A-Level Unit Test: Algebra and Functions Partial Fractions
Transcript of A-Level Unit Test: Algebra and Functions Partial Fractions
A-Level Unit Test: Algebra and Functions
Partial Fractions
1. Express 2π₯π₯β1(π₯π₯β1)(2π₯π₯β3)
in partial fractions (3) 2. Express 2(π₯π₯+5)
8π₯π₯2+10π₯π₯β3 in partial fractions (3)
3. Express π₯π₯+5
(π₯π₯+1)(π₯π₯β3)2 in partial fractions (4)
4. Express 25
π₯π₯2(2π₯π₯+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 xβ β β +
β‘ + + +β β 2x β Β±
find the values of the constants a, b, c, d and e. (3) 6. Express 6π₯π₯2βπ₯π₯+1
(3π₯π₯β1)(π₯π₯+1) = A + π΅π΅
3π₯π₯β1+ πΆπΆ
π₯π₯+1 (4)
7. Given that f(x) = π₯π₯
3+5π₯π₯2β2π₯π₯β19π₯π₯2+7π₯π₯+10
.
Show that f(x) can be written in the form f(x) β‘ x + A + π΅π΅π₯π₯+2
+ πΆπΆπ₯π₯+5
, where A, B and C are integers to be found. (5)
Total marks: 26
Mark Scheme
1. 2π₯π₯β1
(π₯π₯β1)(2π₯π₯β3) β‘ π΄π΄(π₯π₯β1) + π΅π΅
(2π₯π₯β3)2x β 1 = A(2x β 3) + B(x β 1)
M1
x = 1.5 2 = A(0) + B(0.5) B = 4
M1
x = 1 1 = A(-1) + B(0) A = 1
M1
2π₯π₯β1(π₯π₯β1)(2π₯π₯β3) β‘ β1
(π₯π₯β1) + 4(2π₯π₯β3)
2. 2(π₯π₯+5)
8π₯π₯2+10π₯π₯β3 β‘ 2(π₯π₯+5)
(4π₯π₯β1)(2π₯π₯+3)β‘ π΄π΄
(4π₯π₯β1) + π΅π΅(2π₯π₯+3)
2(x + 5) = A(2x + 3) + B(4x β 1) M1
x = ΒΌ 10.5 = 3.5A + B(0) A = 3
M1
x = -1.5 7 = A(0) + B(-7) B = -1
M1
2(π₯π₯+5)8π₯π₯2+10π₯π₯β3
β‘ 3(4π₯π₯β1) - 1
(2π₯π₯+3)
3. π₯π₯+5
(π₯π₯+1)(π₯π₯β3)2 β‘ π΄π΄
(π₯π₯+1) + π΅π΅(π₯π₯β3) + πΆπΆ
(π₯π₯β3)2
x + 5 = A(x β 3)2 + B(x + 1)(x β 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
π₯π₯+5(π₯π₯+1)(π₯π₯β3)2
β‘ 14(π₯π₯ +1) - 1
4(π₯π₯ β3) + 2(π₯π₯β3)2
4.
25π₯π₯2(2π₯π₯+1)
β‘ π΄π΄π₯π₯
+ π΅π΅π₯π₯2
+ πΆπΆ2π₯π₯+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
25π₯π₯2(2π₯π₯+1)
β‘ β50π₯π₯
+ 25π₯π₯2
+ 1002π₯π₯+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. 6π₯π₯2βπ₯π₯+1
(3π₯π₯β1)(π₯π₯+1) β‘ A + π΅π΅
3π₯π₯β1+ πΆπΆ
π₯π₯+1
6x2 β x + 1 = A(3x β 1)(x + 1) + B(x + 1) + C(3x β 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
= π΄π΄(0) + π΅π΅ οΏ½43οΏ½ + πΆπΆ(0)
B = 1
M1
6π₯π₯2βπ₯π₯+1(3π₯π₯β1)(π₯π₯+1)
β‘ 2 + 13π₯π₯β1
β 2π₯π₯+1
7. Long division with π₯π₯
3+5π₯π₯2β2π₯π₯β19π₯π₯2+7π₯π₯+10
M1 A1
f(x) = x β 2 + 2π₯π₯+1π₯π₯2+7π₯π₯+10
2π₯π₯+1
π₯π₯2+7π₯π₯+10 β‘ 2π₯π₯+1
(π₯π₯+5)(π₯π₯+2) β‘ π΄π΄
(π₯π₯+5) + π΅π΅
(π₯π₯+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) β‘ π₯π₯3+5π₯π₯2β2π₯π₯β19π₯π₯2+7π₯π₯+10
β‘ x β 2 + 3 (π₯π₯+5)
- 1 (π₯π₯+2)
A = -2, B = -1, C = 3
A-Level Unit Test: Algebra and Functions
Composite Functions 1. g(x) = π₯π₯
π₯π₯+3 + 3(2π₯π₯+1)
π₯π₯2+π₯π₯β6, x > 3
a. Show that (x) = π₯π₯+1
π₯π₯β2, 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 β€ x β€ 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 β 4+3π₯π₯5βπ₯π₯
, x β β, x β 5 c. Find g-1(x) (3) d. Solve the equation gf(x) = 16 (5)
3. The function f is defined by
f : x β 3β2π₯π₯π₯π₯β5
, x β β, x β 5 a. Find f-1(x). (3) The function g has domain -1 β€ x β€ 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 β 3π₯π₯β5π₯π₯+1
, x β β, x β -1 a. Find f-1(x) (3) b. Show that ff(x) = π₯π₯+1
π₯π₯β1 x β β, x β -1, where a is an integer to be found. (4)
The function g is defined by g : x β x2 β 3x, x β β, 0 β€ x β€ 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 β 2x + ln x, x β β g : x β e2x, x β β
a. Prove that the composite function gf is: (4) gf : x β 4e4x, x β β
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 ππ
πππ₯π₯[ππππ(π₯π₯)] = 3, giving your answer to 3 significant figures. (4)
6. A function g is defined by
g(x) = 3 + βπ₯π₯ + 2 , x β₯ -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 β 2x + 3, x β β g : x β 3 β 4x, x β β
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 β 3x + ln x , x > 0, x β β g : x β πππ₯π₯2, x β β
a. Write down the range of g (1) b. Show that the composite function fg is defined by (2)
fg : β π₯π₯2 + 3πππ₯π₯2, x β β c. Write down the range of fg. (1)
Total marks: 82
Mark Scheme
1a.
π₯π₯π₯π₯+3
+ 3(2π₯π₯+1)π₯π₯2+π₯π₯β6
= π₯π₯π₯π₯+3
+ 3(2π₯π₯+1)(π₯π₯+3)(π₯π₯β2)
M1
= π₯π₯(π₯π₯β2)+3(2π₯π₯+1)(π₯π₯+3)(π₯π₯β2)
M1
= π₯π₯2β2π₯π₯+6π₯π₯+3
(π₯π₯+3)(π₯π₯β2)
= π₯π₯2+4π₯π₯+3(π₯π₯+3)(π₯π₯β2)
M1
= (π₯π₯+3)(π₯π₯+1)(π₯π₯+3)(π₯π₯β2)
= (π₯π₯+1)(π₯π₯β2)
M1
1b.
x > 3, g(x) = (3+1)(3β2)
= 4 M1 g(x) = 1 β asymptote Range, 1 < g(x) β€ 4 M1
1c.
g(x) = π₯π₯+1π₯π₯β2
β y = π₯π₯+1π₯π₯β2
x = π¦π¦+1
π¦π¦β2
x(y β 2) = y + 1 xy β 2x β y = 1
M1
y(x β 1) = 1 + 2x y = 1+2π₯π₯
π₯π₯β1 = g-1(x) M1
g(a) = g-1(a) ππ+1ππβ2
= 1+2ππππβ1
(a + 1)(a β 1) = (1 + 2a)(a β 2) a2 β 1 = 2a2 β 2 β 3a
M1
a2 β 3a β 1 = 0 a = 3 Β± β13
2
As x = a > 3, a = 3+β132
M1
2a.
0 β€ f(x) β€ 10 M1 2b.
ff(0) = f(5) (using symmetry of line) M1 f(5) = ΒΎ way across interval, therefore ff(0) = 3. M1
2c.
y = 4+3π₯π₯5βπ₯π₯
β x = 4+3π¦π¦5βπ¦π¦
x(5 β y) = 4 + 3y 5x β xy = 4 + 3y
M1
3y + xy = 5x β 4 y(3 + x) = 5x β 4 M1
y = 5π₯π₯β43+π₯π₯
β g-1(x) = 5π₯π₯β43+π₯π₯
M1
2d.
g-1gf(x) = g-1(16)
M1
f(x) = g-1(16) = 5(16)β43+(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) = 3β2π₯π₯π₯π₯β5
β x = 3β2π¦π¦π¦π¦β5
M1 x(y β 5) = 3 β 2y xy β 5x = 3 β 2y xy + 2y = 3 + 5x
M1
y(x + 2) = 3 + 5x y = 3+5π₯π₯
π₯π₯+2 β f-1(x) = 3+5π₯π₯
π₯π₯+2 M1
3b.
-9 β€ f(x) β€ 4 M1 3c.
g[g(2)] = g(0) M1 = -6 M1
3d.
f[g(8)] = f(4) M1 f(4) = 3β2(4)
4β5= 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 β€ x β€ 4 M1
4a. y = 3π₯π₯β5
π₯π₯+1 β x= 3π¦π¦β5
π¦π¦+1 M1
x(y + 1) = 3y β 5 xy + x = 3y β 5 3y β xy = x + 5
M1
y(3 β x) = x + 5 y = π₯π₯+5
3βπ₯π₯ β f-1(x) = π₯π₯+5
3βπ₯π₯ M1
4b.
f(x) = 3π₯π₯β5π₯π₯+1
M1
2
2
-6
-6
y = g(x)
y = g-1(x)
y = x
ff(x) = 3(3π₯π₯β5π₯π₯+1 )β5
(3π₯π₯β5π₯π₯+1 )+1 M1
ff(x) = (9π₯π₯β15β5(π₯π₯+1)
π₯π₯+1 )
(3π₯π₯β5+π₯π₯+1π₯π₯+1 ) = 4π₯π₯β20
4π₯π₯β4= π₯π₯β5
π₯π₯β1
a = -5 M1
4c.
g(2) = 22 β 3(2) = -2 M1 f(-2) = 3(β2)β5
(β2)+1 = 11 M1
4d.
g(x) = x2 β 3x g(0) = 0 g(5) = 52 β 3(5) = 25 β 15 = 10
M1
x2 β 3x = (x β 1.5)2 β 2.25 Therefore minimum point of curve = (1.5, -2.25) M1
Range is: -2.25 β€ g(x) β€ 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. πππππ₯π₯
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) β₯ 3 M1
6b.
g(x) = 3 + βπ₯π₯ + 2 β x = 3 + οΏ½π¦π¦ + 2 M1 y = g-1(x) = (x β 3)2 - 2 M1 Domain: x β₯ 3 M1
6c.
3 + βπ₯π₯ + 2 = x βπ₯π₯ + 2 = x β 3
M1
x + 2 = (x β 3)2 x2 β 6x + 9 β x β 2 = 0 x2 β 7x β 7 = 0 x = 7 Β± β21
2
M1
x β₯ 3, therefore x = 7+ β212
M1 6d.
A function and itβs inverse will be equal therefore, g(a) = g-1(a) a = 7+ β21
2
M1
7a.
f(x) β β M1 7b.
g(1) = 3 β (4)(1) = -1 M1 f(-1) = 2(-1) + 3 = 1 M1
7c.
g(x) = 3 β 4x β x = 3 β 4y M1 4y = 3 β x y = 1
4 (3 β x) M1
7d.
gg(x) = 3 β 4(3 β 4x) =3 β 12 + 16 x = 16x β 9 M1 [g(x)] 2 = (3 β 4x)2 = (3 β 4x)(3 β 4x) = 9 + 16x2 β 24x M1 16x β 9 + 9 + 16x2 β 24x = 0 16x2 β 8x = 0 M1
8x(2x β 1) = 0 M1 x = 0 x = 1
2 M1
8a.
g(x) β₯ 1 M1 8b.
3(πππ₯π₯2) + ln (πππ₯π₯2) M1 = 3πππ₯π₯2 + π₯π₯2 M1
8c.
Range: fg(x) β₯ 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 = βf(x β 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 β β Figure 1
The curve passes through the points Q(0, 2) and P(β3, 0) as shown. a. Find the value of ff (β3) (2) On separate diagrams, sketch the curve with equation b. y = f β1(x) (2) c. y = f(|π₯π₯|) β 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 β 3| > 2 β 2x (4)
b. |4x β 3| > 32
β 2x (2)
4. Given that the equation
|2x β 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. |2π₯π₯ β 5| > 7 (2)
b. |2π₯π₯ β 5| > x - 52. Give your answer in set notation. (3)
6a. Given that |π₯π₯| = 3, find the possible values of |2π₯π₯ β 1| (3)
b. Solve the inequality οΏ½π₯π₯ β β2οΏ½ > οΏ½π₯π₯ + 3β2οΏ½ (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 xβ = β or 3 4 2 2x xβ = β 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 = ββ
to give one solution x = ΒΎ
M1 M1
Accept for all values of x except 34x = Or 3
4( , )x xβ β , or 3 34 4,x x< > A1
4. States or uses 8a b+ = B1
Attempts to solve 32 82
x a b xβ + = + in either x or with x = c
32 8 f ( , )2
c a b c kc a bβ + = + β =
M1
Combines f ( , )kc a b= with 8a b+ = 4c aβ = dM1 A1 5a.
(-6, 0)(0, 4)
2x β 5 > 7 2x > 12 x > 6
B1
-(2x β 5) > 7 -2x > 2 x < -1
M1
5b. 2x β 5 > x - 5
2
2x β x > 5 - 52
x > 52
B1
-(2x β 5) > x - 52
-3x > - 152
- x > - 52
x < 52
M1
Set Notation: {x : x < 5
2} U { x : x > 5
2} M1
6a. |π‘π‘| = 3, therefore, t = 3 or t = -3. B1 When t = 3, |2π‘π‘ β 1| = |2(3) β 1| = |5| = 5 M1 When t = -3, |2π‘π‘ β 1| = |2(β3) β 1| = |β7| = 7 M1
6b. (x - β2) 2 > (x + 3β2)2 B1 x2 - 2β2x + 2 > x2 + 6β2x + 18 -8β2x > 16
M1 M1
x < -β2 M1