Statement 1 is the only false one, and it is false twice over, so the correct set is 2, 3 and 4 — option 3.

Why the coefficient is negative. A thermistor is a semiconductor, and in a semiconductor the carrier concentration rises exponentially with temperature as more electrons are excited across the band gap:

\(n\propto e^{-E_{g}/2kT}\)

Conductivity follows, so resistance falls. The standard model is

\(R_{T}=R_{0}\,e^{\beta\left(\frac{1}{T}-\frac{1}{T_{0}}\right)}\)

with \(\beta\) typically 3000 to 4000 K. Statements 2 and 3 are therefore not independent — the second is simply the first expressed as a behaviour.

Why "poor sensitivity" is badly wrong. Sensitivity is exactly a thermistor's strength. Its coefficient is around \(-4\%\) per °C, roughly ten times that of a platinum RTD at \(+0.39\%\) per °C, so a small temperature change produces a large, easily measured resistance change and needs no bridge amplifier of great gain.

Statement 4 explains both. Thermistors are sintered mixtures of the oxides of manganese, nickel, cobalt, copper and iron — semiconducting ceramics, not metals. The exponential law and the large coefficient both follow from that composition.

The price paid is linearity: the exponential response must be linearised in software or by a shunt resistor, and the useful span is only about −50 to +150 °C. An RTD or thermocouple is chosen instead when a wide range or an inherently linear scale matters more than sensitivity.

A note on the exception : PTC thermistors do exist, made from doped barium titanate, and their resistance rises sharply above a Curie point. They are used as resettable fuses and motor-starting devices rather than as thermometers, so statement 1 does not describe them either.

Hence, the correct statements are 2, 3 and 4.