Given that x2018 y2017 = 1/2 and x2016 y2019 = 8, the value of x2+y3 is
- 35/4
- 37/4
- 31/4
- 33/4
The Correct Option is D
Solution and Explanation
Given the equations:
(1) \(x^{2018} y^{2017} = \frac{1}{2}\)
(2) \(x^{2016} y^{2019} = 8\)
We aim to find the value of \(x^2 + y^3\). Let us solve the problem step-by-step:
First, divide equation (1) by equation (2):
\(\frac{x^{2018} y^{2017}}{x^{2016} y^{2019}} = \frac{\frac{1}{2}}{8}\)
Simplifying the left side:
\(x^2 \cdot y^{-2} = \frac{1}{16}\)
This implies:
\(\frac{x^2}{y^2} = \frac{1}{16}\) ⇒ \(x^2 = \frac{y^2}{16}\)
From here, express \(x^2\) in terms of \(y\):
\(x^2 = \frac{y^2}{16}\)
Since we know \(x^2 = \frac{1}{16} y^2\), substitute into one of the equations to find specific values. Consider equation (2):
\((x^2)^{1008} \cdot y^{2016} = 8\)
Substitute \(x^2 = \frac{y^2}{16}\) into the expression:
\(\left(\frac{y^2}{16}\right)^{1008} \cdot y^{2016} = 8\)
This simplifies to:
\(\frac{y^{2016}}{16^{1008}} \cdot y^{2016} = 8\)
So:
\(\frac{y^{4032}}{16^{1008}} = 8\)
Thus, simplifying:
\(y^{4032} = 8 \times 16^{1008} = 2^3 \times (2^4)^{1008}\)
Therefore:
\(y^{4032} = 2^3 \times 2^{4032} = 2^{4035}\)
Solve for \(y\):
\(y = 2\)
Substitute \(y = 2\) into \(x^2 = \frac{y^2}{16}\):
\(x^2 = \frac{4}{16} = \frac{1}{4}\)
Therefore:
\(x = \frac{1}{2}\)
Now compute \(x^2 + y^3\):
\(\frac{1}{4} + 8 = \frac{1}{4} + \frac{32}{4} = \frac{33}{4}\)
The value of \(x^2 + y^3\) is:
\(\boxed{\frac{33}{4}}\)
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