Discrete Mathematics MCQ Number 00944

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 )

Answer

Answer: b
Add constant automatically [Reason:]
Given, ∫∫xye^{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+y² )].
b) x [ytan^(-1) (y)- ¹⁄₂ ln⁡(1+y² )].
c) y [xtan^(-1) (x)- ¹⁄₂ ln⁡(1+x² )].
d) x [ytan^(-1) (y)- ¹⁄₂ ln⁡(1+y² )].

Answer

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)- ¹⁄₂ 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

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) ^ax2⁄₂ – ^x5⁄₃₀
b) ^ax2⁄₂ – ^x3⁄₆
c) ^ax2⁄₂
d) ^ax4⁄₈ – ^x3⁄₆

Answer

Answer: b [Reason:] Add constant automatically
.

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

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 ∫∫₀^x x² + y² dxdy
a) ^x4⁄₆
b) y
c) ^2x3⁄₃y
d) 1

Answer

Answer: c [Reason:] Add constant automatically
Given,f(x)=∫∫₀^x x² + y² dxdy= ∫(^x3⁄₃ + ^x3⁄₃)dxdy = ^2x3⁄₃y.

Answer

Answer: c [Reason:]

8. Find the value of
a) ¹⁶⁄₉₄₆
b) ⁸⁄₉₄₅
c) ¹⁶⁄₉₃₆
d) ¹⁶⁄₉₄₅

Answer

Answer: d [Reason:]

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

Answer

Answer: b [Reason:] Add constant automatically
Given,
f(x) = ^{(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.

Answer

Answer: b [Reason:] Add constant automatically

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

Answer: d [Reason:]Find f_x = 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

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

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

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

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

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

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

Answer

Answer: d [Reason:] Differentiating we have
f_x = 3y + 8xy² = 0
f_y = 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

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

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

Answer

Answer: b [Reason:] Using Euler theorem
xf_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

Answer

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

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

Answer

Answer: c [Reason:] Using Eulers theorem we have
xf_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

Answer

Answer: c [Reason:] Using Euler theorem we have
xf_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

Answer

Answer: d [Reason:] Substituting f_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

Answer

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

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 f_x + f_y at (x,y)=(4,4)
a) 0
b) 78
c) 4² . 3(sin(1) + 1)
d) -12

Answer

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

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

4f_x + 4f_y = 3f(4, 4) = 3⁄4(4³ . sin(1) + 4³)

= 4² . 3(sin(1) + 1).

Discrete Mathematics MCQ Set 4

1. Let f(x) = e^x 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

Answer:b [Reason:] First expanding ^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 + 2ⁿ
b) 0
c) 1
d) None

Answer

Answer: a [Reason:] Assume y = f(x)
Taking ln(x) on both sides The function has to be written in the form ln(y) = e^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 yⁿ 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

Answer

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) ^π⁄₂
c) 45
d) 4

Answer

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) ^π⁄₃₂
c) (π)²
d) cos(1)sinh(1)

Answer

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

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

7. The 7^th 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

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

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

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 ?

Answer

Answer: b [Reason:]

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

Answer

Answer: b [Reason:]

3. Find the differentiation of x⁴ + y⁴ = 0
a) – ^x3⁄_y⁴
b) – ^x4⁄_y³
c) – ^x3⁄_y³
d) ^x3⁄_y³

Answer

Answer: c [Reason:]
x⁴ + y⁴ = 0

4x³ + 4y^{3 dy}⁄_dx = 0

^dy⁄_dx = – ^x3⁄_y³

^dy⁄_dx = Sec² (x)Sec(x) e^x + Sec² (x)Tan(x) e^x + e^x Tan(x)Sec(x)

^dy⁄_dx = Sec² (x) e^x [Sec(x)+Tan(x)] + e^x Tan(x)Sec(x)

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

Answer

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

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

Answer

Answer: c [Reason:]
Tan(y)=Tan(x)

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

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

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

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

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 ^d2y⁄_dx² = ?

Answer

Answer: c [Reason:]

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

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

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.

Answer

Answer: a [Reason:] y(x) = x² + x + 1
Velocity is , v = ^dy⁄_dx = 2x + 1 (not constant)
Acceleration is a = ^dy⁄_dx = 2 (constant).