If $f(x)=\left\{\begin{array}{l}3 x^2+12 x-1, &-1 \leq x \leq 2 \\ 37-x, &2<x \leq 3\end{array}\right.$, then 

all of these

We have,

$\lim\limits_{x \rightarrow 2^{-}} f(x)=\lim\limits_{x \rightarrow 2^{+}} f(x)=f(2)$

So, f(x) is continuous at x = 2.

Hence, f(x) is continuous on [-1, 3]. Thus, options (b) and (c) are correct.

Also,

(LHD at x = 2) = $\left\{\frac{d}{d x}\left(3 x^2+12 x-1\right)\right\}_{\text {at } x=2}=24$

and,

(RHD at x = 2) = $\left\{\frac{d}{d x}(37-x)\right\}_{\text {at } x=2}=-1$

Clearly, f(x) is not differentiable at x = 2. So, f'(2) does not exist.

Now, $f'(x)=\left\{\begin{array}{lr}6 x+12, ~~-1<x<2 \\ -1, ~~~~2<x<3\end{array}\right.$

Clearly, f'(x) > 0 for all $x \in[-1,2]$

So, f(x) is increasing on [-1, 2]

Thus, option (a) is correct.