FPM(2023) January Paper 02R Q2 and solution

Problem and Solution

Find the set of values of $x$ for which:

1. $3x - 8 \lt 5x + 3$
2. $4x^2 - 7x + 1 > 6 - 2x^2$
3. Both $3x - 8 < 5x + 3$ and $4x^2 - 7x + 1 > 6 - 2x^2$

Part (a): Solve $3x - 8 \lt 5x + 3$

Step 1: Rearrange the inequality:

$$ 3x - 8 \lt 5x + 3 \implies -8 - 3 \lt 5x - 3x \implies -11 \lt 2x $$

Step 2: Solve for $x$:

$$ x > -\frac{11}{2} $$

The solution to part (a) is:

$$ x > -\frac{11}{2} $$

Part (b): Solve $4x^2 - 7x + 1 > 6 - 2x^2$

Step 1: Rearrange the inequality:

$$ 4x^2 - 7x + 1 > 6 - 2x^2 $$ $$ \implies 4x^2 + 2x^2 - 7x + 1 - 6 > 0 $$ $$ \implies 6x^2 - 7x - 5 > 0 $$

Step 2: Solve the quadratic inequality:

Let $$ f(x) = 6x^2 - 7x - 5 $$ Solve $$ 6x^2 - 7x - 5 = 0 $$ using the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 6,\; b = -7,\; c = -5 $$

$$ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(6)(-5)}}{2(6)} = \frac{7 \pm \sqrt{49 + 120}}{12} = \frac{7 \pm \sqrt{169}}{12} = \frac{7 \pm 13}{12} $$

Step 3: Find the roots:

$$ x = \frac{7 + 13}{12} = \frac{20}{12} = \frac{5}{3} $$ $$ x = \frac{7 - 13}{12} = \frac{-6}{12} = -\frac{1}{2} $$

Step 4: Test intervals:

The roots divide the $x$-axis into three intervals:

$$ x \lt -\frac{1}{2}, \quad -\frac{1}{2} \lt x \lt \frac{5}{3}, \quad x > \frac{5}{3} $$

Test a point from each interval in $$ f(x) = 6x^2 - 7x - 5 $$

• For $x = -1$: $$ f(-1) = 6(-1)^2 - 7(-1) - 5 = 6 + 7 - 5 = 8 > 0 $$
• For $x = 0$: $$ f(0) = 6(0)^2 - 7(0) - 5 = -5 \lt 0 $$
• For $x = 2$: $$ f(2) = 6(2)^2 - 7(2) - 5 = 24 - 14 - 5 = 5 > 0 $$

Step 5: Determine the solution:

The solution to $$ 6x^2 - 7x - 5 > 0 $$ is:

$$ x \lt -\frac{1}{2} \quad \mbox{or} \quad x > \frac{5}{3} $$

Part (c): Solve both inequalities simultaneously

Combine the solutions:

• From part (a): $$ x > -\frac{11}{2} $$
• From part (b): $$ x \lt -\frac{1}{2} \quad \mbox{or} \quad x > \frac{5}{3} $$

The intersection of these solutions is:

$$ -\frac{11}{2} \lt x \lt -\frac{1}{2} \quad \mbox{or} \quad x > \frac{5}{3} $$

Final Answer

1. $$ x > -\frac{11}{2} $$
2. $$ x \lt -\frac{1}{2} \quad \mbox{or} \quad x > \frac{5}{3} $$
3. $$ -\frac{11}{2} \lt x \lt -\frac{1}{2} \quad \mbox{or} \quad x > \frac{5}{3} $$