Question:hard

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
  • \(1\)
  • \(2\)
  • \(3\)
  • \(m\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Special case check
Try $m = 2$: $x^2 + y^2 = k$ is a circle. The second derivative of a circle is $y'' = -\dfrac{k}{y^3}$.

Step 2: Match
Compare with $\dfrac{a x^b}{y^c}$: $a = -k$, $b = 0$, $c = 3$. Then $a + b + c = -k + 3 = 0$, so $k = 3$.

Step 3: General confirmation
Doing the same with a general $m$, the sum $a + b + c$ factorises as $(m-1)(3-k)$, so the answer 3 holds for every $m \neq 1$.

Step 4: Result
$k = 3$, which is option (C).

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