1. We are given a function $f: X \to Y$ with $X = \{a,b,c,d\}$ and $Y = \{1,2,3\}$, and the mappings $f(a) = 1$, $f(b) = 2$, $f(c) = 3$, $f(d) = 3$. 2. The problem asks to find values of $k$ such that $$ff^{-1}(k^2 - 8) = 2k.$$ Here, $f^{-1}$ denotes the inverse image under $f$, meaning for a value in $Y$, $f^{-1}$ gives the set of elements in $X$ mapped to it. 3. Since $f$ is not one-to-one (both $c$ and $d$ map to 3), $f^{-1}(y)$ for $y \in Y$ is a set of elements in $X$. 4. The expression $ff^{-1}(k^2 - 8)$ means applying $f$ to the set $f^{-1}(k^2 - 8)$. Since $f$ maps elements of $X$ to $Y$, applying $f$ to $f^{-1}(y)$ returns $y$ if $y \in Y$, because $f^{-1}(y)$ is the preimage of $y$. 5. Therefore, for $k^2 - 8 \in Y = \{1,2,3\}$, we have $$ff^{-1}(k^2 - 8) = k^2 - 8.$$ 6. The equation becomes $$k^2 - 8 = 2k.$$ 7. Rearranging gives $$k^2 - 2k - 8 = 0.$$ 8. Solve the quadratic equation using the quadratic formula $$k = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 1 \times (-8)}}{2} = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}.$$ 9. The two solutions are $$k = \frac{2 + 6}{2} = 4$$ and $$k = \frac{2 - 6}{2} = -2.$$ 10. Check if $k^2 - 8$ is in $Y$ for these values: - For $k=4$, $k^2 - 8 = 16 - 8 = 8$, which is not in $Y$. - For $k=-2$, $k^2 - 8 = 4 - 8 = -4$, which is not in $Y$. 11. Since $k^2 - 8$ must be in $Y = \{1,2,3\}$ for $ff^{-1}(k^2 - 8)$ to be defined, check these values: - If $k^2 - 8 = 1$, then $k^2 = 9$, so $k = \pm 3$. - If $k^2 - 8 = 2$, then $k^2 = 10$, so $k = \pm \sqrt{10}$. - If $k^2 - 8 = 3$, then $k^2 = 11$, so $k = \pm \sqrt{11}$. 12. Now check which of these satisfy the original equation $ff^{-1}(k^2 - 8) = 2k$: - For $k^2 - 8 = 1$, $ff^{-1}(1) = 1$, so $1 = 2k \Rightarrow k = \frac{1}{2}$, which contradicts $k = \pm 3$. - For $k^2 - 8 = 2$, $ff^{-1}(2) = 2$, so $2 = 2k \Rightarrow k = 1$, which contradicts $k = \pm \sqrt{10}$. - For $k^2 - 8 = 3$, $ff^{-1}(3) = 3$, so $3 = 2k \Rightarrow k = \frac{3}{2}$, which contradicts $k = \pm \sqrt{11}$. 13. Therefore, no $k$ satisfies both conditions simultaneously. **Final answer:** There are no values of $k$ such that $$ff^{-1}(k^2 - 8) = 2k$$ given the function $f$ and sets $X$ and $Y$ as defined.