1. Given that $$x^{2018}y^{2017}=\frac{1}{2}$$, and $$x^{2016}y^{2019}=8$$, then value of $$x^{2}+y^{3}$$ is





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  • By: anil on 05 May 2019 03.04 pm
    Given that $$x^{2018}y^{2017}=frac{1}{2}$$  ... (1) $$x^{2016}y^{2019}=8$$ ... (2) Equation (2)/ Equation (1) $$dfrac{y^2}{x^2} = dfrac{8}{1/2}$$ $$dfrac{y}{x} = 4$$ or $$-4$$
    Case 1: When $$dfrac{y}{x} = 4$$ $$x^{2018}(4x)^{2017}=dfrac{1}{2}$$
    $$x^{2018+2017}(2)^{4034}=dfrac{1}{2}$$
    $$x^{4035}=dfrac{1}{(2)^{4035}}$$
    $$x=dfrac{1}{2}$$ Since, $$dfrac{y}{x} = 4$$, => y = 2
    Therefore, $$x^{2}+y^{3}$$ = $$dfrac{1}{4}+8$$ = $$dfrac{33}{4}$$ Case 2: When $$dfrac{y}{x} = -4$$ $$x^{2018}(-4x)^{2017}=dfrac{1}{2}$$ $$x^{2018+2017}(2)^{4034}=dfrac{-1}{2}$$ $$x^{4035}=dfrac{1}{(-2)^{4035}}$$ $$x=dfrac{-1}{2}$$ Since, $$dfrac{y}{x} = -4$$, => y = 2 Therefore, $$x^{2}+y^{3}$$ = $$dfrac{1}{4}+8$$ = $$dfrac{33}{4}$$. Hence, option D is the correct answer. 
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