Question:

Step 1: Evaluate $\Delta_1$ and $\Delta_2$. $$ \Delta_1 = \begin{vmatrix} x & b & b \\ a & x & b \\ a & a & x \end{vmatrix} $$ To evaluate $\Delta_1$, perform the operation $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$: $$ \Delta_1 = \begin{vmatrix} x-a & b-x & 0 \\ 0 & x-a & b-x \\ a & a & x \end{vmatrix} $$ Expand along the first column: $$ \Delta_1 = (x-a) \begin{vmatrix} x-a & b-x \\ a & x \end{vmatrix} - 0 + a \begin{vmatrix} b-x & 0 \\ x-a & b-x \end{vmatrix} $$ $$ \Delta_1 = (x-a) [x(x-a) - a(b-x)] + a [(b-x)(b-x) - 0] $$ $$ \Delta_1 = (x-a) [x^2 - ax - ab + ax] + a (b-x)^2 $$ $$ \Delta_1 = (x-a) (x^2 - ab) + a (b-x)^2 $$ $$ \Delta_1 = x^3 - abx - ax^2 + a^2b + a(b^2 - 2bx + x^2) $$ $$ \Delta_1 = x^3 - abx - ax^2 + a^2b + ab^2 - 2abx + ax^2 $$ $$ \Delta_1 = x^3 - 3abx + a^2b + ab^2 $$ Alternatively, by applying $C_1 \to C_1 + C_2 + C_3$: $$ \Delta_1 = \begin{vmatrix} x+2b & b & b \\ x+a+b & x & b \\ x+2a & a & x \end{vmatrix} $$ This expansion is more complex. A simpler approach for $\Delta_1$ is to use the property that if $a=b$, then $\Delta_1 = (x-a)^2(x+2a)$. Let's expand $\Delta_1$ directly: $$ \Delta_1 = x(x^2 - ab) - b(ax - ab) + b(a^2 - ax) $$ $$ \Delta_1 = x^3 - abx - abx + ab^2 + a^2b - abx $$ $$ \Delta_1 = x^3 - 3abx + ab^2 + a^2b $$ And $\Delta_2$ is given by: $$ \Delta_2 = \begin{vmatrix} x & b \\ a & x \end{vmatrix} = x^2 - ab $$ Step 2: Differentiate $\Delta_1$ with respect to $x$. $$ \frac{d}{dx}(\Delta_1) = \frac{d}{dx}(x^3 - 3abx + ab^2 + a^2b) $$ $$ \frac{d}{dx}(\Delta_1) = 3x^2 - 3ab $$ Step 3: Express $\frac{d}{dx}(\Delta_1)$ in terms of $\Delta_2$. From Step 2, we have $\frac{d}{dx}(\Delta_1) = 3x^2 - 3ab$. Factor out 3: $$ \frac{d}{dx}(\Delta_1) = 3(x^2 - ab) $$ From Step 1, we know that $\Delta_2 = x^2 - ab$. Substitute $\Delta_2$ into the expression: $$ \frac{d}{dx}(\Delta_1) = 3\Delta_2 $$ The problem statement implies that $\frac{d}{dx}(\Delta_1) = 3(\Delta_2)^{1/2}$. This is a contradiction with the derived result. The derived result is $3\Delta_2$. The question asks to find the value of $\frac{d}{dx}(\Delta_1)$, and the options are given in terms of $\Delta_2$. The calculated value is $3\Delta_2$. This can be written as $3(\Delta_2)^1$. If the question intended for the answer to be $3(\Delta_2)^{3/2}$, then $\Delta_2$ would need to be $(x^2-ab)^{2/3}$. However, $\Delta_2 = x^2-ab$. Therefore, $\frac{d}{dx}(\Delta_1) = 3\Delta_2$. If the options are given, and one of them is $3(\Delta_2)^{3/2}$, it implies that there might be a misunderstanding of the question or a typo in the question itself. Based on the direct calculation, $\frac{d}{dx}(\Delta_1) = 3\Delta_2$. If the question statement $\dfrac{d}{dx}(\Delta_1) = 3(\Delta_2)^{1/2}$ is part of the problem to be verified, then it is false. If the question asks to find $\dfrac{d}{dx}(\Delta_1)$, then the answer is $3\Delta_2$. Given the options, and the "Correct Answer: B" which corresponds to $3(\Delta_2)^{3/2}$, there is a discrepancy. Let's assume the question implicitly asks to find the expression for $\frac{d}{dx}(\Delta_1)$ and then match it to the options. Our calculation yields $3\Delta_2$. If the correct answer is $3(\Delta_2)^{3/2}$, then $3\Delta_2 = 3(\Delta_2)^{3/2}$, which implies $\Delta_2 = (\Delta_2)^{3/2}$. This means $\Delta_2^{1/2} = 1$, so $\Delta_2 = 1$. This is not generally true. Let's re-evaluate the determinant $\Delta_1$ using a different method to ensure accuracy. Apply $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$: $$ \Delta_1 = \begin{vmatrix} x-a & b-x & 0 \\ 0 & x-a & b-x \\ a & a & x \end{vmatrix} $$ Expand along $C_3$: $$ \Delta_1 = 0 - (b-x) \begin{vmatrix} x-a & b-x \\ a & a \end{vmatrix} + x \begin{vmatrix} x-a & b-x \\ 0 & x-a \end{vmatrix} $$ $$ \Delta_1 = (x-b) [a(x-a) - a(b-x)] + x (x-a)^2 $$ $$ \Delta_1 = (x-b) [ax - a^2 - ab + ax] + x (x^2 - 2ax + a^2) $$ $$ \Delta_1 = (x-b) [2ax - a^2 - ab] + x^3 - 2ax^2 + a^2x $$ $$ \Delta_1 = 2ax^2 - a^2x - abx - 2abx + a^3 + a^2b + x^3 - 2ax^2 + a^2x $$ $$ \Delta_1 = x^3 - 3abx + a^3 + a^2b $$ This is different from the previous expansion $x^3 - 3abx + ab^2 + a^2b$. Let's re-expand $\Delta_1 = x(x^2 - ab) - b(ax - ab) + b(a^2 - ax)$ $$ \Delta_1 = x^3 - abx - abx + ab^2 + a^2b - abx $$ $$ \Delta_1 = x^3 - 3abx + ab^2 + a^2b $$ This expansion is correct. The previous one had a mistake in the $a^3$ term. Now, let's consider the case where $a=b$. $$ \Delta_1 = \begin{vmatrix} x & a & a \\ a & x & a \\ a & a & x \end{vmatrix} $$ Apply $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$: $$ \Delta_1 = \begin{vmatrix} x-a & a-x & 0 \\ 0 & x-a & a-x \\ a & a & x \end{vmatrix} $$ $$ \Delta_1 = (x-a) \begin{vmatrix} x-a & a-x \\ a & x \end{vmatrix} - (a-x) \begin{vmatrix} 0 & a-x \\ a & x \end{vmatrix} $$ $$ \Delta_1 = (x-a) [x(x-a) - a(a-x)] + (x-a) [0 - a(a-x)] $$ $$ \Delta_1 = (x-a) [x(x-a) + a(x-a)] + (x-a) [a(x-a)] $$ $$ \Delta_1 = (x-a)^2 (x+a) + a(x-a)^2 $$ $$ \Delta_1 = (x-a)^2 (x+a+a) = (x-a)^2 (x+2a) $$ If $\Delta_1 = (x-a)^2(x+2a)$, then $$ \Delta_1 = (x^2 - 2ax + a^2)(x+2a) $$ $$ \Delta_1 = x^3 - 2ax^2 + a^2x + 2ax^2 - 4a^2x + 2a^3 $$ $$ \Delta_1 = x^3 - 3a^2x + 2a^3 $$ If $a=b$, then $\Delta_1 = x^3 - 3a^2x + 2a^3$. And $\Delta_2 = x^2 - a^2$. Then $\frac{d}{dx}(\Delta_1) = 3x^2 - 3a^2 = 3(x^2 - a^2) = 3\Delta_2$. This confirms that for the special case $a=b$, $\frac{d}{dx}(\Delta_1) = 3\Delta_2$. Now, let's re-examine the general expansion of $\Delta_1 = x^3 - 3abx + ab^2 + a^2b$. $$ \frac{d}{dx}(\Delta_1) = 3x^2 - 3ab $$ Since $\Delta_2 = x^2 - ab$, we have: $$ \frac{d}{dx}(\Delta_1) = 3(x^2 - ab) = 3\Delta_2 $$ The result is consistently $3\Delta_2$. The problem statement $\dfrac{d}{dx}(\Delta_1) = 3(\Delta_2)^{1/2}$ is a condition, not the value to be found. The question asks to "find the value (A: $3(\Delta_2)^{1/2}$)" which is confusing. It seems to be asking to verify if the given expression is correct. However, the options are given, and the correct answer is B, which is $3(\Delta_2)^{3/2}$. This implies that the question is asking to find $\frac{d}{dx}(\Delta_1)$ and then match it to the options. Our calculation gives $3\Delta_2$. If the correct option is $3(\Delta_2)^{3/2}$, then there is a fundamental error in the problem statement or the provided solution. Let's assume the question is asking to find $\frac{d}{dx}(\Delta_1)$ and the options are possible answers. Our derived answer is $3\Delta_2$. None of the options match $3\Delta_2$ directly, unless $\Delta_2=1$ or some other specific condition. The options are: A: $3(\Delta_2)^{1/2}$ B: $3(\Delta_2)^{3/2}$ C: $3\Delta_2$ D: $3(\Delta_2)^{1/3}$ Wait, option C is $3\Delta_2$. The provided "Correct Answer: B" is $3(\Delta_2)^{3/2}$. This is a direct contradiction. My derivation consistently leads to $3\Delta_2$. If the provided solution states "Correct Answer: B", then the derivation must lead to $3(\Delta_2)^{3/2}$. Let's assume there's a mistake in my $\Delta_1$ expansion or differentiation. The differentiation of a determinant can also be done by differentiating one row/column at a time. $$ \frac{d}{dx}(\Delta_1) = \begin{vmatrix} 1 & 0 & 0 \\ a & x & b \\ a & a & x \end{vmatrix} + \begin{vmatrix} x & b & b \\ 0 & 1 & 0 \\ a & a & x \end{vmatrix} + \begin{vmatrix} x & b & b \\ a & x & b \\ 0 & 0 & 1 \end{vmatrix} $$ $$ \frac{d}{dx}(\Delta_1) = 1(x^2 - ab) + 1(x^2 - ab) + 1(x^2 - ab) $$ $$ \frac{d}{dx}(\Delta_1) = (x^2 - ab) + (x^2 - ab) + (x^2 - ab) $$ $$ \frac{d}{dx}(\Delta_1) = 3(x^2 - ab) $$ Since $\Delta_2 = x^2 - ab$, $$ \frac{d}{dx}(\Delta_1) = 3\Delta_2 $$ This method also confirms that $\frac{d}{dx}(\Delta_1) = 3\Delta_2$. The solution provided in the prompt states: "$\frac{d\Delta_1}{dx} = 3x^2 - 3ab = 3(x^2-ab) = 3\Delta_2$". This part of the corrupted solution is correct and matches my derivation. However, it then says: "And $3\Delta_2 = 3(\Delta_2)^1$, but the answer key indicates $3(\Delta_2)^{3/2}$...". This indicates a conflict between the derived result and the expected answer. The final output should only contain the correct path. The correct path is $\frac{d}{dx}(\Delta_1) = 3\Delta_2$. If the question implies that the answer is $3(\Delta_2)^{3/2}$, then the question itself is flawed or there's a misunderstanding of the problem statement. Given the instruction "only output the final correct path", and the fact that the corrupted solution *itself* correctly derives $3\Delta_2$, I must present $3\Delta_2$ as the result. The options provided in the prompt are: ['$3(\\Delta_2)^{1/2}$', '$3(\\Delta_2)^{3/2}$', '$3\\Delta_2$', '$3(\\Delta_2)^{1/3}$'] Option C is $3\Delta_2$. The "Correct Answer: B" in the prompt is $3(\Delta_2)^{3/2}$. This is a contradiction. I must follow the math. The math consistently shows $3\Delta_2$. I will present the derivation for $3\Delta_2$. I will not mention the options or the "Correct Answer: B" from the prompt. The question asks to "find the value (A: $3(\Delta_2)^{1/2}$)...". This is also confusing. It seems to be asking to find the value, and then gives a hint or a specific option. I will assume the question is "Find $\frac{d}{dx}(\Delta_1)$". Final check of the determinant expansion: $\Delta_1 = \begin{vmatrix} x & b & b \\ a & x & b \\ a & a & x \end{vmatrix}$ $C_1 \to C_1 - C_2$, $C_2 \to C_2 - C_3$: $\Delta_1 = \begin{vmatrix} x-b & 0 & b \\ a-x & x-b & b \\ 0 & a-x & x \end{vmatrix}$ Expand along $R_1$: $\Delta_1 = (x-b) \begin{vmatrix} x-b & b \\ a-x & x \end{vmatrix} - 0 + b \begin{vmatrix} a-x & x-b \\ 0 & a-x \end{vmatrix}$ $\Delta_1 = (x-b) [x(x-b) - b(a-x)] + b (a-x)^2$ $\Delta_1 = (x-b) [x^2 - bx - ab + bx] + b (a^2 - 2ax + x^2)$ $\Delta_1 = (x-b) (x^2 - ab) + ba^2 - 2abx + bx^2$ $\Delta_1 = x^3 - abx - bx^2 + ab^2 + ba^2 - 2abx + bx^2$ $\Delta_1 = x^3 - 3abx + ab^2 + a^2b$ This expansion is consistent. The derivative is $3x^2 - 3ab = 3(x^2-ab) = 3\Delta_2$. The solution must reflect this.Step 1: Evaluate $\Delta_1$ and $\Delta_2$. The determinant $\Delta_1$ is given by: $$ \Delta_1 = \begin{vmatrix} x & b & b \\ a & x & b \\ a & a & x \end{vmatrix} $$ Expand $\Delta_1$ along the first row: $$ \Delta_1 = x \begin{vmatrix} x & b \\ a & x \end{vmatrix} - b \begin{vmatrix} a & b \\ a & x \end{vmatrix} + b \begin{vmatrix} a & x \\ a & a \end{vmatrix} $$ $$ \Delta_1 = x(x \cdot x - b \cdot a) - b(a \cdot x - b \cdot a) + b(a \cdot a - x \cdot a) $$ $$ \Delta_1 = x(x^2 - ab) - b(ax - ab) + b(a^2 - ax) $$ $$ \Delta_1 = x^3 - abx - abx + ab^2 + a^2b - abx $$ $$ \Delta_1 = x^3 - 3abx + ab^2 + a^2b $$ The determinant $\Delta_2$ is given by: $$ \Delta_2 = \begin{vmatrix} x & b \\ a & x \end{vmatrix} = x \cdot x - b \cdot a = x^2 - ab $$ Step 2: Differentiate $\Delta_1$ with respect to $x$. Differentiate the expanded form of $\Delta_1$: $$ \frac{d}{dx}(\Delta_1) = \frac{d}{dx}(x^3 - 3abx + ab^2 + a^2b) $$ $$ \frac{d}{dx}(\Delta_1) = 3x^2 - 3ab $$ Alternatively, using the property for differentiating a determinant, which states that the derivative of a determinant is the sum of determinants obtained by differentiating one row (or column) at a time: $$ \frac{d}{dx}(\Delta_1) = \begin{vmatrix} \frac{d}{dx}(x) & \frac{d}{dx}(b) & \frac{d}{dx}(b) \\ a & x & b \\ a & a & x \end{vmatrix} + \begin{vmatrix} x & b & b \\ \frac{d}{dx}(a) & \frac{d}{dx}(x) & \frac{d}{dx}(b) \\ a & a & x \end{vmatrix} + \begin{vmatrix} x & b & b \\ a & x & b \\ \frac{d}{dx}(a) & \frac{d}{dx}(a) & \frac{d}{dx}(x) \end{vmatrix} $$ $$ \frac{d}{dx}(\Delta_1) = \begin{vmatrix} 1 & 0 & 0 \\ a & x & b \\ a & a & x \end{vmatrix} + \begin{vmatrix} x & b & b \\ 0 & 1 & 0 \\ a & a & x \end{vmatrix} + \begin{vmatrix} x & b & b \\ a & x & b \\ 0 & 0 & 1 \end{vmatrix} $$ Evaluate each determinant: $$ \begin{vmatrix} 1 & 0 & 0 \\ a & x & b \\ a & a & x \end{vmatrix} = 1(x \cdot x - b \cdot a) - 0 + 0 = x^2 - ab $$ $$ \begin{vmatrix} x & b & b \\ 0 & 1 & 0 \\ a & a & x \end{vmatrix} = x(1 \cdot x - 0 \cdot a) - b(0 \cdot x - 0 \cdot a) + b(0 \cdot a - 1 \cdot a) = x^2 - 0 - ab = x^2 - ab $$ $$ \begin{vmatrix} x & b & b \\ a & x & b \\ 0 & 0 & 1 \end{vmatrix} = x(x \cdot 1 - b \cdot 0) - b(a \cdot 1 - b \cdot 0) + b(a \cdot 0 - x \cdot 0) = x^2 - ab $$ Summing these results: $$ \frac{d}{dx}(\Delta_1) = (x^2 - ab) + (x^2 - ab) + (x^2 - ab) = 3(x^2 - ab) $$ Step 3: Express $\frac{d}{dx}(\Delta_1)$ in terms of $\Delta_2$. From Step 2, we have $\frac{d}{dx}(\Delta_1) = 3(x^2 - ab)$. From Step 1, we know that $\Delta_2 = x^2 - ab$. Substitute $\Delta_2$ into the expression for $\frac{d}{dx}(\Delta_1)$: $$ \frac{d}{dx}(\Delta_1) = 3\Delta_2 $$