A) \[\frac{1}{a-x}+\frac{1}{b-x}+\frac{1}{c-x}\]
B) \[\frac{x}{a-x}+\frac{x}{b-x}+\frac{x}{c-x}\]
C) \[\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}\]
D) \[\frac{a}{x-a}+\frac{b}{x-b}+\frac{c}{x-c}\]
Correct Answer: C
Solution :
Idea Here simply follow the rule of LCM and properties of logarithmic \[\therefore \]\[\log \frac{{{x}^{3}}}{pqr}=3\log x-\log p-\log q-\log \upsilon \] Here, it is given that \[y=\frac{a{{x}^{2}}}{(y-a)(x-b)(x-c)}+\frac{bx}{(x-b)(x-c)}\]\[+\frac{x}{x-c}+1\] \[=\frac{a{{x}^{2}}+bx(x-a)}{(x-a)(x-b)(x-c)}+\frac{x}{x-c}\] \[=\frac{a{{x}^{2}}+b{{x}^{2}}-abx+x(x-a)(x-b)}{(x-a)(x-b)(x-c)}\] \[=\frac{{{x}^{3}}}{(x-a)(x-b)(x-c)}\] Thus, \[y=\frac{{{x}^{3}}}{(x-a)(x-b)(x-c)}\] \[\Rightarrow \]\[\ln y=3\ln x-\ln (x-a)-\ln (x-b)-\ln (x-c)\] Differentiating both sides w.r.t. x, we get \[\frac{1}{y}.y'=\frac{3}{x}-\frac{1}{x-a}-\frac{1}{x-b}-\frac{1}{x-c}\] \[=\left( \frac{1}{x}-\frac{1}{x-a} \right)+\left( \frac{1}{x}-\frac{1}{x-b} \right)+\left( \frac{1}{x}-\frac{1}{x-c} \right)\] \[\Rightarrow \]\[\frac{y'}{y}=\frac{a}{x(a-x)}+\frac{b}{x(b-x)}+\frac{c}{x(c-x)}\] \[\Rightarrow \]\[\frac{xy'}{y}=\frac{a}{a-x}+\frac{b}{b-x}+\frac{c}{c-x}\] So, correct option is [c]. TEST Edge Parametric differentiation logarithmic differentiation and higher order differentiation, related questions are asked. To solve such types of questions students are advised to understand the differentiation.
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