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A) Given that (x-1) is a factor of the polynomial f(x) = ax4 - x3 - 12x2 - x + 2. Find the value of the constant a and verify that f(-1)=0. b) Find the value of the constant k for which the equation x

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2 parts
b

k=12 or k=2k = \frac{1}{2} \text{ or } k = 2

b(i)
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sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ=tan⁡θ\frac{\sin\theta + \sin 2\theta}{1 + \cos\theta + \cos 2\theta} = \tan\theta

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Part 1

Part a· 2 steps

Given that (x−1)(x-1) is a factor of the polynomial f(x)=ax4−x3−12x2−x+2f(x) = ax^4 - x^3 - 12x^2 - x + 2. Find the value of the constant aa and verify that f(−1)=0f(-1)=0.

  1. Use the Factor Theorem.

    Since (x−1)(x-1) is a factor of f(x)f(x), we know that 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−12=0a - 12 = 0 a=12a = 12

  2. Verify f(−1)=0f(-1)=0 with a=12a=12.

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

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

    There seems to be a discrepancy with the verification part of the question, as f(−1)≠0f(-1) \neq 0. Assuming the question meant to ask for f(1)=0f(1)=0 or there's a typo in the polynomial or the verification value. Based on the given polynomial and factor, f(−1)=4f(-1)=4.

Part b

k=12 or k=2k = \frac{1}{2} \text{ or } k = 2

Find the value of the constant kk for which the equation x2+(k+1)x+k=0x^2 + (k+1)x + k = 0 has one root, double the other.

  1. Set up relationships between roots and coefficients.

    Let the roots be α\alpha and 2α2\alpha.
    From Vieta's formulas:
    Sum of roots: α+2α=−(k+1)  ⟹  3α=−(k+1)(1)\alpha + 2\alpha = -(k+1) \implies 3\alpha = -(k+1) \quad (1)
    Product of roots: α⋅2α=k  ⟹  2α2=k(2)\alpha \cdot 2\alpha = k \implies 2\alpha^2 = k \quad (2)

  2. Solve for kk.

    From equation (1), α=−k+13\alpha = -\frac{k+1}{3}.
    Substitute this 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 2k2−5k+2=02k^2 - 5k + 2 = 0

    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

    So, 2k−1=02k-1=0 or k−2=0k-2=0.

    k=12ork=2k = \frac{1}{2} \quad \text{or} \quad k = 2

    The values of kk are k=12 or k=2\boxed{k = \frac{1}{2} \text{ or } k = 2}.

Part 2

Part b(i)

sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ=tan⁡θ\frac{\sin\theta + \sin 2\theta}{1 + \cos\theta + \cos 2\theta} = \tan\theta

Show that sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ=tan⁡θ\frac{\sin\theta + \sin 2\theta}{1 + \cos\theta + \cos 2\theta} = \tan\theta.

  1. Apply double angle identities.

    Recall the identities: sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta and cos⁡2θ=2cos⁡2θ−1\cos 2\theta = 2\cos^2\theta - 1.
    Substitute these into the left-hand side (LHS) of the equation:

    LHS=sin⁡θ+2sin⁡θcos⁡θ1+cos⁡θ+(2cos⁡2θ−1)\text{LHS} = \frac{\sin\theta + 2\sin\theta\cos\theta}{1 + \cos\theta + (2\cos^2\theta - 1)}

  2. Simplify the expression.

    LHS=sin⁡θ(1+2cos⁡θ)cos⁡θ+2cos⁡2θ\text{LHS} = \frac{\sin\theta(1 + 2\cos\theta)}{\cos\theta + 2\cos^2\theta}

    Factor out cos⁡θ\cos\theta from the denominator:

    LHS=sin⁡θ(1+2cos⁡θ)cos⁡θ(1+2cos⁡θ)\text{LHS} = \frac{\sin\theta(1 + 2\cos\theta)}{\cos\theta(1 + 2\cos\theta)}

    Cancel the common term (1+2cos⁡θ)(1 + 2\cos\theta), assuming 1+2cos⁡θ≠01 + 2\cos\theta \neq 0:

    LHS=sin⁡θcos⁡θ\text{LHS} = \frac{\sin\theta}{\cos\theta} LHS=tan⁡θ\text{LHS} = \tan\theta

    This matches the right-hand side (RHS).
    Therefore, sin⁡θ+sin⁡2θ1+cos⁡θ+cos⁡2θ=tan⁡θ\boxed{\frac{\sin\theta + \sin 2\theta}{1 + \cos\theta + \cos 2\theta} = \tan\theta}.

Part b(ii)· 3 steps

Find the general solution of the equation sin⁡4x+cos⁡2x=0\sin 4x + \cos 2x = 0.

  1. Use the double angle identity for sin⁡4x\sin 4x.

    Recall sin⁡4x=2sin⁡2xcos⁡2x\sin 4x = 2\sin 2x \cos 2x.
    Substitute this into the equation:

    2sin⁡2xcos⁡2x+cos⁡2x=02\sin 2x \cos 2x + \cos 2x = 0

  2. Factor the equation.

    Factor out cos⁡2x\cos 2x:

    cos⁡2x(2sin⁡2x+1)=0\cos 2x (2\sin 2x + 1) = 0

    This gives two separate equations:

    1. cos⁡2x=0\cos 2x = 0
    2. 2sin⁡2x+1=0  ⟹  sin⁡2x=−122\sin 2x + 1 = 0 \implies \sin 2x = -\frac{1}{2}

  3. Solve for $2x$ in each case.

    For cos⁡2x=0\cos 2x = 0:
    The general solution for cos⁡A=0\cos A = 0 is A=π2+nπA = \frac{\pi}{2} + n\pi, where n∈Zn \in \mathbb{Z}.
    So, 2x=π2+nπ2x = \frac{\pi}{2} + n\pi

    x=π4+nπ2x = \frac{\pi}{4} + \frac{n\pi}{2}

    For sin⁡2x=−12\sin 2x = -\frac{1}{2}:
    The principal values for sin⁡A=−12\sin A = -\frac{1}{2} are A=−π6A = -\frac{\pi}{6} (or 7π6\frac{7\pi}{6}) and A=11π6A = \frac{11\pi}{6}.
    The general solution for $\sin

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