Question

Find all values of $k$ for which every solution of the inequality $x^2+3k^2-1⩾2k(2x-1)$ is a solution of the inequality $x^2-(2x-1)k+k^2⩾0$

Answer 1

1. $x^2+3k^2-1\ge 2k(2x-1)$

$\Rightarrow x^2-4kx+2k+3k^2-1\ge 0$

$\Rightarrow (x-\frac{4k\pm \sqrt{16k^2+4-8k-12k^2}}{2})\ge 0$

$\Rightarrow x-2k\pm \sqrt{k^2-2k+1}\ge 0$

$\Rightarrow (x-2k\pm (k-1\left)\right)\ge 0$

$\Rightarrow (x-k-1)(x-3k+1)\ge 0$

$\Rightarrow x\in (-\infty ,k+1)\cup (3k-1,\infty )$

2. $x^2-(2x-1)k+k^2\ge 0$

$\Rightarrow x^2-2xk+k+k^2\ge 0$

$\Rightarrow x-\frac{2k\pm \sqrt{4k^2-4k-4k^2}}{2}\ge 0$

$\Rightarrow x-k\pm \sqrt{-k}\ge 0$

$\Rightarrow (x-k+\sqrt{-k})(x-k-\sqrt{-k})\ge 0$

Case 1 When $k>0$, for $k>0$, 2 has a negative discriminant, so it is greater than zero ,$\forall x\in R$

So, $3\in R,\forall x\in R$

Values of k, $k\in (0,\infty )$

Case 2 When $k<0$, 2 has a domain, $x\in (-\infty ,k-\sqrt{-k})\cup (\sqrt{-k}+k,\infty )$

So, set 3 becomes , $x\in (-\infty ,3k-1)\cup (k+1,\infty )$ as $k<0$

Now, set in 6 must be a subset of set in 6.

1.$\Rightarrow 3k-1<k-\sqrt{-k}$

$\Rightarrow 2k-1<-\sqrt{-k}$

$\Rightarrow 4k^2-4k+1<-k$

$\Rightarrow 4k^2-3k+1<0$

$\Rightarrow D<0,a>0$, equation must be greater than zero so it can not be negative.

$k\in \varphi$ for $k<0$

$\Rightarrow k\in (0,\infty )$is the condition to get the above statement.