For semiconductors, the charge carrier density $n$ increases exponentially with temperature $T$ according to:

Since electrical resistivity $\rho = \frac{m}{n e^2 \tau}$, as temperature increases, the increase in carrier concentration $n$ heavily dominates over the decrease in relaxation time $\tau$.
Consequently, the resistivity of a semiconductor decreases exponentially with an increase in temperature:

Graph 3 shows this characteristic non-linear, exponential decrease of resistivity with increasing temperature.

Evaluation of statements:

• Doping pure silicon with trivalent impurities (like Boron, Aluminium, Indium) creates excess holes, forming a p-type semiconductor (Correct statement).
• In an n-type semiconductor, doping with pentavalent impurities introduces free conduction electrons as majority carriers, whereas holes are minority carriers. Therefore, the statement claiming majority carriers are holes is incorrect (NOT correct statement).
• In a p-type semiconductor, holes are majority carriers and electrons are minority carriers (Correct statement).
• In intrinsic semiconductors, higher temperature breaks covalent bonds to generate electron-hole pairs, increasing conductivity and thus decreasing resistivity (Correct statement).

Circuit Analysis:

• Let node $B$ be at reference potential $V_B = 0\,\text{V}$.
• The battery maintains the top wire at $24\,\text{V}$.
• The junction between the two vertical $8\,\text{k}\Omega$ resistors is at a positive potential relative to $B$.
• Because the p-side (anode) of the diode is connected to this positive junction and the n-side (cathode) connects to node $A$ (which is connected to $0\,\text{V}$ via the right $8\,\text{k}\Omega$ resistor), the diode is FORWARD BIASED.
• Assuming an ideal diode, it acts as a short circuit with zero resistance.
  6. Consequently, the right branch $8\,\text{k}\Omega$ resistor is connected in parallel with the bottom $8\,\text{k}\Omega$ resistor across terminals $A$ and $B$:
  $$R_{\text{parallel}} = \frac{8\,\text{k}\Omega \times 8\,\text{k}\Omega}{8\,\text{k}\Omega + 8\,\text{k}\Omega} = 4\,\text{k}\Omega$$
  7. The total equivalent circuit resistance is:
  $$R_{\text{total}} = 8\,\text{k}\Omega + R_{\text{parallel}} = 8 + 4 = 12\,\text{k}\Omega$$
  8. Using the voltage divider rule, the potential across the parallel combination (between $A$ and $B$) is:
  $$V_{AB} = 24\,\text{V} \times \frac{4\,\text{k}\Omega}{12\,\text{k}\Omega} = 24 \times \frac{1}{3} = 8\,\text{V}$$