JEE Quadratic#9

Let , $\displaystyle{A={ x }^{ 2 }+x+2\quad \& \quad B={ x }^{ 2 }+x+1}$

Again Also , Let :

$\displaystyle{t=\cfrac { A }{ B } =1+\cfrac { 1 }{ { x }^{ 2 }+x+1 } =1+\cfrac { 1 }{ { (x+\cfrac { 1 }{ 2 } ) }^{ 2 }+\cfrac { 3 }{ 4 } } \\ t\in (1,\cfrac { 7 }{ 3 } ]}$

So required question is analogous to : $Q:\quad { t }^{ 2 }-(a-3)t+a-4=0$ finding values of 'a' for which quadratic 'Q' has at least one real root in the interval $1$ to $7/3$ , in which t=7/3 (achieve when $x=-1/2$ ) is inclusive value and 1 is exclusive( can't achive , since 'x' can't be infinity ) .

Analysing the quadratic ' Q ' , we get it's Discriminate as positive , (discriminant can't be zero , Since if so , then $a=5$ and $t=1$ which is not in our domain ) Since $Q(t):\quad { t }^{ 2 }-(a-3)t+a-4=0$

By analysing some graphs for this quadratic equation 'Q' , we concluded that required Condition is simply nothing but : $\displaystyle{Q(\cfrac { 7 }{ 3 } )\ge 0\\ a\le \cfrac { 19 }{ 3 } =6.33\\ \boxed { { \left\lfloor a \right\rfloor }_{ max }=6 } }$

Consider the equation:

$(x^2+x+2)^2-(a-3)(x^2+x+2)(x^2+x+1)+(a-4)(x^2+x+1)^2=0$

As $x^2+x+1>0$ for all real x, we can divide both sides by $(x^2+x+1)^2$ , resulting in this equation:

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

Let $t = \frac {x^2+x+2} {x^2+x+1} = 1+ \frac {1} {x^2+x+1}$ .

Then $1 < t \le \frac {7} {3}$ .

The original equation therefore becomes: $t^2 - (a-3)t + a-4= 0$

Which can be factored as: $(t-1)[t-(a-4)] = 0$ .

The original equation has one real root if and only if the equation $(t-1)[t-(a-4)] = 0$ has at least one real root that satisfies $1 < t \le \frac {7} {3}$ . Since one of the roots is $t=1$ , the other one must be $\le \frac {7} {3}$ .

Therefore, $a-4 \le \frac {7} {3}$ , or $a \le \frac {19} {3}$ .

The maximum integral $a$ that satisfies this condition is $a = \boxed {6}$ .