JEE Advanced: Quadratic equations & inequations

Suppose
$f(x)=x^{2}-(3k-1)x+2k^{2}-3k-2$

$= x^{2}-[(2k+1)+(k-2)]x+(2k-1)(k-2)$

$=[x-(2k+1)][x-(k-2)]$

So, the inequality simplifies to, $[x-(2k+1)][x-(k-2)] \geq 0$

Also, $x^{2}-1\geq 0$ is true for all x lying in $(\infty,-1)\cup(1,\infty)$ .

We are being asked to find the interval for k such that all values of x which satisfy $f(x)\geq 0$ are a subset of $(\infty,-1)\cup(1,\infty)$

$\Rightarrow$ Both 1 and -1 must lie in between the two distinct roots f(x)=0, which is a quadratic equation 'opening upwards'.

The required necessary and sufficient conditions are:

$1.f(1) \leq 0$

$\Rightarrow 1^{2}-(3k-1)(1)+2k^{2}-3k-2\leq0$

$\Rightarrow 2k^{2}-6k\leq0$

$\Rightarrow k(k-3)\leq 0$

$\Rightarrow k$ lies in [0,3] .

$2.f(-1)\leq0 \Rightarrow (-1)^{2}-(3k-1)(-1)+2k^{2}-3k-2\leq0$

$\Rightarrow 2k^{2}-2\leq0$

$\Rightarrow (k-1)(k+1)\leq 0$

$\Rightarrow k$ lies in [-1,1].

Lastly, for two distinct roots of f(x)=0, we need,

$3.D > 0$ , where $D$ is the discriminant of f(x)=0.

$\Rightarrow (3k-1)^{2}-4(1)(2k^{2}-3k-2)>0$

$\Rightarrow 9k^{2}-6k+1-8k^{2}+12k+8>0$

$\Rightarrow k^{2}+6k+9>0$

$\Rightarrow (k+3)^{2}>0$ , which is true for any real $k$ .

$\Rightarrow k$ lies in $(-\infty,\infty)$

The intersection of the above three intervals of k gives us the required set, which is [0,1].

Hence, $l=0, m=1$

$\therefore l+m = \boxed1$