Question

The set of values of $a$ for which the equation $\left(x^2+x\right)+2{)}^2-(a-3)\left(x^2+x+2\right)\left(x^2+x+1\right)+(a-4)(x^2+(x+1{)}^2=0$ 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
$(z+1{)}^2-(a-3)z(z+1)+(a-4)z^2=0$
$[$ Putting $x^2+x+1=z]$
$\Rightarrow (1+3-a+a-4)z^2+(2+3-a)z+1=0$
$\Rightarrow (5-a)z+1=0$
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 $\ge 0$
$\Rightarrow 1-\frac{4(a-6)}{a-5}\ge 0$
$\Rightarrow \frac{3a-19}{a-5}\le 0$
$\therefore 5<a\le \frac{19}{3}$