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Solution for If Y 1 N + Y − 1 N = 2 X , Then Find ( X 2 − 1 ) Y 2 + X Y 1 = ? - CBSE (Science) Class 12 - Mathematics

ConceptSimple Problems on Applications of Derivatives

Question

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

Solution

$\left( c \right) n^2 y$

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

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

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

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

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

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

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

$\text{ Dividing the above equation by } y_1$

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

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

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

$\text { Now,}$

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

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

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

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

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

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Solution for question: If Y 1 N + Y − 1 N = 2 X , Then Find ( X 2 − 1 ) Y 2 + X Y 1 = ? concept: Simple Problems on Applications of Derivatives. For the courses CBSE (Science), PUC Karnataka Science, CBSE (Arts), CBSE (Commerce)
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