The set of all values of

Solution:

$\left(3 x^{2}+4 x+3\right)^{2}-(k+1)\left(3 x^{2}+4 x+3\right)\left(3 x^{2}+4 x+2\right)$

$+k\left(3 x^{2}+4 x+2\right)^{2}=0$

Let $3 x^{2}+4 x+3=a$

and $3 x^{2}+4 x+2=b \Rightarrow b=a-1$

Given equation becomes

$\Rightarrow a^{2}-(k+1) a b+k b^{2}=0$

$\Rightarrow a(a-k b)-b(a-k b)=0$

$\Rightarrow(a-k b)(a-b)=0 \Rightarrow a=k b$ or $a=b$ (reject)

$\because a=k b$

$\Rightarrow 3 x^{2}+4 x+3=k\left(3 x^{2}+4 x+2\right)$

$\Rightarrow 3(k-1) x^{2}+4(k-1) x+(2 k-3)=0$

$D \geq 0$

$\Rightarrow 16(\mathrm{k}-1)^{2}-4(3(\mathrm{k}-1))(2 \mathrm{k}-3) \geq 0$

$\Rightarrow 4(\mathrm{k}-1)\{4(\mathrm{k}-1)-3(2 \mathrm{k}-3)\} \geq 0$

$\Rightarrow 4(\mathrm{k}-1)\{-2 \mathrm{k}+5\} \geq 0$

$\Rightarrow-4(\mathrm{k}-1)\{2 \mathrm{k}-5\} \geq 0$

$\Rightarrow(\mathrm{k}-1)(2 \mathrm{k}-5) \leq 0$

$\therefore \mathrm{k} \in\left[1, \frac{5}{2}\right]$

$\because \mathrm{k} \neq 1$

$\therefore \mathrm{k} \in\left(1, \frac{5}{2}\right]$ Ans.