JEE MAIN - Mathematics (2020 - 7th January Morning Slot - No. 6)
Let xk + yk = ak, (a, k > 0 ) and $${{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0$$, then k is:
$${1 \over 3}$$
$${2 \over 3}$$
$${4 \over 3}$$
$${3 \over 2}$$
Explanation
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 = $${2 \over 3}$$
$$ \Rightarrow $$ 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 = $${2 \over 3}$$
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