For any quadratic polynomial f(x), it is true | vedclass

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(C) Given $g(x) = 3x^2 + 4x + 7$. By Taylor expansion about $x=2$, we have $g(x) = g(2) + g^{\prime}(2)(x-2) + \frac{g^{\prime \prime}(2)}{2!}(x-2)^2$

Vedclass · Accepted Answer

$g(2)+g^{\prime}(2)+\frac{g^{\prime \prime}(2)}{2!}$

Vedclass · Answer

(C) Given $g(x) = 3x^2 + 4x + 7$. By Taylor expansion about $x=2$, we have $g(x) = g(2) + g^{\prime}(2)(x-2) + \frac{g^{\prime \prime}(2)}{2!}(x-2)^2$.Dividing by $(x-2)^3$ is not applicable here as the degree of $g(x)$ is $2$. However, the equation $\frac{3x^2+4x+7}{(x-2)^3} = \frac{A}{(x-2)^3} + \frac{B}{(x-2)^2} + \frac{C}{(x-2)}$ implies $3x^2+4x+7 = A + B(x-2) + C(x-2)^2$.Comparing this with the Taylor expansion $g(x) = g(2) + g^{\prime}(2)(x-2) + \frac{g^{\prime \prime}(2)}{2!}(x-2)^2$, we identify $A = g(2)$, $B = g^{\prime}(2)$, and $C = \frac{g^{\prime \prime}(2)}{2!}$.Calculating the values: $g(2) = 3(4)+4(2)+7 = 27$, $g^{\prime}(x) = 6x+4 \Rightarrow g^{\prime}(2) = 16$, $g^{\prime \prime}(x) = 6 \Rightarrow \frac{g^{\prime \prime}(2)}{2!} = \frac{6}{2} = 3$.Thus, $A=27, B=16, C=3$.$A+B+C = 27+16+3 = 46$.Checking the options: $g(2)+g^{\prime}(2)+\frac{g^{\prime \prime}(2)}{2!} = 27+16+3 = 46$.Therefore, the correct option is $C$.

For any quadratic polynomial $f(x)$, it is true that $f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2!}(x-a)^2$ where $a$ is any real number. If $\frac{3 x^2+4 x+7}{(x-2)^3}=\frac{A}{(x-2)^3}+\frac{B}{(x-2)^2}+\frac{C}{(x-2)}$ and $g(x)=3 x^2+4 x+7$, then $A+B+C=$

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