Question

Please answer ASAP with explanation.

Answer 1

Step-by-step explanation:

For any function, the maclaurin's series expansion is given by

$f(x) = f(0) + \frac{x}{1!}f^{ \prime} (0) + \frac{ {x}^{2} }{2!} f^{ \prime \prime} (0) + \frac{ {x}^{3} }{3!}f^{ \prime \prime \prime} (0) + ...$

For $f(x)=e^{x}$

$f ^{ \prime} (0) = f ^{ \prime \prime} (0) = f ^{ \prime \prime \prime} (0) = 1$

So,

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

Answer 2

$\large\underline{\sf{Solution-}}$

We have to find the expansion for

$\rm :\longmapsto\:f(x) = {e}^{x}$

So,

$\rm :\longmapsto\:f(0) ={e}^{0} = 1$

Let first find the derivatives of f(x).

$\rm :\longmapsto\:f'(x) ={e}^{x}$

$\rm :\longmapsto\:f''(x) ={e}^{x}$

$\rm :\longmapsto\:f'''(x) ={e}^{x}$

$\rm :\longmapsto\:f''''(x) ={e}^{x}$

Now, Let we find the value of these derivatives at x = 0

$\rm :\longmapsto\:f'(0) ={e}^{0} = 1$

$\rm :\longmapsto\:f''(0) ={e}^{0} = 1$

$\rm :\longmapsto\:f'''(0) ={e}^{0} = 1$

$\rm :\longmapsto\:f''''(0) ={e}^{0} = 1$

Now,

By Maclaurin's Series, we have

$\boxed{ \sf{ \:f(x) = f(0) + xf'(0) + \dfrac{ {x}^{2} }{2!}f''(0) + \dfrac{ {x}^{3} }{3!}f'''(0) + \dfrac{ {x}^{4} }{4!}f''''(0) - - }}$

So, on substituting all the values evaluated above, we get

$\rm :\longmapsto\: {e}^{x} = 1 + x + \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{3} }{3!} + \dfrac{ {x}^{4} }{4!}+ - - -$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{ \underbrace{\boxed{ \bf{ \:Option \: \: \: B \: \: \: is \: \: \: correct}}}}$

Additional Information :-

Let's solve one more problem!!

Question :- Find the expansion of cosx.

Solution :-

Given

$\rm :\longmapsto\:f(x) = cosx$

So,

$\rm :\longmapsto\:f(0) = cos0 = 1$

Let find few derivatives of f(x).

$\rm :\longmapsto\:f'(x) = - sinx$

$\rm :\longmapsto\:f''(x) = - cosx$

$\rm :\longmapsto\:f'''(x) = sinx$

$\rm :\longmapsto\:f''''(x) = cosx$

Now, Let find the value of derivatives at x = 0

$\rm :\longmapsto\:f'(0) = - sin0 = 0$

$\rm :\longmapsto\:f''(0) = - cos0 = - 1$

$\rm :\longmapsto\:f'''(0) = sin0 = 0$

$\rm :\longmapsto\:f''''(0) = cos0 = 1$

Now,

By Maclaurin's Series, we have

$\boxed{ \sf{ \:f(x) = f(0) + xf'(0) + \dfrac{ {x}^{2} }{2!}f''(0) + \dfrac{ {x}^{3} }{3!}f'''(0) + \dfrac{ {x}^{4} }{4!}f''''(0) - - }}$

So, on substituting all the values evaluated above, we get

$\rm :\longmapsto\: cosx = 1 - \dfrac{ {x}^{2} }{2!} + \dfrac{ {x}^{4} }{4!}+ - - -$