Question

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

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

Answer 1

Answer: (A)

${\left(3x^2+4x+3\right)}^2-(k+1)\left(3x^2+4x+3\right)\left(3x^2+4x+2\right)$ $+k{\left(3x^2+4x+2\right)}^2=0$
Let $3x^2+4x+3=a$
and $3x^2+4x+2=b$
$\Rightarrow b=a-1$
Given equation becomes
$\Rightarrow a^2-(k+1)ab+k{b}^2=0$
$\Rightarrow a(a-kb)-b(a-kb)=0$
$\Rightarrow (a-kb)(a-b)=0$
$\Rightarrow a=kb$ or $a=b($ reject $)$
$\because a=kb$
$\Rightarrow 3x^2+4x+3=k\left(3x^2+4x+2\right)$
$\Rightarrow 3(k-1)x^2+4(k-1)x+(2k-3)=0$
for real roots $D\ge 0$
$\Rightarrow 16(k-1{)}^2-4(3(k-1))(2k-3)\ge 0$
$\Rightarrow 4(k-1)\{4(k-1)-3(2k-3)\}\ge 0$
$\Rightarrow 4(k-1)\{-2k+5\}\ge 0$
$\Rightarrow -4(k-1)\{2k-5\}\ge 0$
$\Rightarrow (k-1)(2k-5)\le 0$

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