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 $\frac{3x^2+4x+7}{(x-2{)}^3}=\frac{A}{(x-2{)}^3}+\frac{B}{(x-2{)}^2}+\frac{C}{(x-2)}$ and $g(x)=$ $3x^2+4x+7$ then $A+B+C=$

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

Answer 1

Answer: (C)

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