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
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))
Answer
Answer: b
Add constant automatically [Reason:] 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. 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
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
Answer
Answer: c [Reason:] Add constant automatically
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
Answer: a [Reason:] Since slope of a function is given by dy⁄dx at that point. Hence, when dy⁄dx = 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
Answer: a [Reason:] Since slope of a function is given by dy⁄dx at that point.Hence,when dy⁄dx = ∞ 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
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
Answer
Answer: a [Reason:] Add constant automatically
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
Answer: c [Reason:] Add constant automatically
Given, et = x => dx = et dt,
10. Find the value of ∫ cot3() cosec4 ()
Answer
Answer: c [Reason:] Add constant automatically
Answer
Answer: d [Reason:] Add constant automatically
12. Find the value of 
a) 1⁄8 sin(-1)(x + 1⁄2)
b)1⁄4 tan(-1)(x + 1⁄2)
c) 1⁄8 sec(-1)(x + 1⁄2)
d) 1⁄4 cos(-1)(x + 1⁄2)
Answer
Answer: b [Reason:] Add constant automatically
Answer
Answer: c [Reason:] Add constant automatically
Discrete Mathematics MCQ Set 2
1. Find the value of ∫tan-1(x)dx
a) sec-1 (x) – 1⁄2 ln(1 + x2)
b) xtan-1 (x) – 1⁄2 ln(1 + x2)
c) xsec-1 (x) – 1⁄2 ln(1 + x2)
d) tan-1 (x) – 1⁄2 ln(1 + x2)
Answer
Answer: b [Reason:] Add constant automatically
Given, ∫tan-1(x)dx
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) – 1⁄2 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))
Answer
Answer: b [Reason:] Add constant automatically
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
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
Answer
Answer: c [Reason:] Add constant automatically
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
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
Answer
Answer: a [Reason:] Add constant automatically
7. Find the area of a function f(x) = x2 + xCos(x) from x = 0 to a, where , a>0,
a) a2⁄2 + aSin(a) + Cos(a) – 1
b) a3⁄3 + aSin(a) + Cos(a)
c) a3⁄3 + aSin(a) + Cos(a) – 1
d) a3⁄3 + Cos(a) + Sin(a) – 1
Answer
Answer: c [Reason:] Given, f(x) = x2 + xCos(x)
Hence, F(x) = ∫x2 + xCos(x) dx = x3⁄3 + xSin(x) + Cos(x)
Hence, area inside f(x) is,
F(a) – F(0) = a3⁄3 + aSin(a) + Cos(a) – 1.
8. Find the area ln(x)⁄x from x = x = aeb to a
a) b2⁄2
b) b⁄2
c) b
d) 1
Answer
Answer: a [Reason:]
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)
Answer
Answer: d [Reason:]
10. Find the area inside integral 
from x = 0 to π
a) π
b) 0
c) 1
d) 2
Answer
Answer: b [Reason:]
11. Find the area inside function 
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) = ,
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
12. Find the value of ∫(x4 – 5x2 – 6x)4 4x3 – 10x – 6 dx
Answer
Answer: b [Reason:] Add constant automatically
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
Answer: a [Reason:] Temperature at distance x is ,
T = ∫T(x) dx = ∫x2 + 2x dx = x3⁄3 + x2 + C
At x=0 given T = 40 C
C = T(x = 0) = 40 C
At x= 10,
T(x = 10) = 1000⁄3 + 100 + 43 = 473 C.
14. Find the value of
a) 1⁄8 sin-1(x + 1⁄2)
b) 1⁄8 tan-1(x + 1⁄2)
c) 1⁄8 sec-1(x + 1⁄2)
d) 1⁄4 cos-1(x + 1⁄2)
Answer
Answer: b [Reason:] Add constant automatically
Discrete Mathematics MCQ Set 3
1. Find 
a) 0
b) π⁄8
c) π⁄4
d) 15π⁄96
Answer
Answer: d [Reason:] Using the formula for even n we have
.
2. Find 
a) 1
b) 0
c) 13π⁄1098
d) 21π⁄2048
Answer
Answer: d [Reason:] Rewriting the function as
3. Find 
a) 0
b) 1
c)-1
d) None of these
Answer
Answer: b [Reason:] Using the formula we have
4. Find the value of 
a) 1⁄10!
b) 5!6!⁄11!
c) 10!⁄5!6!
d) 0
Answer
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) 1⁄5 – 1⁄3 + 1⁄1 – π⁄4
Answer
Answer : d [Reason:] Simplifying we have
6. Find 
a) -1
b) 1
c) 0
d) 4((π⁄2)3 – 3π + 1)
Answer
Answer: b [Reason:] Using the formula
7. Find 
a) 1
b) 199
c) -5!
d) 5!
Answer
Answer: c [Reason:] 
Using the formula
8. Find 
a) 0
b) 5
c) 87
d) -16⁄105
Answer
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
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
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
Answer: c [Reason:]The lagrange multiplier can be applied to any number of constraints and the condition is
∇f = £ni=1 μi ∇gi
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
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
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
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
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 £mi=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
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
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
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
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
Answer: a [Reason:] We know that,
5. Find the laplace transform of et Sin(t).
Answer
Answer: b [Reason:]
6. Laplace transform of t2 sin(2t)
Answer
Answer: d [Reason:] We know that,
7. Find the laplace transform of t5⁄2
Answer
Answer: b [Reason:]
8. Value of ∫-∞∞et Sin(t)Cos(t)dt = ?
a) 0.5
b) 0.75
c) 0.2
d) 0.71
Answer
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
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
Answer
Answer: b [Reason:]
11. Find the laplace transform of y(t)=et.t.Sin(t)Cos(t)
Answer
Answer: d [Reason:]
12. Find the value of ∫0∞ tSin(t)Cos(t)
a) s ⁄ s2+22
b) a ⁄ a2+s4
c) 1
d) 0
Answer
Answer: d [Reason:]
13. Find the laplace transform of y(t)=e|t-1| u(t).
Answer
Answer: d [Reason:]
