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IG MATH: A&G AWAY TESTS
📐 SCIMAiQ AWAY TEST 1
⏱️ 30 minutes
📊 Total marks: 18
🧮 Calculator allowed
📌 Topic focus: inequalities, derivatives, turning points
✏️ Answer all questions. Use black/dark blue pen. Show all necessary working.
Numerical answers: 3 significant figures or 1 decimal for angles. π = 3.142 or calculator value.
Name: _________________________ Adm No.: _________________________
1 [4]
(a) Write down the inequality shown on the number line. [2]
Inequality: .............................
(b) Find the integer values of x that satisfy −5 ≤ 3x + 1 < 10. [2]
Integers: .............................
2 [5]
y = 2x3 − 5x2 + 3x − 4
(a) Find dy/dx. [2]
dy/dx = ................................
(b) Calculate the gradient of the curve when x = 2. [1]
Gradient = ................................
(c) Find the coordinates of the point on the curve where the gradient is 1. [2]
( ......... , ......... )
3 [3]
Given y = 4xp − q x3 and dy/dx = 20x4 + 6x2.
Find the value of p and the value of q.
p = ............ , q = ............
4 [6]
A curve has equation y = x3 − 6x2 + 5x.
(a) Find the coordinates of the two turning points. Show all steps clearly. [4]
(b) Determine whether each turning point is a maximum or a minimum. Give reasons. [2]
📘 SOLUTIONS · 30‑minute test
Inequalities · Derivatives · Gradients · Turning points
1 [4]
(a) Number line shows an open circle at 1 and a closed circle at 4. The interval is highlighted between them.
Inequality: 1 < x ≤ 4.
(b) −5 ≤ 3x + 1 < 10
Subtract 1 from all parts: −6 ≤ 3x < 9
Divide by 3: −2 ≤ x < 3
Integer values: −2, −1, 0, 1, 2.
1(a) 1 < x ≤ 4 1(b) −2, −1, 0, 1, 2
2 [5]
y = 2x³ − 5x² + 3x − 4
(a) dy/dx = 6x² − 10x + 3
(b) At x = 2: 6(4) − 10(2) + 3 = 24 − 20 + 3 = 7
(c) Set dy/dx = 1: 6x² − 10x + 3 = 1 → 6x² − 10x + 2 = 0
Divide 2: 3x² − 5x + 1 = 0
Using quadratic formula: x = [5 ± √(25 − 12)] / 6 = [5 ± √13] / 6
Numerical values: x ≈ (5+3.606)/6 = 1.434, x ≈ (5−3.606)/6 = 0.232
Find corresponding y:
For x = 1.434: y = 2(1.434)³ −5(1.434)² +3(1.434)−4 ≈ 2(2.95) −5(2.056) +4.302−4 = 5.90 −10.28 +0.302 ≈ −4.08
For x = 0.232: y = 2(0.0125) −5(0.0538) +0.696−4 ≈ 0.025 −0.269 +0.696−4 = −3.548
Coordinates (to 3 s.f.): (1.43, −4.08) and (0.232, −3.55)
2(a) 6x²−10x+3 2(b) 7 2(c) (1.43,−4.08) and (0.232,−3.55)
3 [3]
y = 4xᵖ − q x³
dy/dx = 4p·xᵖ⁻¹ − 3q x²
Given dy/dx = 20x⁴ + 6x²
Compare powers:
x⁴ term → p−1 = 4 ⇒ p = 5. Then 4p = 4·5 = 20, matches coefficient.
x² term → −3q = 6 ⇒ q = −2.
Thus p = 5, q = −2.
p = 5 , q = −2
4 [6]
y = x³ − 6x² + 5x
(a) dy/dx = 3x² − 12x + 5
Set dy/dx = 0: 3x² − 12x + 5 = 0
Quadratic formula: x = [12 ± √(144 − 60)] / 6 = [12 ± √84] / 6 = [12 ± 2√21] / 6 = 2 ± (√21)/3
√21 ≈ 4.583, so (√21)/3 ≈ 1.528
x₁ = 2 + 1.528 = 3.528, x₂ = 2 − 1.528 = 0.472
Find y‑coordinates:
y(3.528) = (3.528)³ −6(3.528)² +5(3.528)
≈ 43.91 −6(12.45) +17.64 = 43.91 −74.70 +17.64 = −13.15
y(0.472) = (0.472)³ −6(0.472)² +5(0.472)
≈ 0.105 −6(0.223) +2.36 = 0.105 −1.338 +2.36 = 1.127
Turning points: (3.53, −13.2) and (0.472, 1.13) (3 s.f.)
(b) Second derivative: d²y/dx² = 6x − 12
At x = 3.53: 6(3.53)−12 = 21.18−12 = 9.18 > 0 → minimum
At x = 0.472: 6(0.472)−12 = 2.83−12 = −9.17 < 0 → maximum
4(a) (3.53,−13.2) and (0.472,1.13) 4(b) (3.53,−13.2) min, (0.472,1.13) max
📐 SCIMAiQ AWAY TEST
⏱️ 25 minutes
📊 Total marks: 14
🧮 Calculator allowed
📌 Topic focus: Functions (all aspects)
✏️ Answer all questions. Use black/dark blue pen. Show all necessary working.
Numerical answers: 3 significant figures where appropriate.
Name: _________________________ Adm No.: _________________________
1 [5]
f(x) = 4x − 3 g(x) = x2 + 2x h(x) = 12/x
(a) Find f(5). [1]
.......................
(b) Find g(−3). [1]
.......................
(c) Find h(4). [1]
.......................
(d) Find f(2x). [1]
.......................
(e) Find g(x − 1). [1]
.......................
2 [4]
f(x) = 3x + 5 g(x) = (x − 2)/4
(a) Find f−1(x). [2]
.......................
(b) Find g−1(x). [2]
.......................
3 [3]
f(x) = 2x − 1 g(x) = x2 + 3
(a) Simplify f(x) + g(x). [1]
.......................
(b) Simplify f(x) − g(x). [1]
.......................
(c) Simplify 2f(x) − 3g(x). [1]
.......................
4 [2]
h(x) = x2 − 4x + 1
(a) Find h(2a). [1]
.......................
(b) Find h(3 − x). [1]
.......................
📘 SOLUTIONS · Functions focus test
25 minutes · Total marks 14
1 [5]
f(x) = 4x − 3 g(x) = x2 + 2x h(x) = 12/x
(a) f(5) = 4·5 − 3 = 20 − 3 = 17
(b) g(−3) = (−3)2 + 2(−3) = 9 − 6 = 3
(c) h(4) = 12/4 = 3
(d) f(2x) = 4(2x) − 3 = 8x − 3
(e) g(x − 1) = (x − 1)2 + 2(x − 1)
= (x2 − 2x + 1) + (2x − 2) = x2 − 1
1(a) 17 1(b) 3 1(c) 3 1(d) 8x−3 1(e) x²−1
2 [4]
f(x) = 3x + 5 g(x) = (x − 2)/4
(a) Let y = 3x + 5. Swap x and y: x = 3y + 5
3y = x − 5 → y = (x − 5)/3
So f−1(x) = (x − 5)/3
(b) Let y = (x − 2)/4. Swap x and y: x = (y − 2)/4
4x = y − 2 → y = 4x + 2
So g−1(x) = 4x + 2
2(a) (x−5)/3 2(b) 4x+2
3 [3]
f(x) = 2x − 1 g(x) = x2 + 3
(a) f(x) + g(x) = (2x − 1) + (x2 + 3) = x2 + 2x + 2
(b) f(x) − g(x) = (2x − 1) − (x2 + 3) = 2x − 1 − x2 − 3 = −x2 + 2x − 4
(c) 2f(x) − 3g(x) = 2(2x − 1) − 3(x2 + 3) = 4x − 2 − 3x2 − 9 = −3x2 + 4x − 11
3(a) x²+2x+2 3(b) −x²+2x−4 3(c) −3x²+4x−11
4 [2]
h(x) = x2 − 4x + 1
(a) h(2a) = (2a)2 − 4(2a) + 1 = 4a2 − 8a + 1
(b) h(3 − x) = (3 − x)2 − 4(3 − x) + 1
= (9 − 6x + x2) − 12 + 4x + 1
= x2 + (−6x + 4x) + (9 −12 + 1) = x2 − 2x − 2
4(a) 4a²−8a+1 4(b) x²−2x−2