Set of real values of $k$ such that $4\cdot 2^{2x}-4k\cdot 2^x+5-4k\geq 0$ has a real solution

Question

Complete set of real values of $k$ for which

$4\cdot 2^{2x}-4k\cdot 2^x+5-4k\geq 0$ has at least

one real solution .

Try: From $$4k(2^x+1)\leq 4\cdot 2^{2x}+5$$

$$4k\leq \frac{4\cdot 2^{2x}+5}{2^x+1}$$

put $2^{x}=t>1$ for all $x \in R$

$$4k\leq \frac{4t^2+5}{t+1}=4(t-1)+\frac{9}{t+1}$$

let $\displaystyle f(t) = 4(t-1)+\frac{9}{t+1}$ and $\displaystyle f'(t)=4-\frac{9}{(t+1)^2}>0$ for $t>1$

so function $f(t)$ is strictly increasing function

$$4k<\lim_{t\rightarrow \infty}\frac{4t^2+5}{t+1}<\infty\Rightarrow k<\infty$$

could some help me where i am missing

Answer 1

A simpler argument than my original answer: for any given $k \in \mathbb{R}$, we have $\lim_{x \to \infty} 4 \cdot 2^{2x} - 4k \cdot 2^x = \lim_{t \to \infty} 4t^2 - 4kt = \infty$, so the quantity $4 \cdot 2^{2x} - 4k \cdot 2^x$ can be made arbitrarily large by taking $x$ sufficiently large and positive. In particular, it can be made greater than or equal to $4k-5$, whence there is always a real solution to the desired inequality, and so any $k \in \mathbb{R}$ works.

Answer 2

Write it as a quadratic in $t=2^x$ right from the start. So, you're looking at the quadratic $4t^2 + 4kt + 5-4k$. Determine the range of $t \in \mathbb{R}$ in terms of $k$ for which this quadratic is non-negative, and then determine the values of $x$ to which these values of $t$ correspond (if there are any).

Answer 3

Yes, you are right! You got that all real value of $k$ is valid and it gives the answer: $\mathbb R$.

Indeed, you got: $$k\leq\frac{4\cdot2^{2x}+5}{4(2^x+1)},$$ which gives that for all real $k$ there is real $x$, for which this inequality is true.

You proved it!

Answer 4

Though there are a few errors in your try, what you are trying to do looks good to me.

As Sam Streeter pointed out in the comments, $2^x=t\gt 1$ is incorrect. It should be $2^x=t\gt 0$.

Since $$f'(t)=\frac{(2t-1)(2t+5)}{(t+1)^2}$$ we see that $f(t)$ is decreasing for $0\lt t\lt\frac 12$ and is increasing for $t\gt \frac 12$ with $f(\frac 12)=4$ and $\displaystyle\lim_{t\to\infty}f(t)=\infty$. (All we need to know is $\displaystyle\lim_{t\to\infty}f(t)=\infty$.)

(For every positive real $t$, there is $x\in\mathbb R$ such that $2^x=t$, so we want to find all $k$ such that there exists at least one positive real number $t$ satisfying $4k\le f(t)$, so...)

It follows that the answer is $\mathbb R$.