## 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

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^{s}⁄

_{ a2 + s2}

L^{-1} [^{s} ⁄_{ a2 + s2} ] = Cos(at).

2. Inverse Laplace transform of 1/(s+1)(s-1)(s+2) is

a) –^{1}⁄_{2} e^{t} + ^{1}⁄_{6} e^{-t} + ^{1}⁄_{3} e^{2t}

b) –^{1}⁄_{2} e^{-t} + ^{1}⁄_{6} e^{t} + ^{1}⁄_{3} e^{-2t}

c) ^{1}⁄_{2} e^{-t} – ^{1}⁄_{6} e^{t} – ^{1}⁄_{3} e^{-2}

d) –^{1}⁄_{2} e^{-t} + ^{1}⁄_{6} e^{-t} + ^{1}⁄_{3} e^{-2}

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3. Inverse laplace transform of 1/(s-1)^{2} (s+5) is

a) ^{1}⁄_{6} e^{ – t} – ^{1}⁄_{36} e^{t} + ^{1}⁄_{36} e^{-5t}

b) ^{1}⁄_{6} e^{t}t – ^{1}⁄_{36} e^{t} + ^{1}⁄_{36} e^{-5t}

c) ^{1}⁄_{6} e^{-t}t^{2} – ^{1}⁄_{36} e^{-t} + ^{1}⁄_{36} e^{5t}

d) ^{1}⁄_{6} e^{-t} t-^{1}⁄_{36} e^{-t} + ^{1}⁄_{36} e^{5t}

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4. Find the inverse lapalce transform of

a) ^{1}⁄_{12} e^{t} – ^{1}⁄_{13} Cos(-t) – ^{1}⁄_{12} Sin(-t) – ^{1}⁄_{156} e^{-5t}

b) ^{1}⁄_{12} e^{-t} – ^{1}⁄_{13} Cos(t) – ^{1}⁄_{12} Sin(t) – ^{1}⁄_{156} e^{5t}

c) ^{1}⁄_{12} e^{t} – ^{1}⁄_{13} Cos(t) – ^{1}⁄_{12} Sin(t) – ^{1}⁄_{156} e^{-5t}

d) ^{1}⁄_{12} e^{t} + ^{1}⁄_{13} Cos(t) + ^{1}⁄_{12} Sin(t) + ^{1}⁄_{156} e^{-5t}

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5. Find the inverse laplace transform of ^{s}⁄_{(s2 + 4)2}

a) ^{1}⁄_{4} sin(2t)

b) ^{t2}⁄_{4} sin(2t)

c) ^{t}⁄_{4} sin(2t)

d) ^{t}⁄_{4} sin(2t^{2})

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6. Final value theorem states that

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7. Initial value theorem states that

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8. Find the value of x(∞) if

a) 5

b) 4

c) ^{12}⁄_{20}

d) 2

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^{12}⁄

_{20}

9. Find the value of x(0) if

a) 5

b) 4

c) ^{12}⁄_{20}

d) 2

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10. Find the inverse lapace of

a) ^{1}⁄_{3} e^{t} [Cos(t) – Cos(2t)].

b) ^{1}⁄_{3} e^{-t} [Cos(t) + Cos(2t)].

c) ^{1}⁄_{3} e^{t} [Cos(t) + Cos(2t)].

d) ^{1}⁄_{3} e^{-t} [Cos(t) – Cos(2t)].

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11. Find the inverse laplace transform of

a) -e^{-t + 1} + e^{t – 1}

b)-e^{-t + 1} – e^{t + 1}

c) -e^{-t + 1} + e^{t + 1}

d)-e^{-t + 1} – e^{t – 1}

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12. Find the inverse lapalce transform of ^{1}⁄_{s(s-1)(s2+1)}

a) ^{1}⁄_{2} e^{-t} + ^{1}⁄_{2} Sin(-t) – ^{1}⁄_{2} Cos(-t)

b) ^{1}⁄_{2} e^{t} + ^{1}⁄_{2} Sin(t) – ^{1}⁄_{2} Cos(t)

c) ^{1}⁄_{2} e^{t} + ^{1}⁄_{2} Sin(t) + ^{1}⁄_{2} Cos(t)

d) ^{1}⁄_{2} e^{t} – ^{1}⁄_{2} Sin(t) – ^{1}⁄_{2} Cos(t)

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## Discrete Mathematics MCQ Set 2

1. Consider the f(x, y) = x^{2} + y^{2} – 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)

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_{x}= 2x and f

_{y}= 2y There is no a here. Thus, independent of a.

2. The critical point exist for the function f(x, y) = x^{n} + x^{n-1} y +……+y^{n} at (0,0)

a) True

b) False

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3. f(x, y) = sin(x).cos(y) Which of the following is a critical point

a) (^{Π}⁄_{4}, ^{Π}⁄_{4})

b) (- ^{Π}⁄_{4}, ^{Π}⁄_{4})

c) (0, ^{Π}⁄_{2})

d) (0, 0)

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_{x}= cos(x).cos(y) = 0 f

_{y}= – sin(x).sin(y) →(x, y) = (0,

^{Π}⁄

_{2}).

4. The point (0,0) in the domain of f(x, y) = sin(xy) is a point of

a) Saddle

b) Minima

c) Maxima

d) Constant

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_{xx}= -y

^{2}.sin(xy) f

_{yy}= -x

^{2}.sin(xy) f

_{xy}= -yx.sin(xy) Observe that f

_{xx}. f

_{yy}– (f

_{xy})

^{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

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_{xx}= -y

^{2}.sin(xy) f

_{yy}= -x

^{2}.sin(xy) f

_{xy}= -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

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_{x}= f1x g1x (x) d

_{y}= 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

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_{y}= -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

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## 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

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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

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^{∂f}⁄

_{∂x}= 0 &

^{∂f}⁄

_{∂y}= 0.

3. For function f(x,y) to have minimum value at (a,b) value,

a) rt – s^{2}>0 and r<0

b) rt – s^{2}>0 and r>0

c) rt – s^{2}<0 and r<0

d) rt – s^{2}>0 and r>0

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rt – s^{2}>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 – s^{2}>0 and r<0

b) rt – s^{2}>0 and r>0

c) rt – s^{2}<0 and r<0

d) rt – s^{2}>0 and r>0

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rt – s^{2}>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 – s^{2}>0

b) rt – s^{2}<0

c) rt – s^{2} = 0

d) rt – s^{2} ≠ 0

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^{2}< 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)=x^{2} + y^{2} + 6x + 12

a) 3

b) 3

c) -9

d) 9

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^{2}+ y

^{2}+ 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 – s^{2} = 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) = y^{2} + 4xy + 3x^{2} + x^{3}

a) minimum at (0,0)

b) maximum at (0,0)

c) minimum at (2/3, -4/3)

d) maximum at (2/3, -4/3)

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^{2}+ 4xy + 3x

^{2}+ x

^{3}Now,

^{∂f}⁄

_{∂x}= 4y + 6x + 3x

^{2}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 – s

^{2}= 12 – 16<0,hence it has no extremum at this point.

Now,at (^{2}⁄_{3},-^{4}⁄_{3}) r= ^{∂2f}⁄_{∂x2}= 6 + 6x = 10>0 and t= ^{∂2f}⁄_{∂y2} =2>0 and s= ^{∂2f}⁄_{∂x∂y}=4
hence, rt – s^{2} = 20 – 16 > 0 and r>0, hence it has minimum at this point.(^{2}⁄_{3}, –^{4}⁄_{3}).

8.Find the minimum value of xy+a^{3} (^{1}⁄_{x} + ^{1}⁄_{y})

a) 3a^{2}

b) a^{2}

c) a

d) 1

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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

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f = xy + y(120-x-y) + x(120-x-y) = 120x + 120y – xy – x^{2} – y^{2}

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 – s^{2} = 5 > 0
since, r<0 and rt – s^{2} > 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√3}⁄_{8}

b) ^{3√4}⁄_{8}

c) –^{3√3}⁄_{8}

d) ^{π}⁄_{8}

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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-s^{2}= ^{9}⁄_{4∂x}>0

hence, r<0 andrt-s^{2}>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-s^{2}= ^{9}⁄_{4∂x}>0
And this value is ^{3√3}⁄_{8} hence, r>0 and rt-s^{2} >0 hence,f(x,y)or f(A,B)have minimum value at (60,60)
and this value is –^{3√3}⁄_{8}.

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

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## 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

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^{0}⁄

_{0}=

^{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.

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3. If f(x) = x^{2} – 3x + 2 and g(x) = x^{3} – x^{2} + x – 1 than find value of

a) 0.5

b) 1

c) -.5

d) -1

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4. If f(x) = Tan(x)-1 and g(x) = Sin(x) – Cos(x) than find value of lim_{x → π⁄4} ^{f(x)}⁄_{g(x)}

a) -√2

b) √(-2)

c) √2

d) √3

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5. If f(x) = Tan(x) and g(x) = e^{x} – 1 than find value of lim_{x → 0} ^{f(x)}⁄_{g(x)}

a) 1

b) 0

c) -1

d) 2

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6. Find the value of

a) e-1

b) e

c) e + 1

d) 1

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7. If f(x) = x^{3} + 3x^{2} + Sin(x) and g(x) = e^{x} – 1 than find value of

a) e^{6e}

b) e^{(e/6)}

c) e^{6}

d) e^{(6/e)}

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8. If f(x) = Sin(x) and g(x) = x than find value of lim_{x → 0} ^{f(x)}⁄_{g(x)}

a) -1

b) 0

c) 1

d) 2

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_{x → 0}

^{f(x)}⁄

_{g(x)}= lim

_{x → 0}

^{sin(x)}⁄

_{x}=

^{0}⁄

_{0}(Indeterminate forms) By L’Hospital rule lim

_{x → 0}

^{cos(x)}⁄

_{1}= 1.

9. If f(x) = sin(x)cos(x) and g(x) = x^{2} than find value of lim_{x → 0} ^{f(x)}⁄_{g(x)}

a) 2

b) 0

c) -1

d) Cannot be found

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10. If f(x) = e^{x} + xcos(x) and g(x) = Sin(x) than find value of lim_{x → 0} ^{f(x)}⁄_{g(x)}

a) 2

b) 1

c) 3

d) 4

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11. Find the value of is, where {x} is the fractional part of x.

a) ^{2}⁄_{e}

b) ^{1}⁄_{e}

c) 0

d) –^{1}⁄_{e}

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## 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

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2. The shape of the Normal Curve is

a) Bell Shaped

b) Flat

c) Circular

d) Spiked

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3. Normal Distribution is symmetric is about

a) Variance

b) Mean

c) Standard deviation

d) Covariance

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4. For a standard normal variate, the value of mean is

a) ∞

b) 1

c) 0

d) not defined

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5. The area under a standard normal curve is

a) 0

b) 1

c) ∞

d) not defined

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6. The standard normal curve is symmetric about the value

a) 0.5

b) 1

c) ∞

d) 0

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7. For a standard normal variate, the value of Standard Deviation is

a) 0

b) 1

c) ∞

d) not defined

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8. Normal Distribution is also known as

a) Cauchy’s Distribution

b) Laplacian Distribution

c) Gaussian Distribution

d) Lagrangian Distribution

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9. Skewness of Normal distribution is

a) Negative

b) Positive

c) 0

d) Undefined

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10. For a normal distribution its mean, median, mode are equal

a) True

b) False

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11. In Normal distribution, the highest value of ordinate occurs at

a) Mean

b) Variance

c) Extremes

d) Same value occurs at all points

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12. The shape of the normal curve depends on its

a) Mean deviation

b) Standard deviation

c) Quartile deviation

d) Correlation

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13.The value of constant ‘e’ appearing in normal distribution is

a) 2.5185

b) 2.7836

c) 2.1783

d) 2.7183

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14. In Standard normal distribution, the value of mode is

a) 2

b) 1

c) 0

d) Not fixed

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15. In Standard normal distribution, the value of median is

a) 1

b) 0

c) 2

d) Not fixed