Let <math xmlns="http://www.w3.org/1998/Math/MathML"><mfenced open="|" close="|"><mtable><mtr><mtd><mn>1</mn></mtd><mtd><mn>1</mn></mtd><mtd><mn>1</mn></mtd></mtr><mtr><mtd><mn>1</mn></mtd><mtd><mn>2</mn></mtd><mtd><mn>3</mn></mtd></mtr><mtr><mtd><mi>α</mi></mtd><mtd><mi>β</mi></mtd><mtd><mi>γ</mi></mtd></mtr></mtable></mfenced><mo>=</mo><mi>t</mi></math>, where t is an even prime number & α, β, γ are the integral roots of the equation x<sup>3</sup> - 14x<sup>2</sup> + Px - 36 = 0<br>On the basis of above information, answer the following :<br>The value of P is -
Step-by-Step Solution
Key Concept: The determinant evaluates to (γ - β) - (γ - α) + (β - α) = γ - β - γ + α + β - α = 0. Wait, let's re-evaluate: 1(2γ - 3β) - 1(γ - 3α) + 1(β - 2α) = 2γ - 3β - γ + 3α + β - 2α = α - 2β + γ = t. Since t is an even prime, t = 2. Given α, β, γ are roots of x^3 - 14x^2 + Px - 36 = 0, we have α + β + γ = 14, αβ + βγ + γα = P, αβγ = 36. Using α - 2β + γ = 2 and α + β + γ = 14, we get 3β = 12, so β = 4. Then α + γ = 10 and αγ = 36/4 = 9. Solving x^2 - 10x + 9 = 0 gives roots 1 and 9. So {α, β, γ} = {1, 4, 9}. P = αβ + βγ + γα = 1*4 + 4*9 + 9*1 = 4 + 36 + 9 = 49.
Step 1: Evaluate the determinant and determine the value of $t$.
The determinant is given as:
$$ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ \alpha & \beta & \gamma \end{vmatrix} $$
Expanding the determinant:
$$ 1(2\gamma - 3\beta) - 1(\gamma - 3\alpha) + 1(\beta - 2\alpha) $$
$$ = 2\gamma - 3\beta - \gamma + 3\alpha + \beta - 2\alpha $$
$$ = \alpha - 2\beta + \gamma $$
The problem states that this determinant equals $t$, where $t$ is an even prime number. The only even prime number is 2.
Therefore, we have the equation:
$$ \alpha - 2\beta + \gamma = 2 \quad (1) $$
Step 2: Apply Vieta's formulas to the given cubic equation.
The cubic equation is $x^3 - 14x^2 + Px - 36 = 0$, and its integral roots are $\alpha, \beta, \gamma$.
From Vieta's formulas, we have:
Sum of the roots:
$$ \alpha + \beta + \gamma = -(-14)/1 = 14 \quad (2) $$
Product of the roots:
$$ \alpha\beta\gamma = -(-36)/1 = 36 \quad (3) $$
Sum of the products of the roots taken two at a time:
$$ \alpha\beta + \beta\gamma + \gamma\alpha = P $$
Step 3: Solve for the roots $\alpha, \beta, \gamma$.
Subtract equation (1) from equation (2):
$$ (\alpha + \beta + \gamma) - (\alpha - 2\beta + \gamma) = 14 - 2 $$
$$ 3\beta = 12 $$
$$ \beta = 4 $$
Substitute $\beta = 4$ into equation (2):
$$ \alpha + 4 + \gamma = 14 $$
$$ \alpha + \gamma = 10 $$
Substitute $\beta = 4$ into equation (3):
$$ \alpha(4)\gamma = 36 $$
$$ 4\alpha\gamma = 36 $$
$$ \alpha\gamma = 9 $$
Now we have a system for $\alpha$ and $\gamma$:
$$ \alpha + \gamma = 10 $$
$$ \alpha\gamma = 9 $$
Consider a quadratic equation whose roots are $\alpha$ and $\gamma$: $y^2 - (\alpha+\gamma)y + \alpha\gamma = 0$.
$$ y^2 - 10y + 9 = 0 $$
Factoring the quadratic equation:
$$ (y-1)(y-9) = 0 $$
Thus, the values for $\alpha$ and $\gamma$ are $1$ and $9$.
The integral roots of the cubic equation are $\{1, 4, 9\}$.
Step 4: Calculate the value of P.
The value of $P$ is given by the sum of the products of the roots taken two at a time:
$$ P = \alpha\beta + \beta\gamma + \gamma\alpha $$
Using the roots $\{1, 4, 9\}$:
$$ P = (1)(4) + (4)(9) + (9)(1) $$
$$ P = 4 + 36 + 9 $$
$$ P = 49 $$
Step 5: Characterize P.
The value of $P$ is $49$. This is an odd natural number.
Correct Answer: D