Finding all $a$ such that $f(x)=\sin2x-8(a+1)\sin x+(4a^2+8a-14)x$ is increasing and has no critical points

Question

Find the set of all values of the parameter $a$ for which the function, $$f(x)=\sin\left(2x\right)-8\left(a+1\right)\sin\left(x\right)+\left(4a^2+8a-14\right)x$$ increases for all $x\in\Bbb{R}$ and has no critical points for all $x\in\Bbb{R}$.

Obviously, the first thing I did was to find the derivative of this function and simplify it a bit and I got:

$$f'(x)=4\left(\;\cos^2x-2\left(a+1\right)\cos x+\left(a^2+2a-4\right)\;\right)$$ But now how do I proceed further, had it been a simple quadratic in $x$. I would've calculated $D<0$ but this is a quadratic in $\cos x$. Can I do the same here? Why or why not? How should I go ahead?

Not just this, there are many instances where the quadratic is not in x, but in expressions like $e^{x}$. Is there any general approach to solving these quadratics for things like no solutions etc-

Answer 1

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

Now,

$$4(\cos^{2}x-2\left(a+1\right)\cos x+\left(a^{2}+2a-4\right))>0$$

$$\cos^{2}x-2\left(a+1\right)\cos x+\left(a^{2}+2a-4\right)>0$$

$$(\cos^{2}x-2\left(a+1\right)\cos x+a^{2}+2a+1)>5$$

$$((\cos x -(a+1))^2>5$$

$\cos x -(a+1)>√5$ or $\cos x-(a+1)<-√5$

$$a<\cos x-1-√5$$ or $$a>\cos x-1+√5$$

Hence, $a<-2-√5$ or $a>√5$.

Answer 2

Here in as we need to find the parameter $[a]$ if $[f(x) = \sin 2x - 8(a + 1)\sin x + (4{a^2} + 8a - 14)x]$ increases for all $[x \in R]$ and has no critical points for all $[x \in R]$ which means $[f'(x) \geqslant 0]$ because the function increases but it is mentioned that there are no critical points for all $[x \in R]$ so we need to find $[f'(x){\text{ > }}0]$ only as if there is no critical point then $[f'(x) \ne 0]$ and simplify the functions by differentiating.

If you still can't solve it here is the answer.

Answer 3

Since I got the answer: To solve:

$f'\left(x\right)>0$ i.e. $f'(x)=4\left(\;\cos^2x-2\left(a+1\right)\cos x+\left(a^2+2a-4\right)\;\right)$ $>0$

Completing the square:
$4(\left[\cos x-\left(a+1\right)\right]^{2}-5)>0$

Solving this, considering the critical cases according to what value $\cos x$ takes and final answer is:

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

Answer 4

A preliminary exploration indicates that the answer is $a\in(-\infty,p)\cup(q,\infty),$ where $p≈−4.25,q≈2.25.$

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Here's the full solution. We want: $$\forall x\in\mathbb R\quad f'(x)>0.$$

Let $y=\cos x.$ Then we want: $$\forall y\in[-1,1]\quad y^2-2(a+1)y+(a^2+2a-4)>0.$$ 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 $1:$ $$a+1+\sqrt5<-1 \quad\text{or}\quad a+1-\sqrt5>1\\ a\in(-\infty,-2-\sqrt5)\cup(\sqrt5,\infty).$$