Limit Continuity And Differentiability Ques 21
- If $y=\frac{a x^2}{(x-a)(x-b)(x-c)}+\frac{b x}{(x-b)(x-c)}+\frac{c}{(x-c)}+1$, Prove that $\frac{y^{\prime}}{y}=\frac{1}{x}\left(\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}\right)$.
$(1998,8 \mathrm{M})$
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Solution: $y=\frac{a x^2}{(x-a)(x-b)(x-c)}+\frac{b x}{(x-b)(x-c)}+\frac{c}{(x-c)}+1$
$ \begin{aligned} & =\frac{a x^2}{(x-a)(x-b)(x-c)}+\frac{b x}{(x-b)(x-c)}+\frac{x}{(x-c)} \\ & =\frac{a x^2}{(x-a)(x-b)(x-c)}+\frac{x}{(x-c)}\left(\frac{b}{x-b}+1\right) \\ & =\frac{a x^2}{(x-a)(x-b)(x-c)}+\frac{x}{(x-c)} \cdot \frac{x}{(x-b)} \\ & =\frac{x^2}{(x-c)(x-b)}\left(\frac{a}{x-1}+1\right) \Rightarrow y=\frac{x^3}{(x-a)(x-b)(x-c)} \end{aligned} $
$\Rightarrow \quad \log y=\log x^3-\log (x-a)(x-b)(x-c)$
$ \Rightarrow \quad \log y=3 \log x-\log (x-a)-\log (x-b)-\log (x-c) $
On differentiating, we get
$ \begin{aligned} \frac{y^{\prime}}{y} & =\frac{3}{x}-\frac{1}{x-a}-\frac{1}{x-b}-\frac{1}{x-c} \\ \Rightarrow \quad \frac{y^{\prime}}{y} & =\left(\frac{1}{x}-\frac{1}{x-a}\right)+\left(\frac{1}{x}-\frac{1}{x-b}\right)+\left(\frac{1}{x}-\frac{1}{x-c}\right) \\ \Rightarrow \quad \frac{y^{\prime}}{y} & =\frac{-a}{x(x-a)}-\frac{b}{x(x-b)}-\frac{c}{x(x-c)} \\ \Rightarrow \quad \frac{y^{\prime}}{y} & =\frac{a}{x(a-x)}+\frac{b}{x(b-x)}+\frac{c}{x(c-x)} \\ \Rightarrow \quad \frac{y^{\prime}}{y} & =\frac{1}{x}\left(\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}\right) \end{aligned} $
BETA
