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Question

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A.${{(2011)}^{n}}$

B.${{(2012)}^{n}}$

C.${{(2012)}^{n}}-1$

D.$n{{(2011)}^{n}}$

Answer

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Hint: We are given $f(x)={{(2011+x)}^{n}}$ and we have to find the value of $f(0)+f'(0)+\dfrac{f''(0)}{2!}+.....+\dfrac{{{f}^{n-1}}(0)}{(n-1)!}$ . For that, differentiate $f(x)={{(2011+x)}^{n}}$ thrice and substitute the value in the above given equation. Try it and you will get the answer.

Complete step-by-step answer:

In calculus, differentiation is one of the two important concepts apart from integration. It is a method of finding the derivative of a function or instantaneous rate of change in function based on one of its variables. If x is a variable and y is another variable, then the rate of change of x with respect to y is given by dy/dx.

Functions are generally classified in two categories under Calculus, namely:

(i) Linear functions

(ii) Non-linear functions

A linear function varies with a constant rate through its domain. Therefore, the overall rate of change of the function is the same as the rate of change of a function at any point.

However, the rate of change of function varies from point to point in case of non-linear functions. The nature of variation is based on the nature of the function.

The rate of change of a function at a particular point is defined as a derivative of that particular function.

Differentiation formulas list has been provided here for students so that they can refer to these to solve problems based on differential equations. This is one of the most important topics in higher class Mathematics. The general representation of the derivative is d/dx.

This formula list includes derivatives for constant, trigonometric functions, polynomials, hyperbolic, logarithmic functions, exponential, inverse trigonometric functions etc. Based on these, there are a number of examples and problems present in the syllabus of class 11 and 12, for which students can easily write answers.

Differentiation is the essence of Calculus. A derivative is defined as the instantaneous rate of change in function based on one of its variables. It is similar to finding the slope of a tangent to the function at a point.

We are given $f(x)={{(2011+x)}^{n}}$ ,

So now differentiating above we get,

$f'(x)=n{{(2011+x)}^{n-1}}$

Again differentiating we get,

$f''(x)=n(n-1){{(2011+x)}^{n-2}}$ and so on.

So putting $x=0$ in $f(x)={{(2011+x)}^{n}}$ , $f'(x)=n{{(2011+x)}^{n-1}}$ and $f''(x)=n(n-1){{(2011+x)}^{n-2}}$ we get,

$f(0)={{(2011)}^{n}}$

$f'(0)=n{{(2011)}^{n-1}}$

$f''(0)=n(n-1){{(2011)}^{n-2}}$

Now, let $S=f(0)+f'(0)+\dfrac{f''(0)}{2!}+.....+\dfrac{{{f}^{n-1}}(0)}{(n-1)!}$ .

So substituting the above values we get,

$S={{(2011)}^{n}}+n{{(2011)}^{n-1}}+\dfrac{n(n-1){{(2011)}^{n-2}}}{2!}+.....$

Since, there is a missing term $\dfrac{f''(0)}{1!}$ to complete the exact expansion of ${{(1+x)}^{n}}$.

So, $S={{(2011+1)}^{n}}-1$

\[S={{(2012)}^{n}}-1\]

So we get,

$f(0)+f'(0)+\dfrac{f''(0)}{2!}+.....+\dfrac{{{f}^{n-1}}(0)}{(n-1)!}={{(2012)}^{n}}-1$ .

The correct answer is option (C).

Note: Read the question and see what is asked. Your concept regarding differentiation should be clear. A proper assumption should be made. Do not make silly mistakes while substituting. Equate it in a proper manner and don't confuse yourself.

Complete step-by-step answer:

In calculus, differentiation is one of the two important concepts apart from integration. It is a method of finding the derivative of a function or instantaneous rate of change in function based on one of its variables. If x is a variable and y is another variable, then the rate of change of x with respect to y is given by dy/dx.

Functions are generally classified in two categories under Calculus, namely:

(i) Linear functions

(ii) Non-linear functions

A linear function varies with a constant rate through its domain. Therefore, the overall rate of change of the function is the same as the rate of change of a function at any point.

However, the rate of change of function varies from point to point in case of non-linear functions. The nature of variation is based on the nature of the function.

The rate of change of a function at a particular point is defined as a derivative of that particular function.

Differentiation formulas list has been provided here for students so that they can refer to these to solve problems based on differential equations. This is one of the most important topics in higher class Mathematics. The general representation of the derivative is d/dx.

This formula list includes derivatives for constant, trigonometric functions, polynomials, hyperbolic, logarithmic functions, exponential, inverse trigonometric functions etc. Based on these, there are a number of examples and problems present in the syllabus of class 11 and 12, for which students can easily write answers.

Differentiation is the essence of Calculus. A derivative is defined as the instantaneous rate of change in function based on one of its variables. It is similar to finding the slope of a tangent to the function at a point.

We are given $f(x)={{(2011+x)}^{n}}$ ,

So now differentiating above we get,

$f'(x)=n{{(2011+x)}^{n-1}}$

Again differentiating we get,

$f''(x)=n(n-1){{(2011+x)}^{n-2}}$ and so on.

So putting $x=0$ in $f(x)={{(2011+x)}^{n}}$ , $f'(x)=n{{(2011+x)}^{n-1}}$ and $f''(x)=n(n-1){{(2011+x)}^{n-2}}$ we get,

$f(0)={{(2011)}^{n}}$

$f'(0)=n{{(2011)}^{n-1}}$

$f''(0)=n(n-1){{(2011)}^{n-2}}$

Now, let $S=f(0)+f'(0)+\dfrac{f''(0)}{2!}+.....+\dfrac{{{f}^{n-1}}(0)}{(n-1)!}$ .

So substituting the above values we get,

$S={{(2011)}^{n}}+n{{(2011)}^{n-1}}+\dfrac{n(n-1){{(2011)}^{n-2}}}{2!}+.....$

Since, there is a missing term $\dfrac{f''(0)}{1!}$ to complete the exact expansion of ${{(1+x)}^{n}}$.

So, $S={{(2011+1)}^{n}}-1$

\[S={{(2012)}^{n}}-1\]

So we get,

$f(0)+f'(0)+\dfrac{f''(0)}{2!}+.....+\dfrac{{{f}^{n-1}}(0)}{(n-1)!}={{(2012)}^{n}}-1$ .

The correct answer is option (C).

Note: Read the question and see what is asked. Your concept regarding differentiation should be clear. A proper assumption should be made. Do not make silly mistakes while substituting. Equate it in a proper manner and don't confuse yourself.