Question

If $a,b,c,\cdots k,$ are roots of the equation $f\left(x\right)=0$, then the value of $\frac{f\left(x\right)}{x-a}+\frac{f\left(x\right)}{x-b}+\cdots +\frac{f\left(x\right)}{x-k}$, is

• A. $f\left(x\right)$
• B. $2f^{\prime }\left(x\right)$
• C. $2f\left(x\right)$
• D. $f^{\prime }\left(x\right)$

Answer 1

The correct option is D $f^{\prime }\left(x\right)$

$a,b,c...k,arerootsoff\left(x\right)=0.So,f\left(x\right)=(x-a)(x-b)(x-c)...(x-k)\Rightarrow \log f\left(x\right)=\log (x-a)+\log (x-b)+...+\log (x-k)\Rightarrow \frac{1}{f\left(x\right)}{f}^{\prime }\left(x\right)=\frac{1}{x-a}+\frac{1}{x-b}+...\frac{1}{x-k}\Rightarrow f^{\prime }\left(x\right)=\frac{f\left(x\right)}{x-a}+\frac{f\left(x\right)}{x-b}+...\frac{f\left(x\right)}{x-k}$