A) 0
B) 1
C) - 2
D) Cannot be determined
Correct Answer: C
Solution :
Coefficient\[x\]in\[f(x)\]is equal to coefficient of\[x\]in \[\left| \begin{matrix} x & {{(1+x-\frac{{{x}^{3}}}{3!}...)}^{3}} & 1-\frac{{{x}^{2}}}{2!} \\ 1 & x-\frac{{{x}^{2}}}{2} & 2 \\ {{x}^{2}} & 1+{{x}^{2}} & 0 \\ \end{matrix} \right|\] coefficient of\[x\]in\[\left| \begin{matrix} x & 1 & 1 \\ 1 & x & 2 \\ {{x}^{2}} & 1 & 0 \\ \end{matrix} \right|\] = Coefficient of\[x\]in\[[x(0-2)-1(0-2{{x}^{2}})\] \[+1(1-{{x}^{3}})]\] = - 2You need to login to perform this action.
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