1. Recall the frequency dependence of each quantity

For an A.C. circuit:

• Capacitive reactance: $X_C=\frac{1}{\omega C}$ So, as frequency $f$ (or angular frequency $\omega$) increases, $X_C$ decreases hyperbolically.

• Inductive reactance: $X_L=\omega L$ So, as frequency increases, $X_L$ increases linearly from zero.

• Resistance of a pure resistor: $R=\text{constant}$ So its graph is a horizontal line.

• Impedance of an LCR series circuit: $Z=\sqrt{R^2+(X_L-X_C)^2}$ This generally gives a curve with a minimum at resonance, not a simple straight line or hyperbola.

2. Interpret the graph shapes

Although the actual graph image is not shown here, the options indicate that curve $A$ and curve $B$ must be identified from standard shapes.

The only pair among the options that matches the standard frequency behavior of two reactances is:

• one curve decreasing with frequency $\Rightarrow X_C$
• one curve increasing with frequency $\Rightarrow X_L$

Thus: $A=X_C, \qquad B=X_L$

3. Check options

• Option A: $A=X_L, B=R$
  Not possible if one curve is decreasing and the other is horizontal/increasing incorrectly.

• Option B: $A=X_L, B=Z$
  $Z$ is not typically a simple straight line; it has a minimum near resonance.

• Option C: $A=X_C, B=X_L$
  Correct: one decreases as $1/\omega$, the other increases as $\omega$.

• Option D: $A=X_C, B=R$
  $R$ would be constant, not increasing.

4. Final answer

Therefore, the correct representation is: $A=X_C, \quad B=X_L$

So the correct option is C.