Edutest Year 11 Mathematics Practice

Try each question first, then open the worked solution to check your answer and see how to reach it.

1. 1.In triangle ABC, a=5, b=7, and angle C=60°. Find the length of side c.

   • A√39
   • B√40
   • C√41
   • D√42
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   Answer: A (√39)

   Two sides and the included angle are given, so use the cosine rule: c² = a² + b² − 2ab cos C. Substitute: c² = 25 + 49 − 2 × 5 × 7 × cos 60°. Since cos 60° = 0.5, the last term is 35, so c² = 74 − 35 = 39 and c = √39. Pythagoras does not apply, because C is not 90°.

2. 2.Solve for x: log₃(x) = 4.

   • A12
   • B64
   • C81
   • D243
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   Answer: C (81)

   A logarithm asks 'what power?'. log₃(x) = 4 means 3 raised to the power 4 equals x. So x = 3⁴ = 3 × 3 × 3 × 3 = 81. The answer 12 multiplies 3 × 4, and 64 is 4³, which reverses the base and the power. Rewriting a log as an index statement is the key step.

3. 3.Rationalise the denominator of 3 / (√5 - 1).

   • A(3√5 + 3)/4
   • B(3√5 - 3)/4
   • C(√5 + 1)/2
   • D3√5 + 3
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   Answer: A ((3√5 + 3)/4)

   Multiply the top and bottom by the conjugate, √5 + 1, which removes the surd from the denominator. Bottom: (√5 − 1)(√5 + 1) = 5 − 1 = 4. Top: 3(√5 + 1) = 3√5 + 3. So the result is (3√5 + 3)/4. Option B uses the wrong sign, because the conjugate flips the minus to a plus.

4. 4.If P(A) = 0.5, P(B) = 0.4, and P(A and B) = 0.2, find P(A or B).

   • A0.9
   • B0.8
   • C0.7
   • D0.6
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   Answer: C (0.7)

   Use the addition rule: P(A or B) = P(A) + P(B) − P(A and B). That gives 0.5 + 0.4 − 0.2 = 0.7. You subtract P(A and B) because outcomes in both A and B would otherwise be counted twice. The answer 0.9 simply adds the probabilities and forgets the overlap.

5. 5.If f(x) = 2x + 1, find the inverse function f⁻¹(x).

   • A(x - 1)/2
   • Bx/2 - 1
   • C2x - 1
   • D1 - 2x
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   Answer: A ((x - 1)/2)

   To find an inverse, write y = 2x + 1, swap x and y, then solve for y. Swapping gives x = 2y + 1. Subtract 1: x − 1 = 2y. Divide by 2: y = (x − 1)/2. Check: f(3) = 7, and f⁻¹(7) = (7 − 1)/2 = 3. Option B divides before subtracting, which is the wrong order.

6. 6.If sin(θ) = 3/5 and θ is acute, find cos(θ).

   • A3/5
   • B4/5
   • C5/4
   • D5/3
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   Answer: B (4/5)

   If sin θ = 3/5, picture a right triangle with opposite side 3 and hypotenuse 5. By Pythagoras, the adjacent side is √(25 − 9) = 4, the 3-4-5 triangle. So cos θ = adjacent ÷ hypotenuse = 4/5. You can also use sin²θ + cos²θ = 1: cos²θ = 1 − 9/25 = 16/25.

7. 7.What is the remainder when x³ + 2x² - 5x + 1 is divided by (x - 2)?

   • A5
   • B6
   • C7
   • D8
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   Answer: C (7)

   The remainder theorem says dividing a polynomial p(x) by (x − a) leaves remainder p(a). Here a = 2: p(2) = 2³ + 2(2²) − 5(2) + 1 = 8 + 8 − 10 + 1 = 7. This avoids doing long division. Be careful to use a = 2, not −2, because the factor is (x − 2).

8. 8.Solve for x: 4^(x+1) = 32.

   • A1
   • B1.5
   • C2
   • D2.5
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   Answer: B (1.5)

   Write both sides as powers of 2. Since 4 = 2², the left side is (2²)^(x+1) = 2^(2x+2), and 32 = 2⁵. Equal bases mean equal powers: 2x + 2 = 5, so 2x = 3 and x = 1.5. Check: 4^2.5 = (√4)⁵ = 2⁵ = 32.

9. 9.Find the turning point of the parabola y = x² - 6x + 5.

   • A(3, -4)
   • B(3, 5)
   • C(-3, 32)
   • D(6, 5)
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   Answer: A ((3, -4))

   The turning point of y = ax² + bx + c is at x = −b/(2a) = 6/2 = 3. Substitute: y = 9 − 18 + 5 = −4, so the turning point is (3, −4). Completing the square confirms it: y = (x − 3)² − 4. Option B uses the y-intercept, 5, instead of calculating y at x = 3.

10. 10.If x varies directly with y² and x=12 when y=2, find x when y=3.

    • A18
    • B24
    • C27
    • D36
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    Answer: C (27)

    'x varies directly with y²' means x = ky². Use x = 12 when y = 2 to find k: 12 = k × 4, so k = 3. Then for y = 3: x = 3 × 3² = 27. The answer 18 scales x with y instead of y², and 36 squares the wrong value.