3

$$p_k(x) = \frac {\pi_k \frac {1} {\sqrt{2 \pi} \sigma_k} \exp(- \frac {1} {2 \sigma_k^2} (x - \mu_k)^2) } {\sum { \pi_l \frac {1} {\sqrt{2 \pi} \sigma_l} \exp(- \frac {1} {2 \sigma_l^2} (x - \mu_l)^2) }} \\ \log(p_k(x)) = \frac {\log(\pi_k) + \log(\frac {1} {\sqrt{2 \pi} \sigma_k}) + - \frac {1} {2 \sigma_k^2} (x - \mu_k)^2 } {\log(\sum { \pi_l \frac {1} {\sqrt{2 \pi} \sigma_l} \exp(- \frac {1} {2 \sigma_l^2} (x - \mu_l)^2) })} \\ \log(p_k(x)) \log(\sum { \pi_l \frac {1} {\sqrt{2 \pi} \sigma_l} \exp(- \frac {1} {2 \sigma_l^2} (x - \mu_l)^2) }) = \log(\pi_k) + \log(\frac {1} {\sqrt{2 \pi} \sigma_k}) + - \frac {1} {2 \sigma_k^2} (x - \mu_k)^2 \\ \delta(x) = \log(\pi_k) + \log(\frac {1} {\sqrt{2 \pi} \sigma_k}) + - \frac {1} {2 \sigma_k^2} (x - \mu_k)^2$$

As you can see, $\delta(x)$ is a quadratic function of $x$.