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STPM Mathematics 954 study materials
STPM Mathematics 954 study resources

STPM Mathematics 954 practice questions

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Question 1Algebra and Functions

The function f is defined by f(x) = 2x - 5 for x ∈ ℝ. Find f⁻¹(x).

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

Explanation

To invert, write y = 2x - 5, then solve for x: x = (y + 5)/2. Swapping the variable name gives f⁻¹(x) = (x + 5)/2. This is the reflection of f in the line y = x.

Question 2Algebra and Functions

Given that f(x) = x² + 3x and g(x) = x - 1, find the composite function fg(x).

  1. x² - x + 2
  2. x² + 3x - 1
  3. x² + 5x + 4
  4. x² + x - 2 Correct answer.

Explanation

The notation fg(x) means f(g(x)), so substitute g(x) into f. Thus f(x - 1) = (x - 1)² + 3(x - 1) = x² - 2x + 1 + 3x - 3 = x² + x - 2.

Question 3Algebra and Functions

Which of the following relations is a one-to-one function from ℝ to ℝ?

  1. f(x) = |x|
  2. f(x) = x³ Correct answer.
  3. f(x) = x²
  4. f(x) = x⁴ + 1

Explanation

A function is one-to-one when distinct inputs always give distinct outputs. The cube function is strictly increasing on ℝ, so f(a) = f(b) implies a = b. The other options are even functions, so f(a) = f(-a) for a ≠ 0.

Question 4Algebra and Functions

The polynomial P(x) = x³ - 4x² + x + 6 has a factor (x - 2). When P(x) is divided by (x - 2), what is the remainder?

  1. 6
  2. 2
  3. 0 Correct answer.
  4. -4

Explanation

By the factor theorem, if (x - 2) is a factor then P(2) = 0, and by the remainder theorem the remainder on division by (x - 2) equals P(2). Computing P(2) = 8 - 16 + 2 + 6 = 0 confirms the remainder is zero.

Question 5Algebra and Functions

Solve the inequality (x - 1)(x + 3) < 0.

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

Explanation

The quadratic has roots at x = -3 and x = 1 and opens upwards, so it is negative strictly between the roots. Hence the solution set is -3 < x < 1. Testing x = 0 gives (-1)(3) = -3 < 0, confirming this interval.

Question 6Algebra and Functions

Express 2x² + 8x + 5 in the form a(x + h)² + k.

  1. 2(x + 2)² + 5
  2. 2(x + 2)² - 3 Correct answer.
  3. (2x + 4)² - 11
  4. 2(x + 4)² - 27

Explanation

Factor out 2 from the x-terms: 2(x² + 4x) + 5. Completing the square inside gives x² + 4x = (x + 2)² - 4, so the expression equals 2(x + 2)² - 8 + 5 = 2(x + 2)² - 3.

Question 7Algebra and Functions

If log₂ x = 5, find the value of x.

  1. 1/32
  2. 32 Correct answer.
  3. 10
  4. 25

Explanation

The equation log₂ x = 5 means 2⁵ = x by the definition of a logarithm. Since 2⁵ = 32, we obtain x = 32.

Question 8Algebra and Functions

The function f(x) = 2x + 1 has domain {x : 0 ≤ x ≤ 3}. State the range of f.

  1. {y : 1 ≤ y ≤ 7} Correct answer.
  2. {y : 1 ≤ y ≤ 3}
  3. {y : 0 ≤ y ≤ 3}
  4. {y : 2 ≤ y ≤ 6}

Explanation

Since f is increasing, the range is found by evaluating f at the endpoints of the domain. f(0) = 1 and f(3) = 7, so the range is {y : 1 ≤ y ≤ 7}.

Question 9Algebra and Functions

Find the partial fractions of (3x + 5)/((x - 1)(x + 2)).

  1. 3/(x - 1) + 2/(x + 2)
  2. 1/(x - 1) + 2/(x + 2)
  3. 2/(x - 1) - 1/(x + 2)
  4. 8/[3(x - 1)] + 1/[3(x + 2)] Correct answer.

Explanation

Writing the decomposition as A/(x - 1) + B/(x + 2) gives A + B = 3 and 2A - B = 5 by comparing coefficients. Solving yields A = 8/3 and B = 1/3, giving 8/[3(x - 1)] + 1/[3(x + 2)].

Question 10Algebra and Functions

The polynomial P(x) = x³ + ax² - 4x + b is divisible by both (x - 1) and (x + 2). Find the values of a and b.

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

Explanation

By the factor theorem, P(1) = 0 and P(-2) = 0. These give 1 + a - 4 + b = 0, so a + b = 3, and -8 + 4a + 8 + b = 0, so 4a + b = 0. Solving simultaneously gives a = -1 and b = 4, so recheck: subtracting gives 3a = -3, a = -1, b = 4.

Question 11Algebra and Functions

Solve the equation 2^(2x) - 5(2^x) + 4 = 0.

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

Explanation

Let u = 2^x. Then u² - 5u + 4 = 0, so (u - 1)(u - 4) = 0, giving u = 1 or u = 4. Hence 2^x = 1 gives x = 0 and 2^x = 4 gives x = 2.

Question 12Algebra and Functions

Find the set of values of k for which the quadratic equation x² + kx + 9 = 0 has two distinct real roots.

  1. k < -6 only
  2. k > 6 only
  3. k < -6 or k > 6 Correct answer.
  4. -6 < k < 6

Explanation

Two distinct real roots require the discriminant b² - 4ac to be positive. Here k² - 36 > 0, so k² > 36, giving k < -6 or k > 6.

Question 13Algebra and Functions

Given that f(x) = 3x - 2 and g(x) = x² + 1, find the value of f⁻¹g(2).

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

Explanation

First compute g(2) = 4 + 1 = 5. The inverse of f is f⁻¹(x) = (x + 2)/3, so f⁻¹(5) = 7/3. Since 7/3 is not listed, recheck: g(2) = 5 and (5 + 2)/3 = 7/3.

Question 14Algebra and Functions

A geometric progression has first term 3 and common ratio 2. Find the least number of terms required for the sum to exceed 1000.

  1. 8
  2. 9 Correct answer.
  3. 10
  4. 7

Explanation

The sum of n terms is Sₙ = 3(2ⁿ - 1)/(2 - 1) = 3(2ⁿ - 1). Requiring Sₙ > 1000 gives 2ⁿ > 334.33, so n > log₂(334.33) ≈ 8.39. The least integer is therefore 9.

Question 15Algebra and Functions

Find the coefficient of x³ in the expansion of (2 + x)⁵.

  1. 80
  2. 10
  3. 40 Correct answer.
  4. 160

Explanation

The general term is C(5,r)·2^(5-r)·x^r. For x³, r = 3, giving C(5,3)·2² = 10·4 = 40.

Question 16Algebra and Functions

Express 2/(x - 3) - 1/(x + 1) as a single fraction in its simplest form.

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

Explanation

Combine over the common denominator (x - 3)(x + 1). The numerator becomes 2(x + 1) - (x - 3) = 2x + 2 - x + 3 = x + 5.

Question 17Algebra and Functions

The function f(x) = x² - 6x + 5 is defined for x ≥ 3. Find f⁻¹(x).

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

Explanation

Complete the square: f(x) = (x - 3)² - 4. For x ≥ 3, let y = (x - 3)² - 4, so (x - 3)² = y + 4 and x - 3 = √(y + 4) since x ≥ 3. Hence f⁻¹(x) = 3 + √(x + 4).

Question 18Algebra and Functions

Find the range of values of x satisfying |2x - 3| ≤ 5.

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

Explanation

The inequality |2x - 3| ≤ 5 is equivalent to -5 ≤ 2x - 3 ≤ 5. Adding 3 gives -2 ≤ 2x ≤ 8, so -1 ≤ x ≤ 4.

Question 19Algebra and Functions

The roots of the equation 2x² - 7x + 3 = 0 are α and β. Find the value of α² + β².

  1. 13/4
  2. 49/4
  3. 25/4
  4. 37/4 Correct answer.

Explanation

By Vieta's formulas, α + β = 7/2 and αβ = 3/2. Then α² + β² = (α + β)² - 2αβ = 49/4 - 3 = 37/4.

Question 20Algebra and Functions

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

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

Explanation

Using the product law, log₃[x(x + 2)] = 1, so x(x + 2) = 3, giving x² + 2x - 3 = 0 and (x + 3)(x - 1) = 0. Since logarithms require x > 0, only x = 1 is valid.

Question 21Algebra and Functions

A sequence is defined by u₁ = 2 and uₙ₊₁ = 3uₙ - 1. Find u₄.

  1. 40
  2. 13
  3. 41 Correct answer.
  4. 14

Explanation

Compute successively: u₂ = 3(2) - 1 = 5, u₃ = 3(5) - 1 = 14, u₄ = 3(14) - 1 = 41.

Question 22Algebra and Functions

Show that the equation x³ - 3x + 1 = 0 has a root in the interval (1, 2), and determine which of the following is a valid conclusion from the intermediate value theorem.

  1. There is at least one root c with 1 < c < 2 Correct answer.
  2. The root is at x = 1.5
  3. There are no roots in (1, 2)
  4. There is exactly one root in (1, 2)

Explanation

Let f(x) = x³ - 3x + 1. Then f(1) = -1 and f(2) = 3, which have opposite signs. Since f is continuous, the intermediate value theorem guarantees at least one root in (1, 2), but not uniqueness.

Question 23Algebra and Functions

Find the set of values of x for which (x + 1)/(x - 2) ≥ 2.

  1. x ≤ 5
  2. x < 2 or x ≥ 5
  3. 2 < x ≤ 5 Correct answer.
  4. 2 ≤ x ≤ 5

Explanation

Rearrange: (x + 1)/(x - 2) - 2 ≥ 0 gives (x + 1 - 2x + 4)/(x - 2) ≥ 0, that is (5 - x)/(x - 2) ≥ 0. The critical values are x = 2 and x = 5, and a sign analysis gives 2 < x ≤ 5, with x = 2 excluded because the expression is undefined.

Question 24Algebra and Functions

The cubic equation x³ + px² + qx + r = 0 has roots α, β, γ with α + β + γ = 3, αβ + βγ + γα = -4 and αβγ = -12. Find the value of α² + β² + γ².

  1. 13
  2. 17 Correct answer.
  3. 9
  4. 25

Explanation

Use the identity α² + β² + γ² = (α + β + γ)² - 2(αβ + βγ + γα). Substituting gives 9 - 2(-4) = 9 + 8 = 17.

Question 25Algebra and Functions

The function f is defined by f(x) = (2x + 1)/(x - 1) for x ≠ 1. Determine f⁻¹(x) and state its domain.

  1. f⁻¹(x) = (x - 1)/(2x + 1), domain x ≠ -1/2
  2. f⁻¹(x) = (x + 1)/(x - 2), domain x ≠ 2 Correct answer.
  3. f⁻¹(x) = (2x + 1)/(x - 1), domain x ≠ 2
  4. f⁻¹(x) = (x + 1)/(x - 2), domain x ≠ 1

Explanation

Let y = (2x + 1)/(x - 1). Then y(x - 1) = 2x + 1, so x(y - 2) = y + 1 and x = (y + 1)/(y - 2). Hence f⁻¹(x) = (x + 1)/(x - 2), defined for x ≠ 2, which is the value excluded from the range of f.

Question 26Algebra and Functions

The sum of the first n terms of an arithmetic progression is given by Sₙ = 3n² + 2n. Find the 10th term of the progression.

  1. 320
  2. 62
  3. 59 Correct answer.
  4. 56

Explanation

The 10th term equals S₁₀ − S₉. Here S₁₀ = 3(100) + 2(10) = 320 and S₉ = 3(81) + 2(9) = 261, so the tenth term is 320 − 261 = 59. Equivalently, the term formula is 6n − 1, giving 59 when n = 10.

Question 27Algebra and Functions

Given that logₐ 8 = 3/2, find the value of a.

  1. 16
  2. 2
  3. 64
  4. 4 Correct answer.

Explanation

Rewriting the logarithm in exponential form gives a^(3/2) = 8. Squaring both sides yields a³ = 64, so a = 4, which checks since 4^(3/2) = (√4)³ = 8. The base must be positive and not equal to 1, and 4 satisfies this.

Question 28Algebra and Functions

The function f is defined by f(x) = x² + 4x + 7 for x ≥ −2. Find f⁻¹(x) and state its domain.

  1. f⁻¹(x) = 2 + √(x − 3), domain x ≥ 3
  2. f⁻¹(x) = −2 + √(x − 3), domain x ≥ 0
  3. f⁻¹(x) = −2 − √(x − 3), domain x ≥ 3
  4. f⁻¹(x) = −2 + √(x − 3), domain x ≥ 3 Correct answer.

Explanation

Completing the square gives f(x) = (x + 2)² + 3, so the range for x ≥ −2 is f(x) ≥ 3. Solving y = (x + 2)² + 3 gives x + 2 = ±√(y − 3), and since x ≥ −2 we take the positive root, so f⁻¹(x) = −2 + √(x − 3) with domain x ≥ 3, matching the range of f.

Question 29Algebra and Functions

Find the set of values of x for which the inequality (x − 1)/(x + 3) < 2 holds.

  1. x < −3 or x > −7 Correct answer.
  2. x > −3
  3. x < −7 or x > −3
  4. −7 < x < −3

Explanation

Rearrange to (x − 1)/(x + 3) − 2 < 0, which simplifies to (−x − 7)/(x + 3) < 0, that is (x + 7)/(x + 3) > 0. The critical values are −7 and −3, and a sign analysis shows the quotient is positive when x < −7 or x > −3.

Question 30Algebra and Functions

The roots of the cubic equation x³ − 6x² + 11x − 6 = 0 are α, β and γ. Find the value of 1/α + 1/β + 1/γ.

  1. 1
  2. 11/6 Correct answer.
  3. 6/11
  4. 6

Explanation

By Vieta's formulas, α + β + γ = 6, αβ + βγ + γα = 11 and αβγ = 6. Since 1/α + 1/β + 1/γ = (αβ + βγ + γα)/(αβγ), substituting gives 11/6.

Question 31Calculus

Find the derivative of f(x) = (3x² + 1)⁵ with respect to x.

  1. 30x(3x² + 1)⁴ Correct answer.
  2. 30x(3x² + 1)⁵
  3. 5(3x² + 1)⁴
  4. 6x(3x² + 1)⁴

Explanation

Apply the chain rule: differentiate the outer power to get 5(3x² + 1)⁴, then multiply by the derivative of the inner function 3x² + 1, which is 6x. The product is 5(3x² + 1)⁴ · 6x = 30x(3x² + 1)⁴.

Question 32Calculus

Evaluate the definite integral ∫₀² (4x³) dx.

  1. 32
  2. 16 Correct answer.
  3. 8
  4. 4

Explanation

An antiderivative of 4x³ is x⁴. Evaluating from 0 to 2 gives 2⁴ − 0⁴ = 16 − 0 = 16.

Question 33Calculus

What is the limit of (sin 3x)/x as x approaches 0?

  1. 3 Correct answer.
  2. 0
  3. 1
  4. It does not exist

Explanation

Using the standard limit sin(u)/u → 1 as u → 0, write (sin 3x)/x = 3 · (sin 3x)/(3x). As x → 0 the factor (sin 3x)/(3x) → 1, so the limit is 3.

Question 34Calculus

Find the gradient of the tangent to the curve y = x³ − 2x at the point where x = 1.

  1. 3
  2. 5
  3. 1 Correct answer.
  4. −1

Explanation

The gradient function is dy/dx = 3x² − 2. Substituting x = 1 gives 3(1)² − 2 = 1, which is the gradient of the tangent at that point.

Question 35Calculus

Determine ∫ (1/x) dx for x > 0.

  1. ln x + c Correct answer.
  2. −1/x² + c
  3. eˣ + c
  4. x⁻² + c

Explanation

The function 1/x has antiderivative ln x for x > 0, since the derivative of ln x is 1/x. The constant of integration c must be included.

Question 36Calculus

The function f is defined by f(x) = x² − 4x + 7. Find the x-coordinate of the stationary point of the curve y = f(x).

  1. 7
  2. 4
  3. −2
  4. 2 Correct answer.

Explanation

Stationary points satisfy f′(x) = 0. Here f′(x) = 2x − 4, so 2x − 4 = 0 gives x = 2.

Question 37Calculus

Differentiate y = e^(2x) cos x with respect to x.

  1. 2e^(2x) cos x + e^(2x) sin x
  2. e^(2x)(2 cos x − sin x) Correct answer.
  3. e^(2x)(cos x − 2 sin x)
  4. −2e^(2x) sin x

Explanation

Apply the product rule with u = e^(2x) and v = cos x. Then u′ = 2e^(2x) and v′ = −sin x, giving y′ = e^(2x)(2 cos x) + e^(2x)(−sin x) = e^(2x)(2 cos x − sin x).

Question 38Calculus

Find ∫ x e^(x²) dx.

  1. (1/2)e^(x²) + c Correct answer.
  2. e^(x²) + c
  3. 2e^(x²) + c
  4. x²e^(x²) + c

Explanation

Let u = x², so du = 2x dx and x dx = du/2. The integral becomes (1/2)∫e^u du = (1/2)e^u + c = (1/2)e^(x²) + c.

Question 39Calculus

Use the first derivative test to classify the stationary point of y = x³ − 3x at x = 1.

  1. Local minimum
  2. Local maximum Correct answer.
  3. Neither maximum nor minimum, since the gradient does not change sign
  4. Point of inflection with zero gradient

Explanation

Here dy/dx = 3x² − 3 = 3(x − 1)(x + 1). For x slightly less than 1 the gradient is negative, and for x slightly greater than 1 it is positive, so the gradient changes from negative to positive, indicating a local minimum at x = 1.

Question 40Calculus

Find the equation of the normal to the curve y = x² at the point (1, 1).

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

Explanation

The gradient of the curve is dy/dx = 2x, which equals 2 at x = 1. The normal has gradient −1/2, so y − 1 = −(1/2)(x − 1), giving y = −(1/2)x + 3/2.

Question 41Calculus

Evaluate ∫₁^e (1/x) dx.

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

Explanation

An antiderivative of 1/x is ln x. Evaluating from 1 to e gives ln e − ln 1 = 1 − 0 = 1.

Question 42Calculus

A curve is defined parametrically by x = t² + 1 and y = t³ − 3t. Find dy/dx in terms of t.

  1. (3t² − 3)/(t² + 1)
  2. (2t)/(3t² − 3)
  3. 3t² − 3
  4. (3t² − 3)/(2t) Correct answer.

Explanation

For parametric curves, dy/dx = (dy/dt)/(dx/dt). Here dy/dt = 3t² − 3 and dx/dt = 2t, so dy/dx = (3t² − 3)/(2t).

Question 43Calculus

Find the area of the region enclosed between the curve y = x² and the line y = x + 2.

  1. 9/2 Correct answer.
  2. 7/2
  3. 3/2
  4. 5/2

Explanation

The curves meet where x² = x + 2, i.e. x² − x − 2 = 0, giving x = −1 and x = 2. The area is ∫₋₁² (x + 2 − x²) dx = [x²/2 + 2x − x³/3]₋₁² = 10/3 − (−7/6) = 9/2.

Question 44Calculus

Differentiate y = ln(3x + 1) with respect to x.

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

Explanation

By the chain rule, the derivative of ln u is (1/u)(du/dx). With u = 3x + 1, du/dx = 3, so dy/dx = 3/(3x + 1).

Question 45Calculus

A curve has equation y = x³ − 6x² + 9x + 1. Find the x-coordinates of its stationary points.

  1. 3 and 6
  2. 1 and 3 Correct answer.
  3. 0 and 2
  4. −1 and −3

Explanation

Set dy/dx = 3x² − 12x + 9 = 0, dividing by 3 gives x² − 4x + 3 = 0, so (x − 1)(x − 3) = 0 and x = 1 or x = 3.

Question 46Calculus

Evaluate the limit of (1 − cos x)/x² as x approaches 0.

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

Explanation

Multiplying numerator and denominator by (1 + cos x) gives (sin²x)/(x²(1 + cos x)). Since (sin x)/x → 1, the expression tends to 1/(1 + 1) = 1/2.

Question 47Calculus

Find the derivative of y = tan(2x) with respect to x.

  1. −2 cosec²(2x)
  2. 2 sec²(2x) Correct answer.
  3. 2 tan(2x) sec(2x)
  4. sec²(2x)

Explanation

The derivative of tan u is sec²u · du/dx. With u = 2x, du/dx = 2, so dy/dx = 2 sec²(2x).

Question 48Calculus

The curve y = x³ + ax + b has a stationary point at (1, 3). Find the values of a and b.

  1. a = 3, b = −1
  2. a = 1, b = 3
  3. a = −3, b = 5 Correct answer.
  4. a = −3, b = 1

Explanation

dy/dx = 3x² + a, and at x = 1 the gradient is zero, so 3 + a = 0 and a = −3. The point (1, 3) lies on the curve, so 1 + a + b = 3, giving 1 − 3 + b = 3 and b = 5.

Question 49Calculus

Evaluate ∫ x ln x dx for x > 0.

  1. (x²/2) ln x − x²/2 + c
  2. (x²/2) ln x − x²/4 + c Correct answer.
  3. (x²/2) ln x + x²/4 + c
  4. x ln x − x + c

Explanation

Integrate by parts with u = ln x and dv = x dx, so du = dx/x and v = x²/2. Then ∫x ln x dx = (x²/2) ln x − ∫(x²/2)(1/x) dx = (x²/2) ln x − x²/4 + c.

Question 50Calculus

Find the equation of the tangent to the curve x² + y² = 25 at the point (3, 4).

  1. 4x − 3y = 0
  2. 3x − 4y = −7
  3. 4x + 3y = 25
  4. 3x + 4y = 25 Correct answer.

Explanation

Differentiate implicitly: 2x + 2y(dy/dx) = 0, so dy/dx = −x/y. At (3, 4) the gradient is −3/4, giving y − 4 = −(3/4)(x − 3), which simplifies to 3x + 4y = 25.