∂x < 0 and ∂f∂y > 0

Answer: c [Reason:] Point where function f(x,y) either have maximum or minimum value is called saddle point. i.e, ∂f∂x = 0 & ∂f∂y = 0.

3. For function f(x,y) to have minimum value at (a,b) value,
a) rt – s2>0 and r<0
b) rt – s2>0 and r>0
c) rt – s2<0 and r<0
d) rt – s2>0 and r>0

Answer: b [Reason:] For the function f(x,y) to have minimum value at (a,b)

rt – s2>0 and r>0

where, r = 2f∂x2, t=2f∂y2, s=(2f∂x∂y, at (x,y) => (a,b).

4. For function f(x,y) to have maximum value at (a,b),
a) rt – s2>0 and r<0
b) rt – s2>0 and r>0
c) rt – s2<0 and r<0
d) rt – s2>0 and r>0

Answer: a [Reason:] For the function f(x,y) to have maximum value at (a,b)

rt – s2>0 and r<0

where, r = 2f∂x2, t=2f∂y2, s=(2f∂x∂y, at (x,y) => (a,b).

5. For function f(x,y) to have no extremum value at (a,b),
a) rt – s2>0
b) rt – s2<0
c) rt – s2 = 0
d) rt – s2 ≠ 0

Answer: b [Reason:] For the function f(x,y) to have no extremum value at (a,b) rt – s2 < 0 where, r = 2f∂x2, t=2f∂y2, s=(2f∂x∂y, at (x,y) => (a,b).

6. Discuss minimum value of f(x,y)=x2 + y2 + 6x + 12
a) 3
b) 3
c) -9
d) 9

Answer: b [Reason:] Given, f(x, y) = x2 + y2 + 6x + 12 Now, ∂f∂x = 2x + 6 and ∂f∂y = 2

Putting, ∂f∂x and ∂f∂y = 0 we get, (x,y) = (-3,0)

Now, r= 2f∂x2 = 2>0 and t= 2f∂y2 = 2 and s= 2f∂x∂y = 0

hence, rt – s2 = 4>0 and r>0

hence. f(x,y) has minimum value at (-3,0),which is f(x,y) = 12 + 9 – 18 = 3.

7. Discuss maximum or minimum value of f(x,y) = y2 + 4xy + 3x2 + x3
a) minimum at (0,0)
b) maximum at (0,0)
c) minimum at (2/3, -4/3)
d) maximum at (2/3, -4/3)

Answer: c [Reason:] Given,f(x,y) = y2 + 4xy + 3x2 + x3 Now,∂f∂x = 4y + 6x + 3x2 and ∂f∂y = 2y + 4x Putting,∂f∂x and ∂f∂y = 0,and solving two equations,we get, (x,y) = (0,0) or (2/3, -4/3) Now,at (0,0) r= 2f∂x2=6+6x=6>0 and t= 2f∂y2 =2>0 and s= 2f∂x∂y=4 hence, rt – s2 = 12 – 16<0,hence it has no extremum at this point.

Now,at (23,-43) r= 2f∂x2= 6 + 6x = 10>0 and t= 2f∂y2 =2>0 and s= 2f∂x∂y=4 hence, rt – s2 = 20 – 16 > 0 and r>0, hence it has minimum at this point.(23, –43).

8.Find the minimum value of xy+a3 (1x + 1y)
a) 3a2
b) a2
c) a
d) 1

Answer: a [Reason:] 9. Divide 120 into three parts so that the sum of their products taken two at a time is maximum. If x, y, z are two parts, find value of x, y and z
a) x=40, y=40, z=40
b) x=38, y=50, z=32
c) x=50, y=40, z=30
d) x=80, y=30, z=50

Answer: b [Reason:] Now, x + y + z = 120 => z = 120 – x – y f = xy + yz + zx

f = xy + y(120-x-y) + x(120-x-y) = 120x + 120y – xy – x2 – y2

Hence, ∂f∂x = 120 – y – 2x and ∂f∂y = 120 – x – 2y

putting ∂f∂x and ∂f∂y equals to 0 we get, (x, y)=>(40, 40) Now at (40,40), r=2f∂x2 = -2 < 0,s = 2f∂x∂y = -1, and t = 2f∂y2 = -2 hence, rt – s2 = 5 > 0 since, r<0 and rt – s2 > 0 f(x,y)has maixum value at (40,40), Hence, maximum value of f(40,40) = 120 – 40 – 40 = 40, Hence, x = y = z = 40.

10. Find the maximum value of Sin(A)Sin(B)Sin(C) if A, B , C are the angles of triangle.
a) 3√38
b) 3√48
c) –3√38
d) π8

Answer: a [Reason:] Given f(A,B,C)=Sin(A)Sin(B)Sin(c),

Since,A,B,C are the angle of triangle,hence,C = 180 – (A+B),

hence,f(x,y) = Sin(x)Sin(y)Sin(x+y), where A = x and B = y

Hence,∂f∂x = Cos(x)Sin(y)Sin(x+y) + Sin(x)Sin(y)Cos(x+y) = Sin(y)Sin(y+2x) and,∂f∂y = Sin(x)Cos(y)Sin(x+y) + Sin(x)Sin(y)Cos(x+y) = Sin(x)Sin(x+2y)

Hence, putting ∂f∂x and ∂f∂y = 0, we get (x,y)=(60,60), (120,120) Hence, at (x,y) = (60,60)we get,r = -√3, s = -√3/2, t = -√3, hence, rt-s2= 94∂x>0

hence, r<0 andrt-s2>0 hence,f(x,y) or f(A,B)have maximum value at (60,60)

Hence, at (x,y)=(120,120)we get,r=√3,s=√3/2,t=√3,hence,rt-s2= 94∂x>0 And this value is 3√38 hence, r>0 and rt-s2 >0 hence,f(x,y)or f(A,B)have minimum value at (60,60) and this value is –3√38.

11. The drawback of Lagrange’s Method of Maxima and MIinima is
a) Maxima or Minima is not fixed
b) Nature of stationary point is can not be known
c) Accuracy is not good
d) Nature of stationary point is known but can not give maxima or minima

Answer: b [Reason:] In lagrange’s theorem of maxima of minima one can not determine the nature of stationary points.

## Discrete Mathematics MCQ Set 4

1. What are Intermediate Forms ?
a) Forms(f(x)/g(x)) whose limits x tends to ‘a’ can give rational number directly
b) Forms(f(x)/g(x)) whose limits x tends to ‘a’ can give finite number directly
c) Forms(f(x)/g(x)) whose limits x tends to ‘a’ can not be infinite output or cannot be solved directly
d) Forms(f(x)/g(x)) whose limits x tends to ‘a’ can gives finite output

Answer: c [Reason:] = 00 = 0 = 0 = are all intermediate forms. Special Rules are made to solve these forms. They cannot be solved directly.

2. L’Hospital Rule states that
a) If is an ideterminate form than = if has a finite value
b) always equals to c) if an indeterminate form than cannot be solved
d) if an indeterminate form than it is equals to zero.

Answer: a [Reason:] According to L’Hospital Rule, if is indeterminate and has a finite value then = . It is helpful in solving limits of indeterminate forms.

3. If f(x) = x2 – 3x + 2 and g(x) = x3 – x2 + x – 1 than find value of a) 0.5
b) 1
c) -.5
d) -1

Answer: c [Reason:] 4. If f(x) = Tan(x)-1 and g(x) = Sin(x) – Cos(x) than find value of limx → π4 f(x)g(x)
a) -√2
b) √(-2)
c) √2
d) √3

Answer: c [Reason:] 5. If f(x) = Tan(x) and g(x) = ex – 1 than find value of limx → 0⁡ f(x)g(x)
a) 1
b) 0
c) -1
d) 2

Answer: a [Reason:] 6. Find the value of a) e-1
b) e
c) e + 1
d) 1

Answer: b [Reason:] 7. If f(x) = x3 + 3x2 + Sin(x) and g(x) = ex – 1 than find value of a) e6e
b) e(e/6)
c) e6
d) e(6/e)

Answer: c [Reason:] 8. If f(x) = Sin(x) and g(x) = x than find value of limx → 0⁡ f(x)g(x)
a) -1
b) 0
c) 1
d) 2

Answer: c [Reason:] limx → 0⁡ f(x)g(x) = limx → 0⁡ sin(x)x = ⁡00 (Indeterminate forms) By L’Hospital rule limx → 0⁡ cos(x)1 = 1.

9. If f(x) = sin(x)cos(x) and g(x) = x2 than find value of limx → 0⁡ f(x)g(x)
a) 2
b) 0
c) -1
d) Cannot be found

Answer: b [Reason:] 10. If f(x) = ex + xcos(x) and g(x) = Sin(x) than find value of limx → 0⁡ f(x)g(x)
a) 2
b) 1
c) 3
d) 4

Answer: a [Reason:] 11. Find the value of is, where {x} is the fractional part of x.
a) 2e
b) 1e
c) 0
d) –1e

Answer: b [Reason:] ## Discrete Mathematics MCQ Set 5

1. Normal Distribution is applied for
a) Continuous Random Distribution
b) Discrete Random Variable
c) Irregular Random Variable
d) Uncertain Random Variable

Answer: a [Reason:] This is the rule on which Normal distribution is defined, no details on this as of why For more knowledge on this aspect, you can refer to any book or website which speaks on the same.

2. The shape of the Normal Curve is
a) Bell Shaped
b) Flat
c) Circular
d) Spiked

Answer: a [Reason:] Due to the nature of the Probability Mass function, a bell shaped curve is obtained.

3. Normal Distribution is symmetric is about
a) Variance
b) Mean
c) Standard deviation
d) Covariance

Answer: b [Reason:] Due to the very nature of p.m.f of Normal Distribution, the graph appears such that it is symmetric about its mean.

4. For a standard normal variate, the value of mean is
a) ∞
b) 1
c) 0
d) not defined

Answer: c [Reason:] For a normal variate, if its mean = 0 and standard deviation = 1, then its called as Standard Normal Variate. Here, the converse is asked.

5. The area under a standard normal curve is
a) 0
b) 1
c) ∞
d) not defined

Answer: b [Reason:] For any probability distribution, the sum of all probabilities is 1. Area under normal curve refers to sum of all probabilities.

6. The standard normal curve is symmetric about the value
a) 0.5
b) 1
c) ∞
d) 0

Answer: d [Reason:] Normal curve is always symmetric about mean, for standard normal curve or variate mean = 0.

7. For a standard normal variate, the value of Standard Deviation is
a) 0
b) 1
c) ∞
d) not defined

Answer: b [Reason:] If the mean and standard deviation of a normal variate are 0 and 1 respectively, it is called as standard normal variate. Here the converse is asked.

8. Normal Distribution is also known as
a) Cauchy’s Distribution
b) Laplacian Distribution
c) Gaussian Distribution
d) Lagrangian Distribution

Answer: c [Reason:] Named after the one who proposed it. For further details, refer to books or internet.

9. Skewness of Normal distribution is
a) Negative
b) Positive
c) 0
d) Undefined

Answer: c [Reason:] Since the normal curve is symmetric about its mean, its skewness is zero. This is a theoretical explanation for mathematical proofs, you can refer to books or websites that speak on the same in detail.

10. For a normal distribution its mean, median, mode are equal
a) True
b) False

Answer: a [Reason:] It has a theoretical evidence that requires some serious background on several topics For more details you can refer to any book or website that speaks on the same.

11. In Normal distribution, the highest value of ordinate occurs at
a) Mean
b) Variance
c) Extremes
d) Same value occurs at all points

Answer: a [Reason:] This is due the behaviour of the pdf of Normal distribution.

12. The shape of the normal curve depends on its
a) Mean deviation
b) Standard deviation
c) Quartile deviation
d) Correlation

Answer: b [Reason:] This can be seen in the pdf of normal distribution where standard deviation is a variable.

13.The value of constant ‘e’ appearing in normal distribution is
a) 2.5185
b) 2.7836
c) 2.1783
d) 2.7183

Answer: d [Reason:] This is a standard constant.

14. In Standard normal distribution, the value of mode is
a) 2
b) 1
c) 0
d) Not fixed