Question

If \(y^{\frac{1}{m}}+y^{\frac{-1}{m}} = 2x\) , then \((x^2-1)y_1^2 =\)

• A. \(my^2\)
• B. \(m^2y^2\)
• C. \(m^2y\)
• D. \(my\)

Hint:

Solve for y^(1/m) and write y as a power of x + sqrt(x^2 - 1).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Let \(t = y^{1/m}\). The equation \(t + \dfrac1t = 2x\) is a quadratic in \(t\).

Step 2: Solve for y:
\(t^2 - 2xt + 1 = 0\) gives \(t = x + \sqrt{x^2 - 1}\) (taking the root with the plus sign).
\[ y = \left(x + \sqrt{x^2 - 1}\right)^m \]

Step 3: Differentiate:
\[ y_1 = m\left(x + \sqrt{x^2-1}\right)^{m-1}\left(1 + \frac{x}{\sqrt{x^2-1}}\right) = m\left(x + \sqrt{x^2-1}\right)^{m-1}\frac{\sqrt{x^2-1} + x}{\sqrt{x^2-1}} \]
\[ y_1 = \frac{m\,y}{\sqrt{x^2 - 1}} \]

Step 4: Square and rearrange:
\((x^2 - 1)y_1^2 = m^2y^2\). This is option (B). The other options do not carry the factor \(m^2\) with \(y^2\).

Final Answer:
The result is m squared times y squared. \[ \boxed{\text{(B) }m^2y^2} \]