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## Discrete Mathematics MCQ Set 1

1. Find
a) ∞
b) 0
c) Does Not Exist
d) 121

Answer: c [Reason:] Put x = t : y = a.t32 we have By varying a we get different limits along different paths Hence, Does Not Exist is the right answer.

2. Find
a) 0
b)
c) Does Not exist
d) ∞

Answer: c [Reason:] Put Put x = t : y = a.t52 we have By varying a we get different limits along different paths Hence, Does Not exist is the right answer.

3. Find
a) ∞
b) 12
c)
d) 13

Answer: c [Reason:] Treating limits separately we have

lt(x, y)→(0, 0) sin(x)x * lt(x, y)→(0, 0) sec(y) = 1 * 1 = 1.

4. Find
a) 1
b) 0
c) ∞
d) -90

Answer: b [Reason:] Simplifying the expression we have = 0.

5. Find
a) 9
b) 0
c) 6
d) 2

Answer: d [Reason:] Simplifying the expression we have

6. Find
a) 0
b) ∞
c)
d) -1

Answer: a [Reason:] Converting into Polar form we have

7. Find
a)
b) 0
c) ∞
d) Does Not Exist

Answer: d [Reason:] Put x = t : y = at By varying a we get different limits Hence, Does Not Exist is the right answer.

8. Find
a)
b) -1
c) ∞
d) Does not Exist

Answer: d [Reason:] Multiplying and dividing by we have .

9. Find
a) Does Not Exist
b) 0
c)
d) ∞

Answer: b [Reason:] Put x =r.cos(ϴ) : y = r.sin(ϴ)

10. Find
a) 0
b) ∞
c)
d) Does Not Exist

Answer: d [Reason:] Put x = at : y = t By varying n we get different limits

Hence, Does Not Exist is the right answer.

11. Find
a) Does not Exist
b) 0
c) ∞
d)

Answer: a [Reason:] Put x = at : y = t By varying n we get different values of limits.

12. Find
a) ∞
b)
c) 12
d) Does Not Exist

Answer: d [Reason:] Put x = t :y = at By varying the value of a we get different limits.

13. Find
a)
b) ∞
c) 0
d) 990

Answer: a [Reason:] = 1 * 1 * 1 = 1.

14. Find
a)
b) ∞
c) 0
d) 990

Answer: a [Reason:] lt(x, y)→(0, 0) sinh(x)x * lt(x, y)→(0, 0) sinh(y)y = 1 * 1 = 1.

## Discrete Mathematics MCQ Set 2

1. Two men on a surface want to meet each other. They have taken the point (0, 0) as meeting point. The surface is 3-D and its equation Given that they both play this game infinite number of times with their starting point as (908, 908) and (90, 180)
(choosing a different path every time they play the game). Will they always meet?
a) They will not meet every time
b) They will meet every time
c) Insufficient information
d) They meet with probability 12

Answer: a [Reason:] The question is asking us to simply find the limit of the given function exists as the pair (x, y) tends to (0, 0) (The two men meet along different paths taken or not)

Thus, put x = t : y = a(t)34 By putting different values of a we get different limits Thus, there are many paths that do not go to the same place. Hence, They will not meet every time is the right answer.

2. Find
a) 1
b) 0
c) ∞
d) Does Not Exist

Answer: d [Reason:] Put x = t : y = a1 * t34 : z = a2 * t34

3. Find
a) ∞
b) 13
c) 1
d) Does Not Exist

Answer: d [Reason:] Put x = t : y = at : z = t

4. Find
a) ∞
b) 123
c) 9098
d) 8

Answer: d [Reason:] Simplifying the expression yields

5. Find
a) 1990
b) ∞
c) Does Not Exist
d) 0

Answer: c [Reason:] Put x = t : y = a1.t12 : z = a2.t : w = a3.t By changing the values of a1 : a2 : a3 we get different values of limit. Hence, Does Not Exist is the right answer.

6. Find
a) 700
b) 701
c) 699
d) 22

Answer: d [Reason:] Simplifying the expression we have

7. Given that limit exists find
a) 1
b) 35
c) 12
d) 0

Answer: b [Reason:] Given that limit exists we can parameterize the curve Put x = t : y = t : z = t

8. Given that limit exist find
a) 2
b) 1
c) 4
d) 3

Answer: c [Reason:] We can parameterize the curve by x = y = z = t

9. Given that limit exists find
a) 1
b) 12
c) 17
d) 27

Answer: d [Reason:] We can parameterize the curve by

10. Given that limit exists find
a) ∞
b) 1
c) 0
d) ln(45)

Answer: a [Reason:] We can parameterize the curve by x = y = z = t

11. Given that limit exists
a) 99
b) 0
c) 1
d) 100

Answer: c [Reason:] Put x = y = z = t

12. Two men on a 3-D surface want to meet each other. The surface is given by . They make their move horizontally or vertically with the X-Y plane as their reference. It was observed that one man was initially at (200, 400) and the other at (100, 100). Their meet point is decided as (0, 0). Given that they travel in straight lines, will they meet?
a) They will meet
b) They Will not meet
c) They meet with probability 12
d) Insufficient information

Answer: b [Reason:] The problem asks us to find the limit of the function f(x, y) along two lines y = x and y = 2x For the first line (first person)

x = t : y = 2t The limits are different and they will not meet.

13. Two men on a 3-D surface want to meet each other. The surface is given by . They make their move horizontally or vertically with the X-Y plane as their reference. It was observed that one man was initially at (400, 1600) and the other at (897, 897). Their meet point is decided as (0, 0). Given that they travel in straight lines, will they meet?
a) They will meet
b) They will not meet
c) They meet with probability 12
d) Insufficient information

Answer: b [Reason:] The problem asks us to find the limit of the function f(x, y) along two lines y = x and y = 4x For the first line (first person) x = t : y = 4t The limits are different and the will not meet.

14. Observe the figure. It is given that the function has no limit as (x, y) → (0 ,0) along the paths given in the figure. Then which of the following could be f(x, y)

Answer: d [Reason:] The curves in the given graph are parabolic and thus they can be parameterized by x = t : y = at2 Substituting in Option (d) we get By varying a we get different limits

## Discrete Mathematics MCQ Set 3

1. The pth derivative of a qth degree monic polynomial, where p, q are positive integers and 2p4 + 3pq32 = 3q32 + 2qp3 is given by
a) Cannot be generally determined
b) (q – 1)!
c) (q)!
d) (q – 1)! * pq

Answer: c [Reason:] First consider the equation 2p4 + 3pq32 After simplification, we get

(2p3 + 3q12) (p – q) = 0 This gives us two possibilities 2p3 = – 3q12 OR

p = q The first possibility can’t be true as we are dealing with positive integers

Hence, we get

p = q

Thus the pth derivative of any monic polynomial of degree p(p = q) is

p! =q!.

2. The first and second derivatives of a quadratic Polynomial at x = 1 are 1 and 2 respectively. Then the value of f(1) – f(0) Is given by
a) 32
b) 12
c) 1
d) 0

Answer: d [Reason:] Let the quadratic polynomial be f(x) = ax2 + bx + c

The first derivative at x = 1 is given by

2a + b = 1

Now consider the second derivative at x = 1 which is given by

2a = 2

Solving for the coefficients using equations, we get the values as a = 1 and b = -1 Putting these values back in the polynomial yields

f(x) = x2 – x + c

Now the required value can be computed as

f(1) – f(0) = (12 – 1 + c) – (02 – 0 + c)

= (0 + c) – (0 + c) = 0.

3. Let f(x) = sin(x) / x – 54 , then the value of f(100)(54) is given by
a) Undefined
b) 100
c) 10
d) 0

Answer: a [Reason:] The key here is to expand the numerator into a taylor series centered at 54 Doing this gives us the following Observe the first term in the infinite series there is always (x – 54) in the denominator which goes to 0 when we substitute Every derivative also will have this term Hence any derivative of the given function goes to x = 54 as ∞ Hence, the answer is Undefined.

4. Let f(x) = ln(x2 + 5x + 6) then the value of f(30)(1) is given by

f(x) = ln(x2 + 5x + 6) Factorising the inner polynomial we get

f(x) = ln((x + 3) (x + 2)) Now using the rule of logarithms ln (m * n ) = ln(m) + ln(n)

we get

f(x) = ln(x + 3) + ln(x + 2)

Now using the nth derivative of logarithmic function We have

5. f(x) = ∫0 π2 sin(ax)da then the value of f(100)(0) is
a) a(100) sin(a)
b) – a(100) sin(a)
c) a(100) cos(a)
d) 0

Answer: d [Reason:] First solve the integral Observe that every term in the expansion is odd powered Hence even derivative at x = 0 has to be 0.

6. Let f(x) = x9 ex then the ninth derivative of f(x) at x = 0 is given by
a) 9!
b) 9! * e9
c) 10!
d) 21!

Answer: a [Reason:] The key here is to expand ex as a mclaurin series and then multiply it by x9 We have Substituting x = 0 gives us the result f(9)(0) = 9!.

7. The following moves are performed on g(x)
(i) Pick (x0, y0) on g(x) and travel toward the left/right to reach the y = x line. Now travel above/below to reach g(x). Call this point on g(x) as (x1, y1)
(ii) Let the new position of (x0, y0)be (x0, y1)
This is performed for all points on g(x) and a new function is got. Again these steps may be repeated on new function and another function is obtained. It is observed that, of all the functions got, at a certain point (i.e. after finite number of moves) the nth derivatives of the intermediate function are constant, and the curve passes through the origin. Then which of the following functions could be g(x)
a) y = √1 – x2
b) xy32 + y = constant
c) x9 y32 + y6 x32 = constant
d) x7 y 8 + 4y = constant

Answer: d [Reason:] The phrasing is a bit convoluted. It is asking us to convert a given function f(x) into its own composition Now the set of finite number of moves asks us to find the composition of composition i.e. f(f(x)) and f(f(x)) and so on. It also says that the intermediate composition function has the property of all nth derivatives being constant. This is possible only if the intermediate function is of the form k(x) = ax + b (Linear Function)

Given another condition that it has to pass through origin leads to the conclusion that b = 0 So the intermediate function has the form k(x) = ax The first possibility gives rise to k(x) = x Now one has to select from the options as to which curve would give k(x) = f(f(….(x)…)) = x when composed with itself a finite number of times. This can be done by simply interchanging the position of y and x in the options and check whether it preserves its structure.

For y = √1 – x2 we have after changing the position

y = √1 – y2 This has the same structure. None of the other options have this property.

8. The first, second and third derivatives of a cubic polynomial f(x) at x = 1 , are 13, 23 and 33 respectively. Then the value of f(0) + f(1) – 2f(-1) is
a) 76
b) 86
c) 126
d) 41.5

Answer: d [Reason:] Assume the polynomial to be of the form f(x) = ax3 + bx2 + cx + d

Now the first derivative at x = 1 yields the following equation

13 = 1 = 3a + 2b + c

The second derivative at x = 1 yields the following expression

23 = 8 = 6a + 2b

The third derivative at x = 1 yields the following equation

33 = 27 = 6a

Solving for a, b and c simultaneously yields (a, b, c) = (92, -192, 132) Hence the assumed polynomial is f(x) = 9x3 – 19x2 + 13x ⁄ 2 + d

Now the given expression can be evaluated as

f(0) + f(1) – 2f(-1) = (d) + (32 + d) – 2(-20 + d)

= 40 + 32 = 41.5.

9. Let g(x) = ln(x) ⁄ x – 1 Then the hundredth derivative at x = 1 is
a) 100!101
b) 99!101
c) 101100!
d) 199!

Answer: a [Reason:] The key here is to again expand the numerator as a Taylor series centered at x = 1 Hence we have the Taylor series as Substituting x = 1 yields 100!101.

10. Let f(x) = ln(x3 – 3x2 – 16x -12) , then the 1729th derivative at x = 234 is

Answer: a [Reason:] The function can be written as f(x) = ln((x – 6)(x + 1)(x + 2)) Using property of logarithms we have f(x) = ln(x + 1) + ln(x + 2) + ln(x – 6) Using the nth derivative of logarithmic functions

11. Find the value of using nth derivatives
a) – 2 * sin(1)
b) 3 * sin(1)
c) 3 * cos(1)
d) – 3 * cos(1)

Answer: a [Reason:] We have to consider the function f(x) = sin(ex) in order to get the series in some way. Expanding the given function into a Taylor series we have Now observe that our series in question doesn’t have the exponential function, this gives us the hint that some derivative of this function has to be taken at x = 0 Observe that the term (2n – 1)3 has exponent equal to 3 Hence we have to take the third derivative of the function to get the required series To find the value of this series we need to take the third derivative of original function at the required point , this is as follows

f(3)(x) = -2exsin(ex) Substituting x = 0 we get f(3)(0) = -2sin(1).

12. Let .Find the third derivative at x = 0
a) 4
b) 13
c) Undefined
d) 14

Answer: b [Reason:] Again the key here is to expand the given function into appropriate Taylor series. Rewriting the function as f(x) = e-x(ln(1 – x)) and then expanding into Taylor series we have Now the question asks us to find the third derivative at x = 0. Thus, it is enough for us to find the coefficient of the x3 term in the infinite polynomial product above The third degree terms can be grouped apart as follows

= x33x32 + x32 Hence the third derivative at x = 0 is simply the coefficient of the third degree term, which is coefficient(x33) = 13.

## Discrete Mathematics MCQ Set 4

1. The Mclaurin Series expansion of sin(ex) is

Answer: a [Reason:] We know the series expansion for sin(x) is

2. What is the coefficient of x101729 in the series expansion of cos(sin(x))
a) 0
b) 1101729!
c) -1101729!
d) 1

Answer: a [Reason:] We know that the series expansion of cos(x) is Observe that every term has odd powered series raised to an even term Thus, we must have only even powered terms in the above series expansion. The coefficient of any odd powered term is zero.

3. Let τ(X) be the Taylor Series expansion of f(x) = x3 + x2 + 1019 centered at
a = 1019 , then what is the value of the expression 2(τ(1729))2 + τ(1729) * f(1729) – 3(f(1729))2 + 1770
a) 1770
b) 1729
c) 0
d) 1

Answer: a [Reason:] Observe first off that the given function is a polynomial and so any other representation (Taylor Series here) which is continuous and differentiable has to be the same polynomial. This gives us τ(x) = f(x) We now evaluate the expression as follows = 2(f(1729))2 + (f(1729))2 – 3(f(1729))2 + 1770

= 3(f(1729))2 – 3(f(1729))2 +1770

= 1770

4. Find the Taylor series expansion of the function cosh(x) centered at x = 0

Answer: c [Reason:] We know the general expression for the expansion of the Taylor series

5. To find the value of sin(9) the Taylor Series expansion should be expanded with center as
a) 9
b) 8
c) 7
d) None of these.

Answer: d [Reason:] The Taylor series gives accurate results around some point taken as center. As we need the value of 9 the center nearer to the point should be taken. Hence, 9 is the right answer.

6. sin(9) Holds good for some functions f(1) (n) = g(n) (0) and f(x) . Now let the coordinate axes containing graph g(x) be rotated by f(x) degrees clockwise, then the corresponding Taylor series for the transformed is
a) g(x)
b)
d) Such function is not continuous

7. Let τ(f(x)) denote the Taylor series for some function f(x). Then the value of τ(τ(τ(f(1729)))) – 2τ(τ(f(1729))) + τ(f(1729)) is
a) 1729
b) -1
c) 1
d) 0

Answer: d [Reason:] We know that the Mclaurin Series for any given function always yields a polynomial (finite OR infinite). Further the Mclaurin series of this polynomial (i.e.τ(τ(f(x))) ) is also a polynomial. Due to uniqueness of this polynomial, no matter how many nested Mclaurin series we might find, they are all equal. Thus, we have τ(τ…….(f(x))….)) = f(x) Substituting this into our required expression we have = f(1729) – 2f(1729) + f(1729) = 0.

8. Let Mclaurin series of some f(x) be given recursively, where an denotes the coefficient of xnin the expansion. Also given an = an-1 / n and a0 = 1, which of the
following functions could be f(x)
a) ex
b) e2x
c) c + ex
d) No closed form exists

Answer: a [Reason:] Observing the recurrence relation we have

9. A function f(x) which is continuous and differentiable over the real domain exists such that f(n) (x) = [ f(n + 1) (x) ]2 , f(0) = a and f(x)
a) True
b) False

Answer: a [Reason:] Writing out the Mclaurin series we have This is a well defined function.

## Discrete Mathematics MCQ Set 5

1. Let τa [ f (x) ] denote the Taylor series of f (x) centered at a then the value of the expression
[ τ199 (τ121 (τ1729 (f (101729))) ]3 – [τ342 (f (101729)) ]3
a) 101729
b) 0
c) 1
d) -101729

Answer: b [Reason:] The Taylor polynomial of any function is unique at any center. Also observe that Taylor series of any function is some polynomial. Coupling these facts we have

τa1(τa2(…….(f(x))….)) = f(x)

Where a1, a2………….an are real numbers Hence the value of the given expression is = [ τ199(τ121(τ1729(f(101729))) ]3 – [τ342(f(101729)) ]3

= [f(101729)]3 – [f(101729)]3 = 0.

2. Function has the property that f(n) (x) = f(n + 2) (x) : n ≥ 1 : n ∊ N Then which of the following is the expression for f(x) in most general form.

Answer: c [Reason:] Consider the general form of Mclaurin series for f(x) Now writing a0 = c1 and a1 = c2 we have f(x) = c1 * cosh(x) + c2 * sinh(x).

3. The value of f(1) can be deduced using Taylor series
a) True
b) False

Answer: a [Reason:] Though the function f(x) is discontinuous at x = 1 we can still find f(1) as the function is continuous and differentiable in the interval. [0, 1 )

4. To find the value of cosh(23) with good accuracy the Taylor series should be centered at
a) 23
b) 22
c) 21
d) Delta (small) interval around 23

Answer: d [Reason:] To find the Taylor Series according to the theory we have to find some center which is really close to 23 and has determinate value to find the value of cosh(23) with Taylor series.

5. Find the Taylor Series expansion of sinh (x) centered around 5

Answer: a [Reason:] We know the Taylor series does not change the polynomial. Hence, be the polynomial centered at 5 or anywhere else would yield the same polynomial. In this case the polynomial centered at 0 has to be equal to the polynomial centered at 5.

6. Let f(1) (n) = g(n) (0) for some functions f(x) and g(x). Now let the coordinate axes having graph f(x) be rotated by 45 degrees (clockwise). Then the corresponding Mclaurin series of transformed g(x) is

Answer: a [Reason:] The general expansion of Mclaurin is given by .

7. Let τa( f(x)) denote the Taylor series of the polynomial f(x) centered at a. Which of the following exactly happens after the Taylor series is formed.
a) τa( f(x)) = f(x)
b) The Taylor series has the effect of scaling GRAPH OF f(x)
c) The Taylor series transforms the origin
d) Scaled up graph obtained by factor of a

Answer: a [Reason:] The Taylor series has no effect whatsoever on the nature of the polynomial. This is because the Taylor series is itself a polynomial and hence no change takes place. τa( f(x)) = f(x).

8. Let τ(x) be the taylor Series centered at 0. Let the Taylor series be approximated to fifth degree polynomial only, which is the interval over which f(x) = sin(x) can be accurately calculated
a) ( -∞, ∞)
b) [ -2π, 2π].
c) [ -4π, 0].
d) It is accurate for any interval

Answer: b [Reason:] The function sin(x) is known to have a root at every x = nπ The center is at 0. The Taylor polynomial also has five roots, spread symmetrically around the origin. After the polynomial crosses the roots it goes outward to infinity, But the sine function is bounded. Hence, the interval where the roots of sine and polynomial approximately coincide is [ -2π, 2π].

9. The Taylor polynomial of degree 6 is approximated for cos(x). Then the interval in which the function can be accurately calculated using Taylor series (center = 80π)
a) [ -3π, 3π].
b) [ 77.5π, 83.5π].
c) [ -2.5π, 2.5π].
d) [ 77π, 83π].

Answer: b [Reason:] Given the Taylor polynomial has degree 6 we must have 6 roots. The symmetry point is at the center of the Taylor polynomial (which is 80π). The cos(x) function has roots at nπ/2 Hence, the interval over which the roots approximately coincide is [ 77.5π, 83.5π].

## Discrete Mathematics MCQ Set 6

1. In euler theorem x ∂z∂x + y ∂z∂y = nz, here `n` indicates
a) order of z
b) degree of z
c) neither order nor degree
d) constant of z

Answer: a [Reason:] Statement of euler theorem is “if z is an homogeneous function of x and y of order `n` then x ∂z∂x + y ∂z∂y = nz ”.

2. If z = xn f(yx) then
a) y ∂z∂x + x ∂z∂y = nz
b) 1/y ∂z∂x + 1/x ∂z∂y = nz
c) x ∂z∂x + y ∂z∂y = nz
d) 1/x ∂z∂x + 1/y ∂z∂y = nz

Answer: c [Reason:] Since the given function is homogeneous of order n , hence by euler’s theorem x ∂z∂x + y ∂z∂y = nz.

3. Necessary condition of euler’s theorem is
a) z should be homogeneous and of order n
b) z should not be homogeneous but of order n
c) z should be implicit
d) z should be the function of x and y only

Answer: a [Reason:] Answer `a` is correct as statement of euler’s theorem is “if z is an homogeneous function of x and y of order `n` then x ∂z∂x + y ∂z∂y = nz” Answer `b` is incorrect as z should be homogeneous. Answer `c` is incorrect as z should not be implicit. Answer `d` is incorrect as z should be the homogeneous function of x and y not non-homogeneous functions.

4. If
a) 0
b) zln(z)
c) z2 ln⁡(z)
d) z

5. If
a) 2 tan(z)
b) 2 cot(z)
c) tan(z)
d) cot(z)

6. Value of is ,
a) -2.5 u
b) -1.5 u
c) 0
d) -0.5 u

Answer: a [Reason:] Since the function can be written as,

7. If f1(x,y) and f2(x,y) are homogeneous and of order `n` then the function f3(x,y) = f1(x,y) + f2(x,y) satisfies euler’s theorem.
a) True
b) False

Answer: a [Reason:] Since f1(x,y) and f2(x,y) are homogeneous and of order n hence,

9. If z = Sin-1 (xy) + Tan-1 (yx) then x ∂z∂x + y ∂z∂y is
a) 0
b) y
c) 1 + xy Sin-1 (xy)
d) 1 + yx Tan-1 (yx)

Answer: a [Reason:] Given z = Sin-1 (xy) + Tan-1 (yx) Let, u = Sin-1 (xy) and v = Tan-1 (yx) hence z = u + v Now, let u’ = Sin(u) = xy = fxy) hence u’ satisfies euler’s theorem,

10. If f(x,y)is a function satisfying euler’ s theorem then

Answer: a [Reason:] Since f satisfies euler’s theorem,

11. If is
a) Sin(4u) – Cos(2u)
b) Sin(4u) – Sin(2u)
c) Cos(4u) – Sin(2u)
d) Cos(4u) – Cos(2u)

12. If
a) u ln⁡(u)
b) u ln⁡(u)2
c) u [1+ln⁡(u)].
d) 0

## Discrete Mathematics MCQ Set 7

1. f(x, y) = x2 + xyz + z Find fx at (1,1,1)
a) 0
b) 1
c) 3
d) -1

Answer: c [Reason:] fx = 2x + yz

Put (x,y,z) = (1,1,1)

fx = 2 + 1 = 3.

2. f(x, y) = sin(xy) + x2 ln(y) Find fyx at (0, π2)
a) 33
b) 0
c) 3
d) 1

Answer: d [Reason:] fy = xcos(xy) + x2y

fyx = cos(xy) – xysin(xy) + 2xy

Put (x,y) = (0, π2) = 1.

3. f(x, y) = x2 + y3 ; X = t2 + t3; y = t3 + t9 Find dfdt at t=1.
a) 0
b) 1
c)-1
d) 164

Answer: d [Reason:] Using chain rule we have = (2x).(2t + 3t2) + (3y2).(3t2 + 9t8) Put t = 1; we have x = 2; y = 2 = 4.(5) + 12.(12) = 164.

4. f(x, y) = sin(x) + cos(y) + xy2; x = cos(t); y = sin(t) Find dfdt at t = π2
a) 2
b)-2
c) 1
d) 0

Answer: b [Reason:]Using chain rule we have = (cos(x) + y2) . (-sin(t)) + (-sin(y) + 2xy) . (cos(t))

Put t= π2; we have x=0; y=1

= (1 + 1) . (-1) + 0 = -2.

5. f(x, y, z, t) = xy + zt + x2 yzt; x = k3 ; y = k2; z = k; t = √k
Find dfdt at k = 1
a) 34
b) 16
c) 32
d) 61

Answer: b [Reason:] Using Chain rule we have Put k=1; we have x=y=z=t=1 9 + 4 + 2 + 1 = 16.

6. The existence of first order partial derivatives implies continuity
a) True
b) False

Answer: b [Reason:] The mere existence cannot be declared as a condition for contnuity because the second order derivatives should also be continuous.

7. The gradient of a function is parallel to the velocity vector of the level curve
a) True
b) False

Answer: b [Reason:] The gradient is perpendicular and not parallel to the velocity vector of the level curve.

8. f(x, y) = sin(y + yx2) / 1 + x2 Value of fxy at (0,1) is
a) 0
b) 1
c) 67
d) 90

Answer: a [Reason:] First find fy = cos(y + yx2) Hence fyx = fxy = – (2xy).sin(y + yx2) Now put (x,y) = (0,1) = 0.

9. f(x, y) = sin(xy + x3y) / x + x3 Find fxy at (0,1)
a) 2
b) 5
c) 1
d) undefined

Answer: c [Reason:] First find fy = sin(xy + x3y) Hence fyx = fxy = (cos(xy + x3y)) . (y + 3x23y) Now put (x,y) = (0,1) = 1.

## Discrete Mathematics MCQ Set 8

1. In a Poisson Distribution, if ‘n’ is the number of trials and ‘p’ is the probability of success, then the mean value is given by
a) m = np
b) m = (np)2
c) m= np(1-p)
d) m= p

Answer: a [Reason:] For a discrete probability function, the mean value or the expected value is given by For Poisson Distribution substitute in above equation and solve to get µ = m = np.

2. If ‘m’ is the mean of a Poisson Distribution, then variance is given by
a) m2
b) m12
c) m
d) m2

Answer: c [Reason:] For a discrete probability function, the variance is given by Where µ is the mean, substitute , in the above equation and put µ = m to obtain V = m.

3. The p.d.f of Poisson Distribution is given by

Answer: a [Reason:] This is a standard formula for Poisson Distribution, it needs no explanation. Even though if you are interested to know the derivation in detail, you can refer to any of the books or source on internet that speaks of this matter.

4. If ‘m’ is the mean of a Poisson Distribution, the standard deviation is given by
a) m12
b) m2
c) m
d) m2

Answer: a [Reason:] The variance of a Poisson distribution with mean ‘m’ is given by V = m, hence Standard Deviation = (variance)12 = m12.

5. In a Poisson Distribution, the mean and variance are equal
a) True
b) False

Answer: a [Reason:] Mean = m Variance = m ∴ Mean = Variance.

6. In a Poisson Distribution, if mean (m) = e, then P(x) is given by

Answer: b [Reason:] Put m = e, and get correct solution.

7. Poisson distribution is applied for
a) Continuous Random Variable
b) Discrete Random Variable
c) Irregular Random Variable
d) Uncertain Random Variable

Answer: b [Reason:] Poisson Distribution along with Binomial Distribution is applied for Discrete Random variable. Speaking more precisely, Poisson Distribution is an extension of Binomial Distribution for larger values ‘n’. Since Binomial Distribution is of discrete nature, so is its extension Poisson Distribution.

8. If ‘m’ is the mean of Poisson Distribution, the P(0) is given by
a) e-m
b) em
c) e
d) m-e

Answer: a [Reason:] Put x = 0, to obtain e-m.

9. In a Poisson distribution, the mean and standard deviation are equal
a) True
b) False

Answer: b [Reason:] In a Poisson Distribution, Mean = m Standard Deivation = m12 ∴ Mean and Standard deviation are not equal.

10. For a Poisson Distribution, if mean(m) = 1, then P(1) is
a) 1/e
b) e
c) e/2
d) Indeterminate