Question

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

• A. $\frac{\left|\begin{array}{ccc}1& a& \frac{f\left(a\right)}{(x-a)}\\ 1& b& \frac{f\left(b\right)}{(x-b)}\\ 1& c& \frac{f\left(c\right)}{(x-c)}\end{array}\right|}{\left|\begin{array}{ccc}1& a& a^2\\ 1& b& b^2\\ 1& c& c^2\end{array}\right|}$
• B. $\frac{\left|\begin{array}{ccc}1& a& \frac{f\left(a\right)}{(x-a)}\\ 1& b& \frac{f\left(b\right)}{(x-b)}\\ 1& c& \frac{f\left(c\right)}{(x-c)}\end{array}\right|}{\left|\begin{array}{ccc}1& a& a^3\\ 1& b& b^3\\ 1& c& c^3\end{array}\right|}$
• C. $0$
• D. none of these

Answer 1

Answer: (A)

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

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

On 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

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