## Discrete Mathematics MCQ Set 1

1. Find

a) ∞

b) 0

c) Does Not Exist

d) 121

### View Answer

^{3⁄2}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) ∞

### View Answer

^{5⁄2}we have By varying a we get different limits along different paths Hence, Does Not exist is the right answer.

3. Find

a) ∞

b) ^{1}⁄_{2}

c)

d) ^{1}⁄_{3}

### View Answer

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

### View Answer

5. Find

a) 9

b) 0

c) 6

d) 2

### View Answer

6. Find

a) 0

b) ∞

c)

d) -1

### View Answer

7. Find

a)

b) 0

c) ∞

d) Does Not Exist

### View Answer

8. Find

a)

b) -1

c) ∞

d) Does not Exist

### View Answer

9. Find

a) Does Not Exist

b) 0

c)

d) ∞

### View Answer

10. Find

a) 0

b) ∞

c)

d) Does Not Exist

### View Answer

Hence, Does Not Exist is the right answer.

11. Find

a) Does not Exist

b) 0

c) ∞

d)

### View Answer

12. Find

a) ∞

b)

c) ^{1}⁄_{2}

d) Does Not Exist

### View Answer

13. Find

a)

b) ∞

c) 0

d) 990

### View Answer

14. Find

a)

b) ∞

c) 0

d) 990

### View Answer

_{(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 ^{1}⁄_{2}

### View Answer

Thus, put x = t : y = a(t)^{3⁄4} 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

### View Answer

_{1}* t

^{3⁄4}: z = a

_{2}* t

^{3⁄4}

3. Find

a) ∞

b) ^{1}⁄_{3}

c) 1

d) Does Not Exist

### View Answer

4. Find

a) ∞

b) 123

c) 9098

d) 8

### View Answer

5. Find

a) 1990

b) ∞

c) Does Not Exist

d) 0

### View Answer

_{1}.t

^{1⁄2}: z = a

_{2}.t : w = a

_{3}.t By changing the values of a

_{1}: a

_{2}: a

_{3}we get different values of limit. Hence, Does Not Exist is the right answer.

6. Find

a) 700

b) 701

c) 699

d) 22

### View Answer

7. Given that limit exists find

a) 1

b) ^{3}⁄_{5}

c) ^{1}⁄_{2}

d) 0

### View Answer

8. Given that limit exist find

a) 2

b) 1

c) 4

d) 3

### View Answer

9. Given that limit exists find

a) 1

b) ^{1}⁄_{2}

c) ^{1}⁄_{7}

d) ^{2}⁄_{7}

### View Answer

10. Given that limit exists find

a) ∞

b) 1

c) 0

d) ln(^{4}⁄_{5})

### View Answer

11. Given that limit exists

a) 99

b) 0

c) 1

d) 100

### View Answer

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 ^{1}⁄_{2}

d) Insufficient information

### View Answer

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 ^{1}⁄_{2}

d) Insufficient information

### View Answer

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)

### View Answer

^{2}Substituting in Option (d) we get By varying a we get different limits

## Discrete Mathematics MCQ Set 3

1. The p^{th} derivative of a q^{th} degree monic polynomial, where p, q are positive integers and 2p^{4} + 3pq^{3⁄2} = 3q^{3⁄2} + 2qp^{3} is given by

a) Cannot be generally determined

b) (q – 1)!

c) (q)!

d) (q – 1)! * p^{q}

### View Answer

^{4}+ 3pq

^{3⁄2}After simplification, we get

(2p^{3} + 3q^{1⁄2}) (p – q) = 0
This gives us two possibilities
2p^{3} = – 3q^{1⁄2} OR

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

Hence, we get

p = q

Thus the p^{th} 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) ^{3}⁄_{2}

b) ^{1}⁄_{2}

c) 1

d) 0

### View Answer

^{2}+ 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) = x^{2} – x + c

Now the required value can be computed as

f(1) – f(0) = (1^{2} – 1 + c) – (0^{2} – 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

### View Answer

4. Let f(x) = ln(x^{2} + 5x + 6) then the value of f^{(30)}(1) is given by

### View Answer

f(x) = ln(x^{2} + 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 n^{th} derivative of logarithmic function
We have

5. f(x) = ∫_{0} ^{π⁄2} 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

### View Answer

6. Let f(x) = x^{9} e^{x} then the ninth derivative of f(x) at x = 0 is given by

a) 9!

b) 9! * e^{9}

c) 10!

d) 21!

### View Answer

^{x}as a mclaurin series and then multiply it by x

^{9}We have Substituting x = 0 gives us the result f

^{(9)}(0) = 9!.

7. The following moves are performed on g(x)

(i) Pick (x_{0}, y_{0}) 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 (x_{1}, y_{1})

(ii) Let the new position of (x_{0}, y_{0})be (x_{0}, y_{1})

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 n^{th} 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 – x^{2}

b) xy^{3⁄2} + y = constant

c) x^{9} y^{3⁄2} + y^{6} x^{3⁄2} = constant

d) x^{7} y ^{8} + 4y = constant

### View Answer

^{th}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 – x^{2} we have after changing the position

y = √1 – y^{2} 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 1^{3}, 2^{3} and 3^{3} respectively. Then the value of f(0) + f(1) – 2f(-1) is

a) 76

b) 86

c) 126

d) 41.5

### View Answer

^{3}+ bx

^{2}+ cx + d

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

1^{3} = 1 = 3a + 2b + c

The second derivative at x = 1 yields the following expression

2^{3} = 8 = 6a + 2b

The third derivative at x = 1 yields the following equation

3^{3} = 27 = 6a

Solving for a, b and c simultaneously yields
(a, b, c) = (^{9}⁄_{2}, ^{-19}⁄_{2}, ^{13}⁄_{2})
Hence the assumed polynomial is f(x) = 9x^{3} – 19x^{2} + 13x ⁄ 2 + d

Now the given expression can be evaluated as

f(0) + f(1) – 2f(-1) = (d) + (^{3}⁄_{2} + d) – 2(-20 + d)

= 40 + ^{3}⁄_{2} = 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) ^{101}⁄_{100!}

d) ^{1}⁄_{99!}

### View Answer

^{100!}⁄

_{101}.

10. Let f(x) = ln(x^{3} – 3x^{2} – 16x -12) , then the 1729^{th} derivative at x = 234 is

### View Answer

^{th}derivative of logarithmic functions

11. Find the value of using n^{th} derivatives

a) – 2 * sin(1)

b) 3 * sin(1)

c) 3 * cos(1)

d) – 3 * cos(1)

### View Answer

^{x}) 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) = -2e^{x}sin(e^{x})
Substituting x = 0 we get
f^{(3)}(0) = -2sin(1).

12. Let .Find the third derivative at x = 0

a) 4

b) ^{1}⁄_{3}

c) Undefined

d) ^{1}⁄_{4}

### View Answer

^{-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 x

^{3}term in the infinite polynomial product above The third degree terms can be grouped apart as follows

= ^{x3}⁄_{3} – ^{x3}⁄_{2} + ^{x3}⁄_{2} Hence the third derivative at x = 0 is simply the coefficient of the third degree term, which is
coefficient(^{x3}⁄_{3}) = ^{1}⁄_{3}.

## Discrete Mathematics MCQ Set 4

1. The Mclaurin Series expansion of sin(e^{x}) is

### View Answer

2. What is the coefficient of x^{101729} in the series expansion of cos(sin(x))

a) 0

b) ^{1}⁄_{101729!}

c) ^{-1}⁄_{101729!}

d) 1

### View Answer

3. Let τ(X) be the Taylor Series expansion of f(x) = x^{3} + x^{2} + 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

### View Answer

^{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

### View Answer

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.

### View 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)

c) No unique answer exist

d) Such function is not continuous

### View Answer

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

### View Answer

8. Let Mclaurin series of some f(x) be given recursively, where a_{n} denotes the coefficient of x_{n}in the expansion. Also given a_{n} = a_{n-1} / n and a_{0} = 1, which of the

following functions could be f(x)

a) e^{x}

b) e^{2x}

c) c + e^{x}

d) No closed form exists

### View Answer

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

### View Answer

## Discrete Mathematics MCQ Set 5

1. Let τ_{a} [ 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

### View Answer

τa_{1}(τa_{2}(…….(f(x))….)) = f(x)

Where a_{1}, a_{2}………….a_{n} 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.

### View Answer

_{0}= c1 and a

_{1}= 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

### View Answer

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

### View Answer

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

### View Answer

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

### View Answer

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

### View Answer

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

### View Answer

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].

### View Answer

## 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

### View Answer

^{∂z}⁄

_{∂x}+ y

^{∂z}⁄

_{∂y}= nz ”.

2. If z = x^{n} f(^{y}⁄_{x}) 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

### View Answer

^{∂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

### View Answer

^{∂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) z^{2} ln(z)

d) z

### View Answer

5. If

a) 2 tan(z)

b) 2 cot(z)

c) tan(z)

d) cot(z)

### View Answer

6. Value of is ,

a) -2.5 u

b) -1.5 u

c) 0

d) -0.5 u

### View Answer

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

### View Answer

### View Answer

9. If z = Sin^{-1} (^{x}⁄_{y}) + Tan^{-1} (^{y}⁄_{x}) then x ^{∂z}⁄_{∂x} + y ^{∂z}⁄_{∂y} is

a) 0

b) y

c) 1 + ^{x}⁄_{y} Sin^{-1} (^{x}⁄_{y})

d) 1 + ^{y}⁄_{x} Tan^{-1} (^{y}⁄_{x})

### View Answer

^{-1}(

^{x}⁄

_{y}) + Tan

^{-1}(

^{y}⁄

_{x}) Let, u = Sin

^{-1}(

^{x}⁄

_{y}) and v = Tan

^{-1}(

^{y}⁄

_{x}) hence z = u + v Now, let u’ = Sin(u) =

^{x}⁄

_{y}= f

^{x}⁄

_{y}) hence u’ satisfies euler’s theorem,

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

### View Answer

11. If is

a) Sin(4u) – Cos(2u)

b) Sin(4u) – Sin(2u)

c) Cos(4u) – Sin(2u)

d) Cos(4u) – Cos(2u)

### View Answer

12. If

a) u ln(u)

b) u ln(u)^{2}

c) u [1+ln(u)].

d) 0

### View Answer

## Discrete Mathematics MCQ Set 7

1. f(x, y) = x^{2} + xyz + z Find f_{x} at (1,1,1)

a) 0

b) 1

c) 3

d) -1

### View Answer

_{x}= 2x + yz

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

f_{x} = 2 + 1 = 3.

2. f(x, y) = sin(xy) + x^{2} ln(y) Find f_{yx} at (0, ^{π}⁄_{2})

a) 33

b) 0

c) 3

d) 1

### View Answer

_{y}= xcos(xy) +

^{x2}⁄

_{y}

f_{yx} = cos(xy) – xysin(xy) + ^{2x}⁄_{y}

Put (x,y) = (0, ^{π}⁄_{2})
= 1.

3. f(x, y) = x^{2} + y^{3} ; X = t^{2} + t^{3}; y = t^{3} + t^{9} Find ^{df}⁄_{dt} at t=1.

a) 0

b) 1

c)-1

d) 164

### View Answer

^{2}) + (3y

^{2}).(3t

^{2}+ 9t

^{8}) Put t = 1; we have x = 2; y = 2 = 4.(5) + 12.(12) = 164.

4. f(x, y) = sin(x) + cos(y) + xy^{2}; x = cos(t); y = sin(t) Find ^{df}⁄_{dt} at t = ^{π}⁄_{2}

a) 2

b)-2

c) 1

d) 0

### View Answer

^{2}) . (-sin(t)) + (-sin(y) + 2xy) . (cos(t))

Put t= ^{π}⁄_{2}; we have x=0; y=1

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

5. f(x, y, z, t) = xy + zt + x^{2} yzt; x = k^{3} ; y = k^{2}; z = k; t = √k

Find ^{df}⁄_{dt} at k = 1

a) 34

b) 16

c) 32

d) 61

### View Answer

6. The existence of first order partial derivatives implies continuity

a) True

b) False

### View Answer

7. The gradient of a function is parallel to the velocity vector of the level curve

a) True

b) False

### View Answer

8. f(x, y) = sin(y + yx^{2}) / 1 + x^{2} Value of f_{xy} at (0,1) is

a) 0

b) 1

c) 67

d) 90

### View Answer

_{y}= cos(y + yx

^{2}) Hence f

_{yx}= f

_{xy}= – (2xy).sin(y + yx

^{2}) Now put (x,y) = (0,1) = 0.

9. f(x, y) = sin(xy + x^{3}y) / x + x^{3} Find f_{xy} at (0,1)

a) 2

b) 5

c) 1

d) undefined

### View Answer

_{y}= sin(xy + x

^{3}y) Hence f

_{yx}= f

_{xy}= (cos(xy + x

^{3}y)) . (y + 3x

^{23}y) 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

### View Answer

2. If ‘m’ is the mean of a Poisson Distribution, then variance is given by

a) m^{2}

b) m^{1⁄2}

c) m

d) ^{m}⁄_{2}

### View Answer

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

### View Answer

4. If ‘m’ is the mean of a Poisson Distribution, the standard deviation is given by

a) m^{1⁄2}

b) m^{2}

c) m

d) ^{m}⁄_{2}

### View Answer

^{1⁄2}= m

^{1⁄2}.

5. In a Poisson Distribution, the mean and variance are equal

a) True

b) False

### View Answer

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

### View Answer

7. Poisson distribution is applied for

a) Continuous Random Variable

b) Discrete Random Variable

c) Irregular Random Variable

d) Uncertain Random Variable

### View Answer

8. If ‘m’ is the mean of Poisson Distribution, the P(0) is given by

a) e^{-m}

b) e^{m}

c) e

d) m^{-e}

### View Answer

^{-m}.

9. In a Poisson distribution, the mean and standard deviation are equal

a) True

b) False

### View Answer

^{1⁄2}∴ 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

### View Answer

11. The recurrence relation between P(x) and P(x +1) in a Poisson distribution is given by

a) P(x+1) – m P(x) = 0

b) m P(x+1) – P(x) = 0

C) (x+1) P(x+1) – m P(x) = 0

d) (x+1) P(x) – x P(x+1) = 0

### View Answer

^{-1}m

^{x+1}/(x + 1)! Divide P(x+1) by P(x) and rearrange to obtain (x+1) P(x+1) – m P(x) = 0.