A-Level Unit Test: Algebra and Functions

Partial Fractions

1. Express 2𝑥𝑥−1(𝑥𝑥−1)(2𝑥𝑥−3)

in partial fractions (3) 2. Express 2(𝑥𝑥+5)

8𝑥𝑥2+10𝑥𝑥−3 in partial fractions (3)

3. Express 𝑥𝑥+5

(𝑥𝑥+1)(𝑥𝑥−3)2 in partial fractions (4)

4. Express 25

𝑥𝑥2(2𝑥𝑥+1) in partial fractions (4)

5. Given that

4 3 22

2 23 2 5 4

4 4x x x dx eax bx c

x x− − − +

≡ + + +− − 2x ≠ ±

find the values of the constants a, b, c, d and e. (3) 6. Express 6𝑥𝑥2−𝑥𝑥+1

(3𝑥𝑥−1)(𝑥𝑥+1) = A + 𝐵𝐵

3𝑥𝑥−1+ 𝐶𝐶

𝑥𝑥+1 (4)

7. Given that f(x) = 𝑥𝑥

3+5𝑥𝑥2−2𝑥𝑥−19𝑥𝑥2+7𝑥𝑥+10

.

Show that f(x) can be written in the form f(x) ≡ x + A + 𝐵𝐵𝑥𝑥+2

+ 𝐶𝐶𝑥𝑥+5

, where A, B and C are integers to be found. (5)

Total marks: 26

Mark Scheme

1. 2𝑥𝑥−1

(𝑥𝑥−1)(2𝑥𝑥−3) ≡ 𝐴𝐴(𝑥𝑥−1) + 𝐵𝐵

(2𝑥𝑥−3)2x – 1 = A(2x – 3) + B(x – 1)

M1

x = 1.5 2 = A(0) + B(0.5) B = 4

M1

x = 1 1 = A(-1) + B(0) A = 1

M1

2𝑥𝑥−1(𝑥𝑥−1)(2𝑥𝑥−3) ≡ −1

(𝑥𝑥−1) + 4(2𝑥𝑥−3)

2. 2(𝑥𝑥+5)

8𝑥𝑥2+10𝑥𝑥−3 ≡ 2(𝑥𝑥+5)

(4𝑥𝑥−1)(2𝑥𝑥+3)≡ 𝐴𝐴

(4𝑥𝑥−1) + 𝐵𝐵(2𝑥𝑥+3)

2(x + 5) = A(2x + 3) + B(4x – 1) M1

x = ¼ 10.5 = 3.5A + B(0) A = 3

M1

x = -1.5 7 = A(0) + B(-7) B = -1

M1

2(𝑥𝑥+5)8𝑥𝑥2+10𝑥𝑥−3

≡ 3(4𝑥𝑥−1) - 1

(2𝑥𝑥+3)

3. 𝑥𝑥+5

(𝑥𝑥+1)(𝑥𝑥−3)2 ≡ 𝐴𝐴

(𝑥𝑥+1) + 𝐵𝐵(𝑥𝑥−3) + 𝐶𝐶

(𝑥𝑥−3)2

x + 5 = A(x – 3)2 + B(x + 1)(x – 3) + C(x + 1) M1

x = 3 8 = A(0) + B(0) + C(4) C = 2

M1

x = -1 4 = A(16) + B(0) + C(0) A = 0.25

M1

x = 0, A = 0.25, C = 2 5 = 2.25 + B(-3) + 2 0.75 = -3B B = -0.25

M1

𝑥𝑥+5(𝑥𝑥+1)(𝑥𝑥−3)2

≡ 14(𝑥𝑥 +1) - 1

4(𝑥𝑥 −3) + 2(𝑥𝑥−3)2

4.

25𝑥𝑥2(2𝑥𝑥+1)

≡ 𝐴𝐴𝑥𝑥

+ 𝐵𝐵𝑥𝑥2

+ 𝐶𝐶2𝑥𝑥+1

25 = A(x)(2x + 1) + B(2x + 1) + C(x2)

M1

Let x = 0 25 = A(0) + B(1) + C(0) B = 25

M1

Let x = -12

25 = A(0) + B(0) + C(0.25) C = 100

M1

Compare x2 coefficients: 0 = 2A + C 0 = 2A + 100 A = -50

M1

25𝑥𝑥2(2𝑥𝑥+1)

≡ −50𝑥𝑥

+ 25𝑥𝑥2

+ 1002𝑥𝑥+1

5. Long division:

( ) ( )x x

x x x x x x

x x x

x x xx x x

x xx x

x

- ++ - - - + -

+ -- + +- + +

- -+ -- +

2

2 4 3 2

4 3 2

3 2

3 2

2

2

3 2 70 4 3 2 5 0 4

3 0 122 7 02 0 8

7 8 47 0 28

8 24

M1 M1

A = 3, B = -2, C = 7, D = -8, E = 24 M1 M1

6. 6𝑥𝑥2−𝑥𝑥+1

(3𝑥𝑥−1)(𝑥𝑥+1) ≡ A + 𝐵𝐵

3𝑥𝑥−1+ 𝐶𝐶

𝑥𝑥+1

6x2 – x + 1 = A(3x – 1)(x + 1) + B(x + 1) + C(3x – 1) M1

Compare coefficients of x2: 6 = 2A, A = 3 M1 Let x = -1 8 = A(0) + B(0) + C(-4) C = -2

M1

Let x = 13

43

= 𝐴𝐴(0) + 𝐵𝐵 �43� + 𝐶𝐶(0)

B = 1

M1

6𝑥𝑥2−𝑥𝑥+1(3𝑥𝑥−1)(𝑥𝑥+1)

≡ 2 + 13𝑥𝑥−1

− 2𝑥𝑥+1

7. Long division with 𝑥𝑥

3+5𝑥𝑥2−2𝑥𝑥−19𝑥𝑥2+7𝑥𝑥+10

M1 A1

f(x) = x – 2 + 2𝑥𝑥+1𝑥𝑥2+7𝑥𝑥+10

2𝑥𝑥+1

𝑥𝑥2+7𝑥𝑥+10 ≡ 2𝑥𝑥+1

(𝑥𝑥+5)(𝑥𝑥+2) ≡ 𝐴𝐴

(𝑥𝑥+5) + 𝐵𝐵

(𝑥𝑥+2)

2x + 1 = A(x + 2) + B(x + 5)

M1

x = -2 -3 = A(0) + B(3) B = -1

M1

x = -5 -9 = A(-3) + B(0) A = 3

M1

f(x) ≡ 𝑥𝑥3+5𝑥𝑥2−2𝑥𝑥−19𝑥𝑥2+7𝑥𝑥+10

≡ x – 2 + 3 (𝑥𝑥+5)

- 1 (𝑥𝑥+2)

A = -2, B = -1, C = 3

A-Level Unit Test: Algebra and Functions

Composite Functions 1. g(x) = 𝑥𝑥

𝑥𝑥+3 + 3(2𝑥𝑥+1)

𝑥𝑥2+𝑥𝑥−6, x > 3

a. Show that (x) = 𝑥𝑥+1

𝑥𝑥−2, x > 3 (4)

b. Find the range of g (2) c. Find the exact value of a for which g(a) = g-1(a) (4)

2. The function f has domain -2 ≤ x ≤ 6 and is linear from (-2, 10) to (2, 0) and from (2, 0) to (6, 4). A sketch of the graph y = f(x) is shown in figure 1.

Figure 1

a. Write down the range of f b. Find ff(0) (1) A function g is defined by (2)

g : x → 4+3𝑥𝑥5−𝑥𝑥

, x ∈ ℝ, x ≠ 5 c. Find g-1(x) (3) d. Solve the equation gf(x) = 16 (5)

3. The function f is defined by

f : x → 3−2𝑥𝑥𝑥𝑥−5

, x ∈ ℝ, x ≠ 5 a. Find f-1(x). (3) The function g has domain -1 ≤ x ≤ 8 and is linear from (-1, -9) to (2, 0) and from (2, 0) to (8, 4). Figure 2 shows a sketch of the graph y = g(x)

Figure 2

b. Write down the range of g. (1) c. Find gg(2) (2)

d. Find fg(8) (2) e. On the same diagram sketch the graphs y = g(x) and y = g-1(x). Show the coordinates of each point at which the graph meets or cuts the aces.

(3)

f. State the domain of the inverse function g-1. (1)

4. The function f is defined by

f : x → 3𝑥𝑥−5𝑥𝑥+1

, x ∈ ℝ, x ≠ -1 a. Find f-1(x) (3) b. Show that ff(x) = 𝑥𝑥+1

𝑥𝑥−1 x ∈ ℝ, x ≠ -1, where a is an integer to be found. (4)

The function g is defined by g : x → x2 – 3x, x ∈ ℝ, 0 ≤ x ≤ 5

c. Find the value of fg(2) (2) d. Find the range of g (3) e. Explain why the function g does not have an inverse (1)

5. The functions f and g are defined by

f : x → 2x + ln x, x ∈ ℝ g : x → e2x, x ∈ ℝ

a. Prove that the composite function gf is: (4) gf : x → 4e4x, x ∈ ℝ

b. Sketch the curve with equation y = gf(x), and show the coordinates of the point where the curve cuts the y-axis. (3) c. Write down the range of gf (1) d. Find the value of x for which 𝑑𝑑

𝑑𝑑𝑥𝑥[𝑔𝑔𝑔𝑔(𝑥𝑥)] = 3, giving your answer to 3 significant figures. (4)

6. A function g is defined by

g(x) = 3 + √𝑥𝑥 + 2 , x ≥ -2 a. State the range of g (1) b. Find g-1(x) and state its domain. (3) c. Find the exact value of x for which, g(x) = x (4) d. Hence state the value of a for which g(a) = g-1(a) (1) 7. The functions f and g are defined by:

f : x → 2x + 3, x ∈ ℝ g : x → 3 – 4x, x ∈ ℝ

a. State the range of f (2) b. Find fg(1) (2) c. Find g-1, the inverse function of g. (2) d. Solve the equation (5)

gg(x) + [g(x))]2 = 0 8. The functions f and g are defined by:

f : x → 3x + ln x , x > 0, x ∈ ℝ g : x → 𝑒𝑒𝑥𝑥2, x ∈ ℝ

a. Write down the range of g (1) b. Show that the composite function fg is defined by (2)

fg : → 𝑥𝑥2 + 3𝑒𝑒𝑥𝑥2, x ∈ ℝ c. Write down the range of fg. (1)

Total marks: 82

Mark Scheme

1a.

𝑥𝑥𝑥𝑥+3

+ 3(2𝑥𝑥+1)𝑥𝑥2+𝑥𝑥−6

= 𝑥𝑥𝑥𝑥+3

+ 3(2𝑥𝑥+1)(𝑥𝑥+3)(𝑥𝑥−2)

M1

= 𝑥𝑥(𝑥𝑥−2)+3(2𝑥𝑥+1)(𝑥𝑥+3)(𝑥𝑥−2)

M1

= 𝑥𝑥2−2𝑥𝑥+6𝑥𝑥+3

(𝑥𝑥+3)(𝑥𝑥−2)

= 𝑥𝑥2+4𝑥𝑥+3(𝑥𝑥+3)(𝑥𝑥−2)

M1

= (𝑥𝑥+3)(𝑥𝑥+1)(𝑥𝑥+3)(𝑥𝑥−2)

= (𝑥𝑥+1)(𝑥𝑥−2)

M1

1b.

x > 3, g(x) = (3+1)(3−2)

= 4 M1 g(x) = 1 → asymptote Range, 1 < g(x) ≤ 4 M1

1c.

g(x) = 𝑥𝑥+1𝑥𝑥−2

→ y = 𝑥𝑥+1𝑥𝑥−2

x = 𝑦𝑦+1

𝑦𝑦−2

x(y – 2) = y + 1 xy – 2x – y = 1

M1

y(x – 1) = 1 + 2x y = 1+2𝑥𝑥

𝑥𝑥−1 = g-1(x) M1

g(a) = g-1(a) 𝑎𝑎+1𝑎𝑎−2

= 1+2𝑎𝑎𝑎𝑎−1

(a + 1)(a – 1) = (1 + 2a)(a – 2) a2 – 1 = 2a2 – 2 – 3a

M1

a2 – 3a – 1 = 0 a = 3 ± √13

2

As x = a > 3, a = 3+√132

M1

2a.

0 ≤ f(x) ≤ 10 M1 2b.

ff(0) = f(5) (using symmetry of line) M1 f(5) = ¾ way across interval, therefore ff(0) = 3. M1

2c.

y = 4+3𝑥𝑥5−𝑥𝑥

→ x = 4+3𝑦𝑦5−𝑦𝑦

x(5 – y) = 4 + 3y 5x – xy = 4 + 3y

M1

3y + xy = 5x – 4 y(3 + x) = 5x – 4 M1

y = 5𝑥𝑥−43+𝑥𝑥

→ g-1(x) = 5𝑥𝑥−43+𝑥𝑥

M1

2d.

g-1gf(x) = g-1(16)

M1

f(x) = g-1(16) = 5(16)−43+(16)

= 4

M1

Using graph, f(x) = 4 M1 x = 6 or x = 1

2(2) = 2

5

gf(x) = 16 when x = 6 or 25

M1 M1

3a.

f(x) = 3−2𝑥𝑥𝑥𝑥−5

→ x = 3−2𝑦𝑦𝑦𝑦−5

M1 x(y – 5) = 3 – 2y xy – 5x = 3 – 2y xy + 2y = 3 + 5x

M1

y(x + 2) = 3 + 5x y = 3+5𝑥𝑥

𝑥𝑥+2 → f-1(x) = 3+5𝑥𝑥

𝑥𝑥+2 M1

3b.

-9 ≤ f(x) ≤ 4 M1 3c.

g[g(2)] = g(0) M1 = -6 M1

3d.

f[g(8)] = f(4) M1 f(4) = 3−2(4)

4−5= 5 M1

3e.

Attempted reflection in y = x M1 g(x) and g-1(x) drawn correctly M1 All points cutting axes labelled M1

3f.

Domain of g-1(x) = -9 ≤ x ≤ 4 M1

4a. y = 3𝑥𝑥−5

𝑥𝑥+1 → x= 3𝑦𝑦−5

𝑦𝑦+1 M1

x(y + 1) = 3y – 5 xy + x = 3y – 5 3y – xy = x + 5

M1

y(3 – x) = x + 5 y = 𝑥𝑥+5

3−𝑥𝑥 → f-1(x) = 𝑥𝑥+5

3−𝑥𝑥 M1

4b.

f(x) = 3𝑥𝑥−5𝑥𝑥+1

M1

2

2

-6

-6

y = g(x)

y = g-1(x)

y = x

ff(x) = 3(3𝑥𝑥−5𝑥𝑥+1 )−5

(3𝑥𝑥−5𝑥𝑥+1 )+1 M1

ff(x) = (9𝑥𝑥−15−5(𝑥𝑥+1)

𝑥𝑥+1 )

(3𝑥𝑥−5+𝑥𝑥+1𝑥𝑥+1 ) = 4𝑥𝑥−20

4𝑥𝑥−4= 𝑥𝑥−5

𝑥𝑥−1

a = -5 M1

4c.

g(2) = 22 – 3(2) = -2 M1 f(-2) = 3(−2)−5

(−2)+1 = 11 M1

4d.

g(x) = x2 – 3x g(0) = 0 g(5) = 52 – 3(5) = 25 – 15 = 10

M1

x2 – 3x = (x – 1.5)2 – 2.25 Therefore minimum point of curve = (1.5, -2.25) M1

Range is: -2.25 ≤ g(x) ≤ 10 M1 4e.

The function g cannot have an inverse as the function is not one to one and hence the inverse would be one to many, which is not a function. M1

5a.

gf (x) = e2(2x + ln x)

= e4x + 2 ln x

M1

= e4x x e2ln x

M1

= e4x x 2 M1

= 2e4x M1 5b. When x = 0, y = 2e0 = 2 M1 Asymptotes at x = 1, y = 0 M1 M1

5c.

Range: fg(x) > 0 M1 5d. 𝑑𝑑𝑑𝑑𝑥𝑥

gf (x) = 2 x e4x x 4 = 8e4x

M1 M1

8e4x = 3 e4x = 3

8 M1

4x = ln38

x = 14 ln3

8

M1

x

y x = 1

(0, 2)

6a. Range: g(x) ≥ 3 M1

6b.

g(x) = 3 + √𝑥𝑥 + 2 → x = 3 + �𝑦𝑦 + 2 M1 y = g-1(x) = (x – 3)2 - 2 M1 Domain: x ≥ 3 M1

6c.

3 + √𝑥𝑥 + 2 = x √𝑥𝑥 + 2 = x – 3

M1

x + 2 = (x – 3)2 x2 – 6x + 9 – x – 2 = 0 x2 – 7x – 7 = 0 x = 7 ± √21

2

M1

x ≥ 3, therefore x = 7+ √212

M1 6d.

A function and it’s inverse will be equal therefore, g(a) = g-1(a) a = 7+ √21

2

M1

7a.

f(x) ∈ ℝ M1 7b.

g(1) = 3 – (4)(1) = -1 M1 f(-1) = 2(-1) + 3 = 1 M1

7c.

g(x) = 3 – 4x → x = 3 – 4y M1 4y = 3 – x y = 1

4 (3 – x) M1

7d.

gg(x) = 3 – 4(3 – 4x) =3 – 12 + 16 x = 16x – 9 M1 [g(x)] 2 = (3 – 4x)2 = (3 – 4x)(3 – 4x) = 9 + 16x2 – 24x M1 16x – 9 + 9 + 16x2 – 24x = 0 16x2 – 8x = 0 M1

8x(2x – 1) = 0 M1 x = 0 x = 1

2 M1

8a.

g(x) ≥ 1 M1 8b.

3(𝑒𝑒𝑥𝑥2) + ln (𝑒𝑒𝑥𝑥2) M1 = 3𝑒𝑒𝑥𝑥2 + 𝑥𝑥2 M1

8c.

Range: fg(x) ≥ 3 M1

A-Level Unit Test: Algebra and Functions

Modulus Functions

1. Given that f(x) = ln x, x> 0

sketch on separate axes the graphs of, a. y = f(x) (2) b. y = |f(x)| (2) c. y = –f(x – 4) (2) Show, on each diagram, the point where the graph meets or crosses the x-axis. In each case, state the equation of the asymptote. 2. Figure 1 shows part of the curve with equation y = f(x) , x ∈ ℝ Figure 1

The curve passes through the points Q(0, 2) and P(−3, 0) as shown. a. Find the value of ff (−3) (2) On separate diagrams, sketch the curve with equation b. y = f −1(x) (2) c. y = f(|𝑥𝑥|) − 2 (2) d. y = 2f(1

2x) (3)

Indicate clearly on each sketch the coordinates of the points at

which the curve crosses or meets the axes. 3. Find the complete set of values of x for which a. |4x – 3| > 2 – 2x (4)

b. |4x – 3| > 32

– 2x (2)

4. Given that the equation

|2x – a| + b = x + 8

has a solution at x = 0 and a solution at x = c, find c in terms of a. (4)

5. Find the complete set of values of x which satisfy,

a. |2𝑥𝑥 − 5| > 7 (2)

b. |2𝑥𝑥 − 5| > x - 52. Give your answer in set notation. (3)

6a. Given that |𝑥𝑥| = 3, find the possible values of |2𝑥𝑥 − 1| (3)

b. Solve the inequality �𝑥𝑥 − √2� > �𝑥𝑥 + 3√2� (4)

Mark Scheme

1a ln graph crossing x axis at (1,0) and asymptote at x = 0

B1

1b.

Shape including cusp Touches or crosses the x axis at (1,0) Asymptote given as x=0

B1ft

B1ft

B1

1c.

Shape Crosses at (5, 0) Asymptote given as x = 4

B1

B1ft

B1

2a. ff(-3) = f(0) = 2 M1 A1 2b. Shape M1 (0, -3) and (2, 0) labelled A1

2c. Shape M1 (0, 0) labelled A1

(0, -3)(2, 0)

y = f -1(x)

(0,0)

2d. Shape M1 (-6, 0) and (0, 4) labelled A1 3a. Solves 4 3 2 2x x− = − or 3 4 2 2x x− = − to give either value of x

Both 56

x = and 12

x = or 56

x > or 12

x <

M1 M1

12

x < or 56

x > A1

3b.

Draws graph Or solves 121 24 3 xx = −−

to give one solution x = ¾

M1 M1

Accept for all values of x except 34x = Or 3

4( , )x x∈ ≠ , or 3 34 4,x x< > A1

4. States or uses 8a b+ = B1

Attempts to solve 32 82

x a b x− + = + in either x or with x = c

32 8 f ( , )2

c a b c kc a b− + = + ⇒ =

M1

Combines f ( , )kc a b= with 8a b+ = 4c a⇒ = dM1 A1 5a.

(-6, 0)(0, 4)

2x – 5 > 7 2x > 12 x > 6

B1

-(2x – 5) > 7 -2x > 2 x < -1

M1

5b. 2x – 5 > x - 5

2

2x – x > 5 - 52

x > 52

B1

-(2x – 5) > x - 52

-3x > - 152

- x > - 52

x < 52

M1

Set Notation: {x : x < 5

2} U { x : x > 5

2} M1

6a. |𝑡𝑡| = 3, therefore, t = 3 or t = -3. B1 When t = 3, |2𝑡𝑡 − 1| = |2(3) − 1| = |5| = 5 M1 When t = -3, |2𝑡𝑡 − 1| = |2(−3) − 1| = |−7| = 7 M1

6b. (x - √2) 2 > (x + 3√2)2 B1 x2 - 2√2x + 2 > x2 + 6√2x + 18 -8√2x > 16

M1 M1

x < -√2 M1