Question 1 – Evaluating Functions

In this question, we have two functions:

• (a) f(x) = x² – 1

• (b) f(x) = √(2x + 3)

For each function, we have to find four different values.

Part (i) f(–3) means we put –3 in place of x. We just substitute and simplify.

Part (ii) f(0) means we put 0 in place of x. Again, substitute and simplify.

Part (iii) f(x – 2) means wherever we see x in the function, we replace it with (x – 2). Then we simplify the expression.

Part (iv) f(x² + 3) means we replace x with (x² + 3) in the function. Then we simplify.

For function (b), since it has a square root, we need to make sure the expression inside the square root is not negative. But for evaluation, we just substitute and simplify.

Question 2 – Finding the Difference Quotient

In this question, we have four different functions. For each, we have to find (f(a + h) – f(a)) / h and simplify it. This is called the difference quotient.

Part (i) f(x) = 4x + 7
First, we find f(a + h) by putting (a + h) in place of x. We get 4(a + h) + 7 = 4a + 4h + 7.
Then we find f(a) by putting a in place of x. We get 4a + 7.
Then we subtract: f(a + h) – f(a) = (4a + 4h + 7) – (4a + 7) = 4h.
Then we divide by h: 4h / h = 4. So the answer is 4.

Part (ii) f(x) = sin x
We find sin(a + h) – sin(a) and divide by h. We use the trigonometric identity: sin(a + h) – sin(a) = 2 cos(a + h/2) sin(h/2). Then we divide by h and simplify.

Part (iii) f(x) = x³ + x² – 1
We expand (a + h)³ + (a + h)² – 1, then subtract (a³ + a² – 1), then divide by h, and simplify. Many terms will cancel, and we will be left with an expression in a and h.

Part (iv) f(x) = tan x
We use the identity: tan(a + h) – tan(a) = sin(h) / (cos(a + h) cos(a)). Then we divide by h and simplify.

Question 3 – Expressing Quantities as Functions

In this question, we have three parts where we need to express one quantity as a function of another.

Part (a): Area A of a square as a function of its perimeter P.
A square has four equal sides. If side length is s, then perimeter P = 4s, so s = P/4.
Area A = s² = (P/4)² = P²/16. So A(P) = P²/16.

Part (b): Circumference C of a circle as a function of its area A.
A circle has area A = πr², so r = √(A/π).
Circumference C = 2πr = 2π√(A/π) = 2√(πA). So C(A) = 2√(πA).

Part (c): Surface area S of a cube as a function of its volume V.
A cube has side length s. Volume V = s³, so s = V^(1/3).
Surface area S = 6s² = 6(V^(1/3))² = 6V^(2/3). So S(V) = 6V^(2/3).

Question 4 – Finding Domain and Range

In this question, we have five functions. We need to find the domain (all possible x values) and range (all possible y values) for each.

Part (i) g(x) = 5 – x
This is a linear function. Domain is all real numbers because we can put any x. Range is also all real numbers because as x varies, 5 – x takes all real values.

Part (ii) g(x) = √(x + 2)
The square root requires that x + 2 ≥ 0, so x ≥ –2. So domain is [–2, ∞). For range, square root gives non-negative values, so range is [0, ∞).

Part (iii) g(x) = { 6x + 7, x ≤ –2; 4 – 3x, x > –2 }
This is a piecewise function. Domain is all real numbers because both pieces cover all x values. For range, we find values from both pieces. For x ≤ –2, 6x + 7 gives values from –∞ to –5. For x > –2, 4 – 3x gives values from –∞ to 10. So range is all real numbers.

Part (iv) g(x) = |x – 5|
Absolute value function. Domain is all real numbers. Range is [0, ∞) because absolute value is always non-negative.

Part (v) g(x) = (x + 2) / (3 – x)
The denominator cannot be zero, so 3 – x ≠ 0, meaning x ≠ 3. So domain is all real numbers except 3. For range, we solve for x in terms of y: y = (x + 2)/(3 – x). Then y(3 – x) = x + 2, so 3y – xy = x + 2, 3y – 2 = x + xy = x(1 + y), x = (3y – 2)/(y + 1). Denominator y + 1 ≠ 0, so y ≠ –1. So range is all real numbers except –1.

Question 5 – Finding Values of a and b

We are given f(x) = x³ – ax² + bx + 1. Also given that f(2) = –3 and f(–1) = 0.

First, put x = 2: f(2) = 8 – 4a + 2b + 1 = –3. So 9 – 4a + 2b = –3, which gives –4a + 2b = –12. Divide by –2: 2a – b = 6. (Equation 1)

Next, put x = –1: f(–1) = –1 – a – b + 1 = 0. So –a – b = 0, which gives a + b = 0. (Equation 2)

Now solve the two equations: From equation 2, b = –a. Put this in equation 1: 2a – (–a) = 6, so 3a = 6, a = 2. Then b = –2.

So a = 2 and b = –2.

Question 6 – Stone Falling Problem

A stone falls from height 60m. Height after x seconds is h(x) = 40 – 10x².

Part (i): We put the given values of x:
(a) x = 1: h(1) = 40 – 10(1)² = 40 – 10 = 30m.
(b) x = 1.5: h(1.5) = 40 – 10(2.25) = 40 – 22.5 = 17.5m.
(c) x = 1.7: h(1.7) = 40 – 10(2.89) = 40 – 28.9 = 11.1m.

Part (ii): The stone strikes the ground when height is 0. So we solve 40 – 10x² = 0. This gives 10x² = 40, x² = 4, x = 2 seconds (since time cannot be negative).

Question 7 – Analyzing a Linear Function

We have f(x) = 3x – 5.

Part (i): Domain is all real numbers because it is a linear function. Range is also all real numbers because as x varies, 3x – 5 takes all real values.

Part (ii): Is f one-to-one? Yes, because for any two different x values, the outputs are different. If f(a) = f(b), then 3a – 5 = 3b – 5, which gives a = b. So it is one-to-one.

Part (iii): Is f onto if co-domain is all real numbers? Yes. For any y in real numbers, we can find x = (y + 5)/3 such that f(x) = y. So it is onto.

Question 8 – Analyzing a Rational Function

We have f(x) = (2x – 3)/(x + 1) defined from R to R.

Part (i): Domain is all real numbers except x = –1 because denominator cannot be zero. For range, we solve y = (2x – 3)/(x + 1). Then y(x + 1) = 2x – 3, yx + y = 2x – 3, yx – 2x = –y – 3, x(y – 2) = –y – 3, x = (–y – 3)/(y – 2). This is undefined when y = 2. So range is all real numbers except 2.

Part (ii): Is f onto? The co-domain is R, but range is R – {2}. So f is not onto because the value 2 is never achieved.

Part (iii): Is f one-to-one? Yes. If f(a) = f(b), then (2a – 3)/(a + 1) = (2b – 3)/(b + 1). Cross multiply: (2a – 3)(b + 1) = (2b – 3)(a + 1). Expand: 2ab + 2a – 3b – 3 = 2ab + 2b – 3a – 3. Cancel 2ab and –3 from both sides. Then 2a – 3b = 2b – 3a, which gives 5a = 5b, so a = b. So it is one-to-one.

Question 9 – Showing a Function is Bijective

We have f: R⁺ → R⁺ defined by f(x) = e⁻ˣ.

A function is bijective if it is both one-to-one and onto.

First, one-to-one: If f(a) = f(b), then e⁻ᵃ = e⁻ᵇ. Taking natural log on both sides: –a = –b, so a = b. So it is one-to-one.

Second, onto: For any y in R⁺, we need to find x in R⁺ such that e⁻ˣ = y. Then –x = ln y, so x = –ln y. Since y > 0, ln y can be negative, so –ln y is positive. So x is in R⁺. Thus every y has a pre-image. So it is onto.

Since it is both one-to-one and onto, it is bijective.

Question 10 – Analyzing a Cubic Function

We have g(x) = x³ – 3x from R to R.

For injective (one-to-one): We check if g(a) = g(b) implies a = b. g(a) = g(b) means a³ – 3a = b³ – 3b. Rearranging: a³ – b³ – 3a + 3b = 0. Factor: (a – b)(a² + ab + b²) – 3(a – b) = 0. Factor out (a – b): (a – b)(a² + ab + b² – 3) = 0. This gives a = b OR a² + ab + b² = 3. The second equation can have solutions where a ≠ b. For example, a = 1, b = –2 gives 1 – 2 + 4 = 3. So g(1) = 1 – 3 = –2 and g(–2) = –8 + 6 = –2. So g(1) = g(–2) but 1 ≠ –2. So g is not one-to-one.

For surjective (onto): Cubic functions with real coefficients always have range all real numbers because as x → ∞, g(x) → ∞ and as x → –∞, g(x) → –∞, and it is continuous. So it is onto.

So g is surjective but not injective.