Question

The range of a ∈ R for which the function $f(x) = (4a - 3)(x + \log_e 5)+2(a - 7)\cot(\frac{x}{2})\sin^2(\frac{x}{2})$, $x \neq 2n\pi, n \in N$, has critical points, is:

Answer 1

[-4/3, 2]

Answer 2

The function is $f(x) = (4a - 3)(x + \log_e 5) + (a - 7)\sin(x)$. The derivative is $f'(x) = (4a - 3) + (a - 7)\cos(x)$. Critical points exist if $f'(x) = 0$ has a solution. If $a=7$, $f'(x) = 25$, which is never zero. If $a \neq 7$, then $\cos(x) = -\frac{4a-3}{a-7} = \frac{3-4a}{a-7}$. For a solution to exist, we need $-1 \le \frac{3-4a}{a-7} \le 1$. Solving $\frac{3-4a}{a-7} \le 1$ gives $\frac{10-5a}{a-7} \le 0 \implies \frac{a-2}{a-7} \ge 0$, so $a \in (-\infty, 2] \cup (7, \infty)$. Solving $\frac{3-4a}{a-7} \ge -1$ gives $\frac{-3a-4}{a-7} \ge 0 \implies \frac{3a+4}{a-7} \le 0$, so $a \in [-4/3, 7)$. The intersection of these ranges is $a \in [-4/3, 2]$. We need to ensure that the solutions $x$ are in the domain $x \neq 2n\pi$. If $a=2$, $\cos(x)=1$, so $x=2k\pi$. $x=0$ is a solution in the domain. If $a=-4/3$, $\cos(x)=-1$, so $x=(2k+1)\pi$. These are not of the form $2n\pi$. For $a \in (-4/3, 2)$, $\cos(x) \in (-1, 1)$, so $x \neq 2k\pi$. Thus, the range of $a$ is $[-4/3, 2]$.