Question

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

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

Answer 1

Answer: (C)

$0$

$f\left(x\right)=x^n\Rightarrow f\left(1\right)=1$

$f^{\prime }\left(x\right)=n{x}^{n-1}\Rightarrow f^{\prime }\left(1\right)=n$

$f^{\prime \prime }\left(x\right)=n(n-1)x^{n-2}\Rightarrow f^{\prime \prime }\left(1\right)=n(n-1)$

... ... ... ...

... ... ... ...

$f^n\left(x\right)=n(n-1)(n-2)...2.1$

$f^n\left(1\right)=n(n-1)(n-2)...2.1$

$f\left(1\right)-\frac{f^{\prime }\left(1\right)}{1!}+\frac{f^{\prime \prime }\left(1\right)}{2!}-\frac{f^{\prime \prime \prime }\left(1\right)}{3!}+...+\frac{(-1{)}^n{f}^n(1)}{n!}$

$=1-\frac{n}{1!}+\frac{n(n-1)}{2!}-\frac{n(n-1)(n-2)}{3!}+...+\frac{(-1{)}^{n}n(n-1\left)\right(n-2)...2.1}{n!}$

$=(1-1{)}^n$ ........... [Using binomial expansion as $(1-x{)}^n={}^n{C}_0{1}^n(-x{)}^0+{}^n{C}_1{1}^{n-1}(-x{)}^1+.....+{}^n{C}_n(-x{)}^n]$

$=0$