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## Discrete Mathematics MCQ Set 1

1. Rolle’s Theorem tells about the
a) Existence of point c where derivative of a function becomes zero.
b) Existence of point c where derivative of a function is positive.
c) Existence of point c where derivative of a function is negative.
d) Existence of point c where derivative of a function is either positive or negative.
Considering function attains same values at point a, b
Where, c∈(a,b).

Answer: a [Reason:] Statement of Rolle’s Theorem is that, If function f(x) attains same value at point ‘a’ and ‘b’ [ f(a) = f(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) and f’(c) = 0.

2. Rolle’s Theorem is a special case of
a) Lebniz Theorem
b) Mean Value 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. Rolle’s theorem is applicable to the
a) Functions differentiable in closed interval [a, b] and continuous in open interval (a, b) only and having
same value at point ‘a’ and ‘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) only and having same value at point ‘a’ and ‘b’
d) Monotonically Increasing funtions.

Answer: c [Reason:] Statement of Rolle’s Theorem is that, If function f(x) attains same value at point ‘a’ and ‘b’ [ f(a) = f(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) and f’(c) = 0.

4. Find the value of c(a point where slope of a atangent to curve is zero) if f(x) = Sin(x) is continuous over interval [0,π] and differentiable over interval (0, π) and c ∈(0,π)
a) π
b) π2
c) π6
d) π4

Answer: b [Reason:] Given, f(x)=Sin(x) , x ∈ [0,π] Now f(0) = f(π) = 0 f’(c) = Cos(c) = 0 c = π2.

5. Find the value of c if f(x) = x(x-3)e3x , is continuous over interval [0,3] and differentiable over interval (0, 3) and c ∈(0,3)
a) 0.369
b) 2.703
c) 0
d) 3

Answer: b [Reason:] f(0) = 0 f(3) = 0 Hence, By rolle’s Theorem f’(c) = (c-3) e3c + c e3c + 3c(c-3) e3c = 0

Hence, c-3 + c + 3c2 -9c = 0

3c2 – 7c – 3 = 0 c = 2.703 , -0.369 Now c ∈(0,3) , hence , c = 2.703.

6. Find the value of c if f(x) = sin3(x)cos(x), is continuous over interval [0, π2] and differentiable over interval (0, π2) and c ∈(0, π2)
a) 0
b) π6
c) π3
d) π2

Answer: c [Reason:] f(x) = sin3(x)cos(x) f(0) = 0 f(π2) = 0 Hence, f’(c) = 3sin2(c)cos(c)cos(c) – sin4(c) = 0

3sin2(c)cos2(c) – sin4(c) = 0

sin2(c)[3 cos2(c) – sin2(c)] = 0

either, sin2(c)=0 or 3 cos2(c) – sin2(c) = 0

Since sin2(c) cannot be zero because c cannot be 0 Hence, 3 cos2(c) – sin2(c)=0

tan2(c) = 3

tan(c) = √3 c = π3.

7. Find value of c where f(x) = sin(x) ex tan(x) , c ∈(0,∞)
a) Tan-1[-(2+c2)/(1+c2)
b) Tan-1[-(2-c2)/(1+c2)].
c) Tan-1[(2+c2)/(1+c2)].
d) Rolle’s Theorem is not applied, Cannot find the value of c

Answer: d [Reason:] Since, f(x) = exsin(x) tan(x) is not continuous over interval (0,∞), Hnece Rolle’s, theorem is not applied.

8. f(x) = 3Sin(2x), is continuous over interval [0,π] and differentiable over interval (0,π) and c ∈(0,π)
a) π
b) π2
c) π4
d) π8

Answer: b [Reason:] f(x) = 3Sin(2x) f(0)=0 f(π)=0 Hence, f’(c) = 6Cos(2c) = 0 c= π2.

9. Find the value of ‘a’ if f(x) = ax2+32x+4 is continuous over [-4, 0] and differentiable over (-4, 0) and satisfy the Rolle’s theorem. Hence find the point in interval (-2,0) at which its slope of a tangent is zero
a) 2, -2
b) 2, -1
c) 8, -1
d) 8, -2

Answer: d [Reason:] Since it satisfies Rolle’s Theorem, f’(c) = 0 = 2ac+32 …………………. (1) and, f(0) = 4 hence by Rolle’s theorem and, f(-4) = 4 = 16a-128+4 (because f(0)=f(-4) condition of rolle’s theorem) ⇒ a = 8 from, eq.(1) ⇒ c = -2.

10. Find the value of ‘a’ & ‘b’ if f(x) = ax2 + bx + sin(x) is continuous over [0, pi] and differentiable over (0, pi) and satisfy the Rolle’s theorem at point c = π4.
a) 0.45,1.414
b) 0.45,-1.414
c) -0.45,1.414
d) -0.45,-1.414

Answer: b [Reason:] Since function f(x) is continuous over [0,pi] and satisfy rolle’s theorem, ⇒ f(0) = f(pi) = 0 ⇒ f(π) = a π2 + b π=0 ⇒ a π+b=0 ………………….(1) Since it satisfies rolle’s theorem at c= π4 f’(c) = 2ac + b + Cos(c) = 0 ⇒ a(π2) + b + 1√2 = 0 ………………..(2) From eq(1) and eq(2) we get, ⇒ a = 0.45 ⇒ b = -1.414.

11. Find value of c(a point in f(x) where slope of tangent to curve is zero) where

a) π4
b) Rolle’s Theorem is not applied, because function is not continuous in interval [0, π2].
c) Rolle’s Theorem is not applied, because function is not differential in interval (0, π2)
d) Function is both continuous and differentiable but Rolle’s theorem is not applicable as f(0) ≠ f(π2)

Answer: b [Reason:] Continuity Check. Hence function is discontinuous in interval (0, π2). Hence Rolle’s theorem cannot be applied.

12. Find value of c(a point in a curve where slope of tangent to curve is zero) where

a) 1.5
b) Rolle’s Theorem is not applied, because function is not continuous in interval [0,2].
c) Rolle’s Theorem is not applied, because function is not differential in interval (0,2)
d) Function is both continuous and differentiable but Rolle’s theorem is not applicable as f(0) ≠ f(2)

Answer: c [Reason:] Continuity Check Hence function is not differentiable so Rolle’s Theorem cannot be applied.

13. f(x) = ln(10-x2) , x=[-3,3], find the point in interval [-3,3] where slope of a tangent is zero,
a) 0
b) Rolle’s Theorem is not applied, because function is not continuous in interval [-3,3 ].
c) Rolle’s Theorem is not applied, because function is not differential in interval (-3,3)
d) 2

Answer: a [Reason:] Domain of f(x) = [-√10, +√10] Hence given f(x) is continuous in interval [-3,3] f’(x) = -2x/(10-x2) ⇒ x ≠ ±√10 ⇒ Domain of f’(x) = (-∞,∞)- ±√10 ⇒ Hence f(x) is differential in interval (-3,3) f’(c) = -2c/(10-c2) = 0 ⇒ c=0.

## Discrete Mathematics MCQ Set 2

1. Find
a) e
b) 1
c) e2
d) 1e

Answer: a [Reason:] Use the form

lt x→ ∞(1 + f(x))g(x) = elt x→ ∞ f(x) * g(x)

Provided as x → ∞ we must have f(x) → 0 g(x) → ∞ These conditions are met in our question L = e1 = e.

2. Find
a) 0
b) 1
c) Undefined
d) – 135

Answer: d [Reason:] The form here is of 00 Applying L hospitals rule would be really tough to differentiate. Hence we use the concept of Taylor Series

3. Find
a) 1a
b) 1
c) 1 – cos(1)
d) 0

Answer: c [Reason:] We use the concept of limit of a sum which is

4. Find
a) 1
b) -1
c) 0
d) Undefined

5. Find
a) 2
b) 1
c) 0
d) Undefined

6. Find
a) e
b) e-1
c) 0
d) 1

7. Find
a) 1
b) 0
c) ∞
d) 00

Answer: a [Reason:] We have 00 form Now we have the form

8. Find
a) ln(2)
b) ln(4)
c) 3ln(2)
d) 1a

Answer: b [Reason:] Using limit of sum we have

9. Find
a) 1
b) 1a + 1
c) 0
d) Undefined

10. Find
a) 0
b) 1
c) ∞
d) 10204 / 12221

11. Find
a) 0
b) 1
c) -45
d) -1

12. Find
a) 1
b) 2
c) 4
d) 3

Answer: c [Reason:] Expand into Mclaurin series

13. Find
a) 1
b) ∞
c) 0
d) -1

14. Find
a) anbn
b) ∞
c) No general form
d) bnan

## Discrete Mathematics MCQ Set 3

1. Find
a) ∞
b) 0
c) 2
d) -∞

Answer: c [Reason:] First evaluate This is true for the above problem Thus, we can deduce the limit as = 1 Hence, 2 is the right answer.

2. Find
a) 32
b) 0
c) 43
d) –43

Answer: c [Reason:] Form is 0 / 0 Applying L hospitals rule we have

3. Find relation between a and b such that the following limit is got after a single application of L hospitals Rule
a) ba = 2
b) ab = 2
c) a = b
d) a = -b

Answer: d [Reason:] Given differentiation is applied once we get ae0 + be0 = 0 = a + b (numerator → zero) be0 + ae0 = 0 = a + b (denominator → zero) Thus the relation between (a, b) and is a + b = 0 OR a = -b.

4. Find
a) -76
b) -6
c) -7
d) 0

Answer: a [Reason:] Form here 00 Applying L hospitals rule we have

5. Find how many rounds of differentiation are required to have finite limit for given that a ≠ b ≠ c
a) 3
b) 0
c) 2
d) 4

Answer: c [Reason:] Applying L hospitals rule Assume now that a + b + 2c = 0 and a + 2b – 3c = 0 We must have a = c = b but given a ≠ c ≠ b Thus, our assumption is false and a finite limit exists after first round of differentiation. Hence, 2 is the right answer.

6. Find
a) 4!
b) 5!
c) 0
d) ∞

7. Find
a) 0
b) 2
c) ∞
d) 1

8. Find
a) 1cosh(1012345)
b) 90987
c) 1012345
d) ∞

9. Let f o n (f(x)) denote the composition of f(x) with itself n number of times then the value of ltn → ∞ f o n (sin(x)) =
a) -1
b) 2
c) ∞
d) 0

Answer: d [Reason:] Drawing the graph of y = x and y = sin(x) we can write the limit value as 0.

10. Find
a) ∞
b) -1
c) 0
d) 22

Answer: c [Reason:] Expand into Taylor Series

11. Find
a) -33
b) 12
c) 0
d) 3132

12. Find
a) √π
b) ∞
c) √π2
d) 0

Answer: c [Reason:] Using the Gauss definition of the Gamma function we have

13. Find
a) e
b) e – 1
c) 0
d) ∞

## Discrete Mathematics MCQ Set 4

1. For the function f(x) = x2 – 2x + 1
we have Rolles point at x = 1. The coordinate axes are then rotated by 45
degrees in anticlockwise sense. What is the position of new Rolles point with respect to the transformed coordinate axes
a) 32
b) 12
c) 52
d) 1

Answer: a [Reason:]The coordinate axes are rotated by 45 degree then the problem transforms into that of Lagrange mean value theorem where the point in some interval has the slope of tan(45). Hence differentiating the function and equating to tan(45). We have f ‘(x) = tan(45) =2x – 2 2x – 2 = 1 x = 32.

2. For the function f(x) = x3 + x + 1
we do not have any Rolles point. The coordinate axes are transformed by rotating them by 60 degrees in anti-clockwise sense. The new Rolles point is
a) 32
b) The function can never have a Rolles point
c) 1
d)

Answer: d [Reason:] The question is simply asking us to find if there is some open interval in the original function f(x) where we have f'(x) = tan(60) We have f'(x) = 3x2 + 1 = tan(60) 3x2 = √3 – 1

x = .

3. What is the minimum angle by which the coordinate axes have to be rotated in anticlockwise sense (in Degrees), such that the function f(x) = 3x3 + 5x + 1016
has at least one Rolles point
a) π180 tan-1(5)
b) tan-1(5)
c) 180π tan-1(5)
d) – tan-1(5)

Answer: c [Reason:] For the transformed function to have a Rolles point is equivalent to the existing function having a Lagrange point somewhere in the real number domain, we are finding the point in the domain of the original function where we have f'(x) = tan(α) Let the angle to be rotated be α We have f'(x) = 9x2 + 5 = tan(α) 9x2 = tan(α) – 5 For the given function to have a Lagrange point we must have the right hand side be greater than zero, so tan(α) – 5 > 0 tan(α) > 5 α > tan-1(5) In degrees we must have, αdeg > 180π tan-1(5).

4. For a third degree monic polynomial, it is seen that the sum of roots are zero. What is the relation between the minimum angle to be rotated to have a Rolles point (α in Radians) and the cyclic sum of the roots taken two at a time c
a) α = π180 * tan-1(c)
b) Can never have a Rolles point
c) α = tan-1(c)
d) α = 180π tan-1(c).

Answer: d [Reason:] From Vietas formulas we can deduce that the x2 coefficient of the monic polynomial is zero (Sum of roots = zero). Hence, we can rewrite our third degree polynomial as c Now the question asks us to relate α and c Where c is indeed the cyclic sum of two roots taken at a time by Vietas formulae As usual, Rolles point in the rotated domain equals the Lagrange point in the existing domain. Hence, we must have y = tan(α) 3x2 + c = tan(α) To find the minimum angle, we have to find the minimum value of α such that the equation formed above has real roots when solved for x So, we can write tan(α) – c > 0 tan(α) > c α > tan-1(c) Thus, the minimum required angle is α = tan-1(c).

5. For the infinitely defined discontinuous function
exist, such that
a) f (c) = 1
b) 256
c) 512
d) 16

Answer: a [Reason:] To find points such that f'(c) = 1 We need to check points on graph where slope remains the same ( 45 degrees) In every interval of the form [(n – 1)π, nπ] we must have 2n – 1 points Because sine curve there has frequency 2n and the graph is going to meet the graph y = x at 2n points. Hence, in the interval [0, 16π] we have = 1 + 3 + 5…….(16terms) =(16)2 = 256.

6. Let g(x) be periodic and non-constant, with period τ. Also we have g(nτ) = 0 : n ∊ N. The function
f(x) is defined as

How many points c ∊ [0, 18τ] exist such that f'(c) = tan-1(2)
a) 325
b) 323
c) 324
d) 162

Answer: c [Reason:] To find points such that f (c) = f (c) = tan-1(2) We need to check points on the graph such that the slope remains the same ( tan-1(2) radians) In every interval of the form we must have 2n points on the graph because the frequency of periodic function in that interval is 2n and we have g(nτ) = 0 And we have 2n – 1 The total number of such points in the interval [0, 18τ] is given by = 1 + 3 + 5…….(18 terms) = (18)2 = 324.

7. Let . Find a point nearest to such that c
a) f'(c) = 1
b) 1
c) 0
d) 2.3445 * 10-9

Answer: d [Reason:] First find f'(x) Thus we have to choose a point nearer to 0 as our answer which is, 458328.33 * 10-3.

8. A function f(x) with n roots should have n – 1 unique Lagrange points
a) True
b) False

Answer: b [Reason:] If the roots of the polynomial are equal then we will have less than n – 1 Lagrange points that are unique. Hence, the right Option is False.

9. Let . Find a point c ∊ (- ∞, ∞) such that f (c) = 0
a) 1
b) 2
c) 0
d) -1

Answer: c Equating to 0 we have

x99 = 0

Observe that x = 0 satisfies the equation. Neglect other roots as they are complex roots.

## Discrete Mathematics MCQ Set 5

1. Let f(x) = sin(x) / 1+x2 . Let y(n) denote the nth derivative of f(x) at x = 0 then the value of y(100) + 9900y(98) is
a) 0
b) -1
c) 100
d) 1729

Answer: a [Reason:]The key here is a simple manipulation and application of the Leibniz rule. Rewriting the given function as y(1 + x2) = sin(x)…….(1) The Leibniz rule for two functions is given by

2. Let f(x) = ln(x)/x+1 and let y(n) denote the nth derivative of f(x) at x = 1 then the value of 2y(100) + 100y(99)
a) (99)!
b) (-99)!
c) (100)!
d) (-98)!

Answer: b [Reason:]Assume f(x) = y Rewrite the function as y(x + 1 ) = ln(x) Now differentiate both sides up to hundredth derivative in accordance to the Leibniz rule we have

3. Let f(x) = √1-x2 and let y(n) denote the nth derivative of f(x) at x = 0 then the value of 6y (1) y(2) + 2y(3) is
a) -998
b) 0
c) 998
d) -1

Answer: b [Reason:]Assume f(x) = y Rewriting the function as y2 = 1 – x2 Differentiating both sides of the equation up to the third derivative using leibniz rule we have Now substituting x = 0 in both the equations and equating them yields 2y (3) y + 6y (2) y(1) = 0.

4. Let f(x) = tan(x) and let y(n) denote the nth derivative of f(x) then the value of y(9998879879789776) is
a) 908090988
b) 0
c) 989
d) 1729

Answer: b [Reason:]Assume y = f(x) and we also know that Using (1) and the above equation one can conclude that y(2) = 0 This gives the value of second derivative to be zero Similarly for any even value of n all the odd derivatives of y in the expression would have sin(x) as their coefficients and as the values of y(0) and y(2) are zero. Every even derivative of the tan(x) function has to be zero. Thus, we have

y(9998879879789776) = 0.

5. If the first and second derivatives at x = 0 of the function
were 2 and 3 then the value of the third derivative is
a) -3
b) 3
c) 2
d) 1

Answer: b [Reason:] Assume f(x) = y Write the given function as y(x2 – x + 1) = cos(x)

Now applying Leibniz rule up to the third derivative we get Substituting x = 0 gives sin(0) = y(3) + 9 -12 y(3) = 12 – 9 = 3.

6. For the given function the values of first and second derivative at x = 0 are assumed as 0 and 1 respectively. Then the value of the third derivative could be
a) 54√2
b) 2√2
c) √2
d) Indeterminate

Answer: a [Reason:]Rewriting the given function as y2 = x3 + x7 Now applying the Leibniz rule up to the third derivative we have Equating both sides and substituting x = 1 we get y(1) = 0 Now assumed in the question are the values y(2) = 1 and y(1) = √2 We also know

2√2 y(3) = 3! + 210 = 216 Putting them in equation (1) we get

y(3) = 54√2.

7. Let and let the nth derivative at x = 0 be given by y(n) Then the value of the expression for y(n) is given by