## 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).

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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

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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.

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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) ^{π}⁄_{2}

c) ^{π}⁄_{6}

d) ^{π}⁄_{4}

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^{π}⁄

_{2}.

5. Find the value of c if f(x) = x(x-3)e^{3x} , 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

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^{3c}+ c e

^{3c}+ 3c(c-3) e

^{3c}= 0

Hence, c-3 + c + 3c^{2} -9c = 0

3c^{2} – 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) = sin^{3}(x)cos(x), is continuous over interval [0, ^{π}⁄_{2}] and differentiable over interval (0, ^{π}⁄_{2}) and c ∈(0, ^{π}⁄_{2})

a) 0

b) ^{π}⁄_{6}

c) ^{π}⁄_{3}

d) ^{π}⁄_{2}

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^{3}(x)cos(x) f(0) = 0 f(

^{π}⁄

_{2}) = 0 Hence, f’(c) = 3sin

^{2}(c)cos(c)cos(c) – sin

^{4}(c) = 0

3sin^{2}(c)cos^{2}(c) – sin^{4}(c) = 0

sin^{2}(c)[3 cos^{2}(c) – sin^{2}(c)] = 0

either, sin^{2}(c)=0 or 3 cos^{2}(c) – sin^{2}(c) = 0

Since sin^{2}(c) cannot be zero because c cannot be 0
Hence, 3 cos^{2}(c) – sin^{2}(c)=0

tan^{2}(c) = 3

tan(c) = √3
c = ^{π}⁄_{3}.

7. Find value of c where f(x) = sin(x) e^{x} tan(x) , c ∈(0,∞)

a) Tan^{-1}[-(2+c^{2})/(1+c^{2})

b) Tan^{-1}[-(2-c^{2})/(1+c^{2})].

c) Tan^{-1}[(2+c^{2})/(1+c^{2})].

d) Rolle’s Theorem is not applied, Cannot find the value of c

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^{x}sin(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) ^{π}⁄_{2}

c) ^{π}⁄_{4}

d) ^{π}⁄_{8}

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^{π}⁄

_{2}.

9. Find the value of ‘a’ if f(x) = ax^{2}+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

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10. Find the value of ‘a’ & ‘b’ if f(x) = ax^{2} + bx + sin(x) is continuous over [0, pi] and differentiable over (0, pi) and satisfy the Rolle’s theorem at point c = ^{π}⁄_{4}.

a) 0.45,1.414

b) 0.45,-1.414

c) -0.45,1.414

d) -0.45,-1.414

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^{2}+ b π=0 ⇒ a π+b=0 ………………….(1) Since it satisfies rolle’s theorem at c=

^{π}⁄

_{4}f’(c) = 2ac + b + Cos(c) = 0 ⇒ a(

^{π}⁄

_{2}) + 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) ^{π}⁄_{4}

b) Rolle’s Theorem is not applied, because function is not continuous in interval [0, ^{π}⁄_{2}].

c) Rolle’s Theorem is not applied, because function is not differential in interval (0, ^{π}⁄_{2})

d) Function is both continuous and differentiable but Rolle’s theorem is not applicable as f(0) ≠ f(^{π}⁄_{2})

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^{π}⁄

_{2}). 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)

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13. f(x) = ln(10-x^{2}) , 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

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^{2}) = 0 ⇒ c=0.

## Discrete Mathematics MCQ Set 2

1. Find

a) e

b) 1

c) e^{2}

d) ^{1}⁄_{e}

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lt_{ x→ ∞}(1 + f(x))^{g(x)} = e^{lt x→ ∞ f(x) * g(x)}

Provided as x → ∞ we must have
f(x) → 0
g(x) → ∞
These conditions are met in our question L = e^{1} = e.

2. Find

a) 0

b) 1

c) Undefined

d) – ^{1}⁄_{35}

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^{0}⁄

_{0}Applying L hospitals rule would be really tough to differentiate. Hence we use the concept of Taylor Series

3. Find

a) ^{1}⁄_{a}

b) 1

c) 1 – cos(1)

d) 0

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4. Find

a) 1

b) -1

c) 0

d) Undefined

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5. Find

a) 2

b) 1

c) 0

d) Undefined

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6. Find

a) e

b) e^{-1}

c) 0

d) 1

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7. Find

a) 1

b) 0

c) ∞

d) ^{0}⁄_{0}

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^{0}⁄

_{0}form Now we have the form

8. Find

a) ln(2)

b) ln(4)

c) 3ln(2)

d) ^{1}⁄_{a}

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9. Find

a) 1

b) ^{1}⁄_{a + 1}

c) 0

d) Undefined

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10. Find

a) 0

b) 1

c) ∞

d) 10204 / 12221

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11. Find

a) 0

b) 1

c) ^{-4}⁄_{5}

d) -1

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12. Find

a) 1

b) 2

c) 4

d) 3

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13. Find

a) 1

b) ∞

c) 0

d) -1

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14. Find

a) ^{an} ⁄ _{bn}

b) ∞

c) No general form

d) ^{bn} ⁄ _{an}

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## Discrete Mathematics MCQ Set 3

1. Find

a) ∞

b) 0

c) 2

d) -∞

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2. Find

a) ^{3}⁄_{2}

b) 0

c) ^{4}⁄_{3}

d) –^{4}⁄_{3}

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3. Find relation between a and b such that the following limit is got after a single application of L hospitals Rule

a) ^{b}⁄_{a} = 2

b) ^{a}⁄_{b} = 2

c) a = b

d) a = -b

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^{0}+ be

^{0}= 0 = a + b (numerator → zero) be

^{0}+ ae

^{0}= 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

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^{0}⁄

_{0}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

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6. Find

a) 4!

b) 5!

c) 0

d) ∞

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7. Find

a) 0

b) 2

c) ∞

d) 1

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8. Find

a) ^{1}⁄_{cosh(1012345)}

b) 90987

c) 1012345

d) ∞

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9. Let f o ^{n} (f(x)) denote the composition of f(x) with itself n number of times then the value of lt_{n → ∞ } f o ^{n} (sin(x)) =

a) -1

b) 2

c) ∞

d) 0

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10. Find

a) ∞

b) -1

c) 0

d) 2^{2}

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11. Find

a) -33

b) ^{1}⁄_{2}

c) 0

d) ^{31}⁄_{32}

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12. Find

a) √π

b) ∞

c) √^{π}⁄_{2}

d) 0

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13. Find

a) e

b) e – 1

c) 0

d) ∞

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## Discrete Mathematics MCQ Set 4

1. For the function f(x) = x^{2} – 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) ^{3}⁄_{2}

b) ^{1}⁄_{2}

c) ^{5}⁄_{2}

d) 1

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^{3}⁄

_{2}.

2. For the function f(x) = x^{3} + 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) ^{3}⁄_{2}

b) The function can never have a Rolles point

c) 1

d)

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^{2}+ 1 = tan(60) 3x

^{2}= √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) = 3x^{3} + 5x + 1016

has at least one Rolles point

a) ^{π}⁄_{180} tan^{-1}(5)

b) tan^{-1}(5)

c) ^{180}⁄_{π} tan^{-1}(5)

d) – tan^{-1}(5)

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^{2}+ 5 = tan(α) 9x

^{2}= 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, α

^{deg}>

^{180}⁄

_{π}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 = ^{π}⁄_{180} * tan^{-1}(c)

b) Can never have a Rolles point

c) α = tan^{-1}(c)

d) α = ^{180}⁄_{π} tan^{-1}(c).

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^{2}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) 3x

^{2}+ 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

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^{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

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^{‘}(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 2

_{n – 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}

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^{-3}.

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

a) True

b) False

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9. Let . Find a point c ∊ (- ∞, ∞) such that f ^{‘}(c) = 0

a) 1

b) 2

c) 0

d) -1

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x^{99} = 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+x^{2} . Let y^{(n)} denote the n^{th} derivative of f(x) at x = 0 then the value of y^{(100)} + 9900y^{(98)} is

a) 0

b) -1

c) 100

d) 1729

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^{2}) = 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 n^{th} derivative of f(x) at x = 1 then the value of 2y^{(100)} + 100y^{(99)}

a) (99)!

b) (-99)!

c) (100)!

d) (-98)!

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3. Let f(x) = √1-x^{2} and let y^{(n)} denote the n^{th} 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

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^{2}= 1 – x

^{2}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 n^{th} derivative of f(x) then the value of y^{(9998879879789776)} is

a) 908090988

b) 0

c) 989

d) 1729

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^{(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

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^{2}– 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

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^{2}= x

^{3}+ x

^{7}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 n^{th} derivative at x = 0 be given by y^{(n)} Then the value of the expression for y^{(n)} is given by

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8. Let f(x) = e^{x} sinh(x) / x , let y^{(n)} denote the n^{th} derivative of f(x) at x = 0 then the expression for y^{(n)} is given by