If f(x) = x^n,then the value of f(1) - frac{f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $f(x) = x^n$. We know that the $k$-th derivative of $f(x)$ at $x=1$ is $f^k(1) = n(n-1)(n-2)...(n-k+1) = \frac{n!}{(n-k)!}$. Substituting th

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Given $f(x) = x^n$. We know that the $k$-th derivative of $f(x)$ at $x=1$ is $f^k(1) = n(n-1)(n-2)...(n-k+1) = \frac{n!}{(n-k)!}$. Substituting this into the expression: $f(1) - \frac{f'(1)}{1!} + \frac{f''(1)}{2!} - \frac{f'''(1)}{3!} + ...... + \frac{(-1)^n f^n(1)}{n!} = \sum_{k=0}^{n} (-1)^k \frac{f^k(1)}{k!}$. Since $\frac{f^k(1)}{k!} = \frac{n!}{k!(n-k)!} = \binom{n}{k}$,the expression becomes: $\sum_{k=0}^{n} (-1)^k \binom{n}{k} = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - ...... + (-1)^n \binom{n}{n}$. Using the binomial theorem,$(1-x)^n = \sum_{k=0}^{n} \binom{n}{k} (-x)^k$. Setting $x=1$,we get $(1-1)^n = 0^n = 0$ for $n \ge 1$. Thus,the value is $0$.

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

Similar Questions

The sum of the coefficients of all odd degree terms in the expansion of $(x + \sqrt{x^3 - 1})^5 + (x - \sqrt{x^3 - 1})^5$,where $x > 1$,is:

If $r, k, p \in W,$ then $\sum\limits_{r + k + p = 10} {{}^{30}{C_r} \cdot {}^{20}{C_k} \cdot {}^{10}{C_p}} $ is equal to -

The coefficient of $x^{50}$ in the expansion of $(1+x)^{1000} + x(1+x)^{999} + x^2(1+x)^{998} + \ldots + x^{1000}$ is

Find the remainder when $3 \times 7^{22} + 2 \times 10^{22} - 44$ is divided by $18$.

$x^5 + 10x^4a + 40x^3a^2 + 80x^2a^3 + 80xa^4 + 32a^5 = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module