Question

If $f\left(x\right)=x^n$ then the value of $f\left(x\right)$ is $f\left(1\right)-\frac{f\text{'}\left(1\right)}{1!}+\frac{f\text{'}\text{'}\left(1\right)}{2!}-\frac{f\text{'}\text{'}\text{'}\left(1\right)}{3!}+...+\frac{{\left(-1\right)}^n{f}^n\left(1\right)}{n!}$

• A. $2^n$
• B. $2^{n-1}$
• C. $0$
• D. $1$

Answer 1

The correct option is C

$0$

Explanation for the correct option:

Finding the value of the expression:

Given that,

$f\left(x\right)=x^n$,

Differentiate the given function with respect to $x$ at $x=1$,

$\begin{array}{rcl}f\text{'}\left(x\right)& =& n{x}^{n-1}\\ & \Rightarrow & f\text{'}\left(1\right)=n{\left(1\right)}^{n-1}\\ & =& n\end{array}$

Now, differentiate $f\text{'}\left(x\right)$ at $x=1$,

$\begin{array}{rcl}f\text{'}\text{'}\left(x\right)& =& n\left(n-1\right)x^{n-2}\\ & \Rightarrow & f\text{'}\text{'}\left(1\right)=n\left(n-1\right)\end{array}$

Similarly, differentiate the function $k$ times at $x=1$

$\begin{array}{rcl}{f}^k\left(x\right)& =& n\left(n-1\right)\left(n-2\right)...\left(n-k+1\right)x^{n-k}\\ & \Rightarrow & f^k\left(1\right)=n\left(n-1\right)\left(n-2\right)...\end{array}$

Substitute these values in the finding expression,

$\begin{array}{ccc}f\left(1\right)-\frac{f\text{'}\left(1\right)}{1!}+\frac{f\text{'}\text{'}\left(1\right)}{2!}-\frac{f\text{'}\text{'}\text{'}\left(1\right)}{3!}+...+\frac{{\left(-1\right)}^n{f}^n\left(1\right)}{n!}& =& n-\frac{n\left(n-1\right)}{1!}+\frac{n\left(n-1\right)\left(n-2\right)}{2!}-...+\frac{{\left(-1\right)}^{n}n\left(n-1\right)\left(n-2\right)...}{n!}\\ & & {=}^n{C}_0{-}^n{C}_1{+}^n{C}_2{-}^n{C}_3+...+\frac{{\left(-1\right)}^{n}n!}{n!}\mathbf{\left[}\mathbf{\text{by binomial expansion}}\mathbf{\right]}\\ & & {=}^n{C}_0{-}^n{C}_1{+}^n{C}_2{-}^n{C}_3+...+{{\left(-1\right)}^n}^n{C}_n\\ & =& \left({}^n{C}_0{+}^n{C}_2{+}^n{C}_4+....\right)-\left({}^n{C}_1{+}^n{C}_3{+}^n{C}_5+...\right)\\ & =& 2^{n-1}-2^{n-1}\\ & =& 0\end{array}$

Hence, the correct option is C.