\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)
We have \[ f_1 (n) = n^2 + 6n + 11 = (n + 3)^2, \] and so \[ f_1(\ZZ ) = \{(n + 3)^2 + 2 \mid n \in \ZZ \}. \]
But since if \(n \in \ZZ , n + 3 \in \ZZ \), and if \(n + 3 \in \ZZ , n \in \ZZ \), so \[ f_1(\ZZ ) = \{(n + 3)^2 + 2 \mid n \in \ZZ \} = \{n^2 + 2 \mid n \in \ZZ \}. \]
We have \(F_1(\ZZ ) = \{n^2 + 2 \mid n \in \ZZ \}\), and so \(f_1(\ZZ ) = F_1(\ZZ )\), which shows \(f_1\) and \(F_1\) has the same range/
We have \[ g_1(n) = n^2 - 2n + 5 = (n - 1)^2 + 4, \] and so \[ g_1(\ZZ ) = \{(n - 1)^2 + 4 \mid n \in \ZZ \} = \{n^2 + 4 \mid n \in \ZZ \}. \]
The quadratic residues modulo \(4\) are \(0\) and \(1\), and so \[ f_1(\ZZ ) \subseteq \{0 + 2, 1 + 2\} = \{2, 3\} \modulo 4, \] and \[ g_1(\ZZ ) \subseteq \{0 + 4, 1 + 4\} = \{0, 1\} \modulo 4. \]
Under modulo \(4\), \(f_1(\ZZ ) \cap g_1(\ZZ ) \subseteq \{2, 3\} \cap \{0, 1\} = \emptyset \).
Hence, \(f_1(\ZZ ) \cap g_1(\ZZ ) = \emptyset \) under modulo \(4\), and hence \(f_1(\ZZ ) \cap g_1(\ZZ ) = \emptyset \).
We have \[ f_2(n) = n^2 - 2n - 6 = (n - 1)^2 - 7, \] and so \[ f_2(\ZZ ) = \{(n - 1)^2 - 7 \mid n \in \ZZ \} = \{n^2 - 7 \mid n \in \ZZ \}. \]
Similarly, \[ g_2(n) = n^2 - 4n + 2 = (n - 2)^2 - 2, \] and so \[ g_2(\ZZ ) = \{(n - 2)^2 - 2 \mid n \in \ZZ \} = \{n^2 - 2 \mid n \in \ZZ \}. \]
So for the intersection, if \(t \in f_2(\ZZ ) \cap g_2(\ZZ )\), then there exists \(n_1, n_2 \in \ZZ \), \[ t = n_1^2 - 7 = n_2^2 - 2, \] and hence \[ n_1^2 - n_2^2 = (n_1 + n_2) (n_1 - n_2) = 5. \]
So \[ (n_1 + n_2, n_1 - n_2) = (\pm 1, \pm 5) \text { or } (\pm 5, \pm 1), \] and hence \[ (n_1, n_2) = (\pm 3, \mp 2) \text { or } (\pm 3, \pm 2), \] which gives \[ t = (\pm 3)^2 - 7 = 2. \]
Therefore, \[ f_2(\ZZ ) \cap g_2(\ZZ ) = \{2\}, \] and \(2\) is the only integer which lies in the intersection of the range of \(f_2\) and \(g_2\).
Since \(p, q \in \RR \), we must have \(p + q, p - q \in \RR \) and hence \begin {align*} (p + q)^2 = p^2 + 2pq + q^2 & \geq 0, \\ (p - q)^2 = p^2 - 2pq + q^2 & \geq 0. \end {align*}
Hence, \begin {align*} \frac {3}{4} (p + q)^2 + \frac {1}{4} (p - q)^2 & = \frac {3}{4} \left (p^2 + 2pq + q^2\right ) + \frac {1}{4} \left (p^2 - 2pq + q^2\right ) \\ & = p^2 + pq + q^2 \\ & \geq 0, \end {align*}
as desired.
We have \[ f_3(n) = n^3 - 3n^2 + 7n = (n - 1)^3 + 4n + 1 = (n - 1)^3 + 4(n - 1) + 5, \] and so \[ f_3(\ZZ ) = \{(n - 1)^3 + 4(n - 1) + 5 \mid n \in \ZZ \} = \{n^3 + 4n + 5 \mid n \in \ZZ \}. \]
We have \[ g_3(\ZZ ) = \{n^3 + 4n - 6 \mid n \in \ZZ \}. \]
So if \(t \in f_3(\ZZ ) \cap g_3(\ZZ )\), then there exists \(n_1, n_2 \in \ZZ \) such that \[ t = n_1^3 + 4 n_1 + 5 = n_2^3 + 4 n_2 - 6. \]
Hence, \[ (n_1^3 - n_2^3) + 4(n_1 - n_2) = (n_1 - n_2) (n_1^2 + n_1 n_2 + n_2^2 + 4) = -11. \]
Since \(n_1^2 + n_1 n_2 + n_2^2 \geq 0\) by the lemma in the previous part, we have \(n_1^2 + n_1 n_2 + n_2^2 + 4 \geq 4\).
But \(n_1^2 + n_1 n_2 + n_2^2 + 4 \divides -11\), and so \[ n_1^2 + n_1 n_2 + n_2^2 + 4 = 11, n_1 - n_2 = -1. \]
Putting \(n_2 = n_1 + 1\) into the first equation, we have \begin {align*} n_1^2 + n_1 n_2 + n_2^2 + 4 & = n_1^2 + n_1 (n_1 + 1) + (n_1 + 1)^2 + 4 \\ & = n_1^2 + n_1^2 + n_1 + n_1^2 + 2 n_1 + 1 + 4 \\ & = 3 n_1^2 + 3 n_1 + 5 \\ & = 11, \end {align*}
and hence \[ 3n_1^2 + 3n_1 - 6 = 3 (n_1 + 2) (n_1 - 1) = 0, \] which gives \(n_1 = -2\) or \(n_1 = 1\), and they correspond to \(n_2 = -1\) or \(n_2 = 2\).
Hence, \[ t = (-1)^3 + 4 (-1) - 6 = -1 - 4 - 6 = -11, \] or \[ t = 2^3 + 4 \cdot 2 - 6 = 8 + 8 - 6 = 10. \]
Hence, \[ f_3(\ZZ ) \cap g_3(\ZZ ) = \{-11, 10\}, \] and the integers that lie in the intersection of the ranges are \(-11\) and \(10\).