g) Use the truth table to show that the proportion is a tautology \( [p \Lambda(p \rightarrow q] \rightarrow q \). (4mks) h) Prove that \( \forall \mu \in \mathbb{N} n^{3}-n \) is divisible by 3 . i) Given that \( f(x)=3 x^{2}+4 \) find i. \( f(2) \) ii. \( f^{-1}(7) \)

a) To show that the proportion \( [p \wedge (p \rightarrow q)] \rightarrow q \) is a tautology using a truth table, we need to construct a truth table for the proposition and show that the proposition is always true regardless of the truth values of \( p \) and \( q \). Let's construct the truth table for the proposition \( [p \wedge (p \rightarrow q)] \rightarrow q \): | \( p \) | \( q \) | \( p \wedge (p \rightarrow q) \) | \( [p \wedge (p \rightarrow q)] \rightarrow q \) | |--------|--------|----------------------------------|---------------------------------------------| | T | T | T | T | | T | F | F | T | | F | T | F | T | | F | F | T | T | From the truth table, we can see that the proposition \( [p \wedge (p \rightarrow q)] \rightarrow q \) is always true, regardless of the truth values of \( p \) and \( q \). Therefore, the proportion is a tautology. b) To prove that \( \forall \mu \in \mathbb{N} n^{3}-n \) is divisible by 3, we need to show that for any natural number \( n \), \( n^{3}-n \) is divisible by 3. Let's prove this by mathematical induction: Base case: For \( n = 1 \), \( 1^{3}-1 = 0 \), which is divisible by 3. Inductive step: Assume that for some natural number \( k \), \( k^{3}-k \) is divisible by 3. We need to show that \( (k+1)^{3}-(k+1) \) is also divisible by 3. Expanding \( (k+1)^{3}-(k+1) \): \[ (k+1)^{3}-(k+1) = k^{3}+3k^{2}+3k+1-k-1 = k^{3}+3k^{2}+2k \] Since \( k^{3}+3k^{2}+2k \) is divisible by 3 by the inductive hypothesis, we have shown that \( (k+1)^{3}-(k+1) \) is divisible by 3. Therefore, by mathematical induction, \( \forall \mu \in \mathbb{N} n^{3}-n \) is divisible by 3. c) Given that \( f(x)=3x^{2}+4 \), we need to find: i. \( f(2) \) ii. \( f^{-1}(7) \) i. To find \( f(2) \), we substitute \( x = 2 \) into the function \( f(x) \): \[ f(2) = 3(2)^{2}+4 = 3(4)+4 = 12+4 = 16 \] ii. To find \( f^{-1}(7) \), we need to find the value of \( x \) such that \( f(x) = 7 \). This means solving the equation \( 3x^{2}+4 = 7 \) for \( x \). Solving the equation: \[ 3x^{2}+4 = 7 \] \[ 3x^{2} = 3 \] \[ x^{2} = 1 \] \[ x = \pm 1 \] Therefore, \( f^{-1}(7) = \pm 1 \).