Question:

If \(x^m+y^m = k,(m\neq 1)\) and \(y^{''} = \frac{ax^b}{y^c}\) such that \(a+b+c = 0\), then the value of k is...

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Differentiate twice and compare with a x^b / y^c, then use a + b + c = 0.
Updated On: Oct 1, 2026
  • \(1\)
  • \(2\)
  • \(3\)
  • \(m\)
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The Correct Option is C

Solution and Explanation

Step 1: First derivative
From \(x^m + y^m = k\): \(mx^{m-1} + my^{m-1}y' = 0\), so \(y' = -\left(\dfrac{x}{y}\right)^{m-1}\).

Step 2: Second derivative
Differentiating: \(y'' = -(m-1)\left(\dfrac{x}{y}\right)^{m-2}\cdot\dfrac{y - xy'}{y^2}\). Now \(y - xy' = y + \dfrac{x^m}{y^{m-1}} = \dfrac{y^m + x^m}{y^{m-1}} = \dfrac{k}{y^{m-1}}\).

Step 3: Simplify
\[ y'' = -(m-1)\,\frac{x^{m-2}}{y^{m-2}}\cdot\frac{k}{y^{m-1}\,y^2} = \frac{-(m-1)k\,x^{m-2}}{y^{2m-1}} \]

Step 4: Compare and solve
So \(a = -(m-1)k\), \(b = m - 2\), \(c = 2m - 1\). Then \(a + b + c = -(m-1)k + 3m - 3 = (m-1)(3 - k) = 0\). Since \(m \neq 1\), \(k = 3\). Option (C).

Final Answer:
k equals 3. This is option (C). \[ \boxed{\text{(C) }3} \]
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