## Discrete Mathematics MCQ Set 1

1. Find the value of ∫∫xye^{x + y} dxdy

a) ye^{y} (xe^{x}-e^{x} )

b) (ye^{y}-e^{y})(xe^{x}-e^{x})

c) (ye^{y}-e^{y} )xe^{x}

d) (ye^{y}-e^{y} )(xe^{x}+e^{x} )

### View Answer

^{x + y}dxdy ∫∫xye

^{x}e

^{y}dxdy= ∫ye

^{y}dy∫xe

^{x}dx=(ye

^{y}-e

^{y})(xe

^{x}-e

^{x}).

2. Find the value of ∫∫ ^{x}⁄_{x2 + y2} dxdy

a) [ytan^{(-1)} (y)- ^{1}⁄_{2} ln(1+y^{2} )].

b) x [ytan^{(-1)} (y)- ^{1}⁄_{2} ln(1+y^{2} )].

c) y [xtan^{(-1)} (x)- ^{1}⁄_{2} ln(1+x^{2} )].

d) x [ytan^{(-1)} (y)- ^{1}⁄_{2} ln(1+y^{2} )].

### View Answer

^{2}(z)dz,

x∫ zsec^{2} (z)dz

By integration by parts,

x ∫ zsec^{2} (z)dz=x[ztan(z)-log(sec(z) )]= x[ytan^{(-1)} (y)- ^{1}⁄_{2} ln(1+y^{2} )].

3. Find the ∫∫x^{3} y^{3} sin(x)sin(y) dxdy

a) (x^{3} Cos(x) + 3x^{2} Sin(x) + 6xCos(x)-6Sin(x))(y^{3} Cos(y) + 3[y^{2} Sin(y) – 2[-yCos(y) + Sin(y)]])

b) (-x^{3} Cos(x) – 3x^{2} Sin(x) – 6xCos(x)-6Sin(x))(-y^{3} Cos(y) + 3[y^{2} Sin(y) – 2[-yCos(y) + Sin(y)]])

c) (-x^{3} Cos(x) + 3x^{2} Sin(x) + 6xCos(x)-6Sin(x))(-y^{3} Cos(y) + 3y^{2} Sin(y) + 6yCos(y) – 6Sin(y))

d) (–x^{3} Cos(x) + 6xCos(x) – 6Sin(x))(-y^{3} Cos(y))

### View Answer

^{3}Sin(x)dx = -x

^{3}Cos(x) + 3∫x

^{2}Cos(x)dx

∫x^{2} Cos(x)dx = x^{2} Sin(x) – 2∫xSin(x)dx

∫xSin(x)dx = -xCos(x) + ∫Cos(x)dx = -xCos(x) + Sin(x)

=> ∫x^{3} Sin(x)dx = -x^{3} Cos(x) + 3[x^{2} Sin(x) – 2[-xCos(x) + Sin(x)]]

=> ∫x^{3} Sin(x)dx = -x^{3} Cos(x) + 3x^{2} Sin(x) + 6xCos(x) – 6Sin(x)

and,∫y^{3} Sin(y)dy = -y^{3} Cos(x) + 3∫y^{2} Cos(y)dy
∫y^{2} Cos(y)dy = y^{2} Sin(y) – 2∫ySin(y)dy

∫ySin(y)dy = -yCos(y) + ∫Cos(y)dy = -yCos(y) + Sin(y)

=> ∫y^{3} Sin(y)dy = -y^{3} Cos(y) + 3[y^{2} Sin(y) – 2[-yCos(y) + Sin(y)]]

=> ∫y^{3} Sin(y)dy = -y^{3} Cos(y) + 3y^{2} Sin(y) + 6yCos(y) – 6Sin(y)

Hence,∫∫x^{3} y^{3} sin(x) sin(y) dxdy = (∫x^{3} Sin(x)dx)(∫y^{3} Sin(y)dy) = (-x^{3} Cos(x) + 3x^{2} Sin(x)+6xCos(x) – 6Sin(x))(-y^{3} Cos(y) + 3y^{2} Sin(y) + 6yCos(y) – 6Sin(y)).

4. Find the integration of

a) ^{ax2}⁄_{2} – ^{x5}⁄_{30}

b) ^{ax2}⁄_{2} – ^{x3}⁄_{6}

c) ^{ax2}⁄_{2}

d) ^{ax4}⁄_{8} – ^{x3}⁄_{6}

### View Answer

5. Find the value of ∫∫xy^{7} Cos(x)Cos(y) dxdy

a) (7y^{6} Cos(y) + 42y^{5} Sin(y) + 210y^{4} Cos(y) + 840y^{3} Sin(y) + 2520y^{2} Cos(y) + 5040ySin(y) + 5040Cos(y))(7x^{6} Cos(x) + 42x^{5} Sin(x) + 210x^{4} Cos(x) + 840x^{3} Sin(x) + 2520x^{2} Cos(x) + 5040xSin(x) + 5040Cos(x))

b) (y^{7} Sin(y) + 7y^{6} Cos(y) + 42y^{5} Sin(y) + 210y^{4} Cos(y) + 840y^{3} Sin(y) + 2520y^{2} Cos(y) + 5040ySin(y) + 5040Cos(y))(x^{7} Sin(x) + 7x^{6} Cos(x) + 42x^{5} Sin(x) + 210x^{4} Cos(x) + 840x^{3} Sin(x) + 2520x^{2} Cos(x) + 5040xSin(x) + 5040Cos(x))

c) (y^{7} Sin(y) + 42y^{5} Sin(y) + 210y^{4} Cos(y) + 840y^{3} Sin(y) + 2520y^{2} Cos(y) + 5040ySin(y) + 5040Cos(y))(x^{7} Sin(x) + 7x^{6} Cos(x) + 42x^{5} Sin(x) + 210x^{4} Cos(x) + 840x^{3} Sin(x) + 2520x^{2} Cos(x) + 5040xSin(x) + 5040Cos(x))

d) (y^{7} Sin(y) + 7y^{6} Cos(y) + 42y^{5} Sin(y) + 210y^{4} Cos(y) + 840y^{3} Sin(y) + 2520y^{2} Cos(y) + 5040ySin(y) + 5040Cos(y))(x^{7} Sin(x) + 7x^{6} Cos(x) + 42x^{5} Sin(x) + 210x^{4} Cos(x) + 840x^{3} Sin(x) + 2520x^{2} Cos(x) + 5040xSin(x) + 5040Cos(x))

### View Answer

^{7}and v=Cos(x), ∫x

^{7}Cos(x) dx=x

^{7}Sin(x)+7x

^{6}Cos(x)+42x

^{5}Sin(x)+210x

^{4}Cos(x)+840x

^{3}Sin(x)+2520x

^{2}Cos(x)+5040xSin(x)+5040Cos(x)

Similarly,

∫y^{7} Cos(y) dy=y^{7} Sin(y)+7y^{6} Cos(y)+42y^{5} Sin(y)+210y^{4} Cos(y)+840y^{3} Sin(y)+2520y^{2} Cos(y)+5040ySin(y)+5040Cos(y)

Now,

∫∫xy^{7} Cos(x)Cos(y) dxdy=∫y^{7} Cos(y) dy∫x^{7} Cos(x) dx=(y^{7} Sin(y)+7y^{6} Cos(y)+42y^{5} Sin(y)+210y^{4} Cos(y)+840y^{3} Sin(y)+2520y^{2} Cos(y)+5040ySin(y)+5040Cos(y))(x^{7} Sin(x)+7x^{6} Cos(x)+42x^{5} Sin(x)+210x^{4} Cos(x)+840x^{3} Sin(x)+2520x^{2} Cos(x)+5040xSin(x)+5040Cos(x)).

6. Find the integration of ∫∫_{0}^{x} x^{2} + y^{2} dxdy

a) ^{x4}⁄_{6}

b) y

c) ^{2x3}⁄_{3}y

d) 1

### View Answer

_{0}

^{x}x

^{2}+ y

^{2}dxdy= ∫(

^{x3}⁄

_{3}+

^{x3}⁄

_{3})dxdy =

^{2x3}⁄

_{3}y.

### View Answer

8. Find the value of

a) ^{16}⁄_{946}

b) ^{8}⁄_{945}

c) ^{16}⁄_{936}

d) ^{16}⁄_{945}

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9. Find the area inside function ^{(2x3 + 5 x2 – 4)}⁄_{x2} from x = 1 to a

a) ^{a2}⁄_{2} + 5a – 4ln(a)

b) ^{a2}⁄_{2} + 5a – 4ln(a) – ^{11}⁄_{2}

c) ^{a2}⁄_{2} + 4ln(a) – ^{11}⁄_{2}

d) ^{a2}⁄_{2} + 5a – ^{11}⁄_{2}

### View Answer

^{(2x3 + 5 x2 – 4)}⁄

_{x2},

Integrating it we get, F(x) = ^{x2}⁄_{2} + 5x – 4ln(x)

Hence, area under, x = 1 to a, is

F(a) – F(1) = ^{a2}⁄_{2} + 5a – 4ln(a) – ^{1}⁄_{2} – 5 = ^{a2}⁄_{2} + 5a – 4ln(a) – ^{11}⁄_{2}.

### View Answer

## Discrete Mathematics MCQ Set 2

1. Find the critical points of the function

a) (0,0)

b) (0,-90)

c) (90, 0)

d) None exist

### View Answer

_{x}= 10

10 ≠ 0

Hence, no points exist.

2. For function f(x, y) = sin^{-1}(x^{2} + y^{2}) critical points are found. Now a new graph g(x, y) is formed by coupling graphs f(x, y) and f(x, y) = – sin^{-1}(x^{2} + y^{2}). What are the critical points of g(x, y)

a) (0,0)

b) There are infinite such points

c) Only positive (x, y) are critical points

d) (90,-90)

### View Answer

3. Consider the circular region x^{2} + y^{2} = 81, What is the maximum value of the function

f(x, y) = x^{6} + y^{2}(3x^{4} + 1) + x^{2}.(3y^{4} + 1) + y^{6}

a) 90

b) 80

c) 81 + 81^{3}

d) 100

### View Answer

^{2}+ y

^{2}+ (x

^{2}+ y

^{2})

^{3}Put x

^{2}+ y

^{2}= 81 = 81 + 81

^{3}.

4. What is the maximum value of the function

f(x, y) = x^{2}(1 + 3y) + x^{3} + y^{3} + y^{2}(1 + 3x) + 2xy over the region x=0; y=0; x + y=1

a) 0

b) -1

c) Has no maximum value

d) 2

### View Answer

^{2}+ (x + y)

^{3}Put x + y = 1 = 2.

5. If the Hessian matrix of a function is zero then the critical point is

a) It cannot be concluded

b) Always at Origin

c) Depends on Function

d) (100,100)

### View Answer

6. The maximum value of the function

f(x, y) = sin(x).cos(2y).cos(x + 2y) + sin(2y).cos(x + 2y).cos(x) in the region x=0; y=0; x+2y = 3

a) 90

b) cos(1)

c) sin(1).cos(1)

d) sin(3).cos(3)

### View Answer

7. Find the minimum value of the function f(x, y) = x^{2} + y^{2} +199 over the real domain

a) 12

b) 13

c) 0

d) 199

### View Answer

_{x}= 2x f

_{y}= 2y The critical point is x=0 y=0 (0,0) is the critical point Put it back into the function we get z = 0 + 0 + 199 = 199 is the required minimum value.

8. What is the maximum value of the function f(x, y) = 3xy + 4x^{2}y^{2} in the region

x=0; y=0; 2x + y = 2

a) 1

b) 0

c) 100

d) 10

### View Answer

_{x}= 3y + 8xy

^{2}= 0 f

_{y}= 3x + 8x

^{2}y x = 0 y = 0 (0,0) lies in the region Substitute x=0 f(0, y) = 0 Substitute y=0 f(x, 0) = 0 Substitute y = 2 – 2x is the maximum value

## Discrete Mathematics MCQ Set 3

1. f(x, y) = x^{3} + xy^{2} + 901 satisfies the Eulers theorem

a) True

b) False

### View Answer

2. find the value of f_{y} at (x,y) = (0,1)

a) 101

b) -96

c) 210

d) 0

### View Answer

_{x}+ yf

_{y}= n f(x, y) Substituting x = 0; n=-96 and y = 1 we have f

_{y}= -96. f(0, 1) = -96.(1⁄1) = – 96.

3. A non-polynomial function can never agree with eulers theorem

a) True

b) false

### View Answer

4. Find the value of f_{x} at (1,0)

a) 23

b) 16

c) 17(sin(2) + cos(1⁄2) )

d) 90

### View Answer

_{x}+ yf

_{y}= nf(x, y)

Substituting (x,y)=(1,0) we have

f_{x} = 17f(1, 0)
17 (sin(2) + cos(1⁄2)).

5. For a homogenous function if critical points exist the value at critical points is

a) 1

b) equal to its degree

c) 0

d) -1

### View Answer

_{x}+ yf

_{y}= nf(x, y) At critical points f

_{x}= f

_{y}= 0 f(a, b) = 0(a, b) → criticalpoints.

6. For homogenous function with no saddle points we must have the minimum value as

a) 90

b) 1

c) equal to degree

d) 0

### View Answer

_{x}= f

_{y}= 0 At critical points in euler theorem we have nf(a, b) = 0 ⇒ f(a, b) = 0(a, b) → criticalpoints.

7. For homogenous function the linear combination of rates of independent change along x and y axes is

a) Integral multiple of function value

b) no relation to function value

c) real multiple of function value

d) depends if the function is a polynomial

### View Answer

8. A foil is to be put as shield over a cake (circular) in a shape such that the heat is even along any diameter of the cake.

Given that the heat on cake is proportional to the height of foil over cake, the shape of the foil is given by

a) f(x, y) = sin(y⁄x)x^{2} + xy

b) f(x, y) = x^{2} + y^{3}

c) f(x, y) = x^{2}y^{2} + x^{3}y^{3}

d) not possible by any analytical function

### View Answer

9. f(x, y) = sin(y⁄x)x^{3} + x^{2}y find the value of f_{x} + f_{y} at (x,y)=(4,4)

a) 0

b) 78

c) 4^{2} . 3(sin(1) + 1)

d) -12

### View Answer

_{x}+ yf

_{y}= nf(x, y)

Substituting (x,y)=(4,4) we have

4f_{x} + 4f_{y} = 3f(4, 4) = 3⁄4(4^{3} . sin(1) + 4^{3})

= 4^{2} . 3(sin(1) + 1).

## Discrete Mathematics MCQ Set 4

1. Let f(x) = e^{x} sin(x^{2}) ⁄ x Then the value of the fifth derivative at x = 0 is given by

a) 25

b) 21

c) 0

d) 5

### View Answer

^{sin(x2) }⁄

_{x }into a Taylor series we have

2. Let f(x) = e^{ex} assuming all the n^{th} derivatives at x =0 to be 1 the value of the (n + 1)^{th} derivative can be written as

a) e – 1 + 2^{n}

b) 0

c) 1

d) None

### View Answer

^{y}Now computing the first derivative yields y

^{(1)}= y * e

^{x}Now applying the Leibniz rule up to n

^{th}derivative we have

3. Let f(x) = √sin(x) and let y^{n} denote the n^{th} derivative of f(x) at x = 0 then the value of the expression 12y^{(5)} y^{(1)} + 30 y^{(4)} y^{(2)} + 20 (y^{(3)})^{2} is given by

a) 0

b) 655

c) 999

d) 1729

### View Answer

^{2}= sin(x) Now applying Leibniz rule up to the sixth derivative we have (y

^{2})

^{(6)}= c

_{0}

^{6}y

^{(6)}y + c

_{1}

^{6}y

^{(5)}y(1) + ………+ c

_{6}

^{6}y

^{(6)}y

(y^{2})^{(6)} = 2 y^{(6)} y + 12 y^{(6)} y^{(1)} + 30 y^{(4)} y^{(2)} + 20 (y^{(3)})^{2}

(sin(x))^{(6)} = -sin(x)
Now substituting x = 0 and observing that y(0)= 0 we have
sin(0) = 0 = 12 y^{(6)} y^{(1)} + 30 y^{(4)} y^{(2)} + 20 (y^{(3)})^{2}.

4. The fourth derivative of f(x) = sin(x)sinh(x) ⁄ x at x = 0 is given by

a) 0

b) ^{π}⁄_{2}

c) 45

d) 4

### View Answer

5. The third derivative of f(x) = cos(x)sinh(x) ⁄ x at x = 0 is

a) 0

b) ^{π}⁄_{32}

c) (π)^{2}

d) cos(1)sinh(1)

### View Answer

y^{(3)} = 0.

6. Let f(x) = (x^{2} + x + 1)sinh(x) the (1097)^{th} derivative at x = 0 is

a) 1097

b) 1096

c) 0

d) 1202313

### View Answer

7. The 7^{th} derivative of f(x) = (x^{3} + x^{2} + x + 1) sinh(x) at x = 0 is given by

a) 43

b) 7

c) 0

d) 34

### View Answer

8. The (1071729)^{th} derivative of f(x) = (x^{6} + x^{4} + x^{2}) cosh(x) at x = 0 is given by

a) 0

b) 1071

c) 1729

d) ∞

### View Answer

Also note that, there are no odd powered terms and hence we can conclude that
The (1071729)^{th} derivative must be 0.

9. The (17291728)^{th} derivative of f(x) = (x^{2} + 1)tan^{-1} (x) at x = 0 is

a) 0

b) 1729

c) 1728

d) ∞

### View Answer

^{-1}(x) function into Taylor series we have Now observe that the derivative in question is even. Hence, even terms are the only ones to contribute to the derivative at x = 0

Also note that there are no even powered terms in the function. One can conclude that the (17291728)^{th} derivative at x = 0 is 0.

## Discrete Mathematics MCQ Set 5

1. Find the differentiation of x^{3} + y^{3} – 3xy + y^{2} = 0 ?

### View Answer

2. x^{3} Sin(y) + Cos(x) y^{3} = 0 , its differentiation is

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3. Find the differentiation of x^{4} + y^{4} = 0

a) – ^{x3}⁄_{y4}

b) – ^{x4}⁄_{y3}

c) – ^{x3}⁄_{y3}

d) ^{x3}⁄_{y3}

### View Answer

^{4}+ y

^{4}= 0

4x^{3} + 4y^{3} ^{dy}⁄_{dx} = 0

^{dy}⁄_{dx} = – ^{x3}⁄_{y3}

^{dy}⁄_{dx} = Sec^{2} (x)Sec(x) e^{x} + Sec^{2} (x)Tan(x) e^{x} + e^{x} Tan(x)Sec(x)

^{dy}⁄_{dx} = Sec^{2} (x) e^{x} [Sec(x)+Tan(x)] + e^{x} Tan(x)Sec(x)

4. Find differentiation of xSin(x) + ayCos(x) + Tan(y) = 0

### View Answer

5. Find the derivative of Tan(x) = Tan(y)

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6. Implicit functions are those functions

a) Which can be solved for a single variable

b) Which can not be solved for a single variable

c) Which can be eliminated to give zero

d) Which are rational in nature.

### View Answer

^{3}+y

^{3}-3xy = 0.

7. Evaluate y^{4}4 + 3xy^{3} + 6x^{2} y^{2} – 7y + 8 = 0.

### View Answer

^{4}4 + 3xy

^{3}+ 6x

^{2}y

^{2}– 7y + 8 = 0. Differentiating it we get

8. If Sin(y)=Sin^{(-1)} (y) then

a) (1-y^{2} )(1 – Cos^{2} y) = 1

b) (1-y^{2} )(1 – Sin^{2} y) = 1

c) (1-y^{2} )(1 – Siny)=1

d) (1-y^{2} )(1 – Cosy)=1

### View Answer

^{(-1)}(y) Differntiating both sides

9. If Cos(y)=Cos^{(-1)} (y) then

a) (1 – y^{2} )(1 – Cos^{2} (y))=1

b) (1 – y^{2} )(1 – Cos(y))=1

c) (1 – y^{2} )(1 – Sin^{2} (y))=1

d) (1 – y^{2} )(1 – Sin(y))=1

### View Answer

^{(-1)}(y) Differentiating both sides -Sin(y) = -1/√(1-y

^{2}) (1 – y

^{2})(1 – Cos

^{2}(y)) = 1.

10. If y^{2} + xy + x^{2} – 2x = 0 then ^{d2y}⁄_{dx2} = ?

### View Answer

11. If the velocity of car at time t(sec) is directly proportional to the square of its velocity at time (t-1)(sec). Then find the ratio of acceleration at t=10sec to 9sec if proportionality constant is k=10 sec/mt and velocity at t=9sec is 10 mt/sec

a) 100

b) 200

c) 150

d) 250

### View Answer

^{2}(t-1) Differentiating w.r.t time we get

^{dv(t)}⁄

_{dt}= 2kv(t-1)

^{dv(t – 1)}⁄

_{dt}a(t) = 2*10*10 a(t-1)

^{a(t)}⁄

_{a(t – 1)}= 200.

12. If z(x,y) = 2Sin(x)+Cos(y)Sin(x) find ^{d2z(xy)}⁄_{dxdy}= ?

a) –Cos(y)Cos(x)

b) -Sin(y)Sin(x)

c) –Sin(y)Cos(x)

d) -Cos(y)Sin(x)

### View Answer

13. If the car is having a displace from point 1 to point 2 in t sec which is given by equation y(x) = x^{2} + x + 1. Then,

a) Car is moving with constant acceleration

b) Car is moving with constant velocity.

c) Neither acceleration nor velocity is constant.

d) Both aceleration and velocity is contant.

### View Answer

^{2}+ x + 1 Velocity is , v =

^{dy}⁄

_{dx}= 2x + 1 (not constant) Acceleration is a =

^{dy}⁄

_{dx}= 2 (constant).