Question

If $g\left(x\right)=\frac{f\left(x\right)}{(x-a)(x-b)(x-c)}$, where $f\left(x\right)$ is a polynomial of degree $<3$, then

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

Answer 1

Answer: (A), (B)

The correct options are
A $\int g\left(x\right)dx=\left|\begin{array}{ccc}1& a& f\left(a\right)\log |x-a|\\ 1& b& f\left(b\right)\log |x-b|\\ 1& c& f\left(c\right)\log |x-c|\end{array}\right|÷\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|+k$
B $\frac{dg\left(x\right)}{dx}=\left|\begin{array}{ccc}1& a& f\left(a\right){(x-a)}^{-2}\\ 1& b& f\left(b\right){(x-b)}^{-2}\\ 1& c& f\left(c\right){(x-c)}^{-2}\end{array}\right|÷\left|\begin{array}{ccc}{a}^2& a& 1\\ b^2& b& 1\\ c^2& c& 1\end{array}\right|$

$g\left(x\right)=\frac{f\left(x\right)}{(x-a)(x-b)(x-c)}=\frac{A}{(x-a)}+\frac{B}{(x-b)}+\frac{C}{(x-c)}$...........(1)

One comparing the various powers of x, we get

$\Rightarrow \{\begin{array}{c}A=\frac{f\left(a\right)}{(a-b)(a-c)}=-\frac{f\left(a\right)}{(a-b)(c-a)}\\ B=\frac{f\left(b\right)}{(b-a)(b-c)}=-\frac{f\left(b\right)}{(a-b)(b-c)}\\ C=\frac{f\left(c\right)}{(c-a)(c-b)}=-\frac{f\left(c\right)}{(b-c)(c-a)}\end{array}$

Now from (1) we have

$g\left(x\right)=\frac{f\left(x\right)}{(x-a)(x-b)(x-c)}=\frac{(c-b)\frac{f\left(a\right)}{(x-a)}-(c-a)\frac{f\left(b\right)}{(x-b)}+(b-a)\frac{f\left(c\right)}{(x-c)}}{(a-b)(b-c)(c-a)}$ $=\frac{\left|\begin{array}{ccc}1& a& f\left(a\right)/(x-a)\\ 1& b& f\left(b\right)/(x-b)\\ 1& c& f\left(c\right)/(x-c)\end{array}\right|}{\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|}$
Hence option A,B are correct choice.