If y 1 - m - x -1- x 2- then 1- x 2 y 2- xy 1 is equal to-A- m yB- m2 y- m y2D- m2 y2

The correct option is B y=(x+√(1+x^2))^m⇒ y_1=m(x+√(1+x^2))^m 1.{ 1+2x/2√(1+x^2)} ⇒√(1+x^2)y_1=m(x+√(1+x^2))^m⇒√(1+x^2)y_1=my ⇒ (1+x^2)y_2+xy_1=m^2y ...

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Byju's Answer

Standard XII

Mathematics

How to Find the Inverse of a Function

If y 1 / m =[...

Question

If $y^{1 / m} = [x + \sqrt{1 + x^{2}}] t h e n (1 + x^{2}) y_{2} + x y_{1}$ is equal to:

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Solution

The correct option is A
$y = (x + \sqrt{1 + x^{2}})^{m} \Rightarrow y_{1} = m (x + \sqrt{1 + x^{2}})^{m - 1} . {1 + \frac{2 x}{2 \sqrt{1 + x^{2}}}} \Rightarrow \sqrt{1 + x^{2}} y_{1} = m (x + \sqrt{1 + x^{2}})^{m} \Rightarrow \sqrt{1 + x^{2}} y_{1} = m y \Rightarrow (1 + x^{2}) y_{2} + x y_{1} = m^{2} y$

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If y1/m=[x+√1+x2] then (1+x2)y2+xy1 is equal to:

The correct option is A y=(x+√1+x2)m⇒y1=m(x+√1+x2)m−1.{1+2x2√1+x2}⇒√1+x2y1=m(x+√1+x2)m⇒√1+x2y1=my⇒(1+x2)y2+xy1=m2y

If $y^{1 / m} = [x + \sqrt{1 + x^{2}}] t h e n (1 + x^{2}) y_{2} + x y_{1}$ is equal to: