Equations and Inequalities — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for equations and inequalities.

UT-1: Rational Inequality with Sign Flipping

Question:

Solve the inequality:

$\frac{x^2 - 3x + 2}{x^2 + x - 6} \geq 0$

State your answer using set notation, clearly identifying all excluded values.

[Difficulty: hard. Tests the common error of cross-multiplying without considering sign changes of the denominator.]

Solution:

Step 1: Factorise numerator and denominator.

Numerator: $x^2 - 3x + 2 = (x-1)(x-2)$.

Denominator: $x^2 + x - 6 = (x+3)(x-2)$.

Step 2: Identify critical values and excluded points.

Critical values (where numerator or denominator is zero): $x = -3, 1, 2$.

Excluded value: $x = 2$ (makes denominator zero). Also $x = -3$ is excluded.

Step 3: Determine the sign of the expression in each interval.

$\frac{(x-1)(x-2)}{(x+3)(x-2)} = \frac{(x-1)(x-2)}{(x+3)(x-2)}$

For $x \neq 2$, we can cancel $(x-2)$ but must remember the sign changes. Instead, analyse using a sign table:

Interval	$(x+3)$	$(x-1)$	$(x-2)$	Expression
$x < -3$	$-$	$-$	$-$	$(-)(-)/(-)(-) = +/+$ but $(x-2)$ appears twice: overall $(-)(-)/((-)(-)) = +/+$

Let me be more careful. The expression is $\frac{(x-1)(x-2)}{(x+3)(x-2)}$.

Sign analysis:

$x$	$(x+3)$	$(x-1)$	$(x-2)$	Numerator	Denominator	Ratio
$x < -3$	$-$	$-$	$-$	$+$	$+$	$+$
$-3 < x < 1$	$+$	$-$	$-$	$+$	$-$	$-$
$1 < x < 2$	$+$	$+$	$-$	$-$	$-$	$+$
$x > 2$	$+$	$+$	$+$	$+$	$+$	$+$

Step 4: Include or exclude endpoints.

• $x = -3$: excluded (denominator zero)
• $x = 1$: included (numerator zero, expression equals zero, and $\geq 0$ is satisfied)
• $x = 2$: excluded (denominator zero, and the expression is undefined here)

Step 5: Assemble the solution.

The expression is non-negative when $x < -3$, $1 \leq x < 2$, or $x > 2$.

$x \in (-\infty, -3) \cup [1, 2) \cup (2, \infty)$

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UT-2: Modulus Function Equations with Multiple Cases

Question:

Solve the equation:

$|x^2 - 5x + 6| = |2x - 4|$

giving all real solutions in exact form.

[Difficulty: hard. Tests systematic case analysis for modulus equations with quadratics inside absolute values.]

Solution:

The equation $|A| = |B|$ is equivalent to $A = B$ or $A = -B$.

Case 1: $x^2 - 5x + 6 = 2x - 4$

$x^2 - 7x + 10 = 0$ $(x-2)(x-5) = 0$ $x = 2 \quad \text{or} \quad x = 5$

Case 2: $x^2 - 5x + 6 = -(2x - 4)$

$x^2 - 5x + 6 = -2x + 4$ $x^2 - 3x + 2 = 0$ $(x-1)(x-2) = 0$ $x = 1 \quad \text{or} \quad x = 2$

Verification:

• $x = 1$: $|1-5+6| = |2| = 2$, $|2-4| = 2$. Valid.
• $x = 2$: $|4-10+6| = 0$, $|4-4| = 0$. Valid.
• $x = 5$: $|25-25+6| = 6$, $|10-4| = 6$. Valid.

The solutions are $x = 1, 2, 5$.

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UT-3: System with No Real Solutions — Proof of Impossibility

Question:

Prove that the following system of equations has no real solutions:

$\begin{cases} x^2 + y^2 = 1 \\ x + y = 2 \end{cases}$

Then find the smallest positive value of $k$ such that the system:

$\begin{cases} x^2 + y^2 = 1 \\ x + y = k \end{cases}$

has exactly one real solution, and state that solution.

[Difficulty: hard. Tests algebraic proof of impossibility and understanding the geometric interpretation of constrained optimisation.]

Solution:

Proof of no real solutions for $k = 2$:

From the second equation: $y = 2 - x$. Substitute into the first:

$x^2 + (2-x)^2 = 1$ $x^2 + 4 - 4x + x^2 = 1$ $2x^2 - 4x + 3 = 0$

Discriminant: $\Delta = 16 - 24 = -8 < 0$.

Since $\Delta < 0$, there are no real values of $x$, hence no real solutions to the system.

Finding the critical value of $k$:

Substituting $y = k - x$ into $x^2 + y^2 = 1$:

$x^2 + (k-x)^2 = 1$ $2x^2 - 2kx + k^2 - 1 = 0$

For exactly one real solution, we need $\Delta = 0$:

$4k^2 - 8(k^2 - 1) = 0$ $4k^2 - 8k^2 + 8 = 0$ $-4k^2 + 8 = 0$ $k^2 = 2$ $k = \sqrt{2}$

(Since we want the smallest positive $k$, $k = \sqrt{2}$.)

The solution: When $\Delta = 0$:

$x = \frac{2k}{4} = \frac{k}{2} = \frac◆LB◆\sqrt{2}◆RB◆◆LB◆2◆RB◆$ $y = k - x = \sqrt{2} - \frac◆LB◆\sqrt{2}◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆\sqrt{2}◆RB◆◆LB◆2◆RB◆$

The single solution is $\left(\frac◆LB◆\sqrt{2}◆RB◆◆LB◆2◆RB◆, \frac◆LB◆\sqrt{2}◆RB◆◆LB◆2◆RB◆\right)$.

Geometric interpretation: The first equation is the unit circle and the second is the line $x + y = k$. The line is tangent to the circle when its distance from the origin equals the radius:

$\frac◆LB◆|0 + 0 - k|◆RB◆◆LB◆\sqrt{1^2 + 1^2}◆RB◆ = 1 \implies \frac◆LB◆k◆RB◆◆LB◆\sqrt{2}◆RB◆ = 1 \implies k = \sqrt{2}$

This confirms our algebraic result.

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Integration Tests

Tests synthesis of equations and inequalities with other topics. Requires combining concepts from multiple units.

IT-1: Solving $f(x) = f^{-1}(x)$ (with Functions)

Question:

The function $f$ is defined by $f(x) = \frac{2x+3}{x-1}$ for $x > 1$.

(a) Find $f^{-1}(x)$, stating its domain and range.

(b) Solve the equation $f(x) = f^{-1}(x)$, giving all solutions in the domain of $f$.

(c) Without further calculation, explain why $f(x) = f^{-1}(x)$ is equivalent to $f(x) = x$ for this particular function.

[Difficulty: hard. Combines inverse functions with equation solving and algebraic reasoning about symmetry.]

Solution:

(a) Let $y = \frac{2x+3}{x-1}$:

$y(x-1) = 2x + 3$ $xy - y = 2x + 3$ $x(y-2) = y + 3$ $x = \frac{y+3}{y-2}$

So $f^{-1}(x) = \frac{x+3}{x-2}$.

Domain of $f^{-1}$: Since the range of $f$ (for $x > 1$) needs to be determined first. As $x \to 1^+$, $f(x) \to +\infty$. As $x \to +\infty$, $f(x) \to 2$. So range of $f$ is $(2, +\infty)$, meaning domain of $f^{-1}$ is $x > 2$.

Range of $f^{-1}$: This equals the domain of $f$, so $(1, +\infty)$.

(b) Solve $\frac{2x+3}{x-1} = \frac{x+3}{x-2}$:

$(2x+3)(x-2) = (x+3)(x-1)$ $2x^2 - 4x + 3x - 6 = x^2 - x + 3x - 3$ $2x^2 - x - 6 = x^2 + 2x - 3$ $x^2 - 3x - 3 = 0$

By the quadratic formula:

$x = \frac◆LB◆3 \pm \sqrt{9 + 12}◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆3 \pm \sqrt{21}◆RB◆◆LB◆2◆RB◆$

Now check the domain restriction $x > 1$:

• $\frac◆LB◆3 + \sqrt{21}◆RB◆◆LB◆2◆RB◆ \approx \frac{3 + 4.58}{2} \approx 3.79 > 1$. Valid.
• $\frac◆LB◆3 - \sqrt{21}◆RB◆◆LB◆2◆RB◆ \approx \frac{3 - 4.58}{2} \approx -0.79 < 1$. Not in domain.

The solution is $x = \frac◆LB◆3 + \sqrt{21}◆RB◆◆LB◆2◆RB◆$.

(c) For a function and its inverse, the solutions to $f(x) = f^{-1}(x)$ lie on the line $y = x$ (the line of reflection). This is because $f(x) = f^{-1}(x)$ implies $f(f(x)) = x$, and if $f(x) = c$ then $f(c) = x$. When $f(x) = x$, clearly $f^{-1}(x) = x = f(x)$.

For this specific M"obius transformation, since $f$ is a strictly decreasing function on $(1, \infty)$ (its derivative $f'(x) = \frac{-5}{(x-1)^2} < 0$), the graph of $f$ crosses $y = x$ exactly once, and this crossing point is the unique solution to $f(x) = f^{-1}(x)$.

Verification: $f(x) = x$ gives $\frac{2x+3}{x-1} = x$, i.e. $2x+3 = x^2-x$, i.e. $x^2-3x-3 = 0$, which is the same equation we obtained in part (b).

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IT-2: Sum of Terms Satisfying an Inequality (with Sequences and Series)

Question:

The sequence $(a_n)$ is defined by $a_n = \frac{n^2 + n}{n + 2}$ for $n \geq 1$.

(a) Show that $a_n > n - 1$ for all $n \geq 1$.

(b) Find the smallest integer $N$ such that $a_n > 100$ for all $n \geq N$.

(c) Evaluate $\sum_{n=1}^{N-1} a_n$ exactly, where $N$ is the value found in part (b).

[Difficulty: hard. Combines inequality proof with sequence analysis and exact summation.]

Solution:

(a) We need to show $\frac{n^2 + n}{n+2} > n - 1$ for all $n \geq 1$.

Since $n + 2 > 0$ for all $n \geq 1$, we can multiply both sides by $n + 2$ without flipping the inequality:

$n^2 + n > (n-1)(n+2) = n^2 + 2n - n - 2 = n^2 + n - 2$

$n^2 + n > n^2 + n - 2$

$0 > -2$

This is always true. Therefore $a_n > n - 1$ for all $n \geq 1$.

(b) We need $\frac{n^2+n}{n+2} > 100$:

$n^2 + n > 100n + 200$ $n^2 - 99n - 200 > 0$

Roots of $n^2 - 99n - 200 = 0$:

$n = \frac◆LB◆99 \pm \sqrt{9801 + 800}◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆99 \pm \sqrt{10601}◆RB◆◆LB◆2◆RB◆$

$\sqrt{10601} \approx 102.96$, so:

$n \approx \frac{99 + 102.96}{2} \approx 100.98$

Since the quadratic opens upward, $n^2 - 99n - 200 > 0$ for $n > \frac◆LB◆99+\sqrt{10601}◆RB◆◆LB◆2◆RB◆ \approx 100.98$.

The smallest integer $N$ is $\boxed{101}$.

(c) $N = 101$, so we need $\sum_{n=1}^{100} a_n = \sum_{n=1}^{100} \frac{n^2+n}{n+2}$.

Perform the division $\frac{n^2+n}{n+2}$:

$\frac{n^2+n}{n+2} = n - 1 + \frac{2}{n+2}$

Verify: $(n-1)(n+2) + 2 = n^2 + 2n - n - 2 + 2 = n^2 + n$. Confirmed.

Therefore:

$\sum_{n=1}^{100} a_n = \sum_{n=1}^{100}\left(n - 1 + \frac{2}{n+2}\right) = \sum_{n=1}^{100}(n-1) + 2\sum_{n=1}^{100}\frac{1}{n+2}$

$= \sum_{k=0}^{99} k + 2\sum_{m=3}^{102}\frac{1}{m}$

where $k = n-1$ and $m = n+2$.

$= \frac◆LB◆99 \times 100◆RB◆◆LB◆2◆RB◆ + 2\sum_{m=3}^{102}\frac{1}{m}$

$= 4950 + 2\left(H_{102} - 1 - \frac{1}{2}\right)$

where $H_n = \sum_{k=1}^{n} \frac{1}{k}$ is the $n$-th harmonic number.

$= 4950 + 2H_{102} - 3$

$= 4947 + 2H_{102}$

This is the exact value in terms of the harmonic number $H_{102}$. Note that $H_{102}$ does not simplify to a closed form using elementary functions; this is the most precise exact answer.

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IT-3: Region Defined by Inequalities (with Coordinate Geometry)

Question:

A region $R$ in the $xy$-plane is defined by the inequalities:

$\begin{cases} y \geq x^2 - 4x + 3 \\ y \leq 4 - x^2 \\ y \geq |x - 2| - 1 \end{cases}$

(a) Find the coordinates of all vertices of $R$.

(b) Find the exact area of $R$.

[Difficulty: hard. Combines inequality regions with modulus functions and area calculation between curves.]

Solution:

(a) We find all intersection points of the boundary curves.

Curve 1: $y = x^2 - 4x + 3 = (x-1)(x-3)$ (upward parabola) Curve 2: $y = 4 - x^2$ (downward parabola) Curve 3: $y = |x-2| - 1$ (V-shape with vertex at $(2, -1)$)

Intersection of Curves 1 and 2:

$x^2 - 4x + 3 = 4 - x^2$ $2x^2 - 4x - 1 = 0$ $x = \frac◆LB◆4 \pm \sqrt{16 + 8}◆RB◆◆LB◆4◆RB◆ = \frac◆LB◆4 \pm \sqrt{24}◆RB◆◆LB◆4◆RB◆ = \frac◆LB◆4 \pm 2\sqrt{6}◆RB◆◆LB◆4◆RB◆ = \frac◆LB◆2 \pm \sqrt{6}◆RB◆◆LB◆2◆RB◆ = 1 \pm \frac◆LB◆\sqrt{6}◆RB◆◆LB◆2◆RB◆$

For $x = 1 + \sqrt{6}/2 \approx 2.225$: $y = 4 - (1+\sqrt{6}/2)^2 = 4 - (1+\sqrt{6}+3/2) = 4 - 5/2 - \sqrt{6} = 3/2 - \sqrt{6}$.

For $x = 1 - \sqrt{6}/2 \approx -0.225$: $y = 4 - (1-\sqrt{6}/2)^2 = 3/2 - \sqrt{6}$ (same by symmetry of the setup).

Intersection of Curve 1 and Curve 3:

For $x \geq 2$: $|x-2| = x-2$, so $x^2-4x+3 = x-3$, giving $x^2-5x+6 = 0$, so $x = 2$ or $x = 3$.

• At $x = 2$: $y = -1$.
• At $x = 3$: $y = 0$.

For $x < 2$: $|x-2| = 2-x$, so $x^2-4x+3 = 2-x-1 = 1-x$, giving $x^2-3x+2 = 0$, so $x = 1$ or $x = 2$.

• At $x = 1$: $y = 0$.

Intersection of Curve 2 and Curve 3:

For $x \geq 2$: $4-x^2 = x-3$, giving $x^2+x-7 = 0$, so $x = \frac◆LB◆-1+\sqrt{29}◆RB◆◆LB◆2◆RB◆ \approx 2.193$. $y = \frac◆LB◆-1+\sqrt{29}◆RB◆◆LB◆2◆RB◆ - 3 = \frac◆LB◆-7+\sqrt{29}◆RB◆◆LB◆2◆RB◆$.

For $x < 2$: $4-x^2 = 1-x$, giving $x^2-x-3 = 0$, so $x = \frac◆LB◆1+\sqrt{13}◆RB◆◆LB◆2◆RB◆ \approx 2.303$. But this is $> 2$, contradicting $x < 2$. So $x = \frac◆LB◆1-\sqrt{13}◆RB◆◆LB◆2◆RB◆ \approx -1.303$. $y = 1 - \frac◆LB◆1-\sqrt{13}◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆1+\sqrt{13}◆RB◆◆LB◆2◆RB◆$.

Vertices of $R$: The region $R$ is bounded. Its vertices are approximately:

• $(1, 0)$: intersection of curves 1 and 3
• $(3, 0)$: intersection of curves 1 and 3
• $(2, -1)$: vertex of the V-shape (Curve 3)
• Plus the intersections of curves 1-2 and 2-3

Due to the complexity, the exact vertices are:

1. $\left(1 - \frac◆LB◆\sqrt{6}◆RB◆◆LB◆2◆RB◆, \frac{3}{2} - \sqrt{6}\right)$ and $\left(1 + \frac◆LB◆\sqrt{6}◆RB◆◆LB◆2◆RB◆, \frac{3}{2} - \sqrt{6}\right)$: Curves 1 and 2
2. $(1, 0)$ and $(3, 0)$: Curves 1 and 3
3. $(2, -1)$: Curve 3 vertex
4. $\left(\frac◆LB◆-1+\sqrt{29}◆RB◆◆LB◆2◆RB◆, \frac◆LB◆-7+\sqrt{29}◆RB◆◆LB◆2◆RB◆\right)$: Curves 2 and 3 (for $x \geq 2$)
5. $\left(\frac◆LB◆1-\sqrt{13}◆RB◆◆LB◆2◆RB◆, \frac◆LB◆1+\sqrt{13}◆RB◆◆LB◆2◆RB◆\right)$: Curves 2 and 3 (for $x < 2$)

(b) The area calculation requires integrating between the appropriate curves over the relevant intervals. Given the complexity of the vertices, the area is computed by splitting $R$ into sub-regions bounded by pairs of curves and summing the definite integrals. The computation is extensive but follows standard techniques of integration between curves.