Key Concepts and Formulas

• Function Composition: Understanding how to evaluate nested functions, i.e., $f(f(x))$.
• Exponent Rules: Properties of exponents such as $(a^m)^n = a^{mn}$ and $a^0 = 1$.
• Difference of Squares: The algebraic identity $(a-b)(a+b) = a^2 - b^2$.
• Greatest Integer Function: $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$. Key property: $\lfloor x + n \rfloor = \lfloor x \rfloor + n$ for integer $n$.

Step-by-Step Solution

We are given the function $f(x) = {\left( {2\left( {1 - {{{x^{25}}} \over 2}} \right)(2 + {x^{25}})} \right)^{{1 \over {50}}}}$ and $g(x) = f(f(f(x))) + f(f(x))$. We need to find the greatest integer less than or equal to $g(1)$, which is $\lfloor g(1) \rfloor$.

Step 1: Evaluate $f(1)$ To start evaluating $g(1)$, we first need to find the value of the innermost function, $f(1)$. Substitute $x=1$ into the definition of $f(x)$: $f(1) = {\left( {2\left( {1 - {{{1^{25}}} \over 2}} \right)(2 + {1^{25}})} \right)^{{1 \over {50}}}}$ Since $1^{25} = 1$, the expression simplifies to: $f(1) = {\left( {2\left( {1 - {1 \over 2}} \right)(2 + 1)} \right)^{{1 \over {50}}}}$ $f(1) = {\left( {2\left( {{1 \over 2}} \right)(3)} \right)^{{1 \over {50}}}}$ $f(1) = {\left( {1 \cdot 3} \right)^{{1 \over {50}}}}$ $f(1) = {3^{{1 \over {50}}}}$ So, $f(1) = 3^{1/50}$.

Step 2: Evaluate $f(f(1))$ Next, we evaluate $f$ at the result of $f(1)$. Let $y = f(1) = 3^{1/50}$. We need to compute $f(y)$. Substitute $x = 3^{1/50}$ into the definition of $f(x)$: $f\left( {f\left( 1 \right)} \right) = f\left( {{3^{{1 \over {50}}}}} \right) = {\left( {2\left( {1 - {{{{\left( {{3^{{1 \over {50}}}}} \right)}^{25}}} \over 2}} \right)\left( {2 + {{\left( {{3^{{1 \over {50}}}}} \right)}^{25}}} \right)} \right)^{{1 \over {50}}}}$ First, simplify the term inside the parentheses: $\left( {{3^{{1 \over {50}}}}} \right)^{25}$. Using the exponent rule $(a^m)^n = a^{mn}$: $\left( {{3^{{1 \over {50}}}}} \right)^{25} = 3^{{{1 \over {50}}} \times 25} = 3^{{25 \over {50}}} = 3^{{1 \over 2}} = \sqrt{3}$ Now substitute $\sqrt{3}$ back into the expression for $f(f(1))$: $f\left( {f\left( 1 \right)} \right) = {\left( {2\left( {1 - {{\sqrt 3 } \over 2}} \right)\left( {2 + \sqrt 3 } \right)} \right)^{{1 \over {50}}}}$ Let's simplify the expression inside the main parenthesis: $2\left( {1 - {{\sqrt 3 } \over 2}} \right)\left( {2 + \sqrt 3 } \right)$. $2\left( {1 - {{\sqrt 3 } \over 2}} \right)\left( {2 + \sqrt 3 } \right) = 2\left( {{{2 - \sqrt 3 } \over 2}} \right)\left( {2 + \sqrt 3 } \right)$ Cancel out the factor of 2: $= \left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)$ This is a difference of squares, $(a-b)(a+b) = a^2 - b^2$, with $a=2$ and $b=\sqrt{3}$: $= (2^2 - (\sqrt{3})^2) = (4 - 3) = 1$ So, the expression for $f(f(1))$ becomes: $f\left( {f\left( 1 \right)} \right) = {\left( 1 \right)^{{1 \over {50}}}} = 1$ Thus, $f(f(1)) = 1$.

Step 3: Evaluate $f(f(f(1)))$ Now we evaluate $f$ at the result of $f(f(1))$. Let $z = f(f(1)) = 1$. We need to compute $f(z)$, which is $f(1)$. From Step 1, we already found that $f(1) = 3^{1/50}$. Therefore, $f\left( {f\left( {f\left( 1 \right)} \right)} \right) = f(1) = {3^{{1 \over {50}}}}$ So, $f(f(f(1))) = 3^{1/50}$.

Step 4: Calculate $g(1)$ Now we substitute the computed values into the definition of $g(1)$: $g(1) = f\left( {f\left( {f\left( 1 \right)} \right)} \right) + f\left( {f\left( 1 \right)} \right)$ $g(1) = {3^{{1 \over {50}}}} + 1$

Step 5: Find the greatest integer less than or equal to $g(1)$ We need to find $\lfloor g(1) \rfloor = \lfloor {3^{{1 \over {50}}}} + 1 \rfloor$. Using the property $\lfloor x + n \rfloor = \lfloor x \rfloor + n$ for any integer $n$, we get: $\lfloor {3^{{1 \over {50}}}} + 1 \rfloor = \lfloor {3^{{1 \over {50}}}} \rfloor + 1$ Now we need to determine the value of $\lfloor {3^{{1 \over {50}}}} \rfloor$. Consider the value $3^{1/50}$. We know that: $1^{50} = 1$ $2^{50}$ is a very large number. Since $1 < 3$, and raising both sides to the power of $1/50$ preserves the inequality for positive bases: $1^{1/50} < 3^{1/50}$ $1 < 3^{1/50}$ Also, since $3 < 2^{50}$ (as $2^{10} = 1024$, so $2^{50} = (2^{10})^5 = 1024^5$, which is much larger than 3), we have: $3^{1/50} < (2^{50})^{1/50}$ $3^{1/50} < 2$ So, we have established that $1 < 3^{1/50} < 2$. This means that $3^{1/50}$ is a number strictly between 1 and 2. Therefore, the greatest integer less than or equal to $3^{1/50}$ is 1: $\lfloor {3^{{1 \over {50}}}} \rfloor = 1$ Now, substituting this back into the expression for $\lfloor g(1) \rfloor$: $\lfloor g(1) \rfloor = \lfloor {3^{{1 \over {50}}}} \rfloor + 1 = 1 + 1 = 2$

Common Mistakes & Tips

• Algebraic Simplification: Be careful with the algebraic manipulations, especially when dealing with fractions and exponents. The difference of squares identity is a key shortcut here.
• Exponent Rules: Ensure correct application of exponent rules, particularly when raising a power to another power.
• Greatest Integer Function: Remember that $\lfloor x \rfloor$ gives the largest integer less than or equal to $x$. For a number like $1.x$, the floor is 1.

Summary

The problem required evaluating a composite function $g(x)$ at $x=1$, where $g(x)$ is defined in terms of $f(x)$. We systematically evaluated $f(1)$, then $f(f(1))$, and finally $f(f(f(1)))$ by substituting the results back into the function $f(x)$. After finding $g(1) = 3^{1/50} + 1$, we determined the greatest integer less than or equal to $g(1)$ by first estimating the value of $3^{1/50}$ to be between 1 and 2, which allowed us to find its floor and subsequently the floor of $g(1)$.

The final answer is \boxed{2}.