Exponential Function MCQ Quiz - Objective Question with Answer for Exponential Function - Download Free PDF

Concept Used:-

We know that the Taylor expansion of e^x,

\( e^x=1+\frac{x}{1}+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\cdots \infty \)

On putting x = 1 we get,

\(\Rightarrow e=1+1+\frac{1}{2 !}+\frac{1}{3 !}+....\\ \Rightarrow e=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+....\\ \Rightarrow e=\sum_{n=0}^\infty\frac{1}{(n) !}\ \ \ \ \ \ \ \ \ \ \ \ .....(1)\)

Also,

 \(\Rightarrow e=\sum_{n=1}^\infty\frac{1}{(n-1)!} \ \ \ \ \ \ \ \ \ \ \ \ .....(2)\\ \Rightarrow e=\sum_{n=2}^\infty\frac{1}{(n-2) !} \ \ \ \ \ \ \ \ \ \ \ \ .....(3)\)

Explanation:-

Given series is,

\(1+\frac{2^2}{1 !}+\frac{3^2}{2 !}+\frac{4^2}{3 !}+\ldots \ldots.\)

Since, 0! is equal to 1. So, the above sum can be written as,

\(\frac{1}{0!}+\frac{2^2}{1 !}+\frac{3^2}{2 !}+\frac{4^2}{3 !}+\ldots \ldots.\)

Let the sum of the above series is S. Then,

\(S=\frac{1^2}{0!}+\frac{2^2}{1 !}+\frac{3^2}{2 !}+\frac{4^2}{3 !}+\ldots \ldots.\)

The denominator number written with factorial is one less than the numerator number written with square. Thus, in general, this series can be written as,

⇒ \(S=\sum_{n=1}^n\frac{(n+1)^2}{(n)!}\)

Expand the numerator (n+1)2 to solve it further,

\(\Rightarrow S=\sum_{n=0}^n\frac{(n+1)^2}{(n)!}\\ \Rightarrow S=\sum_{n=0}^n\frac{n^2+2n+1}{(n)!}\\ \Rightarrow S=\sum_{n=0}^n\frac{n^2}{(n)!}+\sum_{n=0}^n\frac{2n}{(n)!}+\sum_{n=0}^n\frac{1}{(n)!}\\ \Rightarrow S=\sum_{n=1}^n\frac{n(n-1)+n}{n.(n-1)!}+\sum_{n=1}^n\frac{2n}{n(.n-1)!}+\sum_{n=0}^n\frac{1}{(n)!}\\ \Rightarrow S=\sum_{n=2}^n\frac{1}{(n-2)!}+\sum_{n=1}^n\frac{1}{(n-1)!}+\sum_{n=1}^n\frac{2}{(n-1)!}+\sum_{n=0}^n\frac{1}{(n)!}\\ \Rightarrow S=\sum_{n=2}^n\frac{1}{(n-2)!}+\sum_{n=1}^n\frac{1}{(n-1)!}+2\sum_{n=1}^n\frac{1}{(n-1)!}+\sum_{n=0}^n\frac{1}{(n)!}\)

Now using equation (1), (2) and (3) we get,

⇒ S = e + e + 2e + e

⇒ S = 5e

So, the sum of the given series is 5e.

Hence, the correct option is 5.

Calculation

Given

\(\rm \displaystyle \lim_{x\rightarrow0} \frac{e-(1+2x)^{\frac{1}{2x}}}{x}\)

⇒ \(\operatorname{Lim}_{x \rightarrow 0} \frac{e-e^{\frac{1}{2 x} \ln (1+2 x)}}{x}\)

⇒  \(\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\left(e^{\frac{\ln (1+2 x)}{2 x}-1}-1\right)}{x}\)

⇒  \(\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\ln (1+2 x)-2 x}{2 x^2}\)

⇒  \((-\mathrm{e}) \times(-1) \frac{4}{2 \times 2}=\mathrm{e}\) 

Hence option (1) is correct

Calculation

Given \(\lim _{x \rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e+\frac{1}{2} e x}{x^{2}}\)

Let y = \((1+x)^{\frac{1}{x}}\)

⇒ \(\log y=\frac{1}{x} \log (1+x)=\frac{1}{x}\left[x=\frac{x^{2}}{2}+\frac{x^{3}}{3}-\frac{x^{4}}{4}+\ldots .\right]\)

⇒ logy = \(1-\frac{x}{2}+\frac{x^{2}}{3}-\frac{x^{3}}{4}+\ldots\)

⇒ y = \(e^{\left(1-\frac{x}{2}+\frac{x^{2}}{3}+. .\right)}=e e^{\left(-\frac{x}{2}+\frac{x^{2}}{3}-\right)}\)

⇒ y = \(e\left[1+\left(-\frac{x}{2}+\frac{x^{2}}{3}-\ldots\right)+\frac{1}{2!}\left(-\frac{x}{2}+\frac{x^{2}}{3}-\ldots\right)^{2}+\ldots\right]\)

⇒ \(y-e+\frac{1}{2} ex=e x^{2}\left[\frac{1}{3}+0(x)+\frac{1}{2}\left(-\frac{1}{2}+0(x)\right)^{2}+\ldots\right]\)

[∵ 0(x) in terms containing x]

\(\lim_{x\to0} \frac{y-e+\frac{1}{2} e x}{x^{2}}=e\left[\frac{1}{3}+\frac{1}{8}\right]=\frac{11}{24} e\)

Hence option 1 is correct

Calculation

Given

\(\rm \displaystyle \lim_{x\rightarrow0} \frac{e-(1+2x)^{\frac{1}{2x}}}{x}\)

⇒ \(\operatorname{Lim}_{x \rightarrow 0} \frac{e-e^{\frac{1}{2 x} \ln (1+2 x)}}{x}\)

⇒  \(\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\left(e^{\frac{\ln (1+2 x)}{2 x}-1}-1\right)}{x}\)

⇒  \(\operatorname{Lim}_{x \rightarrow 0}(-e) \frac{\ln (1+2 x)-2 x}{2 x^2}\)

⇒  \((-\mathrm{e}) \times(-1) \frac{4}{2 \times 2}=\mathrm{e}\) 

Hence option (1) is correct

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule ⇔ \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \;\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\) 

Note: We have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{l}} \ne \frac{0}{0}\) where l is a finite value.

Calculation:

\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;\left( {1 + x} \right)}}{{{x^2}}}\)                 [Form 0/0]

Apply L-Hospital Rule,

\( = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;\left( {0 + 1} \right)}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;1}}{{2x}}\)              [Form 0/0]

Again apply L-Hospital Rule,

\( = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;0}}{2} = \frac{{{e^0}}}{2} = \frac{1}{2}\)

Concept:

\(\rm ​​\mathop {\lim }\limits_{x\; \to \;a} \left[ {f\left( x \right) + g\left( x \right)} \right] = \;\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right) + \;\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)\)

\(\rm ​​\mathop {\lim }\limits_{x\; \to \;0} \dfrac { (a^x - 1) }{x} = \log a\)

log mⁿ = n log m

Calculation:

\(\rm \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x + 3^{-x}-2}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1+ 3^{-x}-1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{3^{-x} -1}{x}\\= \displaystyle\lim_{x \rightarrow 0} \dfrac{3^x -1}{x}+\displaystyle\lim_{x \rightarrow 0} \dfrac{(3^{-1})^x -1}{x}\\= \log 3 + \log (3^{-1})\\= \log 3 - \log 3\\=0\)

Concept:

\(\rm \frac{d}{d x} (a^{x})=a^{x} \ln a\)

When there is 0/0 or ∞/∞ form exist we use L-Hospital Rule according to which we differentiate numerator and denominator separately.

Calculation:

\(\mathop {\lim }\limits_{{\rm{x}} \to 0} \frac{{{5^{\rm{x}}} - 1}}{{\rm{x}}}\)

0/0 form, So using L-Hospital rule, we get 

\(\begin{array}{l} \lim _{x\rightarrow 0} \frac{5^{x} \ln 5-0}{1} \\ =\log _{e} 5 \end{array}\)

Hence, option (1) is correct.

Given:

f(x) = \(\rm \displaystyle\lim_{n → ∞} \frac{a^n+b^n}{a^n-b^n}\) and

a > b > 1

Calculation:

We have,

a > b > 1

⇒ \(\frac{a}{b}>1\) or \(\frac{b}{a}<1\)

Given that,

f(x) = \(\rm \displaystyle\lim_{n → ∞} \frac{a^n+b^n}{a^n-b^n}\)    

⇒ f(x) = \(\rm \displaystyle\lim_{n → ∞} \frac{a^n[1+(\frac{b}{a}) ^{n}]}{a^n[1 - (\frac{b}{a}) ^{n}]}\) 

⇒ f(x) = \(\rm \displaystyle\lim_{n → ∞} \frac{[1+(\frac{b}{a}) ^{n}]}{[1 - (\frac{b}{a}) ^{n}]}\) 

Taking limit n→∞

⇒ f(x) = \(\rm \displaystyle\lim_{n → ∞} \frac{[1+(\frac{b^{\infty}}{a^{\infty}}) ]}{[1 - (\frac{b^{\infty}}{a^{\infty}}) ]}\)  

⇒ f(x) = \(\rm \frac{1 + 0}{1 - 0}\)       (∵ \(\frac{b}{a}<1\) )

∴  f(x) = 1

Concept:

L-Hospital Rule: Let f(x) and g(x) be two functions

Suppose that we have one of the following cases,

I. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{0}{0}\)

II. \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{\;}}\frac{\infty }{\infty }\)

Then we can apply L-Hospital Rule ⇔ \(\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}\left( {\bf{x}} \right)}}{{{\bf{g}}\left( {\bf{x}} \right)}} = \;\mathop {\lim }\limits_{{\bf{x}} \to {\bf{a}}} \frac{{{\bf{f}}'\left( {\bf{x}} \right)}}{{{\bf{g}}'\left( {\bf{x}} \right)}}\) 

Note: We have to differentiate both the numerator and denominator with respect to x unless and until \(\mathop {\lim }\limits_{{\rm{x}} \to {\rm{a}}} \frac{{{\rm{f}}\left( {\rm{x}} \right)}}{{{\rm{g}}\left( {\rm{x}} \right)}} = {\rm{l}} \ne \frac{0}{0}\) where l is a finite value.

Calculation:

\(\mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;\left( {1 + x} \right)}}{{{x^2}}}\)                 [Form 0/0]

Apply L-Hospital Rule,

\( = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;\left( {0 + 1} \right)}}{{2x}} = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;1}}{{2x}}\)              [Form 0/0]

Again apply L-Hospital Rule,

\( = \mathop {\lim }\limits_{x \to 0} \frac{{{e^x} - \;0}}{2} = \frac{{{e^0}}}{2} = \frac{1}{2}\)

Given:

The given limiting function is as follows,

\(​\rm\displaystyle\lim _{x \rightarrow 0}\frac{e^x−1−x−x^2}{x^2}\) 

Concept:

Use the L'hospital rule for the limit \(\lim_{x \to a} \frac{f(x)}{g(x)}\) of the form \(\frac{0}{0}\)

\(\lim_{x \to a} \frac{f(x)}{g(x)}=\lim_{x \to a} \frac{f'(x)}{g'(x)}\)

Where,

f'(x) and g'(x) are the derivative of f(x) and g(x).

Solution:

The given limit is as follows,

\(​\rm\displaystyle\lim _{x \rightarrow 0}\frac{e^x−1−x−x^2}{x^2}\) 

\(​\rm\displaystyle\lim _{x \rightarrow 0}\frac{e^x−1−x−x^2}{x^2}=\frac{0}{0}\)

Derivating f(x) and g(x) we will get,

\(​\rm\displaystyle\lim _{x \rightarrow 0}\frac{e^x−1−2x}{2x}=\frac{0}{0}\)

Still, it is of \(\frac{0}{0}\) form.

Derivating again we will get,

\(​\rm\displaystyle\lim _{x \rightarrow 0}\frac{e^x-2}{2}=\frac{1-2}{2}=\frac{-1}{2}\)

Hence, option 3 is correct.

Concept:

Step: 1

Check f(a) is defined. If it is not defined then no need to go further.  The function is not continuous at a. If f(a) is defined then

Step: 2

Check Left-hand limit (LHL) and Right-hand limit (RHL) 

\(LHL = \rm \lim_{x\rightarrow a^{-}}f(x) \)

\(RHL= \lim_{x\rightarrow a^{+}} f(x) \)

If LHL = RHL then limit exist.

Formula used:

1) 1 - cos 2θ = 2sin²θ 

2) \(\rm \displaystyle\lim_ {x\rightarrow 0}\left({\frac{{sinx}}{x}}\right)\) = 1

Calculation: 

We have \(\rm \displaystyle\lim_ {x\rightarrow 1}\left({\frac{√{1-cos2(x-1)}}{(x-1)}}\right)\)

By using the above formula 

\(\rm \displaystyle\lim_ {x\rightarrow 1}\left({\frac{√{2sin^2(x-1)}}{(x-1)}}\right)\)

\(√ 2\rm \displaystyle\lim_ {x\rightarrow 1}\left({\frac{{|sin(x-1)}|}{(x-1)}}\right)\)

Since there is a mod in the expression, we have to check, whether the limit exists or not. 

LHL:

\(√ 2\rm \displaystyle\lim_ {x\rightarrow 1^{-}}\left({\frac{{|sin(x-1)}|}{(x-1)}}\right)\)

\(-√ 2\rm \displaystyle\lim_ {x\rightarrow 1^{-}}\left({\frac{{sin(x-1)}}{(x-1)}}\right)\)

By using the formula (2)

\(-√ 2\rm \displaystyle\lim_ {x-1\rightarrow 0^{}}\left({\frac{{sin(x-1)}}{(x-1)}}\right) \)

LHL = -√2

RHL: 

\(√ 2\rm \displaystyle\lim_ {x\rightarrow 1^{+}}\left({\frac{{|sin(x-1)}|}{(x-1)}}\right) \)

RHL = √2

Since, LHL ≠ RHL

The limit of the expression does not exist.

CONCEPT:

• \(\mathop {\lim }\limits_{x\; \to \;0} \left[ {\frac{{{a^x}\; - \;1}}{x}} \right] = \log a,\;a > 0\)
• \(\mathop {\lim }\limits_{x\; \to \;0} \left[ {\frac{{{e^x} - 1}}{x}} \right] = 1\)
• \(\mathop {\lim }\limits_{x\; \to \;a} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right)}}{{\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)}},\;provided\;\mathop {\lim }\limits_{x\; \to a} g\left( x \right) \ne 0\)

CALCULATION:

Here, we have to find the limit of \(\displaystyle\lim_{x\rightarrow 0} \dfrac{a^x-b^x}{e^x-1}\)

The expression \(\frac{{{a^x} - {b^x}}}{{{e^x} - 1}}\) can be re-written as:

\(\Rightarrow \frac{{{a^x} - {b^x}}}{{{e^x} - 1}} = \left[ {\frac{{\frac{{{a^x} - 1}}{x} - \frac{{{b^x} - 1}}{x}}}{{\frac{{{e^x} - 1}}{x}}}} \right] \)

Now by applying limits on both the sides of the above equation we get

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}} = \;\mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\frac{{{a^x} - 1}}{x} - \frac{{{b^x} - 1}}{x}}}{{\frac{{{e^x} - 1}}{x}}}} \right] \)

As we know that, \(\mathop {\lim }\limits_{x\; \to \;a} \left[ {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right] = \frac{{\mathop {\lim }\limits_{x\; \to \;a} f\left( x \right)}}{{\mathop {\lim }\limits_{x\; \to \;a} g\left( x \right)}},\;provided\;\mathop {\lim }\limits_{x\; \to a} g\left( x \right) \ne 0\)

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \left[ {\frac{{\frac{{{a^x} - 1}}{x} - \frac{{{b^x} - 1}}{x}}}{{\frac{{{e^x} - 1}}{x}}}} \right] = \frac{{\left[ {\mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - 1}}{x}} \right] - \left[ {\mathop {\lim }\limits_{x \to 0} \frac{{{b^x} - 1}}{x}} \right]}}{{\left[ {\mathop {\lim }\limits_{x \to } \frac{{{e^x} - 1}}{x}} \right]}}\)

As we know that, \(\mathop {\lim }\limits_{x\; \to \;0} \left[ {\frac{{{a^x}\; - \;1}}{x}} \right] = \log a,\;a > 0\) and \(\mathop {\lim }\limits_{x\; \to \;0} \left[ {\frac{{{e^x} - 1}}{x}} \right] = 1\)

\(\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{{a^x} - {b^x}}}{{{e^x} - 1}} = \log \frac{a}{b}\)

Hence, option A is true.

Concept Used:-

It is known that the derivative of eˣ is eˣ.

Now when it is written that,

eˣ = a₀ + a₁ x + a₂x² + ... + a_nxⁿ

At, x = 0; a₀ = 1

Differentiating the above equation we get,

eˣ = a₁ + 2a₂x + ...+ na_nx^n-1

Now put x = 0; a₁ = 1

Differentiate it again,

eˣ = 2a₂ +... n (n - 1)x^n-2

Now at, x = 0, 2a₂ = 1

When we, again and again, differentiate it, and put x = 0 to solve for the value of a_n, we get the series:

1 + x + x²/2! + x³/3! + ...+ xⁿ/n! to infinity.

On putting x = 1 we have,

e = 1 + 1/1! + 1/2! + 1/3! + ...+ 1/n!              .......(1)

Explanation:-

Given that the nth term of a series is \( \rm\frac{1}{(n+1)!}\).

\(T_n=\rm\frac{1}{(n+1)!}\)

Here, the series goes from n = 1 to ∞. So, the series will be,

\( S=\frac{1}{2 !}+\frac{1}{3 !}+\ldots-\infty\)

Here, add and subtract 1+1/1! in the right-hand side,

\(\Rightarrow S=\left(1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+--\infty\right)-(1+\frac{1}{1 !})\\ \Rightarrow S=\left(1+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+--\infty\right)-2\)

From equation (1) we get, 

⇒ S = e - 2

So, the sum of the series will be (e - 2)

Concept Used:-

We know that the Taylor expansion of e^x,

\( e^x=1+\frac{x}{1}+\frac{x^2}{2 !}+\frac{x^3}{3 !}+\cdots \infty \)

On putting x = 1 we get,

\(\Rightarrow e=1+1+\frac{1}{2 !}+\frac{1}{3 !}+....\\ \Rightarrow e=\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\frac{1}{3 !}+....\\ \Rightarrow e=\sum_{n=0}^\infty\frac{1}{(n) !}\ \ \ \ \ \ \ \ \ \ \ \ .....(1)\)

Also,

 \(\Rightarrow e=\sum_{n=1}^\infty\frac{1}{(n-1)!} \ \ \ \ \ \ \ \ \ \ \ \ .....(2)\\ \Rightarrow e=\sum_{n=2}^\infty\frac{1}{(n-2) !} \ \ \ \ \ \ \ \ \ \ \ \ .....(3)\)

Explanation:-

Given series is,

\(1+\frac{2^2}{1 !}+\frac{3^2}{2 !}+\frac{4^2}{3 !}+\ldots \ldots.\)

Since, 0! is equal to 1. So, the above sum can be written as,

\(\frac{1}{0!}+\frac{2^2}{1 !}+\frac{3^2}{2 !}+\frac{4^2}{3 !}+\ldots \ldots.\)

Let the sum of the above series is S. Then,

\(S=\frac{1^2}{0!}+\frac{2^2}{1 !}+\frac{3^2}{2 !}+\frac{4^2}{3 !}+\ldots \ldots.\)

The denominator number written with factorial is one less than the numerator number written with square. Thus, in general, this series can be written as,

⇒ \(S=\sum_{n=1}^n\frac{(n+1)^2}{(n)!}\)

Expand the numerator (n+1)2 to solve it further,

\(\Rightarrow S=\sum_{n=0}^n\frac{(n+1)^2}{(n)!}\\ \Rightarrow S=\sum_{n=0}^n\frac{n^2+2n+1}{(n)!}\\ \Rightarrow S=\sum_{n=0}^n\frac{n^2}{(n)!}+\sum_{n=0}^n\frac{2n}{(n)!}+\sum_{n=0}^n\frac{1}{(n)!}\\ \Rightarrow S=\sum_{n=1}^n\frac{n(n-1)+n}{n.(n-1)!}+\sum_{n=1}^n\frac{2n}{n(.n-1)!}+\sum_{n=0}^n\frac{1}{(n)!}\\ \Rightarrow S=\sum_{n=2}^n\frac{1}{(n-2)!}+\sum_{n=1}^n\frac{1}{(n-1)!}+\sum_{n=1}^n\frac{2}{(n-1)!}+\sum_{n=0}^n\frac{1}{(n)!}\\ \Rightarrow S=\sum_{n=2}^n\frac{1}{(n-2)!}+\sum_{n=1}^n\frac{1}{(n-1)!}+2\sum_{n=1}^n\frac{1}{(n-1)!}+\sum_{n=0}^n\frac{1}{(n)!}\)

Now using equation (1), (2) and (3) we get,

⇒ S = e + e + 2e + e

⇒ S = 5e

So, the sum of the given series is 5e.

Hence, the correct option is 1.

\(\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {\frac{1}{n}{e^{\frac{r}{n}}}} \)      ...(A)

If f is integrable on [a, b], then \(\int_a^bf(x) dx=\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n f(x_i)Δ x\)

where, Δx = \(\frac{b-a}{n}\) and xi = a + i Δx

\(\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n { {e^{\frac{r}{n}}\cdot \frac{1}{n}}}\) = \(\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n f(x_i)Δ x\)

Now, Δx = \(\frac{1}{n}\) = \(\frac{1-0}{n}\)

∴ The value of b is 1 and a is 0.

 x_i = a + \(\frac{i}{n}\)

∵ a = 0,  xi = \(\frac{i}{n}\)

a = 0, b = 1 , Δx = \(\frac{1}{n}\) and x_i = \(\frac{i}{n}\)

Now,   \(\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n { {e^{\frac{r}{n}}\cdot \frac{1}{n}}}\) = \(\mathop {\lim }\limits_{n \to \infty } \sum\limits_{i = 1}^n f(\frac{i}{n})\frac{1}{n}\) = \(\int_0^1e^xdx\)

\( = \left[ {{e^x}} \right]_0^1\)

= e¹ - e⁰

= e - 1