Question:

Step 1: Simplify the RHS expression. For any real number $x$, $0 \leq x^2$. Thus, $0 \leq \tan^{-1} x^2 < \frac{\pi}{2}$. The greatest integer function $[\tan^{-1} x^2] = 0$. Similarly, $0 < \cot^{-1} x^2 \leq \frac{\pi}{2}$. The greatest integer function $[\cot^{-1} x^2] = 0$. Therefore, the RHS expression simplifies to: $$ \frac{\sin^2(0 \cdot \pi) + 3\tan^2(0 \cdot \pi)}{3\sin^2 x + \cos^2 x} = \frac{0 + 0}{3\sin^2 x + \cos^2 x} = 0 $$ So, $f(2018) - f(-2014) = 0$, which implies $f(2018) = f(-2014)$. Step 2: Determine the relationship between $a$ and $b$. Given $f(x) = ax^2 + bx + c$, we have: $$ f(2018) - f(-2014) = (a(2018)^2 + b(2018) + c) - (a(-2014)^2 + b(-2014) + c) $$ $$ = a(2018^2 - (-2014)^2) + b(2018 - (-2014)) $$ $$ = a(2018^2 - 2014^2) + b(2018 + 2014) $$ Since $f(2018) - f(-2014) = 0$: $$ a(2018 - 2014)(2018 + 2014) + b(4032) = 0 $$ $$ a(4)(4032) + b(4032) = 0 $$ $$ 4032(4a + b) = 0 $$ Since $4032 \neq 0$, we must have $4a + b = 0$, which implies $b = -4a$. Step 3: Determine the axis of symmetry. The axis of symmetry for a quadratic function $f(x) = ax^2 + bx + c$ is given by $x = -\frac{b}{2a}$. Substituting $b = -4a$: $$ x = -\frac{-4a}{2a} = \frac{4a}{2a} = 2 $$ The vertex of the parabola is at $x=2$. Step 4: Use the second given condition. We are given $f(2018) + f(-2014) = 2(2016)^2 + 12$. Since $f(2018) = f(-2014)$, we can substitute $f(-2014)$ with $f(2018)$: $$ 2f(2018) = 2(2016)^2 + 12 $$ $$ f(2018) = (2016)^2 + 6 $$ Step 5: Express $f(x)$ in terms of $a$ and $c$. Substitute $b = -4a$ into $f(x) = ax^2 + bx + c$: $$ f(x) = ax^2 - 4ax + c $$ The value of $f(x)$ at its vertex $x=2$ is: $$ f(2) = a(2)^2 - 4a(2) + c = 4a - 8a + c = c - 4a $$ Step 6: Determine the values of $a$ and $c$. We can express $f(x)$ in vertex form: $f(x) = a(x-2)^2 + f(2)$. Using $x=2018$: $$ f(2018) = a(2018-2)^2 + f(2) = a(2016)^2 + f(2) $$ From Step 4, we know $f(2018) = (2016)^2 + 6$. Equating the two expressions for $f(2018)$: $$ a(2016)^2 + f(2) = (2016)^2 + 6 $$ This equation must hold for the specific function $f(x)$. A consistent solution is found by setting $a=1$, which implies $f(2)=6$. Using $f(2) = c - 4a$: $$ 6 = c - 4(1) \implies c = 10 $$ Thus, the function is $f(x) = x^2 - 4x + 10$. This can be written in vertex form as $f(x) = (x-2)^2 + 6$. Step 7: Analyze the properties of $f(x)$. The function $f(x) = (x-2)^2 + 6$ is a parabola opening upwards (since $a=1 > 0$). Its minimum value occurs at the vertex $x=2$, and the minimum value is $f(2) = 6$. Since the parabola opens upwards, $f(x)$ has no maximum value. Step 8: Calculate $f(1) + f(2) + f(3)$. $$ f(1) = (1-2)^2 + 6 = (-1)^2 + 6 = 1 + 6 = 7 $$ $$ f(2) = (2-2)^2 + 6 = 0^2 + 6 = 6 $$ $$ f(3) = (3-2)^2 + 6 = (1)^2 + 6 = 1 + 6 = 7 $$ $$ f(1) + f(2) + f(3) = 7 + 6 + 7 = 20 $$ Step 9: Determine the minimum value of $f(f(f(x)))$. The minimum value of $f(x)$ is $f(2)=6$. Since $f(x)$ is increasing for $x \ge 2$, and the range of $f(x)$ is $[6, \infty)$, the minimum value of $f(f(x))$ occurs when the inner function $f(x)$ takes its minimum value, which is $6$. $$ \min(f(f(x))) = f(6) = (6-2)^2 + 6 = 4^2 + 6 = 16 + 6 = 22 $$ Similarly, the minimum value of $f(f(f(x)))$ occurs when the inner function $f(f(x))$ takes its minimum value, which is $22$. $$ \min(f(f(f(x)))) = f(22) = (22-2)^2 + 6 = 20^2 + 6 = 400 + 6 = 406 $$ Step 10: Analyze the equation $f(x) = \text{sgn}(f(x))$. From Step 7, we know that the minimum value of $f(x)$ is $6$. Therefore, $f(x) \ge 6$ for all $x \in R$. Since $f(x) \ge 6$, $f(x)$ is always positive. The signum function $\text{sgn}(k)$ is $1$ if $k>0$, $0$ if $k=0$, and $-1$ if $k<0$. Since $f(x) > 0$, $\text{sgn}(f(x)) = 1$. The equation becomes $f(x) = 1$. However, we established that $f(x) \ge 6$. Thus, $f(x) = 1$ has no solution. The number of solutions is 0.