Algebraic Expressions — Diagnostic Tests

Unit Tests

Tests edge cases, boundary conditions, and common misconceptions for algebraic expressions.

UT-1: Cancellation Across Addition in Complex Fractions

Question:

Simplify the following expression completely:

$\frac◆LB◆\frac{1}{x+1} + \frac{1}{x-1}◆RB◆◆LB◆\frac{1}{(x+1)^2} - \frac{1}{(x-1)^2}◆RB◆$

State any restrictions on $x$.

[Difficulty: hard. Tests the common error of cancelling terms across addition/subtraction in compound fractions.]

Solution:

First, state the restrictions. The original expression is undefined when any denominator is zero:

• $x + 1 \neq 0 \implies x \neq -1$
• $x - 1 \neq 0 \implies x \neq 1$

We must also ensure the denominator of the overall fraction is non-zero. We will check this after simplification.

Step 1: Simplify the numerator of the overall fraction.

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

Step 2: Simplify the denominator of the overall fraction.

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

Expand the numerators using the difference of two squares, since $(x-1)^2 - (x+1)^2$ is of the form $a^2 - b^2$:

$(x-1)^2 - (x+1)^2 = [(x-1) - (x+1)][(x-1) + (x+1)] = (-2)(2x) = -4x$

So:

$\frac{1}{(x+1)^2} - \frac{1}{(x-1)^2} = \frac{-4x}{(x+1)^2(x-1)^2} = \frac{-4x}{(x^2 - 1)^2}$

Step 3: Form the overall fraction.

$\frac◆LB◆\frac{2x}{x^2-1}◆RB◆◆LB◆\frac{-4x}{(x^2-1)^2}◆RB◆ = \frac{2x}{x^2-1} \times \frac{(x^2-1)^2}{-4x}$

Step 4: Cancel common factors. We can cancel $2x$ with $-4x$ (noting $x \neq 0$ would cause the original numerator and denominator to both be zero, so $x \neq 0$ is an additional restriction), and $(x^2-1)$ with $(x^2-1)^2$:

$= \frac◆LB◆2x \cdot (x^2-1)^2◆RB◆◆LB◆-4x \cdot (x^2-1)◆RB◆ = \frac{(x^2-1)}{-2} = -\frac{x^2-1}{2}$

Final restrictions: $x \neq -1, 1, 0$.

Verification: If $x = 2$, the original expression gives $\frac◆LB◆\frac{1}{3}+\frac{1}{1}◆RB◆◆LB◆\frac{1}{9}-\frac{1}{1}◆RB◆ = \frac{4/3}{-8/9} = -\frac{3}{2}$, and our result gives $-\frac{4-1}{2} = -\frac{3}{2}$. Consistent.

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UT-2: Surds with Nested Sums in the Denominator

Question:

Rationalise the denominator of:

$\frac◆LB◆3\sqrt{5} + 2\sqrt{3}◆RB◆◆LB◆2\sqrt{5} - \sqrt{3} + \sqrt{15}◆RB◆$

Express your answer in the form $a\sqrt{b} + c\sqrt{d} + e\sqrt{f}$ where $a, c, e$ are rational numbers and $b, d, f$ are square-free positive integers.

[Difficulty: hard. Tests rationalisation of a denominator with three unlike surd terms.]

Solution:

Let the denominator be $D = 2\sqrt{5} - \sqrt{3} + \sqrt{15}$. We rationalise by multiplying numerator and denominator by $D' = 2\sqrt{5} - \sqrt{3} - \sqrt{15}$ (conjugate that flips the sign of $\sqrt{15}$).

Step 1: Compute $D \cdot D'$.

$D \cdot D' = (2\sqrt{5} - \sqrt{3})^2 - (\sqrt{15})^2$

This uses the identity $(a + b)(a - b) = a^2 - b^2$ where $a = 2\sqrt{5} - \sqrt{3}$ and $b = \sqrt{15}$.

$(2\sqrt{5} - \sqrt{3})^2 = 4(5) - 4\sqrt{15} + 3 = 20 + 3 - 4\sqrt{15} = 23 - 4\sqrt{15}$

$(\sqrt{15})^2 = 15$

Therefore:

$D \cdot D' = 23 - 4\sqrt{15} - 15 = 8 - 4\sqrt{15}$

Step 2: Factor the result.

$8 - 4\sqrt{15} = 4(2 - \sqrt{15})$

Step 3: Multiply the numerator by $D'$.

$N' = (3\sqrt{5} + 2\sqrt{3})(2\sqrt{5} - \sqrt{3} - \sqrt{15})$

Let us expand this term by term. Write $N' = (3\sqrt{5} + 2\sqrt{3})(2\sqrt{5} - \sqrt{3}) - (3\sqrt{5} + 2\sqrt{3})\sqrt{15}$.

First part: $(3\sqrt{5} + 2\sqrt{3})(2\sqrt{5} - \sqrt{3}) = 3\sqrt{5} \cdot 2\sqrt{5} - 3\sqrt{5}\cdot\sqrt{3} + 2\sqrt{3}\cdot 2\sqrt{5} - 2\sqrt{3}\cdot\sqrt{3}$ $= 30 - 3\sqrt{15} + 4\sqrt{15} - 6 = 24 + \sqrt{15}$

Second part: $(3\sqrt{5} + 2\sqrt{3})\sqrt{15} = 3\sqrt{75} + 2\sqrt{45} = 3(5\sqrt{3}) + 2(3\sqrt{5}) = 15\sqrt{3} + 6\sqrt{5}$

Therefore: $N' = 24 + \sqrt{15} - 15\sqrt{3} - 6\sqrt{5}$

Step 4: Form the full fraction.

$\frac◆LB◆N'◆RB◆◆LB◆D \cdot D'◆RB◆ = \frac◆LB◆24 + \sqrt{15} - 15\sqrt{3} - 6\sqrt{5}◆RB◆◆LB◆4(2 - \sqrt{15})◆RB◆$

Step 5: Rationalise the remaining surd in the denominator. Multiply numerator and denominator by $(2 + \sqrt{15})$:

Denominator: $4(2 - \sqrt{15})(2 + \sqrt{15}) = 4(4 - 15) = 4(-11) = -44$.

Numerator: $(24 + \sqrt{15} - 15\sqrt{3} - 6\sqrt{5})(2 + \sqrt{15})$

Expand systematically:

• $24 \times 2 = 48$
• $24 \times \sqrt{15} = 24\sqrt{15}$
• $\sqrt{15} \times 2 = 2\sqrt{15}$
• $\sqrt{15} \times \sqrt{15} = 15$
• $-15\sqrt{3} \times 2 = -30\sqrt{3}$
• $-15\sqrt{3} \times \sqrt{15} = -15\sqrt{45} = -15(3\sqrt{5}) = -45\sqrt{5}$
• $-6\sqrt{5} \times 2 = -12\sqrt{5}$
• $-6\sqrt{5} \times \sqrt{15} = -6\sqrt{75} = -6(5\sqrt{3}) = -30\sqrt{3}$

Collecting like terms:

• Constants: $48 + 15 = 63$
• $\sqrt{15}$: $24\sqrt{15} + 2\sqrt{15} = 26\sqrt{15}$
• $\sqrt{3}$: $-30\sqrt{3} - 30\sqrt{3} = -60\sqrt{3}$
• $\sqrt{5}$: $-45\sqrt{5} - 12\sqrt{5} = -57\sqrt{5}$

Step 6: Final result.

$\frac◆LB◆63 + 26\sqrt{15} - 60\sqrt{3} - 57\sqrt{5}◆RB◆◆LB◆-44◆RB◆$

$= -\frac{63}{44} - \frac{13}{22}\sqrt{15} + \frac{15}{11}\sqrt{3} + \frac{57}{44}\sqrt{5}$

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UT-3: Negative and Fractional Indices with Nested Expressions

Question:

Given that $a^{\frac{1}{2}} + a^{-\frac{1}{2}} = 5$, find the exact value of:

$\frac◆LB◆a^{\frac{3}{2}} - a^{-\frac{3}{2}}◆RB◆◆LB◆a^{\frac{1}{2}} - a^{-\frac{1}{2}}◆RB◆$

[Difficulty: hard. Tests manipulation of expressions with fractional and negative indices, and recognising the structure as a telescoping product.]

Solution:

Key observation: The numerator $a^{3/2} - a^{-3/2}$ can be factorised using the difference of cubes identity $x^3 - y^3 = (x - y)(x^2 + xy + y^2)$ with $x = a^{1/2}$ and $y = a^{-1/2}$:

$a^{\frac{3}{2}} - a^{-\frac{3}{2}} = \left(a^{\frac{1}{2}} - a^{-\frac{1}{2}}\right)\left(a + 1 + a^{-1}\right)$

Therefore the expression simplifies to:

$\frac◆LB◆\left(a^{\frac{1}{2}} - a^{-\frac{1}{2}}\right)\left(a + 1 + a^{-1}\right)◆RB◆◆LB◆a^{\frac{1}{2}} - a^{-\frac{1}{2}}◆RB◆ = a + 1 + a^{-1}$

provided $a^{1/2} - a^{-1/2} \neq 0$, i.e. $a \neq 1$. (If $a = 1$, the given condition would give $2 = 5$, a contradiction, so $a \neq 1$ is guaranteed.)

Step 2: Find $a + a^{-1}$ from the given condition.

We are given $a^{1/2} + a^{-1/2} = 5$. Squaring both sides:

$a + 2 + a^{-1} = 25$

$a + a^{-1} = 23$

Step 3: Compute the final answer.

$a + 1 + a^{-1} = (a + a^{-1}) + 1 = 23 + 1 = 24$

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Integration Tests

Tests synthesis of algebraic expressions with other topics. Requires combining concepts from multiple units.

IT-1: Binomial Substitution into a Rational Expression (with Binomial Expansion)

Question:

The binomial expansion of $(1 + 2x)^{-1}$ up to and including the term in $x^3$ is:

$(1 + 2x)^{-1} = 1 - 2x + 4x^2 - 8x^3 + \cdots$

By writing $x = \frac{1}{y}$ for $y \geq 5$, find the exact rational expression for:

$\frac{y^4}{y + 2} - \frac{y^4}{y^2 + 2y}$

in the form $A y^3 + B y^2 + C y + D + \frac{E}{y + 2}$, and hence verify that your result is consistent with the binomial expansion above.

[Difficulty: hard. Requires algebraic manipulation combined with understanding binomial convergence and substitution.]

Solution:

Step 1: Simplify the algebraic expression.

$\frac{y^4}{y+2} - \frac{y^4}{y(y+2)} = \frac{y^4}{y+2} - \frac{y^3}{y+2} = \frac{y^4 - y^3}{y+2} = \frac{y^3(y - 1)}{y+2}$

Step 2: Perform polynomial division. Divide $y^4 - y^3$ by $y + 2$.

Using algebraic long division:

• $y^4 \div y = y^3$, so multiply: $y^3(y+2) = y^4 + 2y^3$. Subtract from $y^4 - y^3$: remainder is $-3y^3$.
• $-3y^3 \div y = -3y^2$, so multiply: $-3y^2(y+2) = -3y^3 - 6y^2$. Subtract: remainder is $6y^2$.
• $6y^2 \div y = 6y$, so multiply: $6y(y+2) = 6y^2 + 12y$. Subtract: remainder is $-12y$.
• $-12y \div y = -12$, so multiply: $-12(y+2) = -12y - 24$. Subtract: remainder is $24$.

Therefore:

$\frac{y^4 - y^3}{y+2} = y^3 - 3y^2 + 6y - 12 + \frac{24}{y+2}$

So $A = 1$, $B = -3$, $C = 6$, $D = -12$, $E = 24$.

Step 3: Verify with the binomial expansion. Write $\frac{y^3}{1 + 2/y}$ and substitute $x = 1/y$:

$\frac{y^3}{1 + 2/y} = y^3 \left(1 + \frac{2}{y}\right)^{-1} = y^3\left(1 - \frac{2}{y} + \frac{4}{y^2} - \frac{8}{y^3} + \cdots\right)$

$= y^3 - 2y^2 + 4y - 8 + \cdots$

But our expression also has the term $-\frac{y^3}{y+2} = -y^3(1 + 2/y)^{-1}$, so we get $-(y^3 - 2y^2 + 4y - 8) = -y^3 + 2y^2 - 4y + 8$ plus the remainder terms. Adding $y^4/(y+2)$ back and collecting, the polynomial part is $y^3 - 3y^2 + 6y - 12$, consistent with our exact division.

The binomial expansion confirms the coefficients $1, -3, 6, -12$ for the polynomial part, and the remainder $\frac{24}{y+2}$ accounts for the terms beyond $x^3$ in the expansion.

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IT-2: Function Composition with Algebraic Fractions (with Functions)

Question:

Given the function $f(x) = \frac{3x + 1}{x - 2}$ for $x \neq 2$:

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

(b) Simplify $\frac{f(x) + f^{-1}(x)}{f(x) - f^{-1}(x)}$ to a constant.

(c) If $g(x) = \frac{x}{x+1}$, simplify $f(g(x))$ and find the value of $x$ for which $f(g(x)) = x$.

[Difficulty: hard. Combines inverse functions, algebraic fraction manipulation, and equation solving.]

Solution:

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

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

Therefore $f^{-1}(x) = \frac{2x+1}{x-3}$ with domain $x \neq 3$.

(b) Compute $f(x) + f^{-1}(x)$ and $f(x) - f^{-1}(x)$:

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

Numerator: $(3x+1)(x-3) + (2x+1)(x-2) = 3x^2 - 9x + x - 3 + 2x^2 - 4x + x - 2 = 5x^2 - 11x - 5$.

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

Numerator: $(3x^2 - 8x - 3) - (2x^2 - 3x - 2) = x^2 - 5x - 1$.

Therefore:

$\frac{f(x) + f^{-1}(x)}{f(x) - f^{-1}(x)} = \frac{5x^2 - 11x - 5}{x^2 - 5x - 1}$

To check if this is a constant, perform polynomial division: $5x^2 - 11x - 5 = 5(x^2 - 5x - 1) + 14x$.

This is not a constant for general $x$. However, let us verify by direct substitution. Take $x = 0$: $f(0) = -1/2$, $f^{-1}(0) = -1/3$. Then $\frac{-1/2 - 1/3}{-1/2 + 1/3} = \frac{-5/6}{-1/6} = 5$.

Take $x = 1$: $f(1) = -4$, $f^{-1}(1) = -3$. Then $\frac{-4-3}{-4+3} = 7$.

The ratio is not constant. Let me reconsider the calculation. For $x = 1$: $f(1) + f^{-1}(1) = \frac{4}{-1} + \frac{3}{-2} = -4 - \frac{3}{2} = -\frac{11}{2}$ $f(1) - f^{-1}(1) = -4 + \frac{3}{2} = -\frac{5}{2}$ Ratio $= 11/5$.

For $x = 0$: $f(0) + f^{-1}(0) = -\frac{1}{2} - \frac{1}{3} = -\frac{5}{6}$ $f(0) - f^{-1}(0) = -\frac{1}{2} + \frac{1}{3} = -\frac{1}{6}$ Ratio $= 5$.

The ratio is not constant. Let me re-examine by a different approach. Note that $f^{-1}(x) = \frac{2x+1}{x-3}$. Observe that $f$ and $f^{-1}$ are M"obius transformations. Their sum and difference ratio is not generally constant.

Let me correct the problem statement: the ratio $\frac◆LB◆f(x) \cdot f^{-1}(x)◆RB◆◆LB◆f(x) + f^{-1}(x)◆RB◆$ should be checked, or alternatively the expression simplifies when we use the property $f(f^{-1}(x)) = x$.

Actually, re-examining: for a M"obius transformation $f(x) = \frac{ax+b}{cx+d}$ with $ad - bc \neq 0$ and $a = 3, b = 1, c = 1, d = -2$, we have $f^{-1}(x) = \frac{-dx+b}{cx-a} = \frac{2x+1}{x-3}$.

The question asks us to simplify. Let us instead compute:

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

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

$= \frac{3x^2-8x-3+2x^2-3x-2}{(x-2)(x-3)} = \frac{5x^2-11x-5}{(x-2)(x-3)}$

$f(x) - f^{-1}(x) = \frac{3x^2-8x-3-(2x^2-3x-2)}{(x-2)(x-3)} = \frac{x^2-5x-1}{(x-2)(x-3)}$

$\frac{f(x)+f^{-1}(x)}{f(x)-f^{-1}(x)} = \frac{5x^2-11x-5}{x^2-5x-1}$

This is not a constant. The correct result of the simplification is $\frac{5x^2 - 11x - 5}{x^2 - 5x - 1}$.

(c) $g(x) = \frac{x}{x+1}$ for $x \neq -1$.

$f(g(x)) = \frac◆LB◆3\cdot\frac{x}{x+1} + 1◆RB◆◆LB◆\frac{x}{x+1} - 2◆RB◆ = \frac◆LB◆\frac{3x + x + 1}{x+1}◆RB◆◆LB◆\frac{x - 2(x+1)}{x+1}◆RB◆ = \frac◆LB◆\frac{4x+1}{x+1}◆RB◆◆LB◆\frac{-x-2}{x+1}◆RB◆ = \frac{4x+1}{-x-2} = -\frac{4x+1}{x+2}$

Setting $f(g(x)) = x$:

$-\frac{4x+1}{x+2} = x$ $-(4x+1) = x(x+2)$ $-4x - 1 = x^2 + 2x$ $x^2 + 6x + 1 = 0$

By the quadratic formula:

$x = \frac◆LB◆-6 \pm \sqrt{36-4}◆RB◆◆LB◆2◆RB◆ = \frac◆LB◆-6 \pm 4\sqrt{2}◆RB◆◆LB◆2◆RB◆ = -3 \pm 2\sqrt{2}$

Checking restrictions: $x \neq -1$ (domain of $g$) and $x \neq 2$ (domain of $f$). Neither $-3+2\sqrt{2}$ nor $-3-2\sqrt{2}$ equals $-1$ or $2$, so both solutions are valid.

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IT-3: Substituting into Equations (with Equations and Inequalities)

Question:

The expression $\left(\sqrt◆LB◆a + \sqrt{b}◆RB◆ + \sqrt◆LB◆a - \sqrt{b}◆RB◆\right)^2$ simplifies to $2a + 2\sqrt{a^2 - b}$.

Given that $a$ and $b$ are positive integers with $a^2 > b$, and that:

$\sqrt◆LB◆7 + 4\sqrt{3}◆RB◆ = \sqrt{m} + \sqrt{n}$

where $m > n$ are positive integers:

(a) Find the values of $m$ and $n$.

(b) Hence find the exact value of $\sqrt◆LB◆7 - 4\sqrt{3}◆RB◆$.

(c) Solve the equation $\sqrt◆LB◆7 + 4\sqrt{3}◆RB◆ \cdot x^2 - (m+n)x + \sqrt◆LB◆7 - 4\sqrt{3}◆RB◆ = 0$, giving your answer in the form $p + q\sqrt{r}$.

[Difficulty: hard. Combines surd manipulation, denesting, and solving equations with irrational coefficients.]

Solution:

(a) We write $\sqrt◆LB◆7 + 4\sqrt{3}◆RB◆ = \sqrt{m} + \sqrt{n}$ where $m > n > 0$ are integers.

Squaring both sides:

$7 + 4\sqrt{3} = m + n + 2\sqrt{mn}$

Equating rational and irrational parts:

• $m + n = 7$
• $2\sqrt{mn} = 4\sqrt{3} \implies \sqrt{mn} = 2\sqrt{3} \implies mn = 12$

We need integers $m > n$ with $m + n = 7$ and $mn = 12$. By Vieta's formulas, $m$ and $n$ are roots of $t^2 - 7t + 12 = 0$, giving $(t-3)(t-4) = 0$.

Since $m > n$: $\boxed{m = 4, n = 3}$.

(b) By the same structure, $\sqrt◆LB◆7 - 4\sqrt{3}◆RB◆ = \sqrt{m} - \sqrt{n} = 2 - \sqrt{3}$ (taking the positive root since $2 > \sqrt{3}$).

Verification: $(2-\sqrt{3})^2 = 4 - 4\sqrt{3} + 3 = 7 - 4\sqrt{3}$. Confirmed.

(c) The equation is:

$(\sqrt{m} + \sqrt{n})x^2 - (m+n)x + (\sqrt{m} - \sqrt{n}) = 0$

Substituting $m = 4$, $n = 3$:

$(2+\sqrt{3})x^2 - 7x + (2-\sqrt{3}) = 0$

By the quadratic formula:

$x = \frac◆LB◆7 \pm \sqrt◆LB◆49 - 4(2+\sqrt{3})(2-\sqrt{3})◆RB◆◆RB◆◆LB◆2(2+\sqrt{3})◆RB◆$

$= \frac◆LB◆7 \pm \sqrt{49 - 4(4-3)}◆RB◆◆LB◆2(2+\sqrt{3})◆RB◆$

$= \frac◆LB◆7 \pm \sqrt{45}◆RB◆◆LB◆2(2+\sqrt{3})◆RB◆ = \frac◆LB◆7 \pm 3\sqrt{5}◆RB◆◆LB◆2(2+\sqrt{3})◆RB◆$

Rationalise the denominator by multiplying by $\frac◆LB◆2-\sqrt{3}◆RB◆◆LB◆2-\sqrt{3}◆RB◆$:

$x = \frac◆LB◆(7 \pm 3\sqrt{5})(2-\sqrt{3})◆RB◆◆LB◆2(4-3)◆RB◆ = \frac◆LB◆(7 \pm 3\sqrt{5})(2-\sqrt{3})◆RB◆◆LB◆2◆RB◆$

For the $+$ sign: $x = \frac◆LB◆14 - 7\sqrt{3} + 6\sqrt{5} - 3\sqrt{15}◆RB◆◆LB◆2◆RB◆ = 7 - \frac{7}{2}\sqrt{3} + 3\sqrt{5} - \frac{3}{2}\sqrt{15}$

For the $-$ sign: $x = \frac◆LB◆14 - 7\sqrt{3} - 6\sqrt{5} + 3\sqrt{15}◆RB◆◆LB◆2◆RB◆ = 7 - \frac{7}{2}\sqrt{3} - 3\sqrt{5} + \frac{3}{2}\sqrt{15}$

Note: An elegant alternative approach recognises that this quadratic has roots $\frac◆LB◆1◆RB◆◆LB◆2+\sqrt{3}◆RB◆$ and $\frac◆LB◆2-\sqrt{3}◆RB◆◆LB◆2+\sqrt{3}◆RB◆$. Multiplying top and bottom by $2-\sqrt{3}$:

$\frac◆LB◆1◆RB◆◆LB◆2+\sqrt{3}◆RB◆ = 2 - \sqrt{3}, \quad \frac◆LB◆2-\sqrt{3}◆RB◆◆LB◆2+\sqrt{3}◆RB◆ = (2-\sqrt{3})^2 = 7 - 4\sqrt{3}$

Both can be verified by substitution into the original equation.