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प्रश्न
If \[y^\frac{1}{n} + y^{- \frac{1}{n}} = 2x, \text { then find } \left( x^2 - 1 \right) y_2 + x y_1 =\] ?
पर्याय
`-n^2y`
my
`n^2y`
None of these
उत्तर
\[\left( c \right) n^2 y\]
\[ y^\frac{1}{n} + y^{- \frac{1}{n}} = 2x\]
\[\text { Differentiating the above equation with respect to x }\]
\[\left( \frac{1}{n} y^\frac{1}{n} - 1 - \frac{1}{n} y^{- \frac{1}{n} - 1} \right) y_1 = 2\]
\[\frac{1}{ny}\left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right) y_1 = 2\]
\[\left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right) y_1 = 2ny . . . . . \left( 1 \right)\]
\[\left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right) y_2 + y_1 \left( \frac{1}{n} y^\frac{1}{n} - 1 + \frac{1}{n} y^{- \frac{1}{n} - 1} \right) y_1 = 2n y_1 \]
\[ny\left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right) y_2 + {y_1}^2 \left( y^\frac{1}{n} + y^{- \frac{1}{n}} \right) = 2 n^2 y y_1 \]
\[\text{ Dividing the above equation by } y_1 \]
\[\frac{ny}{y_1}\left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right) y_2 + y_1 \left( y^\frac{1}{n} + y^{- \frac{1}{n}} \right) = 2 n^2 y\]
\[\text {Putting y_1 from equation }\left( 1 \right)\]
\[\frac{\left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right)^2}{2} y_2 + y_1 \left( y^\frac{1}{n} + y^{- \frac{1}{n}} \right) = 2 n^2 y . . . . . \left( 2 \right)\]
\[\text { Now,} \]
\[ \left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right)^2 = \left( y^\frac{1}{n} + y^{- \frac{1}{n}} \right)^2 - 4\]
\[ \left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right)^2 = 4 x^2 - 4 . . . . . \left( 3 \right)\]
\[\text { Putting the value of }\left( 3 \right)in\left( 2 \right)\]
\[\frac{4\left( x^2 - 1 \right) y_2}{2} + 2x y_1 = 2 n^2 y\]
\[\left( x^2 - 1 \right) y_2 + x y_1 = n^2 y\]
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