Question

Given that x2018 y2017 = 1/2 and x2016 y2019 = 8, the value of x2+y3 is

• A. 35/4
• B. 37/4
• C. 31/4
• D. 33/4

Answer 1

The Correct Option is D

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:

\(\boxed{\frac{33}{4}}\)