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प्रश्न
If y = xn−1 log x then x2 y2 + (3 − 2n) xy1 is equal to
पर्याय
−(n − 1)2 y
(n − 1)2y
−n2y
n2y
उत्तर
(a) −(n − 1)2 y
Here,
\[y = x^{n - 1} \log x\]
\[ \Rightarrow y_1 = \left( n - 1 \right) x^{n - 2} \log x + \frac{x^{n - 1}}{x}\]
\[ \Rightarrow y_1 = \frac{\left( n - 1 \right) x^{n - 1} \log x + x^{n - 1}}{x}\]
\[ \Rightarrow x y_1 = \left( n - 1 \right)y + x^{n - 1} \]
\[ \Rightarrow x y_2 + y_1 = \left( n - 1 \right) y_1 + \left( n - 1 \right) x^{n - 2} \]
\[ \Rightarrow x y_2 + y_1 = \left( n - 1 \right) y_1 + \frac{\left( n - 1 \right) x^{n - 1}}{x}\]
\[ \Rightarrow x^2 y_2 + x y_1 = x\left( n - 1 \right) y_1 + \left( n - 1 \right) x^{n - 1} \]
\[ \Rightarrow x^2 y_2 + x y_1 = x\left( n - 1 \right) y_1 + \left( n - 1 \right)\left\{ x y_1 - \left( n - 1 \right)y \right\}\]
\[ \Rightarrow x^2 y_2 + x y_1 = x\left( n - 1 \right) y_1 + \left( n - 1 \right)x y_1 - \left( n - 1 \right)^2 y\]
\[ \Rightarrow x^2 y_2 + x y_1 = 2x\left( n - 1 \right) y_1 - \left( n - 1 \right)^2 y\]
\[ \Rightarrow x^2 y_2 + x y_1 - 2x\left( n - 1 \right) y_1 = - \left( n - 1 \right)^2 y\]
\[ \Rightarrow x^2 y_2 + x y_1 \left( 1 - 2n + 2 \right) = - \left( n - 1 \right)^2 y\]
\[ \Rightarrow x^2 y_2 + \left( 3 - 2n \right)x y_1 = - \left( n - 1 \right)^2 y\]
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