Answer :
### Solution
Let's start by addressing each part of the problem systematically.
1. Binary Operation [tex]\(\)[/tex] on Real Numbers [tex]\(R\)[/tex]
The operation defined on the set of real numbers is [tex]\( m n = m + n + 2 \)[/tex].
(a) Identity Element under the Operation :
The identity element [tex]\(e\)[/tex] in an operation [tex]\( \)[/tex] must satisfy the condition that for all elements [tex]\(a\)[/tex] in the set, [tex]\(a e = e a = a\)[/tex].
Let's find [tex]\(e\)[/tex]:
[tex]\[ m * e = m + e + 2 = m \][/tex]
[tex]\[ e + 2 = 0 \][/tex]
[tex]\[ e = -2 \][/tex]
Therefore, the identity element is [tex]\(e = -2\)[/tex].
(b) Inverse of [tex]\( n \)[/tex] under the Operation :
The inverse of an element [tex]\(n\)[/tex] under the operation [tex]\( \)[/tex] must satisfy the condition: [tex]\(n * n^{-1} = -2\)[/tex], where [tex]\(-2\)[/tex] is the identity element.
Let's denote [tex]\(n^{-1}\)[/tex] as [tex]\(m\)[/tex] and solve for [tex]\(m\)[/tex]:
[tex]\[ n * m = n + m + 2 = -2 \][/tex]
[tex]\[ m = -2 - n - 2 \][/tex]
[tex]\[ m = - n - 4 \][/tex]
Therefore, the inverse of [tex]\( n \)[/tex] under the operation [tex]\( * \)[/tex] is [tex]\( - n - 4 \)[/tex].
2. Given that (5, 2), (-4, k), and (2, 1) lie on a straight line, find the value of [tex]\( k \)[/tex]
Points [tex]\((5, 2)\)[/tex], [tex]\((-4, k)\)[/tex], and [tex]\((2, 1)\)[/tex] lie on a straight line, so they must have the same slope between each pair of points.
Slope between [tex]\((5, 2)\)[/tex] and [tex]\((2, 1)\)[/tex]:
[tex]\[ \text{slope}_1 = \frac{1 - 2}{2 - 5} = \frac{-1}{-3} = \frac{1}{3} \][/tex]
Slope between [tex]\((5, 2)\)[/tex] and [tex]\((-4, k)\)[/tex]:
[tex]\[ \text{slope}_2 = \frac{k - 2}{-4 - 5} = \frac{k - 2}{-9} \][/tex]
Since the slopes are equal:
[tex]\[ \frac{1}{3} = \frac{k - 2}{-9} \][/tex]
[tex]\[ 1 \cdot -9 = 3(k - 2) \][/tex]
[tex]\[ -9 = 3k - 6 \][/tex]
[tex]\[ 3k - 6 = -9 \][/tex]
[tex]\[ 3k = -3 \][/tex]
[tex]\[ k = -1 \][/tex]
Therefore, [tex]\(k = -1\)[/tex].
3. If [tex]\( f(x + 2) = 6x^3 + 5x - 8 \)[/tex], find [tex]\( f(5) \)[/tex]
To find [tex]\( f \)[/tex], set [tex]\( x + 2 = 5 \)[/tex]:
[tex]\[ x = 3 \][/tex]
Then:
[tex]\[ f(5) = 6(3)^3 + 5(3) - 8 \][/tex]
[tex]\[ = 6 \cdot 27 + 15 - 8 \][/tex]
[tex]\[ = 162 + 15 - 8 \][/tex]
[tex]\[ = 169 \][/tex]
Therefore, [tex]\( f(5) = 169 \)[/tex].
4. Express [tex]\(\frac{7\sqrt{2} + 3\sqrt{3}}{4\sqrt{2} - 2\sqrt{3}}\)[/tex] in the form [tex]\( p + q\sqrt{r} \)[/tex]
Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator [tex]\(4\sqrt{2} + 2\sqrt{3}\)[/tex].
[tex]\[ \frac{(7\sqrt{2} + 3\sqrt{3})(4\sqrt{2} + 2\sqrt{3})}{(4\sqrt{2} - 2\sqrt{3})(4\sqrt{2} + 2\sqrt{3})} \][/tex]
Calculate the numerator:
[tex]\[ (7\sqrt{2})(4\sqrt{2}) + (7\sqrt{2})(2\sqrt{3}) + (3\sqrt{3})(4\sqrt{2}) + (3\sqrt{3})(2\sqrt{3}) \][/tex]
[tex]\[ = 28 \cdot 2 + 14\sqrt{6} + 12\sqrt{6} + 6 \cdot 3 \][/tex]
[tex]\[ = 56 + 26\sqrt{6} + 18 \][/tex]
[tex]\[ = 74 + 26\sqrt{6} \][/tex]
Calculate the denominator:
[tex]\[ (4\sqrt{2})^2 - (2\sqrt{3})^2 \][/tex]
[tex]\[ = 16 \cdot 2 - 4 \cdot 3 \][/tex]
[tex]\[ = 32 - 12 \][/tex]
[tex]\[ = 20 \][/tex]
Therefore:
[tex]\[ \frac{7\sqrt{2} + 3\sqrt{3}}{4\sqrt{2} - 2\sqrt{3}} = \frac{74 + 26\sqrt{6}}{20} = \frac{74}{20} + \frac{26\sqrt{6}}{20} \][/tex]
[tex]\[ = 3.7 + 1.3\sqrt{6} \][/tex]
Hence, the expression in the required form is [tex]\( 3.7 + 1.3\sqrt{6} \)[/tex].
5. Finding [tex]\( q \)[/tex] in terms of [tex]\( r \)[/tex] and [tex]\( p \)[/tex] for [tex]\( p = \binom{5}{3}, q = \binom{-1}{2}, r = \binom{17}{5} \)[/tex] and [tex]\( r = \alpha p + \beta q \)[/tex]
Given the known values:
[tex]\[ p = \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3! \cdot 2!} = \frac{120}{6 \cdot 2} = 10 \][/tex]
[tex]\[ q = \binom{-1}{2} = \frac{(-1)!}{2!(-3)!} \quad \text{(undefined for negative factorial)} \][/tex]
[tex]\[ r = \binom{17}{5} = \frac{17!}{5!(17-5)!} = \frac{17!}{5! \cdot 12!} = 6188 \][/tex]
Rearranging the given equation [tex]\( r = \alpha p + \beta q \)[/tex], we solve for [tex]\( q \)[/tex]:
[tex]\[ \beta q = r - \alpha p \][/tex]
[tex]\[ q = \frac{r - \alpha p}{\beta} \][/tex]
Therefore, [tex]\( q \)[/tex] in terms of [tex]\( r \)[/tex] and [tex]\( p \)[/tex] is:
[tex]\[ \boxed{q = \frac{r - \alpha p}{\beta}} \][/tex]
### Summary of Calculations
- Identity Element: -2
- Inverse of [tex]\( n \)[/tex]: [tex]\( - n - 4 \)[/tex]
- [tex]\( k = -1 \)[/tex]
- [tex]\( f(5) = 169 \)[/tex]
- [tex]\( \frac{7\sqrt{2} + 3\sqrt{3}}{4\sqrt{2} - 2\sqrt{3}} = 3.7 + 1.3\sqrt{6} \)[/tex]
- [tex]\( q = \frac{r - \alpha p}{\beta} \)[/tex]
Let's start by addressing each part of the problem systematically.
1. Binary Operation [tex]\(\)[/tex] on Real Numbers [tex]\(R\)[/tex]
The operation defined on the set of real numbers is [tex]\( m n = m + n + 2 \)[/tex].
(a) Identity Element under the Operation :
The identity element [tex]\(e\)[/tex] in an operation [tex]\( \)[/tex] must satisfy the condition that for all elements [tex]\(a\)[/tex] in the set, [tex]\(a e = e a = a\)[/tex].
Let's find [tex]\(e\)[/tex]:
[tex]\[ m * e = m + e + 2 = m \][/tex]
[tex]\[ e + 2 = 0 \][/tex]
[tex]\[ e = -2 \][/tex]
Therefore, the identity element is [tex]\(e = -2\)[/tex].
(b) Inverse of [tex]\( n \)[/tex] under the Operation :
The inverse of an element [tex]\(n\)[/tex] under the operation [tex]\( \)[/tex] must satisfy the condition: [tex]\(n * n^{-1} = -2\)[/tex], where [tex]\(-2\)[/tex] is the identity element.
Let's denote [tex]\(n^{-1}\)[/tex] as [tex]\(m\)[/tex] and solve for [tex]\(m\)[/tex]:
[tex]\[ n * m = n + m + 2 = -2 \][/tex]
[tex]\[ m = -2 - n - 2 \][/tex]
[tex]\[ m = - n - 4 \][/tex]
Therefore, the inverse of [tex]\( n \)[/tex] under the operation [tex]\( * \)[/tex] is [tex]\( - n - 4 \)[/tex].
2. Given that (5, 2), (-4, k), and (2, 1) lie on a straight line, find the value of [tex]\( k \)[/tex]
Points [tex]\((5, 2)\)[/tex], [tex]\((-4, k)\)[/tex], and [tex]\((2, 1)\)[/tex] lie on a straight line, so they must have the same slope between each pair of points.
Slope between [tex]\((5, 2)\)[/tex] and [tex]\((2, 1)\)[/tex]:
[tex]\[ \text{slope}_1 = \frac{1 - 2}{2 - 5} = \frac{-1}{-3} = \frac{1}{3} \][/tex]
Slope between [tex]\((5, 2)\)[/tex] and [tex]\((-4, k)\)[/tex]:
[tex]\[ \text{slope}_2 = \frac{k - 2}{-4 - 5} = \frac{k - 2}{-9} \][/tex]
Since the slopes are equal:
[tex]\[ \frac{1}{3} = \frac{k - 2}{-9} \][/tex]
[tex]\[ 1 \cdot -9 = 3(k - 2) \][/tex]
[tex]\[ -9 = 3k - 6 \][/tex]
[tex]\[ 3k - 6 = -9 \][/tex]
[tex]\[ 3k = -3 \][/tex]
[tex]\[ k = -1 \][/tex]
Therefore, [tex]\(k = -1\)[/tex].
3. If [tex]\( f(x + 2) = 6x^3 + 5x - 8 \)[/tex], find [tex]\( f(5) \)[/tex]
To find [tex]\( f \)[/tex], set [tex]\( x + 2 = 5 \)[/tex]:
[tex]\[ x = 3 \][/tex]
Then:
[tex]\[ f(5) = 6(3)^3 + 5(3) - 8 \][/tex]
[tex]\[ = 6 \cdot 27 + 15 - 8 \][/tex]
[tex]\[ = 162 + 15 - 8 \][/tex]
[tex]\[ = 169 \][/tex]
Therefore, [tex]\( f(5) = 169 \)[/tex].
4. Express [tex]\(\frac{7\sqrt{2} + 3\sqrt{3}}{4\sqrt{2} - 2\sqrt{3}}\)[/tex] in the form [tex]\( p + q\sqrt{r} \)[/tex]
Rationalize the denominator by multiplying the numerator and the denominator by the conjugate of the denominator [tex]\(4\sqrt{2} + 2\sqrt{3}\)[/tex].
[tex]\[ \frac{(7\sqrt{2} + 3\sqrt{3})(4\sqrt{2} + 2\sqrt{3})}{(4\sqrt{2} - 2\sqrt{3})(4\sqrt{2} + 2\sqrt{3})} \][/tex]
Calculate the numerator:
[tex]\[ (7\sqrt{2})(4\sqrt{2}) + (7\sqrt{2})(2\sqrt{3}) + (3\sqrt{3})(4\sqrt{2}) + (3\sqrt{3})(2\sqrt{3}) \][/tex]
[tex]\[ = 28 \cdot 2 + 14\sqrt{6} + 12\sqrt{6} + 6 \cdot 3 \][/tex]
[tex]\[ = 56 + 26\sqrt{6} + 18 \][/tex]
[tex]\[ = 74 + 26\sqrt{6} \][/tex]
Calculate the denominator:
[tex]\[ (4\sqrt{2})^2 - (2\sqrt{3})^2 \][/tex]
[tex]\[ = 16 \cdot 2 - 4 \cdot 3 \][/tex]
[tex]\[ = 32 - 12 \][/tex]
[tex]\[ = 20 \][/tex]
Therefore:
[tex]\[ \frac{7\sqrt{2} + 3\sqrt{3}}{4\sqrt{2} - 2\sqrt{3}} = \frac{74 + 26\sqrt{6}}{20} = \frac{74}{20} + \frac{26\sqrt{6}}{20} \][/tex]
[tex]\[ = 3.7 + 1.3\sqrt{6} \][/tex]
Hence, the expression in the required form is [tex]\( 3.7 + 1.3\sqrt{6} \)[/tex].
5. Finding [tex]\( q \)[/tex] in terms of [tex]\( r \)[/tex] and [tex]\( p \)[/tex] for [tex]\( p = \binom{5}{3}, q = \binom{-1}{2}, r = \binom{17}{5} \)[/tex] and [tex]\( r = \alpha p + \beta q \)[/tex]
Given the known values:
[tex]\[ p = \binom{5}{3} = \frac{5!}{3!(5-3)!} = \frac{5!}{3! \cdot 2!} = \frac{120}{6 \cdot 2} = 10 \][/tex]
[tex]\[ q = \binom{-1}{2} = \frac{(-1)!}{2!(-3)!} \quad \text{(undefined for negative factorial)} \][/tex]
[tex]\[ r = \binom{17}{5} = \frac{17!}{5!(17-5)!} = \frac{17!}{5! \cdot 12!} = 6188 \][/tex]
Rearranging the given equation [tex]\( r = \alpha p + \beta q \)[/tex], we solve for [tex]\( q \)[/tex]:
[tex]\[ \beta q = r - \alpha p \][/tex]
[tex]\[ q = \frac{r - \alpha p}{\beta} \][/tex]
Therefore, [tex]\( q \)[/tex] in terms of [tex]\( r \)[/tex] and [tex]\( p \)[/tex] is:
[tex]\[ \boxed{q = \frac{r - \alpha p}{\beta}} \][/tex]
### Summary of Calculations
- Identity Element: -2
- Inverse of [tex]\( n \)[/tex]: [tex]\( - n - 4 \)[/tex]
- [tex]\( k = -1 \)[/tex]
- [tex]\( f(5) = 169 \)[/tex]
- [tex]\( \frac{7\sqrt{2} + 3\sqrt{3}}{4\sqrt{2} - 2\sqrt{3}} = 3.7 + 1.3\sqrt{6} \)[/tex]
- [tex]\( q = \frac{r - \alpha p}{\beta} \)[/tex]
