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

Concept:

The Maclaurin’s series is given by

$f\left(x\right)=f\left(0\right)+x{f}^{\prime }\left(0\right)+{\displaystyle \frac{x^2}{2!}}{f}^{″}\left(0\right)+{\displaystyle \frac{x^3}{3!}}{f}^{‴}\left(0\right)+\dots \mathrm{\infty }$

Calculation:

Given:

$f\left(x\right)=1+x+{\displaystyle \frac{x^2}{2!}}+{\displaystyle \frac{x^3}{3!}}+\dots$

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 e^x

Explanation:

Try to solve options.

Let us take option 3

Then we can say f(x) = e^{sin x}

F(0) = 1

f’(x) = e^{sin x} cos x

f’(0) = 1

f”(x) = e^{sin x} cos² x – e^{sin x} sin x

f”(0) = 1

Similarly f’’’ (0) = 0

f^iv (0) = -3

Now from Maclaurin’s series of expansion –

$f\left(x\right)=f\left(0\right)+\frac{x}{1!}{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{″}\left(0\right)+\frac{x^3}{3!}{f}^{‴}\left(0\right)+\frac{x^4}{4!}{f}^{‴}\left(0\right)$

$=1+x\cdot 1+\frac{x^2}{2!}\left(1\right)+\frac{x^3}{3!}\left(0\right)+\frac{x^4}{4!}(-3)$

Explanation:

Expansion of some important function:

$\sin \theta =\theta -\frac{{\theta }^3}{3!}+\frac{{\theta }^5}{5!}-\frac{{\theta }^7}{7!}+\dots$

$\cos \theta =1-\frac{{\theta }^2}{2!}+\frac{{\theta }^4}{4!}-\frac{{\theta }^6}{6!}+\dots$

$\tan \theta =\theta +\frac{{\theta }^3}{3}+\frac{{2\theta }^5}{15}+\dots$

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots$

The given series is

$\pi -\frac{{\pi }^3}{3!}+\frac{{\pi }^5}{5!}-\frac{{\pi }^7}{7!}+\dots =\sin \pi =0$

Concept:

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

$f\left(x\right)=f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}f\left(0\right)+\frac{x^3}{3!}{f}^{‴}\left(0\right)+\dots$

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

⇒ $f\left(x\right)=x{e}^{-x}=0+x\left(1\right)+\frac{x^2}{2!}\left(-2\right)+\frac{x^3}{3!}\left(3\right)$

$x{e}^{-x}=x-x^2+\frac{x^3}{2}+\dots$

Explanation:

Try to solve options.

Let us take option 3

Then we can say f(x) = e^{sin x}

F(0) = 1

f’(x) = e^{sin x} cos x

f’(0) = 1

f”(x) = e^{sin x} cos² x – e^{sin x} sin x

f”(0) = 1

Similarly f’’’ (0) = 0

f^iv (0) = -3

Now from Maclaurin’s series of expansion –

$f\left(x\right)=f\left(0\right)+\frac{x}{1!}{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{″}\left(0\right)+\frac{x^3}{3!}{f}^{‴}\left(0\right)+\frac{x^4}{4!}{f}^{‴}\left(0\right)$

$=1+x\cdot 1+\frac{x^2}{2!}\left(1\right)+\frac{x^3}{3!}\left(0\right)+\frac{x^4}{4!}(-3)$

Concept:

The Maclaurin’s series is given by

$f\left(x\right)=f\left(0\right)+x{f}^{\prime }\left(0\right)+{\displaystyle \frac{x^2}{2!}}{f}^{″}\left(0\right)+{\displaystyle \frac{x^3}{3!}}{f}^{‴}\left(0\right)+\dots \mathrm{\infty }$

Calculation:

Given:

$f\left(x\right)=1+x+{\displaystyle \frac{x^2}{2!}}+{\displaystyle \frac{x^3}{3!}}+\dots$

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 e^x

Explanation:

Try to solve options.

Let us take option 3

Then we can say f(x) = e^{sin x}

F(0) = 1

f’(x) = e^{sin x} cos x

f’(0) = 1

f”(x) = e^{sin x} cos² x – e^{sin x} sin x

f”(0) = 1

Similarly f’’’ (0) = 0

f^iv (0) = -3

Now from Maclaurin’s series of expansion –

$f\left(x\right)=f\left(0\right)+\frac{x}{1!}{f}^{\prime }\left(0\right)+\frac{x^2}{2!}{f}^{″}\left(0\right)+\frac{x^3}{3!}{f}^{‴}\left(0\right)+\frac{x^4}{4!}{f}^{‴}\left(0\right)$

$=1+x\cdot 1+\frac{x^2}{2!}\left(1\right)+\frac{x^3}{3!}\left(0\right)+\frac{x^4}{4!}(-3)$

Explanation:

Expansion of some important function:

$\sin \theta =\theta -\frac{{\theta }^3}{3!}+\frac{{\theta }^5}{5!}-\frac{{\theta }^7}{7!}+\dots$

$\cos \theta =1-\frac{{\theta }^2}{2!}+\frac{{\theta }^4}{4!}-\frac{{\theta }^6}{6!}+\dots$

$\tan \theta =\theta +\frac{{\theta }^3}{3}+\frac{{2\theta }^5}{15}+\dots$

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\dots$

The given series is

$\pi -\frac{{\pi }^3}{3!}+\frac{{\pi }^5}{5!}-\frac{{\pi }^7}{7!}+\dots =\sin \pi =0$

Concept:

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

$f\left(x\right)=f\left(0\right)+x{f}^{\prime }\left(0\right)+\frac{x^2}{2!}f\left(0\right)+\frac{x^3}{3!}{f}^{‴}\left(0\right)+\dots$

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

⇒ $f\left(x\right)=x{e}^{-x}=0+x\left(1\right)+\frac{x^2}{2!}\left(-2\right)+\frac{x^3}{3!}\left(3\right)$

$x{e}^{-x}=x-x^2+\frac{x^3}{2}+\dots$

Concept:

The Maclaurin’s series is given by

$f\left(x\right)=f\left(0\right)+x{f}^{\prime }\left(0\right)+{\displaystyle \frac{x^2}{2!}}{f}^{″}\left(0\right)+{\displaystyle \frac{x^3}{3!}}{f}^{‴}\left(0\right)+\dots \mathrm{\infty }$

Calculation:

Given:

$f\left(x\right)=1+x+{\displaystyle \frac{x^2}{2!}}+{\displaystyle \frac{x^3}{3!}}+\dots$

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 e^x

$f\left(x\right)=e^{\left(x^2+sinx\right)}$

$f\left(x\right)=e^{x^2}.e^{\sin x}$

So, Maclaurin’s series expansion of

$e^{x^2}=\left(1+x^2+\frac{x^4}{2!}+\frac{x^6}{3!}+\dots \right)$

Also, Maclaurin’s series expansion of e^sinx

$g\left(x\right)=e^{\sin x}$ ........(say)

$g^{\prime }\left(x\right)=e^{\sin x}.\cos x=g\left(x\right).\cos x$

$g^{″}\left(x\right)=g^{\prime }\left(x\right)\cos x-g\left(x\right)\sin x$

$g^{‴}\left(x\right)=g^{″}\left(x\right)\cos x-g^{\prime }\left(x\right)\sin x-g^{\prime }\left(x\right)\sin x-g\left(x\right)\cos x=g^{″}\left(x\right)\cos x-2g^{\prime }\left(x\right)\sin x-g\left(x\right)\cos x$

$g^{{}^{⁗}}\left(x\right)=g^{‴}\left(x\right)\cos x-3g^{″}\left(x\right)\sin x-3g^{\prime }\left(x\right)\cos x+g\left(x\right)\sin x$

g(x = 0) = e^{sin 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

$g^{{}^{⁗}}\left(x=0\right)=g^{‴}\left(x=0\right)\cos 0-3g^{″}\left(x=0\right)\sin 0-3g^{\prime }\left(x=0\right)\cos 0+g\left(x=0\right)\sin 0$

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

Hence,

$e^{\sin x}=1+\left(1\times x\right)+\left(1\times \frac{x^2}{2!}\right)+\left(0\times \frac{x^3}{3!}\right)+\left(-3\right)\times \frac{x^4}{4!}+\dots$

$=1+x+\frac{x^2}{2!}-\frac{3x^4}{4!}\dots$

$=1+x+\frac{x^2}{2!}-\frac{x^4}{8}+\dots$

$\therefore e^{x^2}.e^{\sin x}=\left(1+x^2+\frac{x^4}{2}+\dots \right).\left(1+x+\frac{x^2}{2!}-\frac{x^4}{8}+\dots \right)$

Coefficient of $x^4,A_4=\left(1\times \frac{1}{2}+1\times \frac{1}{2}+1\times \left(-\frac{1}{8}\right)\right)$