$f(x)=\left\{\begin{array}{l}x+1, &x \leq 1 \\ 3-a x^2, &x>1\end{array}\right.$

Value of ‘a’ for which f(x) is continuous, is

1

$f(x)=\left\{\begin{array}{l}x+1 & x \leq 1 \\ 3-ax^2 & x>1\end{array}\right. $

$\lim\limits_{x \rightarrow 1^{-}}(x+1)=2=\lim\limits_{x \rightarrow 1^{+}}\left(3-a x^2\right)=3-a$

3 - a = 2

$\Rightarrow a=3-2 \Rightarrow a=1$

Hence (1) is the correct answer.