Question:
If $f (x) = x^n$, then the value of $f \left(1\right)-\frac{f '\left(1\right)}{1!}+\frac{f ''\left(1\right)}{2!}-\frac{f "'\left(1\right)}{3!}+\dots+\frac{\left(-1\right)^{n}f ^{n}\left(1\right)}{n!}$ is
If $f (x) = x^n$, then the value of $f \left(1\right)-\frac{f '\left(1\right)}{1!}+\frac{f ''\left(1\right)}{2!}-\frac{f "'\left(1\right)}{3!}+\dots+\frac{\left(-1\right)^{n}f ^{n}\left(1\right)}{n!}$ is
Updated On: Jul 27, 2022
- $2^n$
- $2^{n-1}$
- $0$
- $1$
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The Correct Option is C
Solution and Explanation
$LHS = 1-\frac{n}{1!}+\frac{n\left(n-1\right)}{2!}-\frac{n\left(n-1\right)\left(n-2\right)}{3!}+\dots\dots\dots$
$=1-^{n}C_{1}+^{n}C_{2}-\dots\dots\dots$
$=0.$
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.