 Irrerbthist6n

2021-12-26

Finding all a such that $f\left(x\right)=\mathrm{sin}2x-8\left(a+1\right)\mathrm{sin}x+\left(4{a}^{2}+8a-14\right)x$ is increasing and has no critical points
Obviously, the first thing I did was to find the derivative of this function and simplify it a bit and I got:
${f}^{\prime }\left(x\right)=4\left({\mathrm{cos}}^{2}x-2\left(a+1\right)\mathrm{cos}x+\left({a}^{2}+2a-4\right)\right)$
But now how do I proceed further, had it been a simple quadratic in x. Cheryl King

$f\prime \left(x\right)=4\left({\mathrm{cos}}^{2}x-2\left(a+1\right)\mathrm{cos}x+\left({a}^{2}+2a-4\right)\right)$
We need to find a for which f(x) is increasing, which means f'(x)>0.
Now,
$4\left({\mathrm{cos}}^{2}x-2\left(a+1\right)\mathrm{cos}x+\left({a}^{2}+2a-4\right)\right)>0$
${\mathrm{cos}}^{2}x-2\left(a+1\right)\mathrm{cos}x+\left({a}^{2}+2a-4\right)>0$
$\left({\mathrm{cos}}^{2}x-2\left(a+1\right)\mathrm{cos}x+{a}^{2}+2a+1\right)>5$
$\left({\left(\mathrm{cos}x-\left(a+1\right)\right)}^{2}>5$

$a<\mathrm{cos}x-1-\sqrt{5}$
or
$a>\mathrm{cos}x-1+\sqrt{5}$
Hence, kalupunangh

Since I got the answer: To solve:

Completing the square:
$4\left(\left[\mathrm{cos}x-\left(a+1\right){\right]}^{2}-5\right)>0$
Solving this, considering the critical cases according to what value cosx takes and final answer is:
$a\in \left(-\mathrm{\infty },-2-\sqrt{5}\right)\cup \left(\sqrt{5},\mathrm{\infty }\right)$ karton

A preliminary exploration indicates that the answer is $a\in \left(-\mathrm{\infty },p\right)\cup \left(q,\mathrm{\infty }\right)$, where $p\approx -4.25,q\approx 2.25$
Here's the full solution. We want:

Let $y=\mathrm{cos}x$. Then we want:

This function in y opens upwards (${y}^{2}$ has positive coefficient) and has two x−intercepts (discriminant is 5>0). So, we want either its right x−intercept to be less than -1 or its left x−intercept to be greater than 1:

$a\in \left(-\mathrm{\infty },-2-\sqrt{5}\right)\cup \left(\sqrt{5},\mathrm{\infty }\right)$

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