Select Page
Generic selectors
Exact matches only
Search in title
Search in content
Search in posts
Search in pages
Filter by Categories
nmims post
Objective Type Set
Online MCQ Assignment
Question Solution
Solved Question
Uncategorized

Discrete Mathematics MCQ Set 1

1. Find the value of ∫∫xyex + y dxdy
a) yey (xex-ex )
b) (yey-ey)(xex-ex)
c) (yey-ey )xex
d) (yey-ey )(xex+ex )

Answer: b Add constant automatically [Reason:] Given, ∫∫xyex + y dxdy ∫∫xyex ey dxdy= ∫yey dy∫xex dx=(yey-ey)(xex-ex).

2. Find the value of ∫∫ xx2 + y2 dxdy
a) [ytan(-1) (y)- 12 ln⁡(1+y2 )].
b) x [ytan(-1) (y)- 12 ln⁡(1+y2 )].
c) y [xtan(-1) (x)- 12 ln⁡(1+x2 )].
d) x [ytan(-1) (y)- 12 ln⁡(1+y2 )].

Answer: d [Reason:] Add constant automatically Putting, x = tan(z), We get, dz = sec2⁡(z)dz,

x∫ zsec2 (z)dz

By integration by parts,

x ∫ zsec2 (z)dz=x[ztan(z)-log⁡(sec⁡(z) )]= x[ytan(-1) (y)- 12 ln⁡(1+y2 )].

3. Find the ∫∫x3 y3 sin⁡(x)sin⁡(y) dxdy
a) (x3 Cos(x) + 3x2 Sin(x) + 6xCos(x)-6Sin(x))(y3 Cos(y) + 3[y2 Sin(y) – 2[-yCos(y) + Sin(y)]])
b) (-x3 Cos(x) – 3x2 Sin(x) – 6xCos(x)-6Sin(x))(-y3 Cos(y) + 3[y2 Sin(y) – 2[-yCos(y) + Sin(y)]])
c) (-x3 Cos(x) + 3x2 Sin(x) + 6xCos(x)-6Sin(x))(-y3 Cos(y) + 3y2 Sin(y) + 6yCos(y) – 6Sin(y))
d) (–x3 Cos(x) + 6xCos(x) – 6Sin(x))(-y3 Cos(y))

Answer: c [Reason:] Add constant automatically ∫x3 Sin(x)dx = -x3 Cos(x) + 3∫x2 Cos(x)dx

∫x2 Cos(x)dx = x2 Sin(x) – 2∫xSin(x)dx

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

=> ∫x3 Sin(x)dx = -x3 Cos(x) + 3[x2 Sin(x) – 2[-xCos(x) + Sin(x)]]

=> ∫x3 Sin(x)dx = -x3 Cos(x) + 3x2 Sin(x) + 6xCos(x) – 6Sin(x)

and,∫y3 Sin(y)dy = -y3 Cos(x) + 3∫y2 Cos(y)dy ∫y2 Cos(y)dy = y2 Sin(y) – 2∫ySin(y)dy

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

=> ∫y3 Sin(y)dy = -y3 Cos(y) + 3[y2 Sin(y) – 2[-yCos(y) + Sin(y)]]

=> ∫y3 Sin(y)dy = -y3 Cos(y) + 3y2 Sin(y) + 6yCos(y) – 6Sin(y)

Hence,∫∫x3 y3 sin⁡(x) sin⁡(y) dxdy = (∫x3 Sin(x)dx)(∫y3 Sin(y)dy) = (-x3 Cos(x) + 3x2 Sin(x)+6xCos(x) – 6Sin(x))(-y3 Cos(y) + 3y2 Sin(y) + 6yCos(y) – 6Sin(y)).

4. Find the integration of
a) ax22x530
b) ax22x36
c) ax22
d) ax48x36

5. Find the value of ∫∫xy7 Cos(x)Cos(y) dxdy
a) (7y6 Cos(y) + 42y5 Sin(y) + 210y4 Cos(y) + 840y3 Sin(y) + 2520y2 Cos(y) + 5040ySin(y) + 5040Cos(y))(7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x))
b) (y7 Sin(y) + 7y6 Cos(y) + 42y5 Sin(y) + 210y4 Cos(y) + 840y3 Sin(y) + 2520y2 Cos(y) + 5040ySin(y) + 5040Cos(y))(x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x))
c) (y7 Sin(y) + 42y5 Sin(y) + 210y4 Cos(y) + 840y3 Sin(y) + 2520y2 Cos(y) + 5040ySin(y) + 5040Cos(y))(x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x))
d) (y7 Sin(y) + 7y6 Cos(y) + 42y5 Sin(y) + 210y4 Cos(y) + 840y3 Sin(y) + 2520y2 Cos(y) + 5040ySin(y) + 5040Cos(y))(x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x))

Answer: d [Reason:] Add constant automatically Let, u = x7 and v=Cos(x), ∫x7 Cos(x) dx=x7 Sin(x)+7x6 Cos(x)+42x5 Sin(x)+210x4 Cos(x)+840x3 Sin(x)+2520x2 Cos(x)+5040xSin(x)+5040Cos(x)

Similarly,

∫y7 Cos(y) dy=y7 Sin(y)+7y6 Cos(y)+42y5 Sin(y)+210y4 Cos(y)+840y3 Sin(y)+2520y2 Cos(y)+5040ySin(y)+5040Cos(y)

Now,

∫∫xy7 Cos(x)Cos(y) dxdy=∫y7 Cos(y) dy∫x7 Cos(x) dx=(y7 Sin(y)+7y6 Cos(y)+42y5 Sin(y)+210y4 Cos(y)+840y3 Sin(y)+2520y2 Cos(y)+5040ySin(y)+5040Cos(y))(x7 Sin(x)+7x6 Cos(x)+42x5 Sin(x)+210x4 Cos(x)+840x3 Sin(x)+2520x2 Cos(x)+5040xSin(x)+5040Cos(x)).

6. Find the integration of ∫∫0x x2 + y2 dxdy
a) x46
b) y
c) 2x33y
d) 1

Answer: c [Reason:] Add constant automatically Given,f(x)=∫∫0x x2 + y2 dxdy= ∫(x33 + x33)dxdy = 2x33y.

8. Find the value of
a) 16946
b) 8945
c) 16936
d) 16945

9. Find the area inside function (2x3 + 5 x2 – 4)x2 from x = 1 to a
a) a22 + 5a – 4ln(a)
b) a22 + 5a – 4ln(a) – 112
c) a22 + 4ln(a) – 112
d) a22 + 5a – 112

Answer: b [Reason:] Add constant automatically Given, f(x) = (2x3 + 5 x2 – 4)x2 ,

Integrating it we get, F(x) = x22 + 5x – 4ln(x)

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

F(a) – F(1) = a22 + 5a – 4ln(a) – 12 – 5 = a22 + 5a – 4ln(a) – 112.

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

Answer: d [Reason:]Find fx = 10

10 ≠ 0

Hence, no points exist.

2. For function f(x, y) = sin-1(x2 + y2) 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(x2 + y2). 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)

Answer: b [Reason:] The function takes constant values along a circle(observe the function) But it is composed of arc sine function. Hence, we will have critical points at equal intervals Hence, there are infinite such points.

3. Consider the circular region x2 + y2 = 81, What is the maximum value of the function
f(x, y) = x6 + y2(3x4 + 1) + x2.(3y4 + 1) + y6
a) 90
b) 80
c) 81 + 813
d) 100

Answer: c [Reason:] Rewrite the function as f(x, y) = x2 + y2 + (x2 + y2)3 Put x2 + y2 = 81 = 81 + 813.

4. What is the maximum value of the function
f(x, y) = x2(1 + 3y) + x3 + y3 + y2(1 + 3x) + 2xy over the region x=0; y=0; x + y=1
a) 0
b) -1
c) Has no maximum value
d) 2

Answer: d [Reason:] Rewrite the function as f(x, y) = (x + y)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)

Answer: a [Reason:] If the Hessian matrix is zero then the second derivative test fails and nothing can be said about the crtical points.

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)

Answer: d [Reason:]Rewrite the function as f(x, y) = cos(x + 2y) * (sin(x).cos(2y) + cos(x).sin(2y)) f(x, y) = cos(x + 2y).sin(x + 2y) Put x+2y = 3 = sin(3).cos(3).

7. Find the minimum value of the function f(x, y) = x2 + y2 +199 over the real domain
a) 12
b) 13
c) 0
d) 199

Answer: d [Reason:] Find fx = 2x fy = 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 + 4x2y2 in the region
x=0; y=0; 2x + y = 2
a) 1
b) 0
c) 100
d) 10

Answer: d [Reason:] Differentiating we have fx = 3y + 8xy2 = 0 fy = 3x + 8x2y 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) = x3 + xy2 + 901 satisfies the Eulers theorem
a) True
b) False

Answer: b [Reason:] The function is not homogenous and hence does not satisfy the condition posed by eulers theorem.

2. find the value of fy at (x,y) = (0,1)
a) 101
b) -96
c) 210
d) 0

Answer: b [Reason:] Using Euler theorem xfx + yfy = n f(x, y) Substituting x = 0; n=-96 and y = 1 we have fy = -96. f(0, 1) = -96.(1⁄1) = – 96.

3. A non-polynomial function can never agree with eulers theorem
a) True
b) false

Answer: b [Reason:] Counter example is the function .

4. Find the value of fx at (1,0)
a) 23
b) 16
c) 17(sin(2) + cos(1⁄2) )
d) 90

Answer: c [Reason:] Using Eulers theorem we have xfx + yfy = nf(x, y)

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

fx = 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

Answer: c [Reason:] Using Euler theorem we have xfx + yfy = nf(x, y) At critical points fx = fy = 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

Answer: d [Reason:] Substituting fx = fy = 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

Answer: c [Reason:] Eulers theorem is nothing but the linear combination asked here, The degree of the homogenous function can be a real number. Hence, the value is integral multiple of real number.

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)x2 + xy
b) f(x, y) = x2 + y3
c) f(x, y) = x2y2 + x3y3
d) not possible by any analytical function

Answer: b [Reason:]Given that the heat is same along lines we need to choose a homogenous function. Checking options we get that only option satisfies condition for homogenity.

9. f(x, y) = sin(y⁄x)x3 + x2y find the value of fx + fy at (x,y)=(4,4)
a) 0
b) 78
c) 42 . 3(sin(1) + 1)
d) -12

Answer: c [Reason:] Using Euler theorem we have xfx + yfy = nf(x, y)

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

4fx + 4fy = 3f(4, 4) = 3⁄4(43 . sin(1) + 43)

= 42 . 3(sin(1) + 1).

Discrete Mathematics MCQ Set 4

1. Let f(x) = ex sin(x2) ⁄ x Then the value of the fifth derivative at x = 0 is given by
a) 25
b) 21
c) 0
d) 5

Answer:b [Reason:] First expanding sin(x2) x into a Taylor series we have

2. Let f(x) = eex assuming all the nth derivatives at x =0 to be 1 the value of the (n + 1)th derivative can be written as
a) e – 1 + 2n
b) 0
c) 1
d) None

Answer: a [Reason:] Assume y = f(x) Taking ln(x) on both sides The function has to be written in the form ln(y) = ey Now computing the first derivative yields y(1) = y * ex Now applying the Leibniz rule up to nth derivative we have

3. Let f(x) = √sin(x) and let yn denote the nth 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

Answer:a [Reason:] Assume y = f(x) Rewriting the function as y2 = sin(x) Now applying Leibniz rule up to the sixth derivative we have (y2)(6) = c06 y(6) y + c16 y(5) y(1) + ………+ c66 y(6) y

(y2)(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) π2
c) 45
d) 4

Answer: a [Reason:] First convert the function sinh(x)⁄x into its Taylor series expansion

5. The third derivative of f(x) = cos(x)sinh(x) ⁄ x at x = 0 is
a) 0
b) π32
c) (π)2
d) cos(1)sinh(1)

Answer: a [Reason:] Assume y = f(x) Rewriting the part sinh(x)⁄x as infinite series we have Now substituting x = 0 we have

y(3) = 0.

6. Let f(x) = (x2 + x + 1)sinh(x) the (1097)th derivative at x = 0 is
a) 1097
b) 1096
c) 0
d) 1202313

Answer: d [Reason:] Expanding sinh(x) into a taylor series we have

7. The 7th derivative of f(x) = (x3 + x2 + x + 1) sinh(x) at x = 0 is given by
a) 43
b) 7
c) 0
d) 34

Answer: a [Reason:] Expanding sinh(x) into a Taylor series we have

8. The (1071729)th derivative of f(x) = (x6 + x4 + x2) cosh(x) at x = 0 is given by
a) 0
b) 1071
c) 1729
d) ∞

Answer: a [Reason:] Expanding cosh(x) into a Taylor series we have Observe again, that the derivative in question is odd, and hence, only the odd powered terms contribute to the derivative at x = 0

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) = (x2 + 1)tan-1 (x) at x = 0 is
a) 0
b) 1729
c) 1728
d) ∞

Answer: a [Reason:] Expanding the tan-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 x3 + y3 – 3xy + y2 = 0 ?

2. x3 Sin(y) + Cos(x) y3 = 0 , its differentiation is

3. Find the differentiation of x4 + y4 = 0
a) – x3y4
b) – x4y3
c) – x3y3
d) x3y3

Answer: c [Reason:] x4 + y4 = 0

4x3 + 4y3 dydx = 0

dydx = – x3y3

dydx = Sec2 (x)Sec(x) ex + Sec2 (x)Tan(x) ex + ex Tan(x)Sec(x)

dydx = Sec2 (x) ex [Sec(x)+Tan(x)] + ex Tan(x)Sec(x)

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

Answer: c [Reason:] xSin(x) + ayCos(x) + Tan(y) =0 Differentiation of above eqn. is

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

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.

Answer: b [Reason:] Implicit functions are those functions, Which can not be solved for a single variable. For ex, f(x,y) = x3 +y3 -3xy = 0.

7. Evaluate y44 + 3xy3 + 6x2 y2 – 7y + 8 = 0.

Answer: b [Reason:] y44 + 3xy3 + 6x2 y2 – 7y + 8 = 0. Differentiating it we get

8. If Sin(y)=Sin(-1) (y) then
a) (1-y2 )(1 – Cos2 y) = 1
b) (1-y2 )(1 – Sin2 y) = 1
c) (1-y2 )(1 – Siny)=1
d) (1-y2 )(1 – Cosy)=1

Answer: b [Reason:] Sin(y)=Sin(-1) (y) Differntiating both sides

9. If Cos(y)=Cos(-1) (y) then
a) (1 – y2 )(1 – Cos2 (y))=1
b) (1 – y2 )(1 – Cos(y))=1
c) (1 – y2 )(1 – Sin2 (y))=1
d) (1 – y2 )(1 – Sin(y))=1

Answer: a [Reason:] Cos(y)=Cos(-1) (y) Differentiating both sides -Sin(y) = -1/√(1-y2 ) (1 – y2 )(1 – Cos2 (y)) = 1.

10. If y2 + xy + x2 – 2x = 0 then d2ydx2 = ?

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

Answer: b [Reason:] Given,v(t)=kv2 (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)

Answer: c [Reason:] z(x,y) = 2Sin(x) + Cos(y)Sin(x) Hence,

13. If the car is having a displace from point 1 to point 2 in t sec which is given by equation y(x) = x2 + 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.