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## Discrete Mathematics MCQ Set 1

1. Expansion of function f(x) is
a) f(0) + x1! f (0) + x22! f (0)…….+xnn! fn (0)
b) 1 + x1! f (0) + x22! f (0)…….+xnn! fn (0)
c) f(0) – x1! f (0) + x22! f (0)…….+(-1)^n xnn! fn (0)
d) f(1) + x1! f (1) + x22! f (1)…….+xnn! fn (1)

Answer: a [Reason:] By Maclaurin’s series, f(0) + x1! f (0) + x22! f (0)…….+xnn! fn (0)

2. The necessary condition for the maclaurin expansion to be true for function f(x) is
a) f(x) should be continuous
b) f(x) should be differentiable
c) f(x) should exists at every point
d) f(x) should be continuous and differentiable

Answer: d [Reason:] By Maclaurin’s series, f(0) + x1! f (0) + x22! f (0)…….+xnn! fn (0)

Where, f(x) should be continuous and differentiable upto nth derivative.

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

4. The expansion of eSin(x) is
a) 1 + x + x22 + x48 +….
b) 1 + x + x22x48 +….
c) 1 + x – x22 + x48 +….
d) 1 + x + x36x510 +….

Answer: b [Reason:] Now f(x) = eSin(x), f(0) = 1 Hence, f (x)=f(x)Cos(x), f (0) = 1

f (x)=f (x)Cos(x) – f(x)Sin(x),f (0)=1

f”’ (x)=f (x)Cos(x) – 2f (x)Sin(x) – f(x) cos⁡(x),f”’ (x) = 0

f”” (x)=f”’ (x)Cos(x) – 3f (x)Sin(x) – 3f (x) cos⁡(x) + f(x) sin⁡(x), f”” (x) = -3

Hence,

f(x) = eSin(x) = 1 + x + x22x48 +…. (By mclaurin’ sexpansion)

5. Expansion of y = Sin-1(x) is Answer: a [Reason:] Given, y = Sin-1(x), hence at x = 0, y = 0 Now, differentiating it, we get 6. Find the expansion of f(x) = ln⁡(1+ex) Answer: a [Reason:] Given, f(x) = ln⁡(1+ex), f(0) = ln⁡(2) Differentiating it we get 7. Find the expansion of exSin(x) Answer: b [Reason:] Given, f(x) = exSin(x), f(0) = 1 8. Given f(x)= ln⁡(cos⁡(x) ),calculate the value of ln⁡(cos⁡(π2)).
a) -1.741
b) 1.741
c) 1.563
d) -1.563

Answer: a [Reason:] Given f(x) = ln⁡(Cos(x)), f(0) = 0 Differentiating it f'(x) = – tan⁡(x), f'(0) = 0 9. Find the expansion of cos(xsin(t)). Answer: b [Reason:] 10. Find the expansion of Sin(lSin-1 (x)). Answer: b [Reason:] Given, y = f(x) = Sin(lSin-1 (x)) Now, differentiating, 11. Expand (1 + x)1x, gives
a) e [1 + x2 + 11x224 -…..].
b) e [1 – x2 + 11x224 -…..].
c) e [x211x224 -…..].
d) e [x2 + 11x224 -…..].

Answer: b [Reason:] Given, y = (1 + x)1x 12. Find the solution of differential equation, dydx = xy + x2, if y = 1 at x = 0. Answer: d [Reason:] Given dydx = xy + x2

hence, dydy (x=0) = 0

and, d2ydx2 = xy1 + y + 2x

hence, y2 = xy1 + y + 2x

hence, d2ydx2 (x=0)=1

Differentiating it n times we get, ## Discrete Mathematics MCQ Set 2

1. A and B are two events such that P(A) = 0.4 and P(A ∩ B) = 0.2 Then P(A ∩ B) is equal to
a) 0.4
b) 0.2
c) 0.6
d) 0.8

Answer: a [Reason:] P(A ∩ B) = P(A – (A ∩ B)) = P(A) – P(A ∩ B) = 0.6 – 0.2 Using P(A) = 1 – P(A) = 0.4.

2. A problem in mathematics is given to three students A, B and C . If the probability of A solving the problem is 12 and B not solving it is 14 . The whole probability of the problem being solved is 6364 then what is the probability of solving it ?
a) 18
b) 164
c) 78
d) 12

Answer: c [Reason:] Let A be the event of A solving the problem Let B be the event of B solving the problem Let C be the event of C solving the problem Given P(a) = 12, P(~B) = 14 and P(A ∪ B ∪ C) = 63/64

We know P(A ∪ B ∪ C) = 1 – P(A ∪ B ∪ C)

= 1 – P(ABC)

= 1 – P(A) P(B) P(C)

Let P(C) = p ie 6364 = 1 – (12)(14)(p)

= 1 – p8 ⇒ P =1/8 = P(C) ⇒P(C) = 1 – P = 1 – 18 = 78.

3. Let A and B be two events such that P(A) = 15 While P(A or B) = 12. Let P(B) = P. For what values of P are A and B independent?
a) 110 and 310
b) 310 and 45
c) 38 only
d) 310

Answer: c [Reason:] For independent events, P(A ∩ B) = P(A) P(B) P(A ∪ B) = P(A) + P(B) – P(A ∩ B) = P(A) + P(B) – P(A) P(B) = 15 + P (15)P ⇒ 12 = 15 + 45P ⇒ P= 38.

4. If A and B are two mutually exclusive events with P(~A) = 56 and P(b) = 13 then P(A /~B) is equal to
a) 14
b) 12
c) 0, since mutually exclusive
d) 518

Answer: a [Reason:] As A and B are mutually exclusive we have .

5. If A and B are two events such that P(a) = 0.2, P(b) = 0.6 and P(A /B) = 0.2 then the value of P(A /~B) is
a) 0.2
b) 0.5
c) 0.8
d) 13

Answer: a [Reason:] For independent events, P(A /~B) = P(a) = 0.2.

6. If A and B are two mutually exclusive events with P(a) > 0 and P(b) > 0 then it implies they are also independent
a) True
b) False

Answer: b [Reason:] P(A ∩ B) = 0 as (A ∩ B) = ∅ But P(A ∩ B) ≠ 0 , as P(a) > 0 and P(b) > 0 P(A ∩ B) = P(A) P(B), for independent events.

7. Let A and B be two events such that the occurrence of A implies occurrence of B, But not vice-versa, then the correct relation between P(a) and P(b) is
a) P(A) < P(B)
b) P(B) ≥ P(A)
c) P(A) = P(B)
d) P(A) ≥ P(B)

Answer: b [Reason:] Here, according to the given statement A ⊆ B In this case P(B) = P(A ∪ (A ∩ B)) = P(A) + P(A ∩ B) Therefore, P(B) ≥ P(A)

8. In a sample space S, if P(a) = 0, then A is independent of any other event
a) True
b) False

Answer: a [Reason:] P(a) = 0 (impossible event) Hence, A is not dependent on any other event.

9. If then,
a) P(a) > P(b)
b) P(a) > P(b)
c) P(a) = P(b)
d) P(~A) less than P(~B)

Answer: c [Reason:] A ⊂ B, B ⊂ A That means A = B Hence P(a) = P(b).

10. If A is a subset of B then,
a) P(a) is greater than P(b)
b) P(~A) is greater than or equal to P(~B)
c) P(b) is equal to P(a)
d) P(b) is equal to P(~B)

Answer: b [Reason:] A is a subset of B which means (~B) is a subset of (~A) Therefore P(~A) is greater than or equal to P(~B).

11. If A is a perfect subset of B and P(a < Pb), then P(B – A) is equal to
a) P(a) / P(b)
b) P(a)P(b)
c) P(a) + P(b)
d) P(b) – P(a)

Answer: d [Reason:] From Basic Theorem of probability, P(B – A) = P(b) – P(a), this is true only if the condition given in the question is true.

12. What is the probability of an impossible event?
a) 0
b) 1
c) Not defined
d) Insufficient data

Answer: a [Reason:] If the probability of an event is 0, then it is called as an impossible event.

13. If A = A1 ∪ A2……..∪ An, where A1…An are mutually exclusive events then Answer: b [Reason:] A = A1 ∪ A2……..∪ An, where A1…An Since A1…An are mutually exclusive P(a) = P(A1) + P(A2) + … + P(An) Therefore .

## Discrete Mathematics MCQ Set 3

1. If P(BA) = p(b), then P( A and B) =
a) p(b)
b) p(a)
c) p(b)p(a)
d) p(a) + p(b)

Answer: c [Reason:] P(B /A) = p(b) implies A and B are independent events Therefore, P(A and B) = p(a)p(b).

2. Two unbiased coins are tossed. What is the probability of getting at most one head?
a)12
b)13
c)16
d)34

Answer: d [Reason:] Total outcomes = (HH, HT, TH, TT) Favorable outcomes = (TT, HT, TH) At most one head refers to maximum one head, Therefore, Probability = 34.

3. If A and B are two events such that p(a) > 0 and p(b) is not a sure event, then
P(~A /~B) =
a) 1 – P(A /B)
b) P(~A)/P(~B)
c) Not Defined
d) (1 – P(A or B) ) /P(~B)

Answer: d [Reason:] From definition of conditional probability we have 4. If A and B are two events, then the probability of exactly one of them occurs is given by
a) P(A ∩ B) + P( A ∩ B)
b) P(A) + P(B) – 2P(A) P(B)
c) P(A) + P(B) – 2P(A) P(B)
d) P(A) + P(B) – P(A ∩ B)

Answer: a [Reason:] The set corresponding to the required outcome is (A ∩ B) ∪ ( A ∩ B) Hence the required probability is P(A ∩ B) + P( A ∩ B).

5. The probability that at least one of the events M and N occur is 0.6. If M and N have probability of occurring together as 0.2, then P(~M) + P(~N) is
a) 0.4
b) 1.2
c) 0.8
d) Indeterminate

Answer: b [Reason:] Given : P(M or N) = 0.6, P(M and N) = 0.2 P(M or N) + P(M and N) = P(M) + P(N) 2 – ( P(M or N) + P(M and N) ) = 2 – ( P(M) + P(N) ) = ( 1 – P(M) ) + ( 1 – P(N) ) 2 – (0.6 + 0.2) = P(~M) + P(~N) = 1.2.

6. A jar contains ‘y’ blue colored balls and ‘x’ red colored balls. Two balls are pulled from the jar without replacing. What is the probability that the first ball Is blue and second one is red Answer: b [Reason:] Number of blue balls = y Number of Red balls = x Total number of balls = x + y Probability of Blue ball first = y / x + y No. of balls remaining after removing one = x + y – 1 7. A survey determines that in a locality, 33% go to work by Bike, 42% go by Car, and 12% use both. The probability that a random person selected uses neither of them is
a) 0.29
b) 0.37
c) 0.61
d) 0.75

Answer: b [Reason:] Given: p(b) = 0.33, Pc) = 0.42 P(B and C) = 0.12 P(~B and ~C) = ? P(~B and ~C) = 1 – P(B or C) = 1 – p(b) – Pc) + P(B and C) = 1 – 0.22 – 0.42 + 0.12 = 0.37.

8. A coin is biased so that its chances of landing Head is 23 . If the coin is flipped 3 times, the probability that the first 2 flips are heads and the 3rd flip is a tail is
a) 427
b) 827
c) 49
d) 29

Answer: a [Reason:] Required probability = 23 x 23 x 13 = 427.

9. Husband and wife apply for two vacant spots in a company. If the probability of wife getting selected and husband getting selected are 3/7 and 2/3 respectively, what is the probability that neither of them will be selected?
a) 27
b) 57
c) 421
d) 1721

Answer: c [Reason:]Let H be the event of husband getting selected W be the event of wife getting selected Then, the event of neither of them getting selected is = (H ∩ W) P (H ∩ W) = P (H) x P (W) = ( 1 – P (H) ) x ( 1 – P (W) ) = ( 1 – 23 ) x (1 – 37 ) = 421.

10. For two events A and B, if P (B) = 0.5 and P (A ∪ B) = 0.5, then
P (A|B) =
a) 0.5
b) 0
c) 0.25
d) 1

Answer: d [Reason:] We know that, P (A│B) = P(A ∩ B) /P(B) = P((A ∪B) /P(B) ) = ( 1 – P(A ∪ B) ) /P(B) = (1 – 0.5) /0.5 = 1.

11. A fair coin is tossed thrice, what is the probability of getting all 3 same outcomes?
a) 34
b) 14
c) 12
d) 16

Answer: b [Reason:]S= {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} All 3 same outcomes mean either all head or all tail Total outcomes = 8 Favourable outcomes = {HHH, TTT} = 2 ∴ required probability = 28 = 14.

12. A bag contains 5 red and 3 yellow balls. Two balls are picked at random. What is the probability that both are of the same colour? Answer: a [Reason:]Total no.of balls = 5R+3Y = 8 No.of ways in which 2 balls can be picked = 8C2 Probability of picking both balls as red = 5C2 /8C2 Probability of picking both balls as yellow = 3C2 /8C2 ∴ required probability .

## Discrete Mathematics MCQ Set 4

1. nth derivative of Sinh(x) is
a) 0.5(ex – e-x)
b) 0.5(e-x – ex)
c) 0.5(ex – (-1)n e-x)
d) 0.5((-1)-n e-x -ex)

Answer: c [Reason:] Y = Sinh(x) Y = 0.5[ex – e-x]. y1 = 0.5 [ex – (-1)e-x]. y2 = 0.5 [ex – (-1)2 e-x]. Similarly, yn = 0.5 [ex – (-1)n e-x].

2. If y=log⁡(x(x2 – 1)) , then nth derivative of y is ?
a) (-1)(n-1) (n-1)!(x(-n) + (x-1)(-n) + (x+1)(-n) )
b) (-1)n (n)! (x(-n-1) + (x-1)(-n-1) + (x+1)(-n-1) )
c) (-1)(n+1) (n+1)!(x(-n) + (x-1)(-n) + (x+1)(-n) )
d) (-1)n(n)! (x(-n-1) + (x-1)(-n+1) + (x+1)(-n+1) )

Answer: a [Reason:] Y=log(x) – log(x2 – 1) y1 = x(-1)-2x/(x2-1) y1 = x(-1)-(x-1)(-1) + (x+1)(-1) yn = (-1)(n-1) (n-1)!(x(-n)-(x-1)(-n) + (x+1)(-n)).

3. If x = a(Cos(t) + t2) and y = a(Sin(t) + t2 + t3) then dy/dx equals to
a) (Cos(t) + 3t2 + 2t) / (-Sin(t) + 2t)
b) (Sin(t) + 3t2 + 2t) / (-Cos(t) + 2t)
c) (Sin(t) + 3t2 + 2t) / (Cos(t) + 2t)
d) (Cos(t) + 3t2 + 2t) / (Sin(t) + 2t)

Answer: a [Reason:] dx/dt = a(-Sin(t) + 2t) dy/dt = a(Cos(t) + 2t + 3t2) Then, dy/dx = (Cos(t) + 3t2+2t)/(-Sin(t) + 2t).

4. If y=tan(-1)⁡(x) , then which one is correct ?
a) y3 + y12 + 4xy2 y1=0
b) y3 + y12 + xy2 y1=0
c) y3 + 2y12 + xy2 y1=0
d) y2 + 2y12 + 4xy2 y1=0

Answer: d Explanation: y = tan(-1)⁡(x) 5. What is the value of dn (xm)dxn for m<n, m=n, m>n ?
a) 0, n!, mPn x(m-n)
b) mPn x(m-n), n!, 0
c) 0, n!, mCn x(m-n)
d) mCn x(m-n), n!, 0

Answer: a [Reason:] For, m > n 6. Which of the following is true
a) Value of dm (Sin(nx))dxm is always positive for m=0, 1, 4, 5, 8, 9… for 0 < nx < π2 and n<0.
b) Value of dm (Sin(nx))dxm is always positive for m=2, 3, 6, 7, 10, 11… for 0 < nx < π2 and n>0.
c) Value of dm (Sin(nx))dxm is always positive for m=0, 1, 4, 5, 8, 9… for 0 < nx < π2 and n>0.
d) Value of dm (Sin(nx))dxm is always positive for m=2, 3, 6, 7, 10, 11… for 0 < nx < π2 and n<0.

Answer: c [Reason:] 7. If nth derivative of eax sin⁡(bx+c) cos⁡(bx+c) is, eax rn sin⁡(bx+c+2) cos⁡(bx+c+2) then, Answer: b [Reason:] y = eax sin⁡(bx+c) cos⁡(bx+c) y = eax sin⁡2(bx+c)/2 yn = eax rn sin⁡(2(bx+c+nα/2))/2 yn = eax rn sin⁡(bx+c+nα/2) cos⁡(bx+c+nα/2) where r = √(a2+4b2 ) , α = tan(-1)⁡2b/a.

8. If y=x4x2-1 , Then,
a) 0.5*(-1)n (n-1)! [(x-1)-n-1 + (x+1)-n-1 ].
b) 0.5*(-1)n (n-1)! [x– n-1 + (x-1)-n-1 + (x+1)-n-1].
c) 0.5*(-1)n (n-1)! [(x-1)-n + (x+1)-n) ].
d) 0.5*(-1)n (n-1)! [x-n + (x-1)-n + (x+1)-n].

Answer: a [Reason:] 9. If y=sin(-1)⁡(x) then select the true statement
a) y2 = xy13
b) y3 = xy23
c) y2 = xy12
d) y3 = xy12

Answer: a [Reason:] 10. nth derivative of y = sin2x cos3x is
a) 18 cos⁡(x + 2) –116 5n cos⁡(x + 2) – 116 3n cos⁡(3x + 2)
b) 18 sin⁡(x+2) –116 5n cos⁡(x + 2) – 116 3n cos⁡(3x + 2)
c) 18 cos⁡(x+2) –116 5n sin⁡(x + 2) – 116 3n sin⁡(3x + 2)
d) 18 sin⁡(x + 2) –116 5n sin⁡(x + 2) – 116 3n sin⁡(3x + 2)

Answer: a [Reason:] y = sin2x cos2x cos(x) y = 14 sin22x cos2x cos(x) y = 18 (2sin22x) cos(x) y = 18 (1 – cos4x) cos(x) y = 18 (1 – cos4x) cos(x) y = 18 cos(x) – 18 cos4x cos(x) y = 18 cos(x) – 116 (cos5x + cos(3x)) Now, nth derivative is yn = 18 cos⁡(x + 2) – 116 5n cos⁡(x + 2) – 116 3n cos⁡(3x + 2).

11. If , Then value of ‘c’ equals to
a) 1
b) 2
c) 3
d) 4

Answer: a [Reason:] 12. If y = exSin-1(x) and , Then find the value of ‘c’ ?
a) -2
b) 2
c) -0.5
d) 0.5

Answer: a [Reason:] 13. Find nth derivative of y = Sin(x) Cos3(x) Answer: a [Reason:] 14. If nth derivative of then find the value of a and b
a) -1, -2
b) 2, 1
c) 1, 2
d) -2, -1

Answer: c [Reason:] ## Discrete Mathematics MCQ Set 5

1. Mean Value Theorem tells about the
a) Existence of point c in a curve where slope of a tangent to curve is equal to the slope of line joining two points in which curve is continuous and differentiable
b) Existence of point c in a curve where slope of a tangent to curve is equal to zero
c) Existence of point c in a curve where curve meets y axis
d) Existence of point c in a curve where curve meets x axis

Answer: a [Reason:] Statement of Mean Value Theorem is that, If function f(x) is continuous in closed interval [a, b] and differentiable in open interval (a, b), then there exists a point ‘c’ such that c∈(a,b) and f’(c) = [f(b)-f(a)]/(b-a).

2. If f(a) is euquals to f(b) in Mean Value Theorem, then it becomes
a) Lebniz Theorem
b) Rolle’s Theorem
c) Taylor Series of a function
d) Leibnit’x Theorem

Answer: b [Reason:] According to Mean Value Theorem, If a function exist at pt. ‘a’, ‘b’ and continuous in closed interval [a, b] and differentiable in open interval (a, b) then there exists a point ‘c’, such that c∈(a,b), Where, f’(c)= [f(b)-f(a)]/(b-a). Hence, By putting f(b) = f(a) in the statement of Mean Value Theorem, we get f’(c) = [f(b)-f(a)]/(b-a) = 0 . Which is a statement of Rolle’s Theorem.

3. Mean Value theorem is applicable to the
a) Functions differentiable in closed interval [a, b] and continuous in open interval (a, b).
b) Functions continuous in closed interval [a, b] only and having same value at point ‘a’ and ‘b’
c) Functions continuous in closed interval [a, b] and differentiable in open interval (a, b).
d) Functions differentiable in open interval (a, b) only and having same value at point ‘a’ and ‘b’

Answer: c [Reason:] Statement of Mean Value Theorem is that, If function f(x) is continuous in closed interval [a, b] and differentiable in open interval (a, b), then there exists a point ‘c’ such that c∈(a,b) and f’(c) = [f(b)-f(a)]/(b-a).

4. Mean Value theorem is also known as
a) Rolle’s Theorem
b) Lagrange’s Theorem
c) Taylor Expansion
4) Leibnit’s Theorem

Answer: b [Reason:] Statement of Mean Value Theorem is that, If function f(x) is continuous in closed interval [a, b] and differentiable in open interval (a, b), then there exists a point ‘c’ such that c∈(a,b) and f’(c) = [f(b)-f(a)]/(b-a). It is also known as Lgrange’s Theorem.

5. Find the point c in the curve f(x) = x3 + x2 + x + 1 in the interval [0, 1] where slope of a tangent to a curve is equals to the slope of a line joining (0,1)
a) 0.64
b) 0.54
c) 0.44
d) 0.34

Answer: b [Reason:] f(x) = x3 + x2 + x + 1 f(x) is continuous in given interval [0,1]. f’(x) = 3x2+2x+1 Since, value of f’(x) is always finite in interval (0, 1) it is differentiable in interval (0, 1). f(0) = 1 f(1) = 4 By mean value theorem, f’(c) = 3c2 + 2c + 1 = (4-1)/(1-0) = 3 ⇒ c = 0.548,-1.215 Since c belongs to (0, 1) c = 0.54.

6. Find the equation of curve whose roots gives the point which lies in the curve f(x) = xSin(x) in the interval [0, π2] where slope of a tangent to a curve is equals to the slope of a line joining (0, π2)
a) c = -Sec(c) – Tan(c)
b) c = -Sec(c) – Tan(c)
c) c = Sec(c) +Tan(c)
d) c = Sec(c) – Tan(c)

Answer: d [Reason:] f(x) = xSin(x) Since f1(x) = x and f2(x)=Sin(x) both are continuous in interval [0, π2], the curve f(x)=f1(x)f2(x) is also continuous. f’(x) = xCos(x) + Sin(x) f’(x) always have finite value in interval [0, π2] hence it is differentiable in interval (0, π2). f(0) = 0 f(π2) = π2

By mean value theorem, f’(c) = cCos(c) + Sin(c) = (π2 – 0)/(π2 – 0)=1 Hence, c = Sec(c) – Tan(c) is the required curve.

7. Can Mean Value Theorem be applied in the curve a) True
b) False

Answer: b [Reason:] Continuity Check Function f(x) is not continuous hence mean value theorem cannot be applied.

8. Find point c between [2,9] where, the slope of tangent to the function f(x)=1+∛x-1 at point c is equals to the slope of a line joining point (2,f(2)) and (9,f(9)).
(Providing given function is continuous and differentiable in given interval).
a) -2.54
b) 4.56
c) 4.0
d) 4.9

Answer: b [Reason:] Since the given function is continuous and differentiable in a given interval, f(2) = 2

f(9) = 3 Applying mean value theorem, f’(c) = 1/3∛(c-1)2 = [f(9)-f(2)]/(9-2) = 1/7

c = 1 ± (7/3)(3/2)

c = 4.56,-2.54 Since c lies in (2,9), c = 4.56.

9. Find point c between [-1,6] where, the slope of tangent to the function f(x) = x2+3x+2 at point c is equals to the slope of a line joining point (-1,f(-1)) and (6,f(6)).
(Providing given function is continuous and differentiable in given interval).
a) 2.5
b) 0.5
c) -0.5
d) -2.5

Answer: a [Reason:] Since the given function is continuous and differentiable in a given interval, f(-1) = 0

f(6) = 56 Applying mean value theorem,

f’(c) = 2c+3 = [f(6)-f(-1)]/[6-(-1)] = 56/7 = 8 c = 5/2 c = 2.5.

10. If f(x) = Sin(x)Cos(x) is continuous and differentiable in interval (0, x) then 