JEE Main 2019DifferentiationDifferentiationMediumQuestionIf f(x)=xn,f\left( x \right) = {x^n},f(x)=xn, then the value of f(1)−f′(1)1!+f′′(1)2!−f′′′(1)3!+..........(−1)nfn(1)n!f\left( 1 \right) - {{f'\left( 1 \right)} \over {1!}} + {{f''\left( 1 \right)} \over {2!}} - {{f'''\left( 1 \right)} \over {3!}} + ..........{{{{\left( { - 1} \right)}^n}{f^n}\left( 1 \right)} \over {n!}}f(1)−1!f′(1)+2!f′′(1)−3!f′′′(1)+..........n!(−1)nfn(1) isOptionsA111B2n{{2^n}}2nC2n−1{{2^n} - 1}2n−1D000Check AnswerHide SolutionSolutionf(x)=xn⇒f(1)=1f\left( x \right) = {x^n} \Rightarrow f\left( 1 \right) = 1f(x)=xn⇒f(1)=1 f′(x)=nxn−1⇒f′(1)=nf'\left( x \right) = n{x^{n - 1}} \Rightarrow f'\left( 1 \right) = nf′(x)=nxn−1⇒f′(1)=n f′′(x)=n(n−1)xn−2f''\left( x \right) = n\left( {n - 1} \right){x^{n - 2}}f′′(x)=n(n−1)xn−2 ⇒f′′(1)=n(n−1) \Rightarrow f''\left( 1 \right) = n\left( {n - 1} \right)⇒f′′(1)=n(n−1) ∴\therefore∴ fn(x)=n!{f^n}\left( x \right) = n!fn(x)=n! ⇒fn(1)=n! \Rightarrow {f^n}\left( 1 \right) = n!⇒fn(1)=n! =1−n1!+n(n−1)2!n(n−1)(n−2)3! = 1 - {n \over {1!}} + {{n\left( {n - 1} \right)} \over {2!}}{{n\left( {n - 1} \right)\left( {n - 2} \right)} \over {3!}}=1−1!n+2!n(n−1)3!n(n−1)(n−2) +....+(−1)nn!n!\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + .... + {\left( { - 1} \right)^n}{{n!} \over {n!}}+....+(−1)nn!n! =n C0−n C1+n C2−n C3 = {}^n\,{C_0} - {}^n\,{C_1} + {}^n\,{C_2} - {}^n\,{C_3}=nC0−nC1+nC2−nC3 +......+(−1)n nCn=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ...... + {\left( { - 1} \right)^n}\,{}^n{C_n} = 0+......+(−1)nnCn=0