This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “The nth Derivative of Some Elementary Functions – 1”.

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

Explanation: First consider the equation

2p

^{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

Explanation: Let the quadratic polynomial be

f(x) = ax

^{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

Explanation: 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(x^{2} + 5x + 6) then the value of f^{(30)}(1) is given by

View Answer

Explanation: Given function

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

Explanation: 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) = 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

Explanation: The key here is to expand e

^{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

Explanation: 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 n

^{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

Explanation: Assume the polynomial to be of the form f(x) = ax

^{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

Explanation: 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(x^{3} – 3x^{2} – 16x -12) , then the 1729^{th} derivative at x = 234 is

View Answer

Explanation: 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 n

^{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

Explanation: We have to consider the function f(x) = sin(e

^{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

Explanation: 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 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}.

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