Algebra and Graphs Practice Questions

Step 1: Find two numbers that multiply to 6 and add to −5: −2 and −3

Step 2: (x − 2)(x − 3) = 0

Step 3: x = 2 or x = 3

Answer: x = 2 or x = 3

Step 1: a=2, b=3, c=−5

Step 2: x = (−3 ± √(9+40)) / 4

Step 3: x = (−3 ± √49) / 4 = (−3 ± 7) / 4

Step 4: x = 4/4 = 1 or x = −10/4 = −2.5

Answer: x = 1 or x = −2.5

Step 1: Half of 6 is 3, so (x + 3)² = x² + 6x + 9

Step 2: x² + 6x + 2 = (x + 3)² − 9 + 2 = (x + 3)² − 7

Answer: (x + 3)² − 7

Step 1: Substitute y = x+1 into x² + y² = 25

Step 2: x² + (x+1)² = 25

Step 3: x² + x² + 2x + 1 = 25

Step 4: 2x² + 2x − 24 = 0 → x² + x − 12 = 0

Step 5: (x+4)(x−3) = 0 → x = −4 or x = 3

Step 6: y = −3 or y = 4

Answer: (−4, −3) and (3, 4)

Step 1: Factorise numerator: x² − 9 = (x−3)(x+3)

Step 2: Factorise denominator: x²+5x+6 = (x+2)(x+3)

Step 3: Cancel (x+3): (x−3)/(x+2)

Answer: (x − 3)/(x + 2)

Step 1: y(x − 2) = 3x + 1

Step 2: yx − 2y = 3x + 1

Step 3: yx − 3x = 1 + 2y

Step 4: x(y − 3) = 1 + 2y

Step 5: x = (1 + 2y)/(y − 3)

Answer: x = (1 + 2y)/(y − 3)

Step 1: Factorise: (x−5)(x+1) < 0

Step 2: Critical values: x = 5 and x = −1

Step 3: Parabola is negative between roots

Answer: −1 < x < 5

Step 1: Use difference of squares: (a+b)(a−b) = a² − b²

Step 2: = 4 − 3 = 1

Answer: 1

Step 1: g(2) = 4 − 1 = 3, then fg(2) = f(3) = 6 + 3 = 9

Step 2: f(2) = 4 + 3 = 7, then gf(2) = g(7) = 49 − 1 = 48

Answer: fg(2) = 9, gf(2) = 48

Step 1: Let y = (3x−1)/2

Step 2: 2y = 3x − 1

Step 3: 3x = 2y + 1

Step 4: x = (2y + 1)/3

Step 5: f⁻¹(x) = (2x + 1)/3

Answer: f⁻¹(x) = (2x + 1)/3

Step 1: For √(x−4) to exist, x − 4 ≥ 0, so x ≥ 4

Step 2: Output of square root is always ≥ 0

Answer: Domain: x ≥ 4, Range: f(x) ≥ 0

Step 1: fg(x) = f(x/4 + 2) = 3(x/4 + 2) − 2 = 3x/4 + 6 − 2 = 3x/4 + 4

Step 2: Set equal to g(x): 3x/4 + 4 = x/4 + 2

Step 3: 2x/4 = −2

Step 4: x/2 = −2, x = −4

Answer: x = −4

Step 1: ff(x) = f(f(x)) = f(1/x)

Step 2: = 1/(1/x) = x

Step 3: Since ff(x) = x, f is its own inverse

Step 4: Therefore f is self-inverse

Answer: ff(x) = x, so f(x) = 1/x is self-inverse

Step 1: y = 5/(x−2)

Step 2: y(x−2) = 5

Step 3: x − 2 = 5/y

Step 4: x = 5/y + 2 = (5+2y)/y

Step 5: f⁻¹(x) = (5+2x)/x = 5/x + 2

Step 6: Undefined when x = 0

Answer: f⁻¹(x) = 5/x + 2, undefined when x = 0

Step 1: f(x) = x² + 4x = (x+2)² − 4

Step 2: Minimum occurs when (x+2)² = 0, i.e. x = −2

Step 3: Minimum value = −4

Answer: Minimum value = −4 at x = −2

Step 1: gf(x) = g(2x+1) = (2x+1)²

Step 2: (2x+1)² = 25

Step 3: 2x+1 = ±5

Step 4: 2x+1=5 gives x=2; 2x+1=−5 gives x=−3

Answer: gf(x) = (2x+1)²; x = 2 or x = −3

Step 1: First differences: 5, 7, 9, 11

Step 2: Second differences: 2, 2, 2

Step 3: a = 2/2 = 1, so n² term

Step 4: Sequence − n²: 3−1=2, 8−4=4, 15−9=6, 24−16=8

Step 5: Remaining is 2n

Step 6: nth term = n² + 2n

Answer: n² + 2n

Step 1: 8th term = ar^(n−1) = 5 × 2⁷ = 5 × 128 = 640

Step 2: Sum = a(r^n − 1)/(r − 1) = 5(2⁸ − 1)/(2−1)

Step 3: = 5(256 − 1) = 5 × 255 = 1275

Answer: 8th term = 640, sum = 1275

Step 1: n=1: 10, n=2: 13, n=3: 16, n=4: 19

Step 2: 3n + 7 = 100

Step 3: 3n = 93, n = 31

Answer: First 4 terms: 10, 13, 16, 19; 31st term = 100

Step 1: S_n = n/2(2a + (n−1)d)

Step 2: S_20 = 20/2(2(4) + 19(3))

Step 3: = 10(8 + 57) = 10 × 65 = 650

Answer: 650

Step 1: a = 2, r = 3; nth term = 2 × 3^(n−1)

Step 2: 2 × 3^(n−1) > 10000

Step 3: 3^(n−1) > 5000

Step 4: Try: 3⁷ = 2187, 3⁸ = 6561

Step 5: n−1 = 8, so n = 9; term = 2 × 6561 = 13122

Answer: The 9th term (13 122)

Step 1: Triangular numbers: T_n = n(n+1)/2

Step 2: 15th term = 15 × 16 / 2 = 120

Answer: n(n+1)/2; 15th term = 120

Step 1: Continue: 1,1,2,3,5,8,13,21,34,55,89,144

Step 2: The 12th term is 144

Answer: 144

Step 1: Common difference d = 3

Step 2: First term a = 5

Step 3: nth term = a + (n−1)d = 5 + 3(n−1) = 5 + 3n − 3 = 3n + 2

Step 4: Check: n=1 gives 5 ✓, n=2 gives 8 ✓

Answer: nth term = 3n + 2 (proven)

Step 1: 6x − 3 > 4x + 5

Step 2: 2x > 8

Step 3: x > 4

Answer: x > 4

Step 1: Factorise: (x−2)(x−5) ≤ 0

Step 2: Critical values: x = 2 and x = 5

Step 3: Parabola ≤ 0 between roots

Answer: 2 ≤ x ≤ 5

Step 1: Open circle at −1, closed circle at 3, shade between

Step 2: Set notation: {x : −1 < x ≤ 3}

Answer: {x : −1 < x ≤ 3}

Step 1: Region: above y=x, below y=6, right of y-axis

Step 2: x + y = 5 with y ≥ x: need y ≥ x and x ≥ 0

Step 3: If x=0, y=5 ✓ (5≥0)

Step 4: If x=1, y=4 ✓ (4≥1)

Step 5: If x=2, y=3 ✓ (3≥2)

Step 6: If x=3, y=2 ✗ (2<3)

Answer: (0,5), (1,4), (2,3)

Step 1: −3 < 2n − 1 and 2n − 1 ≤ 7

Step 2: −2 < 2n and 2n ≤ 8

Step 3: −1 < n and n ≤ 4

Step 4: Integers: 0, 1, 2, 3, 4

Answer: n = 0, 1, 2, 3, 4

Step 1: x² − 9 > 0 → (x−3)(x+3) > 0

Step 2: Positive when both factors same sign

Step 3: x > 3 or x < −3

Answer: x > 3 or x < −3

Step 1: y = (x−3)(x+1)

Step 2: y = 0 at x = −1 and x = 3

Step 3: y < 0 between the roots

Answer: −1 < x < 3

Step 1: Perimeter: 2x + 2y ≤ 200 → x + y ≤ 100

Step 2: Constraints: x ≥ 40, y ≥ 0

Step 3: y ≤ 100 − x

Step 4: Area = xy = x(100−x) (maximised when y is max)

Step 5: For x ≥ 40: try x=50, y=50, area=2500

Step 6: Check: at x=40, y=60, area=2400; at x=60, y=40, area=2400

Step 7: Maximum at x = y = 50

Answer: Maximum area = 2500 m² when x = y = 50

Step 1: dy/dx = 9x² − 8x + 7

Answer: dy/dx = 9x² − 8x + 7

Step 1: dy/dx = 3x² − 6

Step 2: At x = 2: 3(4) − 6 = 12 − 6 = 6

Answer: Gradient = 6

Step 1: At x=4: y = 16 − 12 + 5 = 9, point is (4, 9)

Step 2: dy/dx = 2x − 3; at x=4, gradient = 5

Step 3: y − 9 = 5(x − 4)

Step 4: y = 5x − 11

Answer: y = 5x − 11

Step 1: dy/dx = 6x² − 18x + 12 = 0

Step 2: x² − 3x + 2 = 0

Step 3: (x−1)(x−2) = 0, x = 1 or x = 2

Step 4: d²y/dx² = 12x − 18

Step 5: At x=1: 12−18 = −6 < 0, maximum; y = 2−9+12 = 5

Step 6: At x=2: 24−18 = 6 > 0, minimum; y = 16−36+24 = 4

Answer: Maximum at (1, 5), minimum at (2, 4)

Step 1: Volume: x²h = 32, so h = 32/x²

Step 2: SA = base + 4 sides = x² + 4xh = x² + 4x(32/x²) = x² + 128/x

Step 3: dA/dx = 2x − 128/x²

Step 4: Set dA/dx = 0: 2x = 128/x²

Step 5: 2x³ = 128, x³ = 64, x = 4

Step 6: d²A/dx² = 2 + 256/x³ > 0, so minimum

Answer: x = 4 cm

Step 1: dy/dx = 3x² − 12

Step 2: Increasing when dy/dx > 0: 3x² − 12 > 0

Step 3: x² > 4, so x > 2 or x < −2

Answer: x > 2 or x < −2

Step 1: At (0,3): c = 3

Step 2: dy/dx = 2ax + b; at x=2: 4a + b = 0

Step 3: At (2,−1): 4a + 2b + 3 = −1, so 4a + 2b = −4

Step 4: From 4a + b = 0: b = −4a

Step 5: 4a + 2(−4a) = −4: 4a − 8a = −4: −4a = −4: a = 1

Step 6: b = −4

Answer: a = 1, b = −4, c = 3

Step 1: dh/dt = 20 − 10t

Step 2: Set dh/dt = 0: 20 − 10t = 0, t = 2

Step 3: d²h/dt² = −10 < 0, so maximum

Step 4: h = 20(2) − 5(4) = 40 − 20 = 20 m

Answer: Maximum height = 20 m at t = 2 seconds