Let $f : R \to [1, \infty )$be defined by$f\left(x\right) = x^2 – 6ax + 3 – 2a + 9a^2$. The integral value of a for which f(x) is surjective, is equal to

$$\begin{gathered} As, f:R\to [1,\infty ], so f\left(x\right)\ge 1\forall x\in R \\ \Rightarrow x^2-6ax+3-2a+9a^2\ge 1\Rightarrow x^2-6ax+2-2a+9a^2\ge 0 \\ For range of g\left(x\right)=x^2-6ax+2-2a+9a^{2}to be[0,\infty ], we must have D=0 \\ \Rightarrow 36a^2-4(2-2a+9a^2)=0\Rightarrow 8a-8=0\Rightarrow a=1 \end{gathered}$$