Answer :
Sure, let's address each problem with a detailed, step-by-step solution.
Problem 6: Find the value of [tex]\( n \)[/tex] such that the equation [tex]\( (5n + 1)x^2 - 8nx + 3n = 0 \)[/tex] is a perfect square.
For a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex] to be a perfect square, the discriminant [tex]\(\Delta\)[/tex] must be zero.
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For the equation [tex]\( (5n + 1)x^2 - 8nx + 3n = 0 \)[/tex], we can identify:
[tex]\[ a = 5n + 1, \quad b = -8n, \quad c = 3n \][/tex]
Thus, the discriminant [tex]\(\Delta\)[/tex] is:
[tex]\[ \Delta = (-8n)^2 - 4(5n + 1)(3n) \][/tex]
Calculating it step-by-step:
[tex]\[ \Delta = 64n^2 - 4 \cdot (5n + 1) \cdot 3n \][/tex]
[tex]\[ = 64n^2 - 4 \cdot (15n^2 + 3n) \][/tex]
[tex]\[ = 64n^2 - 60n^2 - 12n \][/tex]
[tex]\[ = 4n^2 - 12n \][/tex]
For the quadratic to be a perfect square, [tex]\(\Delta = 0\)[/tex]:
[tex]\[ 4n^2 - 12n = 0 \][/tex]
Factor the left-hand side:
[tex]\[ 4n(n - 3) = 0 \][/tex]
This gives two solutions:
[tex]\[ n = 0 \quad \text{or} \quad n = 3 \][/tex]
However, [tex]\( n = 0 \)[/tex] would make the equation trivial. Thus, the non-trivial solution is:
[tex]\[ n = 3 \][/tex]
Problem 7: Prove that [tex]\( \cosh^2 x = 1 + 2\sinh^2 x \)[/tex]
Recall the definitions of hyperbolic functions:
[tex]\[ \cosh x = \frac{e^x + e^{-x}}{2}, \quad \sinh x = \frac{e^x - e^{-x}}{2} \][/tex]
First, express [tex]\(\cosh^2 x\)[/tex] in terms of [tex]\(e^x\)[/tex] and [tex]\(e^{-x}\)[/tex]:
[tex]\[ \cosh^2 x = \left( \frac{e^x + e^{-x}}{2} \right)^2 \][/tex]
[tex]\[ = \frac{(e^x + e^{-x})^2}{4} \][/tex]
[tex]\[ = \frac{e^{2x} + 2e^0 + e^{-2x}}{4} \][/tex]
[tex]\[ = \frac{e^{2x} + 2 + e^{-2x}}{4} \][/tex]
Now, express [tex]\(\sinh^2 x\)[/tex]:
[tex]\[ \sinh^2 x = \left( \frac{e^x - e^{-x}}{2} \right)^2 \][/tex]
[tex]\[ = \frac{(e^x - e^{-x})^2}{4} \][/tex]
[tex]\[ = \frac{e^{2x} - 2e^0 + e^{-2x}}{4} \][/tex]
[tex]\[ = \frac{e^{2x} - 2 + e^{-2x}}{4} \][/tex]
Multiply [tex]\(\sinh^2 x\)[/tex] by 2:
[tex]\[ 2\sinh^2 x = 2 \cdot \frac{e^{2x} - 2 + e^{-2x}}{4} \][/tex]
[tex]\[ = \frac{e^{2x} - 2 + e^{-2x}}{2} \][/tex]
Add 1 to both sides:
[tex]\[ 1 + 2\sinh^2 x = 1 + \frac{e^{2x} - 2 + e^{-2x}}{2} \][/tex]
[tex]\[ = \frac{2 + e^{2x} - 2 + e^{-2x}}{2} \][/tex]
[tex]\[ = \frac{e^{2x} + 2 + e^{-2x}}{2} \][/tex]
Thus,
[tex]\[ 1 + 2\sinh^2 x = \cosh^2 x \][/tex]
Problem 8: Prove that [tex]\(\sqrt{5}\)[/tex] is irrational
To prove that [tex]\(\sqrt{5}\)[/tex] is irrational, we use a proof by contradiction.
Assume, for the sake of contradiction, that [tex]\(\sqrt{5}\)[/tex] is rational. This means that [tex]\(\sqrt{5}\)[/tex] can be expressed as a fraction of two integers [tex]\( \frac{a}{b} \)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are integers with no common factors (i.e., the fraction is in its simplest form), and [tex]\(b \neq 0\)[/tex].
[tex]\[ \sqrt{5} = \frac{a}{b} \][/tex]
Squaring both sides:
[tex]\[ 5 = \frac{a^2}{b^2} \][/tex]
Rearranging gives:
[tex]\[ a^2 = 5b^2 \][/tex]
This equation implies that [tex]\(a^2\)[/tex] is divisible by 5. Therefore, [tex]\(a\)[/tex] must also be divisible by 5 (because 5 is a prime number, if [tex]\(a^2\)[/tex] is divisible by 5, then [tex]\(a\)[/tex] must be).
Let [tex]\( a = 5k \)[/tex] for some integer [tex]\( k \)[/tex].
Substitute back into the equation:
[tex]\[ (5k)^2 = 5b^2 \][/tex]
[tex]\[ 25k^2 = 5b^2 \][/tex]
[tex]\[ 5k^2 = b^2 \][/tex]
This implies that [tex]\(b^2\)[/tex] is also divisible by 5, and hence [tex]\(b\)[/tex] must be divisible by 5.
Since both [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are divisible by 5, they have a common factor of 5, which contradicts our initial assumption that [tex]\( a \)[/tex] and [tex]\( b \)[/tex] have no common factors (i.e., the fraction [tex]\( \frac{a}{b} \)[/tex] is in simplest form). Therefore, our initial assumption that [tex]\(\sqrt{5}\)[/tex] is rational must be false.
Thus, [tex]\(\sqrt{5}\)[/tex] is irrational.
Problem 6: Find the value of [tex]\( n \)[/tex] such that the equation [tex]\( (5n + 1)x^2 - 8nx + 3n = 0 \)[/tex] is a perfect square.
For a quadratic equation of the form [tex]\( ax^2 + bx + c = 0 \)[/tex] to be a perfect square, the discriminant [tex]\(\Delta\)[/tex] must be zero.
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
For the equation [tex]\( (5n + 1)x^2 - 8nx + 3n = 0 \)[/tex], we can identify:
[tex]\[ a = 5n + 1, \quad b = -8n, \quad c = 3n \][/tex]
Thus, the discriminant [tex]\(\Delta\)[/tex] is:
[tex]\[ \Delta = (-8n)^2 - 4(5n + 1)(3n) \][/tex]
Calculating it step-by-step:
[tex]\[ \Delta = 64n^2 - 4 \cdot (5n + 1) \cdot 3n \][/tex]
[tex]\[ = 64n^2 - 4 \cdot (15n^2 + 3n) \][/tex]
[tex]\[ = 64n^2 - 60n^2 - 12n \][/tex]
[tex]\[ = 4n^2 - 12n \][/tex]
For the quadratic to be a perfect square, [tex]\(\Delta = 0\)[/tex]:
[tex]\[ 4n^2 - 12n = 0 \][/tex]
Factor the left-hand side:
[tex]\[ 4n(n - 3) = 0 \][/tex]
This gives two solutions:
[tex]\[ n = 0 \quad \text{or} \quad n = 3 \][/tex]
However, [tex]\( n = 0 \)[/tex] would make the equation trivial. Thus, the non-trivial solution is:
[tex]\[ n = 3 \][/tex]
Problem 7: Prove that [tex]\( \cosh^2 x = 1 + 2\sinh^2 x \)[/tex]
Recall the definitions of hyperbolic functions:
[tex]\[ \cosh x = \frac{e^x + e^{-x}}{2}, \quad \sinh x = \frac{e^x - e^{-x}}{2} \][/tex]
First, express [tex]\(\cosh^2 x\)[/tex] in terms of [tex]\(e^x\)[/tex] and [tex]\(e^{-x}\)[/tex]:
[tex]\[ \cosh^2 x = \left( \frac{e^x + e^{-x}}{2} \right)^2 \][/tex]
[tex]\[ = \frac{(e^x + e^{-x})^2}{4} \][/tex]
[tex]\[ = \frac{e^{2x} + 2e^0 + e^{-2x}}{4} \][/tex]
[tex]\[ = \frac{e^{2x} + 2 + e^{-2x}}{4} \][/tex]
Now, express [tex]\(\sinh^2 x\)[/tex]:
[tex]\[ \sinh^2 x = \left( \frac{e^x - e^{-x}}{2} \right)^2 \][/tex]
[tex]\[ = \frac{(e^x - e^{-x})^2}{4} \][/tex]
[tex]\[ = \frac{e^{2x} - 2e^0 + e^{-2x}}{4} \][/tex]
[tex]\[ = \frac{e^{2x} - 2 + e^{-2x}}{4} \][/tex]
Multiply [tex]\(\sinh^2 x\)[/tex] by 2:
[tex]\[ 2\sinh^2 x = 2 \cdot \frac{e^{2x} - 2 + e^{-2x}}{4} \][/tex]
[tex]\[ = \frac{e^{2x} - 2 + e^{-2x}}{2} \][/tex]
Add 1 to both sides:
[tex]\[ 1 + 2\sinh^2 x = 1 + \frac{e^{2x} - 2 + e^{-2x}}{2} \][/tex]
[tex]\[ = \frac{2 + e^{2x} - 2 + e^{-2x}}{2} \][/tex]
[tex]\[ = \frac{e^{2x} + 2 + e^{-2x}}{2} \][/tex]
Thus,
[tex]\[ 1 + 2\sinh^2 x = \cosh^2 x \][/tex]
Problem 8: Prove that [tex]\(\sqrt{5}\)[/tex] is irrational
To prove that [tex]\(\sqrt{5}\)[/tex] is irrational, we use a proof by contradiction.
Assume, for the sake of contradiction, that [tex]\(\sqrt{5}\)[/tex] is rational. This means that [tex]\(\sqrt{5}\)[/tex] can be expressed as a fraction of two integers [tex]\( \frac{a}{b} \)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are integers with no common factors (i.e., the fraction is in its simplest form), and [tex]\(b \neq 0\)[/tex].
[tex]\[ \sqrt{5} = \frac{a}{b} \][/tex]
Squaring both sides:
[tex]\[ 5 = \frac{a^2}{b^2} \][/tex]
Rearranging gives:
[tex]\[ a^2 = 5b^2 \][/tex]
This equation implies that [tex]\(a^2\)[/tex] is divisible by 5. Therefore, [tex]\(a\)[/tex] must also be divisible by 5 (because 5 is a prime number, if [tex]\(a^2\)[/tex] is divisible by 5, then [tex]\(a\)[/tex] must be).
Let [tex]\( a = 5k \)[/tex] for some integer [tex]\( k \)[/tex].
Substitute back into the equation:
[tex]\[ (5k)^2 = 5b^2 \][/tex]
[tex]\[ 25k^2 = 5b^2 \][/tex]
[tex]\[ 5k^2 = b^2 \][/tex]
This implies that [tex]\(b^2\)[/tex] is also divisible by 5, and hence [tex]\(b\)[/tex] must be divisible by 5.
Since both [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are divisible by 5, they have a common factor of 5, which contradicts our initial assumption that [tex]\( a \)[/tex] and [tex]\( b \)[/tex] have no common factors (i.e., the fraction [tex]\( \frac{a}{b} \)[/tex] is in simplest form). Therefore, our initial assumption that [tex]\(\sqrt{5}\)[/tex] is rational must be false.
Thus, [tex]\(\sqrt{5}\)[/tex] is irrational.
