Class 12 Continuity And Differentiability · Long Answer3 Marksimplicit_differentiation

Class 12 Maths · Continuity And Differentiability · Long Answer #77

Determine all real solutions of x¹³y⁷=(x+y)²⁰ and assess the source claim dy/dx=y/x.

✅ Step-by-Step Solution

Final Answer: Only (0,0); no differentiable graph branch exists.

If xy<0, x¹³y⁷ is negative while (x+y)²⁰ is nonnegative, so equality is impossible.
If xy>0, x,y have the same sign. Let u=|x|,v=|y|>0. Since u<u+v and v<u+v, u¹³v⁷<(u+v)²⁰=|x+y|²⁰, again excluding equality.
If xy=0, the left side is zero and equality forces x+y=0; thus both coordinates are zero.
The only real solution is (0,0). It is an isolated point, so the claimed graph derivative y/x is undefined and cannot be proved from this equation.

[0.5 marks] If xy<0, x¹³y⁷ is negative while (x+y)²⁰ is nonnegative, so equality is impossible.
[1 mark] If xy>0, x,y have the same sign. Let u=|x|,v=|y|>0. Since u<u+v and v<u+v, u¹³v⁷<(u+v)²⁰=|x+y|²⁰, again excluding equality.
[1 mark] If xy=0, the left side is zero and equality forces x+y=0; thus both coordinates are zero.
[0.5 marks] The only real solution is (0,0). It is an isolated point, so the claimed graph derivative y/x is undefined and cannot be proved from this equation.

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