Question

Let $g\left(x\right)=\frac{f\left(x\right)}{(x-a)(x-b)(x-c)},$ where $f\left(x\right)$ is a polynomial of degree less than $3.$ If $(a-b)(b-c)(c-a)=k,$ where $k$ is a non-zero constant, then

• A. $\int g\left(x\right)dx=\frac{1}{k}\left|\begin{array}{ccc}1& a& f\left(a\right)\ln |x-a|\\ 1& b& f\left(b\right)\ln |x-b|\\ 1& c& f\left(c\right)\ln |x-c|\end{array}\right|+C$
• B. $\int g\left(x\right)dx=-\frac{1}{k}\left|\begin{array}{ccc}1& a& f\left(a\right)\ln |x-a|\\ 1& b& f\left(b\right)\ln |x-b|\\ 1& c& f\left(c\right)\ln |x-c|\end{array}\right|+C$
• C. $\frac{dg\left(x\right)}{dx}=\frac{1}{k}\left|\begin{array}{ccc}1& a& f\left(a\right)(x-a{)}^{-2}\\ 1& b& f\left(b\right)(x-b{)}^{-2}\\ 1& c& f\left(c\right)(x-c{)}^{-2}\end{array}\right|$
• D. $\frac{dg\left(x\right)}{dx}=-\frac{1}{k}\left|\begin{array}{ccc}1& a& f\left(a\right)(x-a{)}^{-2}\\ 1& b& f\left(b\right)(x-b{)}^{-2}\\ 1& c& f\left(c\right)(x-c{)}^{-2}\end{array}\right|$

Answer 1

Answer: (A), (D)

$\frac{dg\left(x\right)}{dx}=-\frac{1}{k}\left|\begin{array}{ccc}1& a& f\left(a\right)(x-a{)}^{-2}\\ 1& b& f\left(b\right)(x-b{)}^{-2}\\ 1& c& f\left(c\right)(x-c{)}^{-2}\end{array}\right|$

$g\left(x\right)=\frac{f\left(x\right)}{(x-a)(x-b)(x-c)}$
By partial fractions, we have
$g\left(x\right)=\frac{f\left(a\right)}{(x-a)(a-b)(a-c)}+\frac{f\left(b\right)}{(b-a)(x-b)(b-c)}+\frac{f\left(c\right)}{(c-a)(c-b)(x-c)}$
$\Rightarrow g\left(x\right)=\left[\frac{1}{(a-b)(b-c)(c-a)}\right]\times [\frac{f\left(a\right)(c-b)}{x-a}+\frac{f\left(b\right)(a-c)}{x-b}+\frac{f\left(c\right)(b-a)}{x-c}]$

$\because k=(a-b)(b-c)(c-a)$
$\Rightarrow g\left(x\right)=\frac{1}{k}\times [\frac{f\left(a\right)(c-b)}{x-a}+\frac{f\left(b\right)(a-c)}{x-b}+\frac{f\left(c\right)(b-a)}{x-c}]$

$\Rightarrow g\left(x\right)=\frac{1}{k}\left|\begin{array}{ccc}1& a& \frac{f\left(a\right)}{(x-a)}\\ 1& b& \frac{f\left(b\right)}{(x-b)}\\ 1& c& \frac{f\left(c\right)}{(x-c)}\end{array}\right|$

$\therefore \int g\left(x\right)dx=\frac{1}{k}\left|\begin{array}{ccc}1& a& f\left(a\right)\ln |x-a|\\ 1& b& f\left(b\right)\ln |x-b|\\ 1& c& f\left(c\right)\ln |x-c|\end{array}\right|+C$
and
$\frac{dg\left(x\right)}{dx}=\frac{1}{k}\left|\begin{array}{ccc}1& a& -f\left(a\right)(x-a{)}^{-2}\\ 1& b& -f\left(b\right)(x-b{)}^{-2}\\ 1& c& -f\left(c\right)(x-c{)}^{-2}\end{array}\right|$
$\Rightarrow \frac{dg\left(x\right)}{dx}=-\frac{1}{k}\left|\begin{array}{ccc}1& a& f\left(a\right)(x-a{)}^{-2}\\ 1& b& f\left(b\right)(x-b{)}^{-2}\\ 1& c& f\left(c\right)(x-c{)}^{-2}\end{array}\right|$