determine the value of a for which the function f has no critical number:
f(x)= (a^{2}+a-6)cos2x+(a-2)x+cos1
cos(1) is a constant, so its derivative is zero. I would write the derivative as:
$\displaystyle f'(x)=-2(a^2+a-6)\sin(2x)+(a-2)=-2(a+3)(a-2)\sin(2x)+(a-2)=$
$\displaystyle (a-2)(-2(a+3)\sin(2x)+1)=0$
What condition on $\displaystyle a$ causes the second factor to have no root?
A root is a value that causes an expression to be zero. Anyway, consider:
$\displaystyle -1\le\sin(2x)\le1$
Now, equate the second factor to zero, solve for $\displaystyle \sin(2x)$ and substitute into the above inequality. This will tell the the values of $\displaystyle a$ for which there are critical numbers.
When a=2, your function is constant, so there will be no critical values.
As for the second part, solve for sin(2x), then as an alternate to what I suggested before, set the magnitude of this to greater than 1. You will have 2 cases to consider:
i) $\displaystyle \frac{1}{2(a+3)}>1$
ii) $\displaystyle \frac{1}{2(a+3)}<-1$
We found:
$\displaystyle \sin(2x)=\frac{1}{2(a+3)}$
We therefore know that if $\displaystyle \left|\frac{1}{2(a+3)} \right|>1$ then there is no solution, that is, there is no real value of $\displaystyle a$ which will give us a critical value.