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## Discrete Mathematics MCQ Set 1

1. d(uvw)dx is where u ,v, w are the functions of x
a) u’vw + uv’w + uvw’
b) uvw + uv’w’ + u’v’w’
c) u’v’w + uv’w’ + u’vw’
d) uv’w’ + u’v’w’ + uvw

2. d(u ⁄ v)dx is where u, v are the functions of x
a) v’u’ – uvv2
b) vu’ – uv’v2
c) vu – u’v’v2
d) 0

3. If , find dydx .
a) Sec2 (x) ex [1 + Tan(x)] + ex Tan(x)Sec(x)
b) Sec2 (x) ex [Sec(x) + Tan(x)] + ex Tan(x)Sec(x)
c) Sec2 (x) e2x [Sec(x) + Tan(x)] + ex Tan(x)Sec(x)
d) Sec(x) ex [Sec(x) + Tan(x)] + ex Tan(x)Sec(x)

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

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. Value of ddx⁡ [(1 + xex)/(1-Cos(x))].

5. Find the derivative of Sin(x)Tan(x) w.r.t ex Tan(x)

6.

Answer: b [Reason:] None.

7. Evaluate d/dx xx ln(x)
a) x(x-1) + x2x ln⁡(x) + xx [ln⁡(x) ]2
b) x(x-1) + xx ln⁡(x) + xx [ln⁡(x) ]2
c) x(x-1) + xx ln⁡(x) + xx ln⁡(x)
d) xx + xx ln⁡(x) + xx [ln⁡(x) ]2

8. Evaluate the differentiation of
a) tan-1⁡x
b) 1
c) 0
d) -1

9. If y = Tan(x)Tan(x) then dydx⁡ = ?
a) Tan(x) [1 + lnTan(x)] Tan(x)Tan(x)
b) Tan2 (x) [1 + lnTan(x)] Tan(x)Tan(x)
c) Sec2 (x) [1 + lnTan(x)] Tan(x)Tan(x)
d) Sec(x) [1 + lnTan(x)] Tan(x)Tan(x)

Answer: c [Reason:] y = Tan(x)Tan(x) Taking ln on both side ln y = Tan(x)lnTan(x) Differentiating w.r.t x

10. Evaluate ddx Cot(x)Cosec(x)
a) – Cosec2 (x)- Cosec2 (x)Cot(x)
b) – Cosec3 (x)- 2 (x)Cot(x)
c) – Cosec(x) – Cosec2 (x)Cot(x)
d) – Cosec3 (x)- Cosec(x)Cot2 (x)

Answer: b [Reason:] ddx Cot(x)Cosec(x) = – Cosec3 (x)- 2 (x)Cot(x).

11. Evaluate differentiation of x2 Sin(x) w.r.t Tan(x)Cosec(x)

12. If z = ex Sin(Cos(x))Cos(Sin(x)) Then find dzdx
a) [exSin(Cos(x))Cos(Sin(x))-exCos(x)Cos(Cos(x))Cos(Sin(x))-exSin(x)Sin(Cos(x))Sin(Sin(x))].
b) [exSin(Cos(x))Cos(Sin(x))-exSin(x)Cos(Cos(x))Cos(Sin(x))-exCos(x)Sin(Cos(x))Sin(Sin(x))].
c) [exCos(Cos(x))Sin(Sin(x))-exSin(x)Cos(Cos(x))Cos(Sin(x))-exCos(x)Sin(Cos(x))Sin(Sin(x))].
d) [exSin(Cos(x))Cos(Sin(x))-exCos(x)Cos(Cos(x))Cos(Sin(x))-exSin(x)Sin(Cos(x))Sin(Sin(x))].

Answer: b [Reason:] dzdx = ddx ex Sin(Cos(x))Cos(Sin(x)) = [(ex Sin(Cos(x))Cos(Sin(x)) – ex Sin(x)Cos(Cos(x))Cos(Sin(x)) – ex Cos(x)Sin(Cos(x))Sin(Sin(x)) )].

13. If F(x) = f(x)g(x)h(x) and F’(x) = 10F(x) and f’(x) = 10f(x) , g’(x) = 10g(x) and h’(x) = 10kh(x), then find value of k.
a) 0
b) 1
c) -1
d) 2

Answer: c [Reason:] Given F(x) = f(x)g(x)h(x) Differentiating, F’(x) = f’(x)g(x)h(x) + f(x)g’(x)h(x) + f(x)g(x)h’(x)

Putting value of F’(x), f’(x), g’(x), h’(x)

We get 10 = 10 + 10 + 10k K = -1.

## Discrete Mathematics MCQ Set 2

1. limx → 1⁡ (x-1)Tan(πx2) is
a) 0
b) –1π
c) –2π
d) 2π

2. Value of limit always be in the range of function?
a) True
b) False

Answer: b [Reason:] Because the range of f(x) = {x} is [0,1) and it value at limx → 1⁡ – f(x) is 1 which is not in its range.

3. Necessary Conditions of Sandwich rule is
a) All function must have common domain.
b) All function must have common range.
c) All function must have common domain and range both..
d) Function must not have common domain and range.

Answer: a [Reason:] Statement of sandwich theorem is, If Functions f(x) ,g(x) and h(x) 1. have Common Domain, 2. and, satisfy f(x) ≤ g(x) ≤ h(x) ∀ x ∈ D Then if f(x) = h(x) = L => g(x) = L .

4. The value of limx → 0⁡⁡ [x]Cos(x), [x] denotes the greatest integer function
a) lies between 0 and 1
b) lies between -1 and 0
c) lies between 0 and 2
d) lies between -2 and 0

Answer: b [Reason:] limx → 0⁡⁡ [x]Cos(x) We know that, x-1 < [x] < x

Multiplying by Cos(x), we get (x-1)Cos(x) < [x]Cos(x) < xCos(x)

Taking limits, we get limx → 0 [(x-1)Cos(x)] < limx → 0 [x]Cos(x) < limx → 0[xCos(x)] => -1 < limx → 0 [x]Cos(x) < 0.

5. Value of limx → 0[(1+xex )/(1 – Cos(x))]
a) e
b) 1
c) 2
d) Can not be solved

Answer: c [Reason:] =>limx → 0[(1+xex )/(1 – Cos(x))] = 10 (Indeterminate) => By L’Hospital rule => limx → 0[(1+xex ) / (Sin(x))] = 10 (Again indeterminate) => By L’ Hospital rule => limx → 0[((2+x)ex )/ (Cos(x))] = 2.

6. The value of , [x] denotes the greatest integer function
a) 0
b) 1
c) ∞
d) – ∞

7. Evaluate limx → 0(1+Tan(x))Cot(x)
a) 1
b) e
c) ln(2)
d) e2

Answer: b [Reason:] limx → 0(1+Tan(x))Cot(x) = limtan(x) → 0 (1+Tan(x))1Tan(x) = limt → 0 (1 + t)1t = e.

8. Evaluate limx → 1[(-xx + 1) / (xlog(x)) ]
a) ee
b) e
c) -1
d) e2

9. Find domain of n for which limx → 0enxCot(nx) , has non zero value.
a) n ∈ (0,∞) ∩ (1,5)
b) n ∈ (-∞,∞) ∩ (1,5)
c) n ∈ (-∞,∞)
d) n ∈ (-∞,∞) ~ 5

Answer: c [Reason:] Hence domain of n is n ∈ (-∞,∞).

10. Value of (dSin(x)Cos(x)) / dx is
a) Cos(2x)
b) Sin(2x)
c) Cos2(2x)
d) Sin2(2x)

Answer: a [Reason:] (dSin(x)Cos(x)) / dx=Cos(x) dSin(x)/dx + Sin(x) dCos(x)/dx = Cos2(x) – Sin2(x) = Cos(2x).

11. Evaluate
a) 1
b) e
c) 0
d) e2

12. If , then find the value of a and b.
a) 2.5, -1.5
b) -2.5, -1.5
c) -2.5, 1.5
d) 2.5, 1.5

Answer: b [Reason:] Since, given limit is finite, hence coefficients of powers of x should be zero and x3 should be 1 ⇒ 1 + a – b=0 ⇒ b6a2 = 1 ⇒ Solving the above two equations we get, a = -2.5, b = -1.5.

13. , then find the value of a, b and c.
a) 1.37, -4.13, 4.13
b) 1.37, 4.13, -4.13
c) -1.37, 4.13, 4.13
d) 1.37, 4.13, 4.13

Answer: b [Reason:] Now, coefficient of x and x^3 should be zero and that of x5 should be 1, then ⇒ B + c = 0 ⇒ b6 + c2 = a ⇒ b120 + c24 = 1 ⇒ By solving these 3 equations, a = 1.37, b = 4.13, c = -4.13.

## Discrete Mathematics MCQ Set 3

1. Which of the following mentioned standard Probability density functions is applicable to discrete Random Variables ?
a) Gaussian Distribution
b) Poisson Distribution
c) Rayleigh Distribution
d) Exponential Distribution

Answer: b [Reason:] None.

2. What is the area under a conditional Cumulative density function ?
a) 0
b) Infinity
c) 1
d) Changes with CDF

Answer: c [Reason:] Area under any conditional CDF is 1.

3. When do the conditional density functions get converted into the marginally density functions ?
a) Only if random variables exhibit statistical dependency
b) Only if random variables exhibit statistical independency
c) Only if random variables exhibit deviation from its mean value
d) If random variables do not exhibit deviation from its mean value

Answer: b [Reason:] None.

4. Mutually Exclusive events
a) Contain all sample points
b) Contain all common sample points
c) Does not contain any sample point
d) Does not contain any common sample point

Answer: d [Reason:] Events are said to be mutually exclusive if they do not have any common sample point.

5. What would be the probability of an event ‘G’ if H denotes its complement, according to the axioms of probability?
a) P (G) = 1 / P (H)
b) P (G) = 1 – P (H)
c) P (G) = 1 + P (H)
d) P (G) = P (H)

Answer: b [Reason:] According to the given statement P(G) + P(H) = 1.

6. A table with all possible value of a random variable and its corresponding probabilities is called
a) Probability Mass Function
b) Probability Density Function
c) Cumulative distribution function
d) Probability Distribution

Answer: d [Reason:] The given statement is the definition of a probability distribution.

7. A variable that can assume any value between two given points is called
a) Continuous random variable
b) Discrete random variable
c) Irregular random variable
d) Uncertain random variable

Answer: a [Reason:] This is the definition of a continuous random variable.

8. If a variable can certain integer values between two given points is called
a) Continuous random variable
b) Discrete random variable
c) Irregular random variable
d) Uncertain random variable

Answer: b [Reason:] This is the definition of a Discrete random variable.

9. The expected value of a discrete random variable ‘x’ is given by
a) P(x)
b) ∑ P(x)
c) ∑ x P(x)
d) 1

Answer: c [Reason:] Expected value refers to mean which is given by http://mathurl.com/zqymzn7 in case of discrete probability distribution.

10. If ‘X’ is a continuous random variable, then the expected value is given by
a) P(X)
b) ∑ x P(x)
c) ∫ X P(X)
d) No value such as expected value

Answer: c [Reason:] Since X is a continuous random variable, its expected value is given by c.

11. Out of the following values, which one is not possible in probability ?
a) P(x) = 1
b) ∑ x P(x) = 3
c) P(x) = 0.5
d) P(x) = – 0.5

Answer: d [Reason:] In probability P(x) is always greater than or equal to zero.

12. If E(x) = 2 and E(z) = 4, then E(z – x) =
a) 2
b) 6
c) 0
d) Insufficient data

Answer: a [Reason:] E(z – x) = E(z) – E(x) = 4 – 2 = 2.

## Discrete Mathematics MCQ Set 4

1. The expansion of f(x), about x = a is

Answer: a [Reason:] By taylor expansion, f(a+h) = f(a) + h1! f’ (a) + h22! f (a)…….

2. Find the expansion of ex in terms of x + m, m > 0.

Answer: d [Reason:] Let, h = x + m = > f(x) = f(h-m) = e(h-m) By taylor theorem, putting a = -m , we get,

3. Expand ln(x) in the power of (x-m).

Answer: a [Reason:] where, h = x-m Let, h = x – m => f(x) = f(h+m) = e(h+m) By taylor theorem, putting a = m , we get,

4. Find the value of √10
a) 3.1633
b) 3.1623
c) 3.1632
d) 3.1645

5. Expand f(x) = 1x about x = 1.
a) 1 – (x-1) + (x-1)2 – (x-1)3 +⋯….
b) 1 + (x-1) + (x-1)2 + (x-1)3 +⋯….
c) 1 + (x-1) – (x-1)2 + (x-1)3 +⋯….
d) 1 – (x+1) + (x+1)2 – (x+1)3 +⋯….

Answer: a [Reason:] Given f(x) = 1x Let, x – 1 = h

Hence, x = 1 + h

Hence, f(x) = f(1 + h) = f(1) + h1! f’ (1) + h22! f (1) +h33! f”’ (1)+⋯…

Now, f(1) = 1, f'(1) = -1, f”(1) = 2 ,f”'(1) = -6,…….

Hence, f(1 + h) = 1 – h + h2 – h3+⋯…

hence, 1 – (x-1) + (x-1)2 – (x-1)3 +⋯….

6. Find the expansion of f(x) = ex1+ex, given ∫f(x)dx = ln⁡(2), for x = 0

7. Find the value of eπ4√2
a) 1.74
b) 1.84
c) 1.94
d) 1.64

Answer: a [Reason:] Let,f(x) = exSin(x), f(0) = 1

8. Find the value of ln(sin(31o)) if ln(2) = 0.69315
a) -0.653
b) -0.663
c) -0.764
d) -0.662

Answer: b [Reason:] Let, f(x) = ln⁡(sin⁡(x+h)) Then, f(x) = ln⁡(sin⁡(x)), if h=0

9. The expansion of f(x,y), is

Answer: b [Reason:] By taylor expansion,

10. The expansion of f(x, y)=ex Sin(y), is
a) x + xy + ……..
b) y + y2 x + ……..
c) x + x2 y + ……..
d) y + xy + ……..

Answer: d [Reason:] Now, f(x, y)=ex Sin(y), f(0,0) = 0 Therefore, fx (x,y) = ex Sin(y), hence fx (0,0) = 0

fy (x,y) = ex Cos(y), hence fy (0,0) = 1

fxx (x,y) = ex Sin(y), hence fxx (0,0) = 0

fyy (x,y) = -ex Sin(y), hence fyy (0,0) = 0

fxy (x,y) = ex Cos(y), hence fxy (0,0) = 1 By taylor expansion,

f(x,y) = 0 + 0 + y + 12! [0 + 2xy + 0] +⋯. f(x,y) = y + xy + ……..

11. The expansion of f(x, y) = ex ln(1 + y), is
a) f(x,y)= y + xy – y22 +…….
b) f(x,y)= y – xy + y22 -…….
c) f(x,y)= y + x – y22 +……..
d) f(x,y)= x + y – x22 +……..

Answer: a [Reason:] Now, f(x, y) = ex ln(1 + y) , f(0,0) = 0 Therefore,

## Discrete Mathematics MCQ Set 5

1. For y = -x2 + 2x there exist a c in the interval [ – 19765, 19767] Such that f'(c) = 0
a) True
b) False

Answer: a [Reason:] The key here is to rewrite the function as y = -(x – 1)2 + 1 Observe here that on substituting – 19765 and 19767 in the equation we get (- 19766)2 + 1 and (- 19766)2 respectively. As we are dealing with their squared values they have to be equal We have f(- 19765) = f(19767) Polynomial functions are continuous and differentiable over the whole domain and hence by Rolles Theorem we must have a c such that f'(c) = 0 in the interval [-19765, 19767] Hence, the claim is true.

2. For the function f(x) = sin(x)⁄x2 How many points exist in the interval [0, 7π] Such that f’ (c) = 0
a) 8
b) 0
c) 7
d) 6

Answer: d [Reason:] We know that sine is a periodic function and it is divided by x2. Observe that the sine takes the value of zero at integral arguments, hence at every interval of the form We have f(nπ) = f((n + 1) π) The sine and the polynomial combination is continuous and differentiable at every point except x = 0 Every such interval has a point such that f’ (c) = 0 Hence, by Rolles theorem, in every interval of the form [nπ, (n + 1)π] we must have a point such that Leaving the interval [0, π] we are left with six such intervals from 0 to 7π.

3. f(x) = sin(x)⁄x, How many points exist such that f’ (c) = 0 in the interval [0, 18π]
a) 18
b) 17
c) 8
d) 9

Answer: a [Reason:] We have the sine function that takes the value of zero at Integral multiples, But for sin(x)⁄x we have the exceptional value of limx→0 sin(x)⁄x reaching one. So leaving the first interval [0, π], for every other interval of the form [nπ (n + 1)π ] we must have f(nπ) = f((n + 1)π) By Rolles theorem we have f’ (c) = 0 For every interval of the form [nπ (n + 1)π ] There are 17 such intervals.

4. Let f(x) = x + sin(x) Every point on the graph is rotated by 45 degree with respect to the origin along the radius equal to the radius vector at that point. How many c that belong to [0, 11π] exist Such that f'(c) = 0
a) 10
b) 11
c) 110
d) 9

Answer: b [Reason:] The function has the beautiful property that the sine function is built over the line y = x In other words the axis for the sine graph is no more on the x axis but is the line y = x . When this graph is rotated by 45 degrees which Is equal to the slope of the line y = x we must get the original graph sin(x) again. The number of c in the interval [0, 11π] is then obviously equal to eleven.

5. A Function f(x) has the property f(a) = f(b) for ∀a,b…∊….I and a + b = 20 then which of the following even degree polynomials could be f(x)
a) x44 – 10x3 + 150x2 – 1000x + 10131729
b) x2 + 5x + 6
c) x2 + x + 1
d) Polynomial functions are inadequate representations

Answer: a [Reason:] Given that f(a) = f(b) for all integral values of a, b, one additional constraint is that a + b⁄2 = 10 This means that the graph is symmetric about the line x = 10 This means that the even degree polynomial has a Rolle point(the only Rolle point) at x = 10 We need to differentiate the functions given in the options and observe which of these has a single(OR repeated root), which is equal to 10. The first Option when differentiated yields f'(x) = x3 – 30x2 + 300x – 1000 Equating to zero, we see that x = 10 satisfies the equation (10)3 – 30(10)2 + 300(10) – 1000 = 0.

6. For some function f(x) we have f(a) = f(b) for ∀a,b…∊….I and a + b = 2 then which of the following even degree polynomials could f(x) be
a) x2 + 3x +1
b) 5x22 – 5x + 101
c) x2 + 2x + 1
d) Even degree polynomials of such kind cannot exist

Answer: b [Reason:] Given that f(a) = f(b) for all integral values of a, b with the condition a + b ⁄ 2 = 1 We know that the function is symmetric about the line x = 1 Thus we must have a Rolle point at 1 for any function that satisfies these conditions. Differentiating the given functions in the options we have for a 2x + 3 = 0 x = -32 ≠ we get 5x – 5 = 0 x =1.

7. For all second degree polynomials with y = ax2 + bx + k, it is seen that the Rolles’ point is at c = 0 . Also the value of k is zero. Then what is the value of b
a) 0
b) 1
c) -1
d) 56

Answer: a [Reason:] Given that k = 0 we must rewrite the function to get y = ax2 + bx Now differentiating the function yields y’ = 2ax + b = 0 Equating it to zero we get the Rolle point which is also zero 2a(0) + b = 0 b = 0 .

8. For second degree polynomial it is seen that the roots are equal. Then what is the relation between the Rolles point c and the root x
a) c = x
b) c = x2
c) They are independent
d) c = sin(x)

Answer: a [Reason:] For a polynomial with equal roots we can write the polynomial as f(x) = (x – a)2 Where a is the repeated root. Differentiating it and equating the function to zero we get the Rolles point, f'(x) = 2(x – a) = 0 x = a = c.

9. For any second degree polynomial with two real unequal roots. The relation between Rolles point r1 and the two roots r2 is
a) They are independent
b) c = r1 – r2
c) c = r1 * 1r2
d) c = r1 + r22