Question

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 $A+B+C=$

• A. $g(2)+g^{\prime}(2)+g^{\prime \prime}(2)$
• B. $g^{\prime \prime}(2)+2 g(2)+\frac{g \prime(1)}{2 !}$
• C. $g(2)+g^{\prime}(2)+\frac{g \prime(2)}{2 !}$
• D. $2 g(2)+2 g^{\prime}(2)+\frac{g \prime \prime(2)}{2 !}$

Answer 1

Answer: (C)

$3 x^2+4 x+7=A+B(x-2)+C(x-2)^2$
Given, $f(x)=f(a)+f^{\prime}(a)(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^2$ and $g(x)=3 x^2+4 x+7$
$ \therefore a=2, A=g(2), B=g^{\prime}(2), C=\frac{g^{\prime \prime}(2)}{2 !} $
$ \Rightarrow A+B+C=g(2)+g^{\prime}(2)+\frac{g^{\prime \prime}(2)}{2 !} $