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Given that (x-1) is a factor of f(x) = ax⁴ + x³ - 12x² - x + 2, by the Factor Theorem, f(1) = 0.

Mathematics

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1. (i)

  1. Use the Factor Theorem.

    Given that (x−1)(x-1) is a factor of f(x)=ax4+x3−12x2−x+2f(x) = ax^4 + x^3 - 12x^2 - x + 2, by the Factor Theorem, f(1)=0f(1) = 0.
    Substitute x=1x=1 into f(x)f(x):

    f(1)=a(1)4+(1)3−12(1)2−(1)+2=0f(1) = a(1)^4 + (1)^3 - 12(1)^2 - (1) + 2 = 0 a+1−12−1+2=0a + 1 - 12 - 1 + 2 = 0 a−10=0a - 10 = 0 a=10a = 10

    The value of the constant aa is 10\boxed{10}.

  2. Verify that f(−1)=0f(-1) = 0.

    Now, substitute a=10a=10 into f(x)f(x), so f(x)=10x4+x3−12x2−x+2f(x) = 10x^4 + x^3 - 12x^2 - x + 2.
    Substitute x=−1x=-1 into f(x)f(x):

    f(−1)=10(−1)4+(−1)3−12(−1)2−(−1)+2f(-1) = 10(-1)^4 + (-1)^3 - 12(-1)^2 - (-1) + 2 f(−1)=10(1)+(−1)−12(1)+1+2f(-1) = 10(1) + (-1) - 12(1) + 1 + 2 f(−1)=10−1−12+1+2f(-1) = 10 - 1 - 12 + 1 + 2 f(−1)=9−12+1+2f(-1) = 9 - 12 + 1 + 2 f(−1)=−3+1+2f(-1) = -3 + 1 + 2 f(−1)=0f(-1) = 0

    This verifies that f(−1)=0f(-1) = 0.

    1. (ii)

  3. Define the roots.

    Let the roots of the equation x2+(k+1)x+k=0x^2 + (k+1)x + k = 0 be α\alpha and 2α2\alpha.

  4. Apply Vieta's formulas.

    For a quadratic equation Ax2+Bx+C=0Ax^2 + Bx + C = 0, the sum of roots is −BA-\frac{B}{A} and the product of roots is CA\frac{C}{A}.
    In this equation, A=1A=1, B=(k+1)B=(k+1), C=kC=k.

    Sum of roots:

    α+2α=−(k+1)\alpha + 2\alpha = -(k+1) 3α=−(k+1)(1)3\alpha = -(k+1) \quad (1)

    Product of roots:

    α⋅2α=k\alpha \cdot 2\alpha = k 2α2=k(2)2\alpha^2 = k \quad (2)
  5. Solve the system of equations for kk.

    From equation (1), express α\alpha in terms of kk:

    α=−k+13\alpha = -\frac{k+1}{3}

    Substitute this expression for α\alpha into equation (2):

    2(−k+13)2=k2\left(-\frac{k+1}{3}\right)^2 = k 2(k+1)29=k2\frac{(k+1)^2}{9} = k 2(k2+2k+1)=9k2(k^2 + 2k + 1) = 9k 2k2+4k+2=9k2k^2 + 4k + 2 = 9k 2k2+4k−9k+2=02k^2 + 4k - 9k + 2 = 0 2k2−5k+2=02k^2 - 5k + 2 = 0
  6. Solve the quadratic equation for kk.

    Factor the quadratic equation:

    2k2−4k−k+2=02k^2 - 4k - k + 2 = 0 2k(k−2)−1(k−2)=02k(k - 2) - 1(k - 2) = 0 (2k−1)(k−2)=0(2k - 1)(k - 2) = 0

    This gives two possible values for kk:

    2k−1=0  ⟹  k=122k - 1 = 0 \implies k = \frac{1}{2} k−2=0  ⟹  k=2k - 2 = 0 \implies k = 2

    The values of the constant kk are 12 and 2\boxed{\frac{1}{2} \text{ and } 2}.

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