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## Discrete Mathematics MCQ Set 1

1. Integration of function is same as the
a) Joining many small entities to create a large entity.
b) Indefinetly small difference of a function
c) Multiplication of two function with very small change in value
d) Point where function neither have maximujm value nor minimum value

Answer: a [Reason:] Integration of function is same as the Joining many small entities to create a large entity.

2. Integration of (Sin(x) + Cos(x))ex is
a) ex Cos(x)
b) ex Sin(x)
c) ex Tan(x)
d) ex (Sin(x)+Cos(x))

∫ ex Sin(x)dx = ex Sin(x) – ∫ ex Cos(x)dx

∫ ex Sin(x)dx + ∫ ex Cos(x)dx = ∫ ex [Cos(x)+Sin(x)]dx = ex Sin(x).

3. Integration of (Sin(x) – Cos(x))ex is
a) -ex Cos(x)
b) ex Cos(x)
c) -ex Sin(x)
d) ex Sin(x)

Answer: a [Reason:] Add constant automatically Let f(x) = ex Sin(x) ∫ ex Sin(x)dx = -ex Cos(x) + ∫ ex Cos(x)dx ∫ ex Sin(x)d-∫ ex Cos(x)dx = ∫ ex [Sin(x)-Cos(x)]dx = -ex Cos(x).

4. Value of ∫ Cos2 (x) Sin2 (x)dx

5. If differentiation of any function is zero at any point and constant at other points then it means
a) Function is parallel to x-axis at that point
b) Function is parallel to y-axis at that point
c) Function is constant
d) Function is discontinuous at that point

Answer: a [Reason:] Since slope of a function is given by dydx at that point. Hence, when dydx = 0 means slope of a function is zero i.e, parallel to x axis. Function is not a constant function since it has finite value at other points.

6. If differentiation of any function is infinite at any point and constant at other points then it means
a) Function is parallel to x-axis at that point
b) Function is parallel to y-axis at that point
c) Function is constant
d) Function is discontinuous at that point

Answer: a [Reason:] Since slope of a function is given by dydx at that point.Hence,when dydx = ∞ means slope of a function is 90 degree i.e,parallel to y axis.

7. Integration of function y = f(x) from limit x1 < x < x2 , y1 < y < y2, gives
a) Area of f(x) within x1 < x < x2
b) Volume of f(x) within x1 < x < x2
c) Slope of f(x) within x1 < x < x2
d) Maximum value of f(x) within x1 < x < x2

Answer: a [Reason:] Integration of function y=f(x) from limit x1 < x < x2 , y1 < y < y2, gives area of f(x) within x1 < x < x2.

8. Find the value of ∫ ln⁡(x)x dx
a) 3a2
b) a2
c) a
d) 1

9. Find the value of ∫t(t+3)(t+2) dt, is
a) 2 ln⁡(t+3)-3 ln⁡(t+2)
b) 2 ln⁡(t+3)+3 ln⁡(t+2)
c) 3 ln⁡(t+3)-2 ln⁡(t+2)
d) 3 ln⁡(t+3)+2ln⁡(t+2)

Answer: c [Reason:] Add constant automatically Given, et = x => dx = et dt,

10. Find the value of ∫ cot3() cosec4 ()

12. Find the value of
a) 18 sin(-1)⁡(x + 12)
b)14 tan(-1)⁡(x + 12)
c) 18 sec(-1)⁡(x + 12)
d) 14 cos(-1)⁡(x + 12)

## Discrete Mathematics MCQ Set 2

1. Find the value of ∫tan-1⁡(x)dx
a) sec-1 (x) – 12 ln⁡(1 + x2)
b) xtan-1 (x) – 12 ln⁡(1 + x2)
c) xsec-1 (x) – 12 ln⁡(1 + x2)
d) tan-1 (x) – 12 ln⁡(1 + x2)

Putting, x = tan(y),

We get, dy = sec2(y)dy,

∫ysec2(y)dy

By integration by parts,

ytan(y) – log⁡(sec⁡(y)) = xtan-1 (x) – 12 ln⁡(1 + x2.

2. Integration of (Sin(x) + Cos(x))ex is
a) ex Cos(x)
b) ex Sin(x)
c) ex Tan(x)
d) ex (Sin(x) + Cos(x))

Let f(x) = ex Sin(x) ∫ex Sin(x)dx = ex Sin(x) – ∫ex Cos(x)dx

∫ex Sin(x)dx + ∫ex Cos(x)dx = ∫ex [Cos(x) + Sin(x)]dx = ex Sin(x).

3. Find the value of ∫x3 Sin(x)dx
a) x3 Cos(x) + 3x2 Sin(x) + 6xCos(x) – 6Sin(x)
b) – x3 Cos(x) + 3x2 Sin(x) – 6Sin(x)
c) – x3 Cos(x) – 3x2 Sin(x) + 6xCos(x) – 6Sin(x)
d) – x3 Cos(x) + 3x2 Sin(x) + 6xCos(x) – 6Sin(x)

Answer: d [Reason:] Add constant automatically Let f(x) = x3 Sin(x) ∫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).

4. Value of ∫uv dx,where u and v are function of x

5. Find the value of ∫x7 Cos(x) dx
a) x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x)
b) x7 Sin(x) – 7x6 Cos(x) + 42x5 Sin(x) – 210x4 Cos(x) + 840x3 Sin(x) – 2520x2 Cos(x) + 5040xSin(x) – 5040Cos(x)
c) x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 5040Cos(x)
d) x7 Sin(x) + 7x6 Cos(x) + 42x5 Sin(x) + 210x4 Cos(x) + 840x3 Sin(x) + 2520x2 Cos(x) + 5040xSin(x) + 10080Cos(x)

Answer: a [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)

6. Find the value of ∫x3 ex e2x e3x……..enx dx

7. Find the area of a function f(x) = x2 + xCos(x) from x = 0 to a, where , a>0,
a) a22 + aSin(a) + Cos(a) – 1
b) a33 + aSin(a) + Cos(a)
c) a33 + aSin(a) + Cos(a) – 1
d) a33 + Cos(a) + Sin(a) – 1

Answer: c [Reason:] Given, f(x) = x2 + xCos(x)

Hence, F(x) = ∫x2 + xCos(x) dx = x33 + xSin(x) + Cos(x)

Hence, area inside f(x) is,

F(a) – F(0) = a33 + aSin(a) + Cos(a) – 1.

8. Find the area ln(x)x from x = x = aeb to a
a) b22
b) b2
c) b
d) 1

9. Find the area inside a function f(t) = t/(t+3)(t+2) from t = -1 to 0
a) 4 ln⁡(3) – 5ln⁡(2)
b) 3 ln⁡(3)
c)3 ln⁡(3) – 4ln⁡(2)
d) 3 ln⁡(3) – 5 ln⁡(2)

10. Find the area inside integral from x = 0 to π
a) π
b) 0
c) 1
d) 2

11. Find the area inside function 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) = , 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) – 1/2 – 5=a22 + 5a – 4ln(a) – 112

12. Find the value of ∫(x4 – 5x2 – 6x)4 4x3 – 10x – 6 dx

13. Temperature of a rod is increased by moving x distance from origin and is given by equation T(x) = x2 + 2x , where x is the distance and T(x) is change of temperature w.r.t distance.If,at x = 0,temperature is 40 C,find temperature at,x=10 .
a) 473 C
b) 472 C
c) 474 C
d) 475 C

Answer: a [Reason:] Temperature at distance x is , T = ∫T(x) dx = ∫x2 + 2x dx = x33 + x2 + C At x=0 given T = 40 C C = T(x = 0) = 40 C At x= 10, T(x = 10) = 10003 + 100 + 43 = 473 C.

14. Find the value of
a) 18 sin-1(x + 12)
b) 18 tan-1(x + 12)
c) 18 sec-1(x + 12)
d) 14 cos-1(x + 12)

## Discrete Mathematics MCQ Set 3

1. Find
a) 0
b) π8
c) π4
d) 15π96

Answer: d [Reason:] Using the formula for even n we have .

2. Find
a) 1
b) 0
c) 13π1098
d) 21π2048

Answer: d [Reason:] Rewriting the function as

3. Find
a) 0
b) 1
c)-1
d) None of these

Answer: b [Reason:] Using the formula we have

4. Find the value of
a) 110!
b) 5!6!11!
c) 10!5!6!
d) 0

Answer: b [Reason:] Using the definition of beta function we see that the integral is equal to the beta function at (6,5) Now using the relation between the Beta and the Gamma function we have

5. Find
a)-1
b) 1
c) 0
d) 1513 + 11π4

Answer : d [Reason:] Simplifying we have

6. Find
a) -1
b) 1
c) 0
d) 4((π2)3 – 3π + 1)

Answer: b [Reason:] Using the formula

7. Find
a) 1
b) 199
c) -5!
d) 5!

Answer: c [Reason:] Using the formula

8. Find
a) 0
b) 5
c) 87
d) -16105

Answer: d [Reason:] Rewriting the function as

## Discrete Mathematics MCQ Set 4

1. In a simple one-constraint Lagrange multiplier setup, the constraint has to be always one dimension lesser than the objective function.
a) True
b) False

Answer: b [Reason:] This condition is not always necessary because the lesser dimension curve can still be treated as a higher dimension curve.

2. Maximize the function x + y – z = 1 with respect to the constraint xy=36.
a) 0
b) -8
c) 8
d) No Maxima exists

Answer: d [Reason:] Geometrically, we can see that the level curves can go further the origin along the curve xy=36 infinitely and still not reach its maximum value. What the Lagrange multiplier predicts in this case is the minimum value.

3. Which one of these is the right formula for the Lagrange multiplier with more than one constraint.
a) ∇f = (μ)2 * ∇g1 + ∇g2
b) Cannot be applied to more than one constraint function.
c) ∇f = μ * ∇g1 + λ * ∇g2
d) ∇f = μ * ∇g1 +∇g2

Answer: c [Reason:]The lagrange multiplier can be applied to any number of constraints and the condition is ∇f = Σni=1 μigi Where μi, μ2 ……..μn are appropriate constraints(scalar multiples).

4. Maximum value of a 3-d plane is to be found over a circular region. Which of the following happens if we increase the radius of the circular region.
a) Maximum value is invariant
b) Maximum value decreases
c) Maximum value increases and minimum value goes lesser
d) minimum value goes higher

Answer: c [Reason:]Consider the level curves of the plane. These are the set of straight lines with equal slope and unequal intercepts. Now as the radius of the circular region is increased, we see that the Lagrange condition(i.e. the level curves to be tangent to the circular boundary) happens to occur further away form the origin. Thus the maximum value is pushed further and the minimum value is decreased further.

5. Find the points on the plane x + y + z = 9 which are closest to origin.
a) (3,3,3)
b) (2,1,3)
c) (2,2,2)
d) (3,4,1)

Answer: a [Reason:] The objective function is f(x,y,z) = x2 + y2 + z2 compute gradient ∇f = 2x i + 2y j + 2z k Now compute gradient of the function x + y + z = 9 which is = i + j + k Using Lagrange condition we have ∇f = λ . ∇g 2x i + 2y j + 2z k = λ * (i + j + k) ⇒ x = y =z Put this back into constraint function we get 3x = 9 ⇒ (x,y,z) = (3,3,3).

6. Consider the points closest to the origin on the planes x + y + z =a.
a) The closest point travels farther as a is increased
b) The closest point travels nearer as a is increased
c) The closest point is independent of a as a is not there in the expression of the gradient.
d) Varies as a2, away from the origin.

Answer: a [Reason:] The intercept of the planes increase as we increase the a value. Hence, we may conclude that the closest point of lower a value plane would be closer to the origin. The Lagrange multiplier set up can be used to verified this.

7. The span of a Astroid is increased along both the x and y axes equally. Then the maximum value of: z = x + y along the Astroid
a) Increases
b) Decreases
c) Invariant
d) The scaling of Astroid is irrelevant

Answer: a [Reason:] Calculating the gradients considering the general form of Astroid as x2/3 + y2/3 = a2/3 and then equating them by Lagrange condition. we can conclude that the maximum value increases.

8. The extreme value of the function With respect to the constraint Σmi=1 (xi)2 = 1 where m always stays lesser than n and as m,n tends to infinity is:
a) 1
b) 2/3√3
c) 2
d) 1 ⁄ 2

Answer: b [Reason:] First consider these functions as infinite dimension vectors. Given the constraint dimension is always less than the objective we can apply the Lagrange condition. We now have

## Discrete Mathematics MCQ Set 5

1. Laplace of function f(t) is given by

Answer: a [Reason:] Laplace of function f(t) is given by

2. Laplace transform any function changes it domain to s-domain.
a) True
b) False

Answer: a [Reason:] Laplace of function f(t) is given by ,hence it changes domain of function from one domain to s-domain.

3. Laplace transform if sin⁡(at)u(t) is
a) s ⁄ a2+s2
b) a ⁄ a2+s2
c) s2 ⁄ a2+s2
d) a2 ⁄ a2+s2

Answer: b [Reason:] We know that,

4. Laplace transform if cos⁡(at)u(t) is
a) s ⁄ a2+s2
b) a ⁄ a2+s2
c) s2 ⁄ a2+s2
d) a2 ⁄ a2+s2

Answer: a [Reason:] We know that,

5. Find the laplace transform of et Sin(t).

6. Laplace transform of t2 sin⁡(2t)

Answer: d [Reason:] We know that,

7. Find the laplace transform of t52

8. Value of ∫-∞et Sin(t)Cos(t)dt = ?
a) 0.5
b) 0.75
c) 0.2
d) 0.71

Answer: c [Reason:] L(Sin(2t) = ∫-∞e-st Sin(2t)dt = 2/(s2 + 4) Putting s=-1 ∫-∞et Sin(2t)dt = 0.4 hence, ∫-∞e-st Sin(t)Cos(t)dt = 0.2.

9. Value of ∫-∞et Sin(t) dt = ?
a) 0.50
b) 0.25
c) 0.17
d) 0.12

Answer: a [Reason:] L(Sin(2t) = ∫-∞e-st Sin(t)dt = 1/(s2 + 1) Putting s = -1 ∫-∞et Sin(t)dt = 0.5.

10. Value of ∫-∞et log(1+t)dt = ?
a) Sum of infinite integers
b) Sum of infinite factorials
c) Sum of squares of Integers
d) Sum of square of factorials

11. Find the laplace transform of y(t)=et.t.Sin(t)Cos(t)

12. Find the value of ∫0 tSin(t)Cos(t)
a) s ⁄ s2+22
b) a ⁄ a2+s4
c) 1
d) 0