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प्रश्न
If \[y = \log_e \left( \frac{x}{a + bx} \right)^x\] then x3 y2 =
विकल्प
(xy1 − y)2
(1 + y)2
\[\left( \frac{y - x y_1}{y_1} \right)^2\]
none of these
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उत्तर
(a) (xy1 − y)2
Here,
\[y = \log_e \left( \frac{x}{a + bx} \right)^x \]
\[ \Rightarrow y = x \log_e \left( \frac{x}{a + bx} \right) \]
\[ \Rightarrow y_1 = \log_e \left( \frac{x}{a + bx} \right) + x \times \frac{a + bx}{x}\left( \frac{1}{a + bx} - \frac{bx}{\left( a + bx \right)^2} \right)\]
\[ \Rightarrow y_1 = \log_e \left( \frac{x}{a + bx} \right) + \left( \frac{a}{a + bx} \right) . . . \left( 1 \right)\]
\[ \Rightarrow y_1 = \frac{y}{x} + \left( \frac{a}{a + bx} \right) \left[ \because y = x \log_e \left( \frac{x}{a + bx} \right) \right]\]
\[ \Rightarrow \frac{x y_1 - y}{x} = \frac{a}{a + bx} . . . \left( 2 \right)\]
\[\text{Differentiating } \left( 1 \right) \text { we get }, \]
\[ y_2 = \frac{a + bx}{x}\left( \frac{a + bx - bx}{\left( a + bx \right)^2} \right) - \frac{ba}{\left( a + bx \right)^2}\]
\[ \Rightarrow y_2 = \frac{a}{x\left( a + bx \right)} - \frac{ba}{\left( a + bx \right)^2}\]
\[ \Rightarrow y_2 = \frac{a\left( a + bx \right) - abx}{x \left( a + bx \right)^2}\]
\[ \Rightarrow y_2 = \frac{a^2}{x \left( a + bx \right)^2}\]
\[ \Rightarrow y_2 = \frac{\left( x y_1 - y \right)^2}{x^3} \left[ \text { Using }
\left( 2 \right) \right]\]
\[ \Rightarrow x^3 y_2 = \left( x y_1 - y \right)^2 \]
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