If f(x) = begin{cases} frac{a sin x - b x + c | vedclass

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(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x 	o 0} f(x) = f(0) = 0$. The Taylor expansion of the numerator $N(x)$ is: $N(x) = a(x -

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$a=b$

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(B) For $f(x)$ to be continuous at $x=0$,we must have $\lim_{x 	o 0} f(x) = f(0) = 0$. The Taylor expansion of the numerator $N(x)$ is: $N(x) = a(x - \frac{x^3}{6} + O(x^5)) - bx + cx^2 + x^3 = (a-b)x + cx^2 + (1 - \frac{a}{6})x^3 + O(x^4)$. The Taylor expansion of the denominator $D(x)$ is: $D(x) = 2(x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + O(x^5)) - 2x + x^2 - \frac{2}{3}x^3 = -\frac{1}{2}x^4 + O(x^5)$. For the limit to exist and equal $0$,the coefficients of $x, x^2,$ and $x^3$ in the numerator must be zero. Thus,$a-b = 0 \implies a=b$,$c=0$,and $1 - \frac{a}{6} = 0 \implies a=6$. Since the question asks for the relation between $a, b, c$ based on the provided options,$a=b$ is the correct relation.

If $f(x) = \begin{cases} \frac{a \sin x - b x + c x^2 + x^3}{2 \log(1+x) - 2x + x^2 - \frac{2}{3}x^3} &, x \neq 0 \\ 0 &, x=0 \end{cases}$ is continuous at $x=0$,then find the relation between $a, b, c$.

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