QUESTION 1 1.1 Solve for $$ x $$ 1.1.1 $$ \quad x=\frac{4}{x}+3 $$ 1.1.2 $$ 2 x^{2}=-x+5 $$ ( Are the roots real or Unreal) 1.1.3 $$ \quad x^{2}-x+60 $$ 1.3 Simplify: $$ \frac{2^{x}-2^{-x}}{4^{x}-1} $$

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

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Sure, let's tackle each part of **Question 1** step by step. --- ## **1.1 Solve for $$ x $$** ### **1.1.1 $$ x = \frac{4}{x} + 3 $$** **Solution:** 1. **Multiply both sides by $$ x $$** to eliminate the denominator: $$ x^2 = 4 + 3x $$ 2. **Rearrange the equation** to form a quadratic equation: $$ x^2 - 3x - 4 = 0 $$ 3. **Solve the quadratic equation** using the quadratic formula $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$: $$ a = 1,\quad b = -3,\quad c = -4 $$ $$ x = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2} $$ 4. **Calculate the roots**: $$ x = \frac{3 + 5}{2} = 4 \quad \text{and} \quad x = \frac{3 - 5}{2} = -1 $$ **Answer:** $$ x = 4 $$ and $$ x = -1 $$ --- ### **1.1.2 $$ 2x^{2} = -x + 5 $$** **Determine if the roots are Real or Unreal.** **Solution:** 1. **Rearrange the equation** to standard quadratic form: $$ 2x^2 + x - 5 = 0 $$ 2. **Calculate the discriminant $$ D $$**: $$ D = b^2 - 4ac = 1^2 - 4 \times 2 \times (-5) = 1 + 40 = 41 $$ 3. **Analyze the discriminant**: - If $$ D > 0 $$: Two distinct real roots. - If $$ D = 0 $$: One real root (repeated). - If $$ D < 0 $$: Complex (unreal) roots. Here, $$ D = 41 > 0 $$, so **the roots are real and distinct**. **Answer:** The equation has two distinct real roots. --- ### **1.1.3 $$ x^{2} - x + 6 < 2(x + 2) $$** **Solution:** 1. **Expand the right side**: $$ x^2 - x + 6 < 2x + 4 $$ 2. **Bring all terms to the left side**: $$ x^2 - x + 6 - 2x - 4 < 0 \implies x^2 - 3x + 2 < 0 $$ 3. **Factor the quadratic inequality**: $$ x^2 - 3x + 2 = (x - 1)(x - 2) < 0 $$ 4. **Determine the intervals** where the product is negative: - **Roots:** $$ x = 1 $$ and $$ x = 2 $$ - **Test intervals:** - $$ x < 1 $$: Positive $$(+)\times(+)=+$$ - $$ 1 < x < 2 $$: Negative $$(+)\times(-)=-$$ - $$ x > 2 $$: Positive $$(+)\times(+)=+$$ **Solution:** $$ 1 < x < 2 $$ **Answer:** All real numbers $$ x $$ such that $$ 1 < x < 2 $$ --- ## **1.2 Analyze the Function $$ f(x) = \frac{\sqrt{x + 2}}{3 - 3x^{2}} $$** ### **1.2.2 $$ f(x) $$ is Non-Real** **Solution:** 1. **Determine when the numerator is non-real:** - $$ \sqrt{x + 2} $$ is real if $$ x + 2 \geq 0 $$ - **Non-real when:** $$ x + 2 < 0 \implies x < -2 $$ 2. **Denominator's Reality:** - The denominator $$ 3 - 3x^2 $$ is always real for real $$ x $$. **Answer:** $$ f(x) $$ is non-real for all real numbers $$ x $$ where $$ x < -2 $$ --- ### **1.2.3 $$ f(x) $$ is Undefined** **Solution:** 1. **Identify when the denominator is zero:** $$ 3 - 3x^2 = 0 \implies x^2 = 1 \implies x = \pm 1 $$ 2. **Additionally, ensure the numerator is defined:** - $$ \sqrt{x + 2} $$ requires $$ x \geq -2 $$ **Note:** When $$ x < -2 $$, the function is non-real, but undefined typically refers to points where the function cannot be evaluated within the real number system. **Answer:** $$ f(x) $$ is undefined at $$ x = 1 $$ and $$ x = -1 $$ --- ### **1.2.4 $$ f(x) > 0 $$** **Solution:** 1. **Condition for $$ f(x) > 0 $$:** $$ \frac{\sqrt{x + 2}}{3 - 3x^2} > 0 $$ - **Numerator $$ \sqrt{x + 2} > 0 $$:** $$ x > -2 $$ - **Denominator $$ 3 - 3x^2 > 0 $$:** $$ 3(1 - x^2) > 0 \implies 1 - x^2 > 0 \implies -1 < x < 1 $$ 2. **Combine the conditions:** $$ x > -2 \quad \text{and} \quad -1 < x < 1 \implies -1 < x < 1 $$ **Answer:** $$ f(x) > 0 $$ for all real numbers $$ x $$ such that $$ -1 < x < 1 $$ --- ## **1.3 Simplify:** $$ \frac{2^{x} - 2^{-x}}{4^{x} - 1} $$ **Solution:** 1. **Express all terms with base 2:** $$ 4^x = (2^2)^x = 2^{2x} $$ So, the expression becomes: $$ \frac{2^{x} - 2^{-x}}{2^{2x} - 1} $$ 2. **Simplify the numerator:** $$ 2^{x} - 2^{-x} = 2^{x} - \frac{1}{2^{x}} = \frac{2^{2x} - 1}{2^{x}} $$ 3. **Substitute back into the original expression:** $$ \frac{\frac{2^{2x} - 1}{2^{x}}}{2^{2x} - 1} = \frac{2^{2x} - 1}{2^{x}(2^{2x} - 1)} = \frac{1}{2^{x}} $$ 4. **Express the simplified form:** $$ \frac{1}{2^{x}} = 2^{-x} $$ **Answer:** The expression simplifies to $$ 2^{-x} $$ --- **Solutions:** 1. **1.1.1** $$ x = 4 $$ and $$ x = -1 $$ 2. **1.1.2** The roots are real and distinct. 3. **1.1.3** $$ 1 < x < 2 $$ 4. **1.2.2** $$ f(x) $$ is non-real for $$ x < -2 $$ 5. **1.2.3** $$ f(x) $$ is undefined at $$ x = 1 $$ and $$ x = -1 $$ 6. **1.2.4** $$ f(x) > 0 $$ for $$ -1 < x < 1 $$ 7. **1.3** $$ 2^{-x} $$

QUESTION 1
1.1 Solve for
1.1.1
1.1.2 ( Are the roots real or Unreal)
1.1.3
1.2 If , for which values of is
1.2.2 non real.
1.2.3 undefined
1.2.4
1.3 Simplify:

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QUESTION 1 1.1 Solve for 1.1.1 1.1.2 ( Are the roots real or Unreal) 1.1.3 1.2 If , for which values of is 1.2.2 non real. 1.2.3 undefined 1.2.4 1.3 Simplify: