Question

The set of all values of $k>-1,$ for which the equation $(3x^2+4x+3{)}^2-(k+1\left)\right(3x^2+4x+3\left)\right(3x^2+4x+2)+k(3x^2+4x+2{)}^2=0$ has real roots, is

• A. $(\frac{1}{2},\frac{3}{1}]-\left\{1\right\}$
• B. $[-\frac{1}{2},1)$
• C. $[2,3)$
• D. $(1,\frac{5}{2}]$

Answer 1

The correct option is D $(1,\frac{5}{2}]$

$3x^2+4x+2>0$ $\forall x\in R(\because D<0)$
$(3x^2+4x+3{)}^2–(k+1\left)\right(3x^2+4x+3\left)\right(3x^2+4x+2)+k(3x^2+4x+2{)}^2=0$
$\Rightarrow {\left(\frac{3x^2+4x+3}{3x^2+4x+2}\right)}^2-(k+1)\left(\frac{3x^2+4x+3}{3x^2+4x+2}\right)+k=0\cdots \left(i\right)$
Let $\frac{3x^2+4x+3}{3x^2+4x+2}=t$
$\Rightarrow t=\frac{3x^2+4x+2+1}{3x^2+4x+2}=1+\frac{1}{3x^2+4x+2}$

$3x^2+4x+2\in [\frac{2}{3},\infty )$
$\Rightarrow \frac{1}{3x^2+4x+2}\in (0,\frac{3}{2}]$
$\Rightarrow t=1+\frac{1}{3x^2+4x+2}\in (1,\frac{5}{2}]$
$\Rightarrow t^2-(k+1)t+k=0\text{where}t\in (1,\frac{5}{2}]\cdots \left(ii\right)$
$\left(ii\right)$ should have at least one root in $(1,\frac{5}{2}]$
$\Rightarrow (t-1)(t-k)=0$
$\Rightarrow t=1,t=k$
$\therefore k\in (1,\frac{5}{2}]$