QUESTION 1 1.1 Solve for \( x \) \[ x=\frac{4}{x}+3 \] 1.1.1 \( \quad x=\frac{4}{x}+3 \) \[ \frac{x-\frac{4}{x}=3 \quad x^{2}+4-3}{1}=3 \] 1.1.2 \( 2 x^{2}=-x+5 \) (Are the roots real or Unreal) 1.1.3 \( x^{2}-x+6<2(x+2) \) \[ \begin{array}{ll} \text { T } x \\ 4+x^{2}=3 & \left(x^{2}-1\right) \\ x & (x+1)(x \end{array} \] 1.2 If \( f(x)=\frac{\sqrt{x+2}}{3-3 x^{2}} \), for which values of \( x \) is 1.2.2 \( \quad f(x) \) non real. 1.2.3 \( \quad f(x) \) undefined (4) 1.2.4 \( f(x)>0 \) (4) 1.3 Simplify: \[ \frac{2^{x}-2^{-x}}{4^{x}-1} \] [24]

Let's solve the problem step by step. ### 1.1 Solve for \( x \) #### 1.1.1 Solve the equation \( x = \frac{4}{x} + 3 \) To eliminate the fraction, we can multiply both sides by \( x \) (assuming \( x \neq 0 \)): \[ x^2 = 4 + 3x \] Rearranging gives us: \[ x^2 - 3x - 4 = 0 \] Now, we can solve this quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = -3, c = -4 \). Calculating the discriminant: \[ b^2 - 4ac = (-3)^2 - 4(1)(-4) = 9 + 16 = 25 \] Since the discriminant is positive, the roots are real. Now, substituting into the quadratic formula: \[ x = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2} \] This gives us two solutions: \[ x_1 = \frac{8}{2} = 4, \quad x_2 = \frac{-2}{2} = -1 \] #### 1.1.2 Check if the roots are real or unreal The roots we found are \( x = 4 \) and \( x = -1 \), both of which are real. #### 1.1.3 Solve the inequality \( x^2 - x + 6 < 2(x + 2) \) First, simplify the inequality: \[ x^2 - x + 6 < 2x + 4 \] Rearranging gives: \[ x^2 - 3x + 2 < 0 \] Factoring the quadratic: \[ (x - 1)(x - 2) < 0 \] The critical points are \( x = 1 \) and \( x = 2 \). We can test intervals to find where the product is negative: - For \( x < 1 \): Choose \( x = 0 \) → \( (0 - 1)(0 - 2) = 2 > 0 \) - For \( 1 < x < 2 \): Choose \( x = 1.5 \) → \( (1.5 - 1)(1.5 - 2) = -0.25 < 0 \) - For \( x > 2 \): Choose \( x = 3 \) → \( (3 - 1)(3 - 2) = 2 > 0 \) Thus, the solution to the inequality is: \[ 1 < x < 2 \] ### 1.2 Analyze the function \( f(x) = \frac{\sqrt{x+2}}{3 - 3x^2} \) #### 1.2.2 For which values of \( x \) is \( f(x) \) non-real? The function \( f(x) \) is non-real when the square root is negative: \[ x + 2 \geq 0 \implies x \geq -2 \] Thus, \( f(x) \) is real for \( x \geq -2 \). #### 1.2.3 For which values of \( x \) is \( f(x) \) undefined? The function is undefined when the denominator is zero: \[ 3 - 3x^2 = 0 \implies 3x^2 = 3 \implies x^2 = 1 \implies x = 1 \text{ or } x = -1 \] Thus, \( f(x) \) is undefined at \( x = 1 \) and \( x = -1 \). #### 1.2.4 For which values of \( x \) is \( f(x) > 0 \)? For \( f(x) > 0 \), both the numerator and denominator must be positive: 1. \( \sqrt{x + 2} > 0 \) implies \( x + 2 > 0 \) or \( x > -2 \). 2. \( 3 - 3x^2 > 0 \) implies \( 3x^2 < 3 \) or \( x^2 < 1 \) which gives \( -1 < x < 1 \). Combining these conditions: - From \( x > -2 \) and \( -1 < x < 1 \), we find: \[ -1 < x < 1 \] ### 1.3 Simplify \( \frac{2^x - 2^{-x}}{4^x - 1} \) We can rewrite \( 4^x \) as \( (2^2)^x = (2^x)^2 \): \[ \frac{2^x - 2^{-x}}{(2^x)^2 - 1} \] The denominator can be factored using the difference of squares: \[ (2^x - 1)(2^x + 1) \] Thus, we have: \[ \frac{2^x - 2^{-x}}{(2^x - 1)(2^x + 1)} \] Now, rewrite \( 2^{-x} \) as \( \frac{1}{2^x} \): \[ \frac{2^x - \frac{1}{2^x}}{(2^x - 1)(2^x + 1)} = \frac{2^{2x} - 1}{2^x(2^x - 1)(2^x + 1)} \] This simplifies to: \[ \frac{(2^x - 1)(2^x + 1)}{2^x(2^x - 1)(2^x + 1)} = \frac{1}{2^x} \] Thus, the final simplified expression is: \[ \frac{1}{2^x} \] ### Summary of Results 1.1.1 Solutions: \( x = 4, -1 \) (both real) 1.1.3 Inequality solution: \( 1 < x < 2 \) 1.2.2 Non-real values: \( x < -2 \) 1.2.3 Undefined values: \( x = 1, -1 \) 1.2.4 Positive values: \( -1 < x < 1 \) 1.3 Simplified expression: \( \frac{1}{2^x} \)