Algebra & Functions

The quadratic equation 2x² + kx + 8 = 0 has equal roots, where k > 0. Find the value of k.

• Ak = 4
• Bk = 8✓
• Ck = 2
• Dk = 16

Answer: Equal roots require the discriminant b² − 4ac = 0, so k² − 4(2)(8) = k² − 64 = 0, giving k² = 64 and the positive root k = 8.

Simplify fully (3 + √5)(3 − √5).

• A9 − √5
• B14
• C6
• D4✓

Answer: This is a difference of two squares: (3)² − (√5)² = 9 − 5 = 4.

Given f(x) = x² − 6x + 11, express f(x) in completed-square form a(x + p)² + q, where p and q may be negative.

• A(x − 6)² + 11
• B(x + 3)² + 2
• C(x − 3)² + 2✓
• D(x − 3)² − 2

Answer: Completing the square: x² − 6x + 11 = (x − 3)² − 9 + 11 = (x − 3)² + 2.

Evaluate 27^(2/3) without a calculator.

• A6
• B81
• C18
• D9✓

Answer: 27^(2/3) = (27^(1/3))² = 3² = 9, since the cube root of 27 is 3.

The equation x² + kx + 9 = 0 has two distinct real roots. Find the complete set of values of k.

• A−6 < k < 6
• Bk > 6 only
• Ck < −6 or k > 6✓
• Dk < −3 or k > 3

Answer: Two distinct real roots require discriminant b² − 4ac > 0, so k² − 4(1)(9) = k² − 36 > 0, giving k² > 36 and hence k < −6 or k > 6.

Express 2x² + 12x + 5 in the form a(x + p)² + q.

• A2(x + 3)² − 13✓
• B2(x + 3)² + 5
• C2(x + 6)² − 13
• D(x + 6)² − 31

Answer: Factor 2 from the x-terms: 2(x² + 6x) + 5 = 2[(x + 3)² − 9] + 5 = 2(x + 3)² − 18 + 5 = 2(x + 3)² − 13.

Solve the simultaneous equations y = x + 1 and x² + y² = 25. Which pair gives both solution points?

• A(4, 3) and (−3, −4)
• B(3, 4) and (−4, −3)✓
• C(3, 4) only
• D(0, 1) and (5, 6)

Answer: Substitute y = x + 1: x² + (x + 1)² = 25 → 2x² + 2x + 1 = 25 → 2x² + 2x − 24 = 0 → x² + x − 12 = 0 → (x + 4)(x − 3) = 0. So x = 3 (y = 4) or x = −4 (y = −3), giving (3, 4) and (−4, −3).

Solve the inequality x² − x − 6 < 0.

• Ax < −2 or x > 3
• B−3 < x < 2
• C−2 < x < 3✓
• Dx < −3 or x > 2

Answer: Factorise: x² − x − 6 = (x − 3)(x + 2). The parabola opens upwards and is below the x-axis between its roots, so (x − 3)(x + 2) < 0 gives −2 < x < 3.

Express (5x + 1)/((x + 1)(x − 2)) as partial fractions A/(x + 1) + B/(x − 2). Find A and B.

• AA = 11/3, B = 4/3
• BA = 4/3, B = 11/3✓
• CA = −4/3, B = 11/3
• DA = 4, B = 11

Answer: Write 5x + 1 = A(x − 2) + B(x + 1). Let x = 2: 11 = 3B, so B = 11/3. Let x = −1: −4 = −3A, so A = 4/3.

Solve the modulus equation |2x − 3| = 5.

• Ax = 4 or x = 1
• Bx = 4 only
• Cx = −4 or x = 1
• Dx = 4 or x = −1✓

Answer: Either 2x − 3 = 5, giving 2x = 8 and x = 4, or 2x − 3 = −5, giving 2x = −2 and x = −1. Both satisfy the equation.

The functions are f(x) = 2x + 1 and g(x) = x². Find the value of the composite fg(3), that is f(g(3)).

• A49
• B19✓
• C37
• D13

Answer: fg(3) means apply g first: g(3) = 3² = 9, then f(9) = 2(9) + 1 = 19. Applying the functions in the opposite order gives gf(3) = (2·3 + 1)² = 49, a common order mix-up.

The function f is defined by f(x) = 3x − 2. Find its inverse f⁻¹(x).

• A(x − 2)/3
• B1/(3x − 2)
• C(x + 2)/3✓
• D3x + 2

Answer: Let y = 3x − 2 and make x the subject: y + 2 = 3x, so x = (y + 2)/3. Replacing y with x gives f⁻¹(x) = (x + 2)/3.

Given f(x) = 2x³ + 3x² − x + 4, use the remainder theorem to find the remainder when f(x) is divided by (x − 2).

• A0
• B26
• C34
• D30✓

Answer: The remainder on dividing by (x − 2) is f(2) = 2(8) + 3(4) − 2 + 4 = 16 + 12 − 2 + 4 = 30.

The equation 3x² + kx + 12 = 0 has equal roots, where k > 0. Find k.

• Ak = 12✓
• Bk = 144
• Ck = 6
• Dk = 24√(3)

Answer: Equal roots means the discriminant is zero: b² − 4ac = 0, so k² = 4(3)(12) = 144. Taking the positive root, k = 12. Setting b² − 4ac = 0 and then forgetting to square-root gives 144, which is the most common slip here.

The equation 5x² + kx + 20 = 0 has equal roots, where k > 0. Find k.

• Ak = 40√(5)
• Bk = 400
• Ck = 10
• Dk = 20✓

Answer: Equal roots means the discriminant is zero: b² − 4ac = 0, so k² = 4(5)(20) = 400. Taking the positive root, k = 20. Setting b² − 4ac = 0 and then forgetting to square-root gives 400, which is the most common slip here.

The equation 2x² + kx + 18 = 0 has equal roots, where k > 0. Find k.

• Ak = 6
• Bk = 144
• Ck = 12✓
• Dk = 36√(2)

Answer: Equal roots means the discriminant is zero: b² − 4ac = 0, so k² = 4(2)(18) = 144. Taking the positive root, k = 12. Setting b² − 4ac = 0 and then forgetting to square-root gives 144, which is the most common slip here.

Write x² + 6x + 5 in the form (x + a)² + b.

• A(x + 6)² − 4
• B(x + 3)² + 5
• C(x + 3)² − 4✓
• D(x + 3)² + 4

Answer: Halve the coefficient of x: a = 6/2 = 3. Then (x + 3)² = x² + 6x + 9, so you must subtract 9 to keep the expression equal: b = 5 − 9 = −4.

Write x² + 8x + 12 in the form (x + a)² + b.

• A(x + 4)² − 4✓
• B(x + 4)² + 12
• C(x + 8)² − 4
• D(x + 4)² + 4

Answer: Halve the coefficient of x: a = 8/2 = 4. Then (x + 4)² = x² + 8x + 16, so you must subtract 16 to keep the expression equal: b = 12 − 16 = −4.

Write x² + 4x − 21 in the form (x + a)² + b.

• A(x + 2)² − 25✓
• B(x + 2)² − 21
• C(x + 4)² − 25
• D(x + 2)² + 25

Answer: Halve the coefficient of x: a = 4/2 = 2. Then (x + 2)² = x² + 4x + 4, so you must subtract 4 to keep the expression equal: b = −21 − 4 = −25.

f(x) = 2x + 3. Find f⁻¹(x).

• A(x − 3)/2✓
• B(x + 3)/2
• C2x − 3
• D1/(2x + 3)

Answer: Write y = 2x + 3, swap x and y, then make y the subject: x = 2y + 3, so y = (x − 3)/2. Writing 1/f(x) is a different thing entirely — the notation f⁻¹ means inverse function, not reciprocal.

f(x) = x + 4. Find f⁻¹(x).

• A1/(x + 4)
• Bx + 4
• C4 − x
• Dx − 4✓

Answer: Write y = x + 4, swap x and y, then make y the subject: x = y + 4, so y = x − 4. Writing 1/f(x) is a different thing entirely — the notation f⁻¹ means inverse function, not reciprocal.

f(x) = 3x + 2. Find f⁻¹(x).

• A(x − 2)/3✓
• B(x + 2)/3
• C3x − 2
• D1/(3x + 2)

Answer: Write y = 3x + 2, swap x and y, then make y the subject: x = 3y + 2, so y = (x − 2)/3. Writing 1/f(x) is a different thing entirely — the notation f⁻¹ means inverse function, not reciprocal.

State the range of f(x) = x² + 3 for x ∈ ℝ.

• AAll real numbers, because x can take any value whatsoever
• Bf(x) > 3 for every real value of x with no minimum attained
• Cf(x) ≤ 3, since the parabola opens downwards from its vertex
• Df(x) ≥ 3✓

Answer: x² is never negative and reaches 0 at x = 0, so the smallest output is 0 + 3 = 3 and the curve rises from there. The range is the set of OUTPUTS; the domain here is all reals, which is the trap on offer.

Why is f(x) = x² not invertible over the domain x ∈ ℝ?

• ABecause its graph is a parabola rather than a straight line through the origin
• BIt is not one-to-one✓
• CBecause it is not continuous everywhere on the real number line
• DBecause its range does not include any of the negative real numbers

Answer: Both x = 2 and x = −2 map to 4, so a would-be inverse would not know which to return. Restricting the domain to x ≥ 0 makes it one-to-one and the inverse √x then exists. Continuity is not the issue — x² is continuous everywhere.

Solve 3x − 12 = 0.

• Ax = 1/4, obtained by dividing three by twelve rather than the other way round
• Bx = −4, obtained by moving the twelve across without changing its sign
• Cx = 36, obtained by multiplying the two numbers instead of dividing
• Dx = 4✓

Answer: Add 12 to both sides to get 3x = 12, then divide by 3 to get x = 4. Each wrong option corresponds to one specific slip, so checking by substitution catches all of them: 3(4) − 12 = 0. ✓