Question

If \[y^\frac{1}{n} + y^{- \frac{1}{n}} = 2x, \text { then find } \left( x^2 - 1 \right) y_2 + x y_1 =\] ?

विकल्प

• `-n^2y`

• my

• `n^2y`

• None of these 

Answer 1

\[\left( c \right) n^2 y\]

\[ y^\frac{1}{n} + y^{- \frac{1}{n}} = 2x\]

\[\text { Differentiating the above equation with respect to x }\]

\[\left( \frac{1}{n} y^\frac{1}{n} - 1 - \frac{1}{n} y^{- \frac{1}{n} - 1} \right) y_1 = 2\]

\[\frac{1}{ny}\left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right) y_1 = 2\]

\[\left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right) y_1 = 2ny . . . . . \left( 1 \right)\]

\[\left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right) y_2 + y_1 \left( \frac{1}{n} y^\frac{1}{n} - 1 + \frac{1}{n} y^{- \frac{1}{n} - 1} \right) y_1 = 2n y_1 \]

\[ny\left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right) y_2 + {y_1}^2 \left( y^\frac{1}{n} + y^{- \frac{1}{n}} \right) = 2 n^2 y y_1 \]

\[\text{ Dividing the above equation by } y_1 \]

\[\frac{ny}{y_1}\left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right) y_2 + y_1 \left( y^\frac{1}{n} + y^{- \frac{1}{n}} \right) = 2 n^2 y\]

\[\text {Putting y_1 from equation }\left( 1 \right)\]

\[\frac{\left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right)^2}{2} y_2 + y_1 \left( y^\frac{1}{n} + y^{- \frac{1}{n}} \right) = 2 n^2 y . . . . . \left( 2 \right)\]

\[\text { Now,} \]

\[ \left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right)^2 = \left( y^\frac{1}{n} + y^{- \frac{1}{n}} \right)^2 - 4\]

\[ \left( y^\frac{1}{n} - y^{- \frac{1}{n}} \right)^2 = 4 x^2 - 4 . . . . . \left( 3 \right)\]

\[\text { Putting the value of }\left( 3 \right)in\left( 2 \right)\]

\[\frac{4\left( x^2 - 1 \right) y_2}{2} + 2x y_1 = 2 n^2 y\]

\[\left( x^2 - 1 \right) y_2 + x y_1 = n^2 y\]