This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Taylor Mclaurin Series – 1”.

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

View Answer

Explanation: We know the series expansion for sin(x) is

Hence, Option (a) is the right 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

Explanation: We know that the series expansion of cos(x) is

Observe that every term has odd powered series raised to an even term

Thus, we must have only even powered terms in the above series expansion. The coefficient of any odd powered term is zero.

Hence, Option (a) is the right 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

Explanation: Observe first off that the given function is a polynomial and so any other representation (Taylor Series here) which is continuous and differentiable has to be the

same polynomial. This gives us

τ(x) = f(x)

We now evaluate the expression as follows

= 2(f(1729))

^{2}+ (f(1729))

^{2}– 3(f(1729))

^{2}+ 1770

= 3(f(1729))^{2} – 3(f(1729))^{2} +1770

= 1770

Hence, Option (a) is the right answer.

4. Find the Taylor series expansion of the function cosh(x) centered at x = 0

View Answer

Explanation: We know the general expression for the expansion of the Taylor series

Hence, Option (c) is the right 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

Explanation: The Taylor series gives accurate results around some point taken as center. As we need the value of 9 the center nearer to the point should be taken. Hence, 9 is the right 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

Explanation: We know that the Mclaurin Series for any given function always yields a polynomial (finite OR infinite).

Further the Mclaurin series of this polynomial (i.e.τ(τ(f(x))) ) is also a polynomial. Due to uniqueness of this polynomial, no matter how many nested Mclaurin series we might find, they are all equal. Thus, we have

τ(τ…….(f(x))….)) = f(x)

Substituting this into our required expression we have

= f(1729) – 2f(1729) + f(1729)

= 0

Hence, 0 is the right 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

Explanation: Observing the recurrence relation we have

Hence, Option e

^{x}is the right 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

Explanation: Writing out the Mclaurin series we have

This is a well defined function.

**Sanfoundry Global Education & Learning Series – Engineering Mathematics.**

To practice all areas of Engineering Mathematics, __here is complete set of 1000+ Multiple Choice Questions and Answers__.