Question

If $y=\frac{a{x}^2}{(x-a)(x-b)(x-c)}+\frac{bx}{(x-b)(x-c)}+\frac{c}{x-c}+1,$ then $\frac{y^{\prime }}{y}$ is equal to
$(\text{Here},y^{\prime }=\frac{dy}{dx})$

• A. $\frac{1}{x}(\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x})$
• B. $\frac{1}{x}(\frac{b}{a-x}+\frac{c}{b-x}+\frac{a}{c-x})$
• C. $\frac{abc}{x}(\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x})$
• D. $\frac{1}{x}(\frac{bc}{a-x}+\frac{ac}{b-x}+\frac{ab}{c-x})$

Answer 1

The correct option is A $\frac{1}{x}(\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x})$

$y=\frac{a{x}^2}{(x-a)(x-b)(x-c)}+\frac{bx}{(x-b)(x-c)}+\frac{c}{x-c}+1\Rightarrow y=\frac{a{x}^2}{(x-a)(x-b)(x-c)}+\frac{bx}{(x-b)(x-c)}+\frac{x}{x-c}\Rightarrow y=\frac{a{x}^2}{(x-a)(x-b)(x-c)}+\frac{x^2}{(x-b)(x-c)}\Rightarrow y=\frac{x^3}{(x-a)(x-b)(x-c)}$

Taking $\log$ on both sides, we get
$\log y=3\log x-\left[\log \right(x-a)+\log (x-b)+\log (x-c\left)\right]$
On differentiating with respect to $x$, we get
$\frac{1}{y}\frac{dy}{dx}=\frac{3}{x}-[\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c}]\Rightarrow \frac{y^{\prime }}{y}=(\frac{1}{x}-\frac{1}{x-a})+(\frac{1}{x}-\frac{1}{x-b})+(\frac{1}{x}-\frac{1}{x-c})\Rightarrow \frac{y^{\prime }}{y}=-\left(\frac{a}{x(x-a)}\right)-\left(\frac{b}{x(x-b)}\right)-\left(\frac{c}{x(x-c)}\right)\therefore \frac{y^{\prime }}{y}=\frac{1}{x}(\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x})$