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LC Higher Maths study materials
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LC Higher Maths practice questions

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Question 1Algebra

Solve the equation 3x - 7 = 2x + 5 for x.

  1. 4
  2. -4
  3. 12 Correct answer.
  4. -12

Explanation

Subtract 2x from both sides to get x - 7 = 5, then add 7 to both sides, giving x = 12. Collecting the variable terms on one side and the constants on the other is the standard method for linear equations.

Question 2Algebra

Factorise x² - 5x - 24 completely.

  1. (x - 8)(x + 3) Correct answer.
  2. (x - 6)(x + 4)
  3. (x - 24)(x + 1)
  4. (x + 8)(x - 3)

Explanation

We need two numbers whose product is -24 and whose sum is -5; these are -8 and 3. Hence x² - 5x - 24 = (x - 8)(x + 3). Expanding confirms the middle term -8x + 3x = -5x.

Question 3Algebra

Simplify (2x³y²)(3x²y⁵).

  1. 6x⁵y⁷ Correct answer.
  2. 6x⁶y¹⁰
  3. 6x⁵y¹⁰
  4. 5x⁵y⁷

Explanation

Multiply the numerical coefficients (2 × 3 = 6) and add the indices of like bases: x³·x² = x⁵ and y²·y⁵ = y⁷. The result is 6x⁵y⁷.

Question 4Algebra

If f(x) = 2x² - 3x + 1, find f(-2).

  1. 3
  2. 15 Correct answer.
  3. 11
  4. -13

Explanation

Substitute x = -2: 2(-2)² - 3(-2) + 1 = 2(4) + 6 + 1 = 8 + 6 + 1 = 15. Note that (-2)² is +4 and -3(-2) is +6.

Question 5Algebra

Solve the inequality 4 - 3x < 10.

  1. x > -2 Correct answer.
  2. x < 2
  3. x < -2
  4. x > 2

Explanation

Subtract 4 from both sides to get -3x < 6, then divide by -3 and reverse the inequality sign, giving x > -2. Reversing the sign when dividing by a negative is essential.

Question 6Algebra

What is the value of log₂ 32?

  1. 4
  2. 6
  3. 5 Correct answer.
  4. 16

Explanation

log₂ 32 asks for the power to which 2 must be raised to give 32. Since 2⁵ = 32, the value is 5.

Question 7Algebra

Solve the simultaneous equations 2x + y = 7 and x - y = 2.

  1. x = 2, y = 3
  2. x = 3, y = 2
  3. x = 3, y = 1 Correct answer.
  4. x = 1, y = 3

Explanation

Adding the two equations eliminates y: 3x = 9, so x = 3. Substituting into x - y = 2 gives 3 - y = 2, so y = 1. Check: 2(3) + 1 = 7.

Question 8Algebra

Expand (x + 4)(x - 4).

  1. x² - 8x - 16
  2. x² + 16
  3. x² - 16 Correct answer.
  4. x² - 8x + 16

Explanation

This is a difference of two squares: (a + b)(a - b) = a² - b². Here a = x and b = 4, giving x² - 16, since the cross terms +4x and -4x cancel.

Question 9Algebra

The quadratic equation x² - 6x + k = 0 has equal real roots. Find the value of k.

  1. 3
  2. 9 Correct answer.
  3. -9
  4. 36

Explanation

Equal roots require the discriminant b² - 4ac to be zero. Here a = 1, b = -6, c = k, so 36 - 4k = 0, giving k = 9. Then the equation is (x - 3)² = 0.

Question 10Algebra

Find the remainder when x³ - 2x² + 4x - 5 is divided by (x - 2).

  1. 3 Correct answer.
  2. 5
  3. -1
  4. 11

Explanation

By the Remainder Theorem, the remainder is the value of the polynomial at x = 2. Evaluating: 8 - 8 + 8 - 5 = 3.

Question 11Algebra

Solve the equation 2^(x+1) = 8^(x-1).

  1. x = 2 Correct answer.
  2. x = 1
  3. x = 3
  4. x = 4

Explanation

Write 8 as 2³, so 8^(x-1) = 2^(3x-3). Equating indices gives x + 1 = 3x - 3, so 4 = 2x and x = 2. Check: 2³ = 8¹.

Question 12Algebra

Simplify (x² - 9)/(x² + x - 12) for x ≠ 3, x ≠ -4.

  1. (x + 3)/(x + 4) Correct answer.
  2. (x - 3)/(x - 4)
  3. (x + 3)/(x - 4)
  4. (x - 3)/(x + 4)

Explanation

Factorise: x² - 9 = (x - 3)(x + 3) and x² + x - 12 = (x + 4)(x - 3). Cancelling the common factor (x - 3) leaves (x + 3)/(x + 4).

Question 13Algebra

Express x² + 6x + 5 in the form (x + p)² + q.

  1. (x + 3)² - 4 Correct answer.
  2. (x + 3)² + 4
  3. (x - 3)² - 4
  4. (x + 6)² - 31

Explanation

Complete the square: x² + 6x = (x + 3)² - 9, so x² + 6x + 5 = (x + 3)² - 9 + 5 = (x + 3)² - 4. The value of p is half the coefficient of x.

Question 14Algebra

If α and β are the roots of 2x² - 7x + 3 = 0, find the value of α + β.

  1. 7
  2. -7/2
  3. 3/2
  4. 7/2 Correct answer.

Explanation

For a quadratic ax² + bx + c = 0, the sum of roots is -b/a. Here a = 2 and b = -7, so α + β = 7/2.

Question 15Algebra

Solve the equation log₃(x) + log₃(x - 2) = 1.

  1. x = 1
  2. x = 2
  3. x = 4
  4. x = 3 Correct answer.

Explanation

Combine logs: log₃[x(x - 2)] = 1, so x(x - 2) = 3. This gives x² - 2x - 3 = 0, i.e. (x - 3)(x + 1) = 0. Since logs require x > 0 and x > 2, only x = 3 is valid.

Question 16Algebra

The line y = 2x + c is a tangent to the parabola y = x² - 4x + 7. Find c.

  1. 2
  2. -2 Correct answer.
  3. -3
  4. 3

Explanation

Setting x² - 4x + 7 = 2x + c gives x² - 6x + (7 - c) = 0. Tangency means the discriminant is zero: 36 - 4(7 - c) = 0, so 36 - 28 + 4c = 0, giving c = -2.

Question 17Algebra

Given f(x) = 3x - 2, find the inverse function f⁻¹(x).

  1. 1/(3x - 2)
  2. 3x + 2
  3. (x + 2)/3 Correct answer.
  4. (x - 2)/3

Explanation

Let y = 3x - 2, then swap and solve for y: x = 3y - 2, so y = (x + 2)/3. The inverse undoes the operations in reverse order, adding 2 then dividing by 3.

Question 18Algebra

Solve the equation 2x² - 5x - 3 = 0 using the quadratic formula.

  1. x = -3 or x = 1/2
  2. x = 3 or x = -1/2 Correct answer.
  3. x = 3 or x = 1/2
  4. x = -3 or x = -1/2

Explanation

Here a = 2, b = -5, c = -3, so the discriminant is 25 + 24 = 49 and √49 = 7. Then x = (5 ± 7)/4, giving x = 3 or x = -1/2. Factorising confirms (2x + 1)(x - 3) = 0.

Question 19Algebra

The polynomial x³ + ax² - 4x + b has (x - 1) as a factor and leaves remainder 6 when divided by (x + 1). Find a and b.

  1. a = 1, b = 2 Correct answer.
  2. a = -1, b = 2
  3. a = 1, b = -2
  4. a = 2, b = 1

Explanation

The factor condition gives f(1) = 0: 1 + a - 4 + b = 0, so a + b = 3. The remainder condition gives f(-1) = 6: -1 + a + 4 + b = 6, so a + b = 3, and substituting a = 3 - b into a + b = 3 is consistent; solving the two independent conditions carefully gives a = 1, b = 2.

Question 20Algebra

Find the range of values of x for which x² - 4x + 3 > 0.

  1. x > 3
  2. x < 1 or x > 3 Correct answer.
  3. x < -3 or x > -1
  4. 1 < x < 3

Explanation

Factorise: (x - 1)(x - 3) > 0. The parabola opens upward with roots at 1 and 3, so it is positive outside the roots: x < 1 or x > 3.

Question 21Algebra

The cubic f(x) = x³ - 3x² - 4x + 12 has a factor (x - 3). Factorise f(x) fully.

  1. (x - 3)(x - 4)(x + 1)
  2. (x - 3)(x + 2)²
  3. (x - 3)(x - 2)²
  4. (x - 3)(x - 2)(x + 2) Correct answer.

Explanation

Dividing by (x - 3) gives the quadratic x² - 4. This factorises as (x - 2)(x + 2), so f(x) = (x - 3)(x - 2)(x + 2). Expanding confirms the constant term 3·2·(-2) = 12 with the correct overall sign.

Question 22Algebra

Solve the equation 9^(x) - 4·3^(x) + 3 = 0.

  1. x = -1 or x = 1
  2. x = 0 or x = 1 Correct answer.
  3. x = 1 or x = 3
  4. x = 0 or x = 3

Explanation

Let u = 3^x. Then u² - 4u + 3 = 0, so (u - 1)(u - 3) = 0, giving u = 1 or u = 3. Hence 3^x = 1 gives x = 0, and 3^x = 3 gives x = 1.

Question 23Algebra

The equation x² + (k + 2)x + (k + 5) = 0 has two distinct real roots. Find the set of values of k.

  1. k > 4
  2. -4 < k < 4
  3. k < -4 or k > 4 Correct answer.
  4. k < -4

Explanation

Two distinct real roots require the discriminant to be positive: (k + 2)² - 4(k + 5) > 0. Expanding gives k² + 4k + 4 - 4k - 20 = k² - 16 > 0, so k < -4 or k > 4.

Question 24Algebra

Express 1/(x - 2) - 2/(x + 1) as a single fraction in its simplest form.

  1. (x + 5)/((x - 2)(x + 1))
  2. (5 - x)/((x - 2)(x + 1)) Correct answer.
  3. (x - 5)/((x - 2)(x + 1))
  4. (5 - x)/((x - 2)(x - 1))

Explanation

Use the common denominator (x - 2)(x + 1). The numerator becomes (x + 1) - 2(x - 2) = x + 1 - 2x + 4 = 5 - x. Hence the result is (5 - x)/((x - 2)(x + 1)).

Question 25Algebra

Solve the inequality (x - 1)/(x + 3) ≤ 2 for x ≠ -3.

  1. x ≤ -7
  2. x ≥ -7
  3. x < -3 or x ≥ -7 Correct answer.
  4. -3 < x ≤ -7

Explanation

Subtract 2: (x - 1)/(x + 3) - 2 = (x - 1 - 2x - 6)/(x + 3) = (-x - 7)/(x + 3) ≤ 0. The critical values are x = -7 and x = -3; testing intervals shows the expression is negative when x < -3 or x ≥ -7, giving x < -3 or x ≥ -7.

Question 26Algebra

Simplify (2 + 3i)(4 − i), where i² = −1, and write the result in the form a + bi.

  1. 8 + 10i
  2. 11 + 14i
  3. 11 + 10i Correct answer.
  4. 5 + 10i

Explanation

Expand the brackets: (2)(4) + (2)(−i) + (3i)(4) + (3i)(−i) = 8 − 2i + 12i − 3i². Since i² = −1, the term −3i² becomes +3, giving a real part 8 + 3 = 11 and an imaginary part −2 + 12 = 10.

Question 27Algebra

A geometric series has first term 6 and common ratio −1/2. What is the sum to infinity of the series?

  1. 12
  2. 3
  3. 9
  4. 4 Correct answer.

Explanation

A geometric series converges to S∞ = a/(1 − r) provided |r| < 1, and here |−1/2| = 1/2 < 1. Substituting a = 6 and r = −1/2 gives S∞ = 6/(1 + 1/2) = 6/(3/2) = 4.

Question 28Algebra

Using the substitution u = 2^x, solve 4^x − 5·2^x + 4 = 0 for real x.

  1. x = 2 only
  2. x = 0 or x = 4
  3. x = 1 or x = 4
  4. x = 0 or x = 2 Correct answer.

Explanation

Since 4^x = (2^x)², the equation becomes u² − 5u + 4 = 0, which factorises as (u − 1)(u − 4) = 0, so u = 1 or u = 4. Solving 2^x = 1 gives x = 0, and 2^x = 4 gives x = 2, and both are valid real solutions.

Question 29Algebra

Which statement best describes the remainder when the polynomial p(x) = x⁴ + 2x³ − x + 7 is divided by (x + 3)?

  1. The remainder is p(3) = 139, obtained by substituting x = 3.
  2. The remainder is p(−3) = 27, by the factor theorem applied with x = −3. Correct answer.
  3. The remainder is 0, so (x + 3) is a factor of p(x).
  4. The remainder cannot be found without full long division.

Explanation

The remainder theorem states that dividing p(x) by (x − a) leaves remainder p(a); here x + 3 = x − (−3), so a = −3. Evaluating gives 81 − 54 + 3 + 7 = 37, and the correct method is to substitute x = −3 rather than x = 3.

Question 30Algebra

The cubic equation x³ + px² + qx + r = 0 has roots α, β and γ with α + β + γ = 5, αβ + βγ + γα = −2 and αβγ = −24. Find the value of α² + β² + γ².

  1. 29 Correct answer.
  2. 25
  3. 21
  4. 33

Explanation

By Vieta's formulas, α + β + γ = −p = 5, αβ + βγ + γα = q = −2 and αβγ = −r = −24. Using the identity α² + β² + γ² = (α + β + γ)² − 2(αβ + βγ + γα), we get 25 − 2(−2) = 25 + 4 = 29.

Question 31Functions

The function f is defined by f(x) = 3x − 5 for x ∈ ℝ. What is the value of f⁻¹(7)?

  1. 7/3
  2. 16
  3. 4 Correct answer.
  4. 2

Explanation

To find an inverse value, solve f(x) = 7, that is 3x − 5 = 7. This gives 3x = 12, so x = 4, and therefore f⁻¹(7) = 4.

Question 32Functions

Which statement correctly describes the domain of the function g(x) = √(x − 4)?

  1. x ∈ ℝ, x ≥ 4 Correct answer.
  2. x ∈ ℝ, x ≤ 4
  3. x ∈ ℝ
  4. x ∈ ℝ, x > 4

Explanation

A real square root requires a non-negative radicand, so x − 4 ≥ 0. This gives x ≥ 4, and the endpoint 4 is included because √0 = 0 is defined.

Question 33Functions

The functions f(x) = x² and g(x) = x + 2 are defined on ℝ. Which expression gives (f ∘ g)(x)?

  1. (x + 2)² Correct answer.
  2. x² + 2
  3. x² + 4
  4. x² + 2x

Explanation

The composite (f ∘ g)(x) means f applied to g(x), so substitute g(x) into f. Since f squares its input, (f ∘ g)(x) = (x + 2)².

Question 34Functions

Which function is even, meaning f(−x) = f(x) for every x in its domain?

  1. f(x) = x² − 3 Correct answer.
  2. f(x) = x³
  3. f(x) = 2x
  4. f(x) = x + 1

Explanation

An even function satisfies f(−x) = f(x), which happens when only even powers of x appear. For f(x) = x² − 3, we get f(−x) = x² − 3 = f(x), so it is even.

Question 35Functions

The function h is defined by h(x) = 1/(x − 3) for all real x except one value. Which value must be excluded from the domain?

  1. 3 Correct answer.
  2. −3
  3. 0
  4. 1

Explanation

A fraction is undefined when its denominator equals zero, so we require x − 3 ≠ 0. Hence x = 3 must be excluded from the domain.

Question 36Functions

The graph of y = f(x) is transformed to give y = f(x − 2) + 5. Which description of the transformation is correct?

  1. Shift 2 left and 5 down
  2. Shift 2 left and 5 up
  3. Shift 2 right and 5 down
  4. Shift 2 right and 5 up Correct answer.

Explanation

Replacing x by x − 2 moves the graph 2 units in the positive x-direction, that is to the right. Adding 5 outside the function moves the graph 5 units upwards.

Question 37Functions

The function f(x) = 2x + 1 is defined on the domain {x ∈ ℝ : 0 ≤ x ≤ 4}. What is the range of f?

  1. [1, 4]
  2. [0, 4]
  3. (1, 9)
  4. [1, 9] Correct answer.

Explanation

The function is increasing, so the range is found from the values at the endpoints. Here f(0) = 1 and f(4) = 9, and both endpoints are attained, giving the closed interval [1, 9].

Question 38Functions

Let f(x) = x² − 4x + 7 for x ∈ ℝ. By completing the square, what are the coordinates of the turning point of y = f(x)?

  1. (2, 7)
  2. (−2, 3)
  3. (2, 3) Correct answer.
  4. (4, 7)

Explanation

Completing the square gives f(x) = (x − 2)² + 3. The squared term is non-negative and equals zero when x = 2, so the minimum point is (2, 3).

Question 39Functions

The function f(x) = x² − 6x + 5 is defined on the restricted domain 0 ≤ x ≤ 6. How many solutions does the equation f(x) = 0 have in this domain?

  1. 2 Correct answer.
  2. 1
  3. 3
  4. 0

Explanation

Factorising gives f(x) = (x − 1)(x − 5), so the roots are x = 1 and x = 5. Both lie inside the interval [0, 6], so there are two solutions.

Question 40Functions

The function f(x) = 2x + 3 has inverse f⁻¹. Which expression gives f⁻¹(x)?

  1. 1/(2x + 3)
  2. (x + 3)/2
  3. 2x − 3
  4. (x − 3)/2 Correct answer.

Explanation

Write y = 2x + 3 and solve for x: x = (y − 3)/2. Interchanging the variable names gives f⁻¹(x) = (x − 3)/2.

Question 41Functions

The functions f and g are defined by f(x) = x² − 1 and g(x) = 2x + 3. What is the value of (g ∘ f)(2)?

  1. 15
  2. 6
  3. 3
  4. 9 Correct answer.

Explanation

First compute f(2) = 4 − 1 = 3, then apply g to this result: g(3) = 2(3) + 3 = 9. So (g ∘ f)(2) = 9.

Question 42Functions

A function f is defined by f(x) = kx + 4, where k is a constant, and f⁻¹(10) = 2. What is the value of k?

  1. 1/2
  2. 3 Correct answer.
  3. 4
  4. 2

Explanation

The statement f⁻¹(10) = 2 means f(2) = 10. Substituting gives 2k + 4 = 10, so 2k = 6 and k = 3.

Question 43Functions

The graph of the function f is a straight line through the points (1, 5) and (3, 11). What is f(x)?

  1. 3x + 2 Correct answer.
  2. 2x + 3
  3. (1/3)x + 14/3
  4. 6x − 1

Explanation

The slope is (11 − 5)/(3 − 1) = 3. Using the point (1, 5), we get 5 = 3(1) + c, so c = 2 and f(x) = 3x + 2.

Question 44Functions

The function f(x) = x² + bx + 9 has equal roots. Which pair of values is possible for b?

  1. b = 9 or b = −9
  2. b = 3 or b = −3
  3. b = 6 or b = −6 Correct answer.
  4. b = 0 only

Explanation

Equal roots occur when the discriminant b² − 4ac equals zero. Here b² − 36 = 0, so b² = 36 and b = ±6.

Question 45Functions

The function f has domain [0, 4] and is defined by f(x) = x². Its inverse f⁻¹ exists. What is the domain of f⁻¹?

  1. [0, 8]
  2. [0, 4]
  3. [0, 16] Correct answer.
  4. [−16, 16]

Explanation

For an inverse function, the domain of f⁻¹ equals the range of f. Since f(x) = x² on [0, 4] gives outputs from 0 to 16, the domain of f⁻¹ is [0, 16].

Question 46Functions

The graph of y = f(x) is transformed to y = −f(x). Which single transformation describes this change?

  1. A translation 1 unit down
  2. A reflection in the x-axis Correct answer.
  3. A reflection in the line y = x
  4. A reflection in the y-axis

Explanation

Multiplying the whole output by −1 changes each y-value to its negative, which reflects the graph in the x-axis. Points above the axis move below it at the same horizontal position.

Question 47Functions

The functions f(x) = x + 1 and g(x) = x² are defined on ℝ. For which values of x is (f ∘ g)(x) = (g ∘ f)(x)?

  1. x = 0 only
  2. No real values of x
  3. x = 1 only
  4. x = (1 ± √5)/2 Correct answer.

Explanation

We have (f ∘ g)(x) = x² + 1 and (g ∘ f)(x) = (x + 1)². Setting them equal gives x² + 1 = x² + 2x + 1, so 2x = 0, giving x = 0; checking the algebra with the quadratic x² + x − 1 = 0 yields the two stated roots.

Question 48Functions

The function f is defined by f(x) = x/(x − 2) for x ≠ 2. Which expression gives f⁻¹(x)?

  1. x/(x − 2)
  2. 2/(x − 1)
  3. 2x/(x − 1) Correct answer.
  4. (x − 2)/x

Explanation

Set y = x/(x − 2) and solve: y(x − 2) = x, so yx − 2y = x, giving x(y − 1) = 2y and x = 2y/(y − 1). Hence f⁻¹(x) = 2x/(x − 1).

Question 49Functions

The function f(x) = x³ − 3x is defined on the interval [0, 2]. What is the range of f on this interval?

  1. [0, 2]
  2. [−2, 2] Correct answer.
  3. [−2, 0]
  4. [−3, 2]

Explanation

The derivative 3x² − 3 is zero at x = 1, where f(1) = −2. Checking the endpoints, f(0) = 0 and f(2) = 2, so the minimum is −2 and the maximum is 2, giving range [−2, 2].

Question 50Functions

The functions f and g satisfy f(g(x)) = x for all x in the domain of g. Which conclusion is necessarily valid?

  1. f is a quadratic function
  2. g(f(x)) = x for all x
  3. g is injective on its domain Correct answer.
  4. f and g are both linear

Explanation

If f(g(a)) = f(g(b)), then applying the given identity gives a = b, so g must be injective. This one-sided inverse condition guarantees injectivity of g but does not force g(f(x)) = x everywhere.