Let xk + yk = ak, (a, k 0 ) and {{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0, then k is:

Question

Let xk + yk = ak, (a, k > 0 ) and {{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0, then k is:

Accepted Answer

Answer: B. {2 \over 3} xk + yk = ak \Rightarrow kxk - 1 + kyk - 1{{{dy} \over {dx}}} = 0 \Rightarrow {{{dy} \over {dx}} + {{\left( {{x \over y}} \right)}^{k - 1}}} = 0 ...(1) Given {{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0 ...(2) Comparing (1) and (2), we get k - 1 = - {1 \over 3} \Rightarrow k = {2 \over 3}

Let xk + yk = ak, (a, k > 0 ) and $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$, then k is:

Answer

Explanation

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