Maclaurin Theorem MCQ Quiz - Objective Question with Answer for Maclaurin Theorem - Download Free PDF

Last updated on Mar 20, 2025

Latest Maclaurin Theorem MCQ Objective Questions

Maclaurin Theorem Question 1:

The infinite series 1+x+x22!+x33!+x44!+... corresponds to

  1. sec x
  2. ex
  3. cos x
  4. 1 + sin2x

Answer (Detailed Solution Below)

Option 2 : ex

Maclaurin Theorem Question 1 Detailed Solution

Concept:

The Maclaurin’s series is given by

f(x)=f(0)+xf(0)+x22!f(0)+x33!f(0)+

Calculation:

Given:

f(x)=1+x+x22!+x33!+

Comparing the given series with the standard series, we get

f (0) = 0, f’(0) = 0, f’’(0) = 0

The only option satisfying this condition is ex

Maclaurin Theorem Question 2:

1 + x + x2/2 – x4/8 – x5/15 + … =

  1. etan x
  2. ecos x
  3. esin x
  4. ex sin x

Answer (Detailed Solution Below)

Option 3 : esin x

Maclaurin Theorem Question 2 Detailed Solution

Explanation:

Try to solve options.

Let us take option 3

Then we can say f(x) = esin x

F(0) = 1

f’(x) = esin x cos x

f’(0) = 1

f”(x) = esin x cos2 x – esin x sin x

f”(0) = 1

Similarly f’’’ (0) = 0

fiv (0) = -3

Now from Maclaurin’s series of expansion –

f(x)=f(0)+x1! f(0)+x22! f(0)+x33! f(0)+x44!f(0)

=1+x1+x22! (1)+x33! (0)+x44!(3)

=1+x+x22x48

Maclaurin Theorem Question 3:

By using a suitable Maclaurin series, find the sum to infinity of the following infinite series

ππ33!+π55!π77!+

Answer (Detailed Solution Below) 0

Maclaurin Theorem Question 3 Detailed Solution

Explanation:

Expansion of some important function:

sinθ=θθ33!+θ55!θ77!+

cosθ=1θ22!+θ44!θ66!+

tanθ=θ+θ33+2θ515+

ex=1+x+x22!+x33!+x44!+

The given series is

ππ33!+π55!π77!+=sinπ=0

Maclaurin Theorem Question 4:

Maclaurin’s series expansion for the function f(x)=xex is

  1. 1+xx2+x32+
  2. x+x2+x33+
  3. xx2+x32+
  4. 1x+x2+x32+

Answer (Detailed Solution Below)

Option 3 : xx2+x32+

Maclaurin Theorem Question 4 Detailed Solution

Concept:

Maclaurin's series expansion for a function f(x) is given by

f(x)=f(0)+xf(0)+x22!f(0)+x33!f(0)+

which is nothing but Taylor series expansion about x = 0

Now 

f(x) = xe-x  ⇒ f (0) = 0

f’(x) = e-x + x(-1)e-x = e-x(1 - x) ⇒  f’(0) = 1

f’’(x) = e-x (-1) + (1 - x)(-1) e-x = e-x(x - 2) ⇒  f’’(0) = -2

f’’’(x) = e-x (1) + (x - 2)(-1) e-x = e-x(3 - x) ⇒  f’’’(0) = 3

⇒ f(x)=xex=0+x(1)+x22!(2)+x33!(3)

xex=xx2+x32+

Maclaurin Theorem Question 5:

Let S=n=0nαn where |α|<1. the value of α in the range 0<α<1, such that S=2α is  _______.

Answer (Detailed Solution Below) 0.28 - 0.31

Maclaurin Theorem Question 5 Detailed Solution

s=α+2α2+3α3+.αs=α2+2α3+3α4+.s(1α)=α+α2+α3+.=α1αs=α(1α)2=2α(1α)2=121α=12α=0.2929

Top Maclaurin Theorem MCQ Objective Questions

1 + x + x2/2 – x4/8 – x5/15 + … =

  1. etan x
  2. ecos x
  3. esin x
  4. ex sin x

Answer (Detailed Solution Below)

Option 3 : esin x

Maclaurin Theorem Question 6 Detailed Solution

Download Solution PDF

Explanation:

Try to solve options.

Let us take option 3

Then we can say f(x) = esin x

F(0) = 1

f’(x) = esin x cos x

f’(0) = 1

f”(x) = esin x cos2 x – esin x sin x

f”(0) = 1

Similarly f’’’ (0) = 0

fiv (0) = -3

Now from Maclaurin’s series of expansion –

f(x)=f(0)+x1! f(0)+x22! f(0)+x33! f(0)+x44!f(0)

=1+x1+x22! (1)+x33! (0)+x44!(3)

=1+x+x22x48

Let S=n=0nαn where |α|<1. the value of α in the range 0<α<1, such that S=2α is  _______.

Answer (Detailed Solution Below) 0.28 - 0.31

Maclaurin Theorem Question 7 Detailed Solution

Download Solution PDF
s=α+2α2+3α3+.αs=α2+2α3+3α4+.s(1α)=α+α2+α3+.=α1αs=α(1α)2=2α(1α)2=121α=12α=0.2929

The infinite series 1+x+x22!+x33!+x44!+... corresponds to

  1. sec x
  2. ex
  3. cos x
  4. 1 + sin2x

Answer (Detailed Solution Below)

Option 2 : ex

Maclaurin Theorem Question 8 Detailed Solution

Download Solution PDF

Concept:

The Maclaurin’s series is given by

f(x)=f(0)+xf(0)+x22!f(0)+x33!f(0)+

Calculation:

Given:

f(x)=1+x+x22!+x33!+

Comparing the given series with the standard series, we get

f (0) = 0, f’(0) = 0, f’’(0) = 0

The only option satisfying this condition is ex

Maclaurin Theorem Question 9:

1 + x + x2/2 – x4/8 – x5/15 + … =

  1. etan x
  2. ecos x
  3. esin x
  4. ex sin x

Answer (Detailed Solution Below)

Option 3 : esin x

Maclaurin Theorem Question 9 Detailed Solution

Explanation:

Try to solve options.

Let us take option 3

Then we can say f(x) = esin x

F(0) = 1

f’(x) = esin x cos x

f’(0) = 1

f”(x) = esin x cos2 x – esin x sin x

f”(0) = 1

Similarly f’’’ (0) = 0

fiv (0) = -3

Now from Maclaurin’s series of expansion –

f(x)=f(0)+x1! f(0)+x22! f(0)+x33! f(0)+x44!f(0)

=1+x1+x22! (1)+x33! (0)+x44!(3)

=1+x+x22x48

Maclaurin Theorem Question 10:

By using a suitable Maclaurin series, find the sum to infinity of the following infinite series

ππ33!+π55!π77!+

Answer (Detailed Solution Below) 0

Maclaurin Theorem Question 10 Detailed Solution

Explanation:

Expansion of some important function:

sinθ=θθ33!+θ55!θ77!+

cosθ=1θ22!+θ44!θ66!+

tanθ=θ+θ33+2θ515+

ex=1+x+x22!+x33!+x44!+

The given series is

ππ33!+π55!π77!+=sinπ=0

Maclaurin Theorem Question 11:

Let S=n=0nαn where |α|<1. the value of α in the range 0<α<1, such that S=2α is  _______.

Answer (Detailed Solution Below) 0.28 - 0.31

Maclaurin Theorem Question 11 Detailed Solution

s=α+2α2+3α3+.αs=α2+2α3+3α4+.s(1α)=α+α2+α3+.=α1αs=α(1α)2=2α(1α)2=121α=12α=0.2929

Maclaurin Theorem Question 12:

Maclaurin’s series expansion for the function f(x)=xex is

  1. 1+xx2+x32+
  2. x+x2+x33+
  3. xx2+x32+
  4. 1x+x2+x32+

Answer (Detailed Solution Below)

Option 3 : xx2+x32+

Maclaurin Theorem Question 12 Detailed Solution

Concept:

Maclaurin's series expansion for a function f(x) is given by

f(x)=f(0)+xf(0)+x22!f(0)+x33!f(0)+

which is nothing but Taylor series expansion about x = 0

Now 

f(x) = xe-x  ⇒ f (0) = 0

f’(x) = e-x + x(-1)e-x = e-x(1 - x) ⇒  f’(0) = 1

f’’(x) = e-x (-1) + (1 - x)(-1) e-x = e-x(x - 2) ⇒  f’’(0) = -2

f’’’(x) = e-x (1) + (x - 2)(-1) e-x = e-x(3 - x) ⇒  f’’’(0) = 3

⇒ f(x)=xex=0+x(1)+x22!(2)+x33!(3)

xex=xx2+x32+

Maclaurin Theorem Question 13:

The infinite series 1+x+x22!+x33!+x44!+... corresponds to

  1. sec x
  2. ex
  3. cos x
  4. 1 + sin2x

Answer (Detailed Solution Below)

Option 2 : ex

Maclaurin Theorem Question 13 Detailed Solution

Concept:

The Maclaurin’s series is given by

f(x)=f(0)+xf(0)+x22!f(0)+x33!f(0)+

Calculation:

Given:

f(x)=1+x+x22!+x33!+

Comparing the given series with the standard series, we get

f (0) = 0, f’(0) = 0, f’’(0) = 0

The only option satisfying this condition is ex

Maclaurin Theorem Question 14:

Maclaurin‘s series expansion of f(x)=ex2+sinx has the form f(x) = A0 + A1x + A2x2 + A3x3 + ….

The coefficient A4 (correct to two decimal place) is equal to _____.

Answer (Detailed Solution Below) 0.87 - 0.88

Maclaurin Theorem Question 14 Detailed Solution

f(x)=e(x2+sinx)

f(x)=ex2.esinx

So, Maclaurin’s series expansion of

ex2=(1+x2+x42!+x63!+)

Also, Maclaurin’s series expansion of esinx

g(x)=esinx ........(say)

g(x)=esinx.cosx=g(x).cosx

g(x)=g(x)cosxg(x)sinx

g(x)=g(x)cosxg(x)sinxg(x)sinxg(x)cosx=g(x)cosx2g(x)sinxg(x)cosx

 

g(x)=g(x)cosx3g(x)sinx3g(x)cosx+g(x)sinx

g(x = 0) = esin 0 = e0 = 1

g'(x = 0) = g(x = 0) × cos (0) = 1

g''(x = 0) = g'(x = 0) cos (0) – g(0) sin (0)

= 1 × 1 – 1 × 0

 = 1

g'''(x = 0) = g''(x = 0) cos (0) – 2g'(x = 0) sin (0) – g(x = 0) cos (0)

= 1 × 1 – 2 × 1 × 0 – 1 × 1

= 0

g(x=0)=g(x=0)cos03g(x=0)sin03g(x=0)cos0+g(x=0)sin0

g''''(x = 0) = -3

Hence,

esinx=1+(1×x)+(1×x22!)+(0×x33!)+(3)×x44!+

=1+x+x22!3x44!

=1+x+x22!x48+

ex2.esinx=(1+x2+x42+).(1+x+x22!x48+)

Coefficient of x4,A4=(1×12+1×12+1×(18))

=(118)=78
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