You are given:

• Parabola: \(y = 2x^{2}\)
• Circle: \(x^{2} + y^{2} = 4\)

Points:

• \(P = \left(\frac{3}{2},\frac{1}{2}\right)\)
• \(Q = \left(\frac{1}{2},\frac{3}{2}\right)\)
• \(R = \left(\frac{3}{2},\frac{11}{2}\right)\)
• \(S = \left(\frac{11}{2},\frac{3}{2}\right)\)

A point \((x,y)\) is:

• Above the parabola \(y = 2x^{2}\) if \(y > 2x^{2}\).
• Inside the circle \(x^{2} + y^{2} = 4\) if \(x^{2} + y^{2} < 4\).

Conditions to Check

For each point \((x,y)\):

1. Compute \(y_{\text{parabola}} = 2x^{2}\).
2. Check if actual \(y\) satisfies \(y > 2x^{2}\) (above parabola).
3. Compute \(x^{2} + y^{2}\).
4. Check if \(x^{2} + y^{2} < 4\) (inside circle).

Only the point that satisfies both inequalities is the correct option.

Testing Point \(P\left(\frac{3}{2},\frac{1}{2}\right)\)

Here \(x = \frac{3}{2},\ y = \frac{1}{2}\).

1. Above parabola?
\[
2x^{2} = 2\left(\frac{3}{2}\right)^{2} = 2\cdot\frac{9}{4} = \frac{9}{2}
\]
Compare \(y\) with \(2x^{2}\): \(y = \frac{1}{2},\ 2x^{2} = \frac{9}{2}\).
Since \(\frac{1}{2} < \frac{9}{2}\), point \(P\) lies below the parabola.

2. Inside circle?
\[
x^{2} + y^{2} = \left(\frac{3}{2}\right)^{2} + \left(\frac{1}{2}\right)^{2}
= \frac{9}{4} + \frac{1}{4} = \frac{10}{4} = \frac{5}{2} = 2.5
\]
Since \(2.5 < 4\), point \(P\) is inside the circle.

Conclusion for P: Inside the circle but below the parabola, so it does not satisfy the required condition.

Testing Point \(Q\left(\frac{1}{2},\frac{3}{2}\right)\)

Here \(x = \frac{1}{2},\ y = \frac{3}{2}\).

1. Above parabola?
\[
2x^{2} = 2\left(\frac{1}{2}\right)^{2} = 2\cdot\frac{1}{4} = \frac{1}{2}
\]
Compare: \(y = \frac{3}{2},\ 2x^{2} = \frac{1}{2}\).
Since \(\frac{3}{2} > \frac{1}{2}\), point \(Q\) is above the parabola.

2. Inside circle?
\[
x^{2} + y^{2} = \left(\frac{1}{2}\right)^{2} + \left(\frac{3}{2}\right)^{2}
= \frac{1}{4} + \frac{9}{4} = \frac{10}{4} = \frac{5}{2} = 2.5
\]
Since \(2.5 < 4\), point \(Q\) is inside the circle.

Conclusion for Q: Above the parabola and inside the circle, so \(Q\) satisfies both required conditions.

Testing Point \(R\left(\frac{3}{2},\frac{11}{2}\right)\)

Here \(x = \frac{3}{2},\ y = \frac{11}{2}\).

1. Above parabola?
As earlier, \[
2x^{2} = 2\left(\frac{3}{2}\right)^{2} = \frac{9}{2}
\]
Compare: \(y = \frac{11}{2},\ 2x^{2} = \frac{9}{2}\).
Since \(\frac{11}{2} > \frac{9}{2}\), point \(R\) is above the parabola.

2. Inside circle?
\[
x^{2} + y^{2} = \left(\frac{3}{2}\right)^{2} + \left(\frac{11}{2}\right)^{2}
= \frac{9}{4} + \frac{121}{4} = \frac{130}{4} = 32.5
\]
Since \(32.5 > 4\), point \(R\) is outside the circle.

Conclusion for R: Above the parabola but outside the circle, so it does not satisfy the condition.

Testing Point \(S\left(\frac{11}{2},\frac{3}{2}\right)\)

Here \(x = \frac{11}{2},\ y = \frac{3}{2}\).

1. Above parabola?
\[
2x^{2} = 2\left(\frac{11}{2}\right)^{2} = 2\cdot\frac{121}{4} = \frac{121}{2} = 60.5
\]
Compare: \(y = \frac{3}{2},\ 2x^{2} = \frac{121}{2}\).
Since \(\frac{3}{2} < \frac{121}{2}\), point \(S\) lies below the parabola.

2. Inside circle?
\[
x^{2} + y^{2} = \left(\frac{11}{2}\right)^{2} + \left(\frac{3}{2}\right)^{2}
= \frac{121}{4} + \frac{9}{4} = \frac{130}{4} = 32.5
\]
Since \(32.5 > 4\), point \(S\) is outside the circle.

Conclusion for S: Below the parabola and outside the circle, so it clearly does not satisfy the condition.

Final Answer and Summary Table

Only point \(Q\left(\frac12,\frac32\right)\) lies above the parabola \(y = 2x^{2}\) and inside the circle \(x^{2}+y^{2}=4\).
Therefore, the correct option is (B) Q.

Position of Each Point