Question

The set of values of $a$ for which the equation $\left(x^{2}+x\right)+2)^{2}-(a-3)\left(x^{2}+x+2\right)\left(x^{2}+x+1\right)+(a-4)\left(x^{2}+\right.(x+1)^{2}=0$ has at least one real root is

• A. $\left(5, \frac{19}{3}\right)$
• B. $\left[5, \frac{19}{3}\right]$
• C. $\left[5, \frac{19}{3}\right)$
• D. $\left(5, \frac{19}{3}\right]$

Answer 1

Answer: (D)

The given equation can be written as
$(z+1)^{2}-(a-3) z(z+1)+(a-4) z^{2}=0 $
$\left[\right.$ Putting $\left.x^{2}+x+1=z\right]$
$\Rightarrow (1+3-a+a-4) z^{2}+(2+3-a) z+1=0 $
$\Rightarrow (5-a) z+1=0 $
or $ z=\frac{1}{a-5}$
$\Rightarrow x^{2}+x+1-\frac{1}{a-5}=0 $
$\Rightarrow x^{2}+x+\frac{a-6}{a-5}=0$
whose roots will be real if discriminant $\geq 0$
$\Rightarrow 1-\frac{4(a-6)}{a-5} \geq 0$
$\Rightarrow \frac{3 a-19}{a-5} \leq 0 $
$\therefore 5< a \leq \frac{19}{3}$