Coefficient of linear expansion of material of resistor is $\alpha$. Its temperature coefficient of resistivity and resistance are $\alpha_\rho$ and $\alpha_R$ respectively, then correct relation is
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we will find the relation for small temperature changes.

resistance $R =\rho L / A$

coefficient of linear expansion $= \alpha$

length of conductor: $L = L _0(1+ a \Delta T ) \quad \Delta L = a L _0 \Delta T$

$\beta=$ coefficient of expansion in area : $=2 \alpha$

Area of cross section: $A=A_0(1+2 \alpha \Delta T)$

$\Delta A=2 \alpha A_0 \Delta T$

Resistivity $\rho=\rho_0(1+\alpha_p \Delta T)$

$\Delta \rho=\rho_0 \alpha_p \Delta T$

Resistance $R = R _0(1+\alpha_R \Delta T )$

$\Delta R =\alpha_R R _0 \Delta T$

If $\Delta A =2 a \Delta T$ is very small then, and for small $\Delta T$,

$R_0=\rho_0 L_0 / A_0$

$R =\rho_0(1+ \alpha_p \Delta T ) L _0(1+ a \Delta T ) /\left[ A _0(1+2 \alpha \Delta T )\right]$

$=\left(\rho_0 L_0 / A_0\right)(1+\alpha_p \Delta T)(1+\alpha \Delta T)(1-2 \alpha \Delta T)$

$=R_0(1+\alpha_p \Delta T)(1-a \Delta T) \quad$ ignoring the $2 a^2 \Delta T^2$ term

$=R_0[1+(\alpha_p-\alpha) \Delta T] \quad$ ignoring the $A \rho a \Delta T^2$ term

$\alpha_R=\alpha_P-\alpha$

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