Key Concepts and Formulas

1. Determinant Evaluation: For a $3 \times 3$ matrix, the determinant can be expanded along any row or column. For a matrix $\begin{vmatrix} p & q & r \\ s & t & u \\ v & w & x \end{vmatrix}$, expanding along the first row gives $p(tx - uw) - q(sx - uv) + r(sw - tv)$. Choosing a row or column with zero elements simplifies calculations.
2. Differentiation Rules:
   • Constant Multiple Rule: $\frac{d}{dx}[c \cdot h(x)] = c \cdot h'(x)$, where $c$ is a constant with respect to $x$.
   • Chain Rule: $\frac{d}{dx}[g(x)]^n = n[g(x)]^{n-1} \cdot g'(x)$. This is crucial for differentiating composite functions like $(x+a)^2$.
3. Solving Quadratic Equations and Vieta's Formulas: For a quadratic equation $Ax^2 + Bx + C = 0$ with roots $\alpha$ and $\beta$, the sum of roots $\alpha + \beta = -\frac{B}{A}$ and the product of roots $\alpha \beta = \frac{C}{A}$. The sum of squares of roots is given by $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$.

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Step-by-Step Solution

Step 1: Evaluate the Determinant to Express $f(x)$ as a Polynomial

The first step is to simplify the given determinant into a more manageable polynomial form of $f(x)$. This algebraic expression is necessary for subsequent differentiation.

The given determinant is: $f(x) = \left| {\begin{matrix} a & { - 1} & 0 \\ {ax} & a & { - 1} \\ {a{x^2}} & {ax} & a \\ \end{matrix} } \right|$

To simplify, we expand the determinant along the first row ($R_1$) because it contains a zero, which eliminates one term in the expansion. $f(x) = a \cdot \begin{vmatrix} a & -1 \\ ax & a \end{vmatrix} - (-1) \cdot \begin{vmatrix} ax & -1 \\ a{x^2} & a \end{vmatrix} + 0 \cdot \begin{vmatrix} ax & a \\ a{x^2} & ax \end{vmatrix}$

Now, we evaluate the $2 \times 2$ determinants:

• The first term: $a \cdot (a \cdot a - (-1) \cdot ax) = a(a^2 + ax)$
• The second term: $+1 \cdot (ax \cdot a - (-1) \cdot a{x^2}) = 1 \cdot (a^2x + a{x^2})$
• The third term is $0$, so its contribution is zero.

Combine these results: $f(x) = a(a^2 + ax) + (a^2x + a{x^2})$ $f(x) = a^3 + a^2x + a^2x + a{x^2}$ $f(x) = a^3 + 2a^2x + a{x^2}$

We can factor out $a$ from the expression: $f(x) = a(a^2 + 2ax + x^2)$ Recognizing the perfect square trinomial $(x^2 + 2ax + a^2) = (x+a)^2$, we simplify $f(x)$ to: $f(x) = a(x+a)^2$

Step 2: Differentiate $f(x)$ to Find $f'(x)$

The problem involves $f'(10)$ and $f'(5)$, so we need to find the derivative of $f(x)$ with respect to $x$. Remember that $a$ is treated as a constant here.

Given $f(x) = a(x+a)^2$. Using the constant multiple rule and the chain rule: $f'(x) = a \cdot \frac{d}{dx}((x+a)^2)$ Let $u = x+a$. Then $\frac{du}{dx} = 1$. Applying the chain rule $\frac{d}{dx}(u^2) = 2u \cdot \frac{du}{dx}$: $f'(x) = a \cdot 2(x+a)^{2-1} \cdot \frac{d}{dx}(x+a)$ $f'(x) = a \cdot 2(x+a) \cdot 1$ $f'(x) = 2a(x+a)$

Step 3: Substitute $f'(x)$ into the Given Condition

The problem states the condition $2f'(10) - f'(5) + 100 = 0$. We will evaluate $f'(x)$ at $x=10$ and $x=5$ and substitute these into the equation.

First, evaluate $f'(10)$: $f'(10) = 2a(10+a)$

Next, evaluate $f'(5)$: $f'(5) = 2a(5+a)$

Substitute these expressions into the given condition: $2 \cdot [2a(10+a)] - [2a(5+a)] + 100 = 0$ $4a(10+a) - 2a(5+a) + 100 = 0$

Step 4: Solve the Resulting Quadratic Equation for 'a'

Now, we simplify the equation from Step 3 and solve for the values of $a$. $4a(10+a) - 2a(5+a) + 100 = 0$ Distribute the terms: $(40a + 4a^2) - (10a + 2a^2) + 100 = 0$ Remove parentheses, being careful with the negative sign: $40a + 4a^2 - 10a - 2a^2 + 100 = 0$ Combine like terms: $(4a^2 - 2a^2) + (40a - 10a) + 100 = 0$ $2a^2 + 30a + 100 = 0$ Divide the entire equation by $2$ to simplify: $a^2 + 15a + 50 = 0$ This is a quadratic equation. We can solve it by factorization. We look for two numbers that multiply to $50$ and add up to $15$. These numbers are $10$ and $5$. $(a+10)(a+5) = 0$ This yields two possible values for $a$: $a+10 = 0 \Rightarrow a = -10$ $a+5 = 0 \Rightarrow a = -5$ So, the values of $a$ are $-10$ and $-5$.

Step 5: Calculate the Sum of the Squares of all Values of $a$

The question asks for the sum of the squares of all the values of $a$. Let $a_1 = -10$ and $a_2 = -5$. Sum of squares $= a_1^2 + a_2^2$ $= (-10)^2 + (-5)^2$ $= 100 + 25$ $= 125$

Alternatively, using Vieta's formulas for $a^2 + 15a + 50 = 0$: Here, $A=1$, $B=15$, $C=50$. Sum of roots: $a_1 + a_2 = -\frac{B}{A} = -\frac{15}{1} = -15$ Product of roots: $a_1 a_2 = \frac{C}{A} = \frac{50}{1} = 50$ Sum of squares: $a_1^2 + a_2^2 = (a_1 + a_2)^2 - 2a_1 a_2$ $= (-15)^2 - 2(50)$ $= 225 - 100$ $= 125$ Both methods confirm the result.

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Common Mistakes & Tips

• Determinant Expansion Errors: Be careful with signs when expanding determinants, especially the negative sign for the middle term in a $3 \times 3$ expansion.
• Differentiation of Constants: Remember that $a$ is a constant with respect to $x$. Its derivative with respect to $x$ is $0$, and it acts as a constant multiplier. Do not treat $a$ as a variable when differentiating $f(x)$.
• Chain Rule Application: Ensure the chain rule is correctly applied when differentiating $(x+a)^2$. It's $2(x+a)$ multiplied by the derivative of $(x+a)$, which is $1$.
• Algebraic Simplification: Simplify the expression for $f(x)$ as much as possible, e.g., recognizing $(x^2+2ax+a^2)$ as $(x+a)^2$. This makes differentiation much easier.

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Summary

This problem required a systematic application of determinant evaluation, differentiation, and solving quadratic equations. We first simplified the determinant to obtain $f(x) = a(x+a)^2$. Then, we found its derivative $f'(x) = 2a(x+a)$. Substituting $f'(10)$ and $f'(5)$ into the given condition $2f'(10) - f'(5) + 100 = 0$ led to a quadratic equation $a^2 + 15a + 50 = 0$. Solving this equation yielded $a = -10$ and $a = -5$. Finally, we calculated the sum of the squares of these values, $(-10)^2 + (-5)^2 = 100 + 25 = 125$.

The final answer is $\boxed{125}$, which corresponds to option (C).