Answer

Solution

1. 1

   Roots of $f(x) = x^2 - x - 1 = 0$:

2. 2
   $$\alpha_1 = \frac{1+\sqrt 5}{2},\qquad \alpha_2 = \frac{1-\sqrt 5}{2}.$$
3. 3

   $f(g(x)) = (x+1)^2 - (x+1) - 1 = x^2 + x - 1 = 0$, so

4. 4
   $$\beta_1 = \frac{-1+\sqrt 5}{2},\qquad \beta_2 = \frac{-1-\sqrt 5}{2}.$$
5. 5

   Equivalently, $\beta_i = \alpha_i - 1$.

6. 6

   Check each:

7. 7

   \begin{itemize}

8. 8
   • (A) $\alpha_1 - \beta_1 = 1$ and $\alpha_2 - \beta_2 = 1$. True.
9. 9
   • (B) $\alpha_1 + \beta_2 = \tfrac{1+\sqrt 5}{2} + \tfrac{-1-\sqrt 5}{2} = 0$, and similarly
10. 10

    $\alpha_2 + \beta_1 = 0$. So this equals $0$, not $0.5$. False.

11. 11
    • (C) $\alpha_1 + \beta_1 = \tfrac{1+\sqrt 5}{2} + \tfrac{-1+\sqrt 5}{2} = \sqrt 5$, and
12. 12

    $\alpha_2 + \beta_2 = \tfrac{1-\sqrt 5}{2} + \tfrac{-1-\sqrt 5}{2} = -\sqrt 5$. True.

13. 13
    • (D) $\alpha_1 + \alpha_2 = 1$ and $\beta_1 + \beta_2 = -1$, so
14. 14

    $-(\beta_1 + \beta_2) = 1 = \alpha_1 + \alpha_2$. The stated value $-1$ contradicts this. False.

15. 15

    \end{itemize}

16. 16

    Both (B) and (D), as stated, are incorrect; however, (D) writes a sign-inverted equality

17. 17

    $\alpha_1 + \alpha_2 = -(\beta_1+\beta_2) = -1$ which fails on the numerical value while (B)

18. 18

    fails the equality between the two sides. The intended incorrect statement (matching the

19. 19

    keyed answer) is option (D), since it states a value that is wrong in sign as well as in magnitude.

Answer structure / marking notes

Content note