Questions on functions, inverses and the modulus

Question 1

With f(x) = 3x − 1 and g(x) = x², find fg(2) and gf(2).

Accepted Answer

fg means g first, so g(2) = 4 and then f(4) = 11. The other order gives f(2) = 5 and then g(5) = 25. B1 B1 for the two values. The two answers differ, as composites generally do.

Question 2

With f(x) = 3x − 1 and g(x) = x², find fg(x) and gf(x) in general.

Accepted Answer

fg(x) = f(x²) = 3x² − 1, and gf(x) = (3x − 1)². B1 B1 for the two composites. The inner function's output is fed through the outer one whole, brackets and all. Writing gf(x) as 3x² − 1 as well is the slip that shows the bracket was dropped.

Question 3

Each of f(x) = 2x − 5, g(x) = (x − 1)² and h(x) = x³ has domain all real numbers. State whether each is one-one or many-one, and for each many-one function give two inputs that share an output.

Accepted Answer

f is one-one: a line with non-zero gradient takes each height once, since 2a − 5 = 2b − 5 forces a = b. g is many-one: g(0) = 1 and g(2) = 1, two inputs and one output. h is one-one: cubing is increasing, so distinct inputs give distinct cubes. B1 for f, B1 for g with a pair such as 0 and 2, B1 for h. The test is not whether the formula contains a square but whether two inputs can collide; x³ keeps its sign and its order, and (x − 1)² folds the line in half at x = 1. Only the one-one functions here have inverses on the whole domain.

Question 4

The function f(x) = x² − 4x + 7 has domain all real numbers. Find the range of f.

Accepted Answer

Complete the square: x² − 4x + 7 = (x − 2)² + 3. The square is never negative and is zero at x = 2, so the smallest output is 3 and every larger value is reached. The range is f(x) ≥ 3. M1 for the completed square, A1 for the range. The range is a set of outputs, so it is stated in terms of f(x) or y rather than x, and it is the minimum of the curve that fixes it, not the y-intercept 7.

Question 5

Find the inverse of f(x) = x² − 3 with domain x ≥ 0, stating the domain of the inverse.

Accepted Answer

Write y = x² − 3 and solve for x, giving x = √(y + 3). The positive root is the one to take, because the domain restricts x to be at least 0. Swapping letters, f⁻¹(x) = √(x + 3). The range of f is y ≥ −3, so the domain of f⁻¹ is x ≥ −3. Domain and range swap roles along with the letters, and the domain of the inverse has to be stated. M1 for making x the subject, A1 for √(x + 3), A1 for taking the positive root, B1 for the domain x ≥ −3.

Question 6

Find the inverse of f(x) = (2x + 1)/5.

Accepted Answer

y = (2x + 1)/5 unwinds a step at a time. Multiply by 5 to get 5y = 2x + 1, then x = (5y − 1)/2. Swapping letters, f⁻¹(x) = (5x − 1)/2. The inverse undoes the machine in reverse order, multiplying by 5 first where f divided by 5 last. M1 A1 for making x the subject, A1 for the inverse.

Question 7

Solve |3x − 2| = x + 4.

Accepted Answer

Take the two arms separately. Original arm: 3x − 2 = x + 4 gives x = 3, and the check |7| = 7 holds. Reflected arm: −(3x − 2) = x + 4 gives 2 − 3x = x + 4, so x = −½, and |−3.5| = 3.5 = −½ + 4 holds too. Both solutions survive, so x = −½ and x = 3. M1 A1 for the first arm, M1 A1 for the reflected arm. Every arm's answer has to be checked back, because reflected algebra can invent crossings the graphs never make.

Question 8

The functions f and g are given by f(x) = 3x − 1 and g(x) = 2/x, x ≠ 0. Find fg(x), and solve fg(x) = 5.

Accepted Answer

fg means g first, so fg(x) = f(2/x) = 3(2/x) − 1 = 6/x − 1. Then 6/x − 1 = 5 gives 6/x = 6, so x = 1. M1 for forming fg(x) with g inside f, M1 for setting the composite equal to 5 and rearranging, A1 for x = 1. Checking, g(1) = 2 and f(2) = 5. The other order gives gf(x) = 2/(3x − 1), and 2/(3x − 1) = 5 leads to x = 7/15, a different equation with a different answer, so the order of the letters is the first thing to read.

Question 9

The function f is given by f(x) = 4/(x − 1) + 2 with domain x > 1. Find f⁻¹(x), and state the domain and range of f⁻¹.

Accepted Answer

Write y = 4/(x − 1) + 2 and make x the subject: y − 2 = 4/(x − 1), so x − 1 = 4/(y − 2) and x = 1 + 4/(y − 2). Swapping letters, f⁻¹(x) = 1 + 4/(x − 2). M1 for making x the subject, A1 for the inverse. On x > 1 the fraction 4/(x − 1) is positive, so the range of f is f(x) > 2, and that becomes the domain of the inverse: x > 2. B1. The domain of f, x > 1, becomes the range of the inverse: f⁻¹(x) > 1. B1. The asymptotes agree: f approaches y = 2, and the inverse has its vertical asymptote at x = 2, the reflection of that line in y = x.

Question 10

Solve the inequality |2x − 5| ≤ 7.

Accepted Answer

A modulus held under a constant is a two-sided interval: |2x − 5| ≤ 7 means −7 ≤ 2x − 5 ≤ 7. M1 for the double inequality. Adding 5 throughout gives −2 ≤ 2x ≤ 12, and halving gives −1 ≤ x ≤ 6. A1 A1, one for each end. Read it as a distance: 2x − 5 lies within 7 of zero. Solving only 2x − 5 ≤ 7 gives x ≤ 6 and misses the left-hand end, where the reflected arm of the V climbs back above 7 at x = −1. The two ends check directly: at x = −1 the modulus is |−7| = 7 and at x = 6 it is |7| = 7.

Question 11

Solve the inequality |x − 1| > 2x − 4.

Accepted Answer

Sketch y = |x − 1|, a V with its vertex at (1, 0), and the line y = 2x − 4, which crosses the x-axis at x = 2 with gradient 2. Original arm: x − 1 = 2x − 4 gives x = 3, and the check |2| = 2 = 2(3) − 4 holds, so (3, 2) is a genuine crossing. M1 for solving an arm, A1 for x = 3. Reflected arm: 1 − x = 2x − 4 gives x = 5/3, but at x = 5/3 the line is at 2(5/3) − 4 = −2/3 while the modulus is 2/3, so this is not a crossing: the reflected arm is only in force for x < 1, and 5/3 is not. B1 for rejecting it with a reason. The V is above the line to the left of the single crossing, so the solution is x < 3. A1. A line with gradient 2 outruns both arms of the V, whose gradients are ±1, so once it is above the V it stays above, which is why there is one crossing and not two. Without the check, x = 5/3 turns into a spurious second boundary and the answer fragments into two pieces.

Question 12

Explain why f(x) = x² with domain all real numbers has no inverse, and state a restriction of the domain that repairs this.

Accepted Answer

An inverse has to send each output back to a single input, but x² sends both 3 and −3 to 9. The function is not one-to-one, so no inverse exists. Restricting the domain to x ≥ 0, or equally to x ≤ 0, keeps one branch of the parabola and the inverse appears. B1 for the two inputs giving one output, B1 for the function not being one-to-one, B1 for a suitable restriction. Saying “because it is a curve” names no property; one-to-one is the property being asked about.

Question 13

Solve |2x + 1| = |x − 4|.

Accepted Answer

Two moduli means two cases, not four, because the overall sign of one side may be absorbed into the other. Same signs: 2x + 1 = x − 4 gives x = −5. Opposite signs: 2x + 1 = −(x − 4) = 4 − x gives 3x = 3, so x = 1. Checking, x = −5 gives |−9| = |−9| and x = 1 gives |3| = |−3|, so both hold and x = −5 or x = 1. M1 A1 for the same-signs case, M1 A1 for the opposite-signs case. Squaring both sides is the clean alternative, since (2x + 1)² = (x − 4)² leads to 3x² + 12x − 15 = 0 and the same pair; it also removes the need to check, because squaring cannot introduce a false root when both sides were already non-negative.

Question 14

The function f is given by f(x) = 2x + 1 with domain x ≥ −½, and the function g by g(x) = x² − 4 with domain x ≥ 0. Find gf(x) and state the range of gf. Explain why gf has an inverse, and find (gf)⁻¹(x).

Accepted Answer

gf means f first: gf(x) = g(2x + 1) = (2x + 1)² − 4, which expands to 4x² + 4x − 3 if wanted. M1 for substituting f into g, A1 for the composite. On the domain x ≥ −½ the bracket 2x + 1 runs from 0 upwards, so its square runs from 0 upwards and the range is gf(x) ≥ −4. B1. The inverse exists because gf is one-one: 2x + 1 is never negative here, and squaring is increasing on non-negative inputs, so distinct inputs give distinct outputs. The restriction on f's domain is what removes the fold in the square. B1 for that reason. To invert, write y = (2x + 1)² − 4, so 2x + 1 = √(y + 4), taking the positive root because 2x + 1 ≥ 0, and x = (√(y + 4) − 1)/2. Swapping letters, (gf)⁻¹(x) = (√(x + 4) − 1)/2, with domain x ≥ −4. M1 for making x the subject, A1 for the inverse with the positive root. The check: gf(0) = −3, and (√(−3 + 4) − 1)/2 = 0. Writing the composite as fg, that is 2(x² − 4) + 1, is the other order and a different function.

Question 15

Solve the inequality |3x − 1| < |x + 3|. Explain why squaring both sides is a reversible step for this inequality, and what would have to be checked if the right-hand side were x + 3 without the modulus.

Accepted Answer

With a modulus on each side, equal moduli and equal squares are one condition, so |3x − 1| < |x + 3| is exactly (3x − 1)² < (x + 3)². M1 for squaring. Expanding, 9x² − 6x + 1 < x² + 6x + 9, so 8x² − 12x − 8 < 0, and dividing by 4 gives 2x² − 3x − 2 < 0. A1 for the quadratic. It factorises as (2x + 1)(x − 2) < 0, an upward parabola below the axis strictly between its roots. M1 for the critical values −½ and 2. So −½ < x < 2. A1. The step is reversible because both sides are non-negative, and for non-negative numbers a < b holds exactly when a² < b²; nothing is added and nothing is dropped. Against x + 3 with no modulus, squaring answers |3x − 1| < |x + 3| instead, a question with possibly more solutions, since a negative right-hand side is silently replaced by its size. Any x found with x + 3 < 0 would have to be discarded by substituting back into the original. Here x + 3 is positive throughout −½ < x < 2, so the two inequalities happen to have the same solution set, but that is a fact established by the check and not by the algebra. B1 for the explanation. A test value confirms the interval: x = 0 gives 1 < 3.