Mathematics 9709/12 February/March 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 1 Pure Mathematics 1, February/March 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
(a) Find the first three terms in the expansion, in ascending powers of x, of 1 + x5 . [1]
(b) Find the first three terms in the expansion, in ascending powers of x, of 1 − 2x6 . [2]
(c) Hence find the coefficient of x2 in the expansion of 1 + x5 1 − 2x6 . [2]
Mark scheme
1(a) 1 + 5 x + 10 x 2 B1
1
1(b) 1 − 12 x + 60 x 2 B2, 1, 0 B2 all correct, B1 for two correct components.
2
1(c) (1 + 5 x + 10 x )(1 − 12 x + 60 x ) leading to 60 ‒ 60 + 10
2 2 M1 3 products required
10 A1 Allow 10x2
2
By using a suitable substitution, solve the equation
4
2x − 32 − − 3 = 0. [4]
2x − 32
tan 1 + 2 sin 1
Mark scheme
2 u = 2x − 3 leading to u 4 − 3u 2 − 4 [ = 0 ] M1 Or u = ( 2 x − 3)2 leading to u 2 − 3u − 4 = 0
[ ]
( u − 4 )( u + 1) [ = 0]
2 2 M1 Or ( u − 4 )( u + 1) [ = 0]
2 x − 3 = [ ±] 2 A1
1 5 A1
x= , only
2 2
4
A line has equation y = 3x + k and a curve has equation y = x2 + kx + 6, where k is a constant.
Find the set of values of k for which the line and curve have two distinct points of intersection. [5]
Mark scheme
4 x + kx + 6 = 3x + k leading to x 2 + x ( k − 3 ) + ( 6 − k ) [ = 0 ]
2 M1 Eliminate y and form 3-term quadratic.
( k − 3) − 4 ( 6 − k ) [ > 0]
2 M1 OE. Apply b 2 − 4 ac .
k − 2k − 15[ > 0] A1 Form 3-term quadratic.
2
( k + 3)( k − 5) [ > 0] A1 Or k = − 3, 5 from use of formula or completing
square.
k < − 3, k > 5 A1 FT Or any correct alternative notation, do not allow
, .
FT for their outside regions.
5
y
6
5
4
3
2 y = f x
1
x
O 1 2 3 4 5 6
In the diagram, the graph of y = f x is shown with solid lines. The graph shown with broken lines is a transformation of y = f x.
(a) Describe fully the two single transformations of y = f x that have been combined to give the resulting transformation. [4]
(b) State in terms of y, f and x, the equation of the graph shown with broken lines. [2]
dy 6
Mark scheme
5(a) (Stretch) (factor 3 in y direction or parallel to the y-axis) B1 B1
4 B1 B1 Allow Translation 4 (units) in x direction.
(Translation) N.B. Transformations can be given in either order.
0
4
5(b) [y =] 3f(x ‒ 4) B1 B1 B1 for 3 , B1 for (x ‒ 4) with no extra terms.
2
A curve is such that = and A 1, −3 lies on the curve. A point is moving along the curve dx 3x − 23 and at A the y-coordinate of the point is increasing at 3 units per second.
(a) Find the rate of increase at A of the x-coordinate of the point. [3]
(b) Find the equation of the curve. [4]
Mark scheme
6(a) dy B1
At x = 1, =6
dx
dx dx dy 1 1 M1 A1 Chain rule used correctly.
= × = ×3= Allow alternative and minimal notation.
dt dy dt 6 2
3
6(b) 6 ( 3 x − 2 ) −2 B1 B1
[ y = ] ÷ ( 3 ) [ + c ]
− 2
−3 = −1 + c M1 Substitute x = 1, y = − 3. c must be present.
y = − ( 3x − 2 ) − 2
−2 A1 OE. Allow f(x)=
4
Functions f and g are defined as follows:
f : x → x2 + 2x + 3 for x ≤ −1, g : x → 2x + 1 for x ≥ −1.
(a) Express f x in the form x + a2 + b and state the range of f. [3]
(b) Find an expression for f −1 x. [2]
(c) Solve the equation gf x = 13. [3]
Mark scheme
7(a) f ( x ) = ( x + 1) + 2
2
B1 B1 Accept a = 1, b = 2 .
Range [of f is ( y )] 2 B1FT OE. Do not allow x 2 , FT on their b.
3
7(b) y = ( x + 1) + 2 leading to x = [ ± ] y − 2 − 1
2 M1 Or by using the formula. Allow one sign error.
f −1 ( x ) = − x − 2 − 1 A1
2
7(c) ( )
2 x 2 + 2 x + 3 + 1 = 13 B1 Or using a correct completed square form of f(x).
2x + 4x − 6[ = 0] leading to ( 2 )( x − 1)( x + 3) [ = 0] B1 Or x = 1, x = − 3 using formula or completing
2
square. Must reach 2 solutions.
x = −3 only B1
3
The points A 7, 1, B 7, 9 and C 1, 9 are on the circumference of a circle.
(a) Find an equation of the circle. [5]
(b) Find an equation of the tangent to the circle at B. [2]
Mark scheme
8(a) Centre of circle is (4, 5) B1 B1
r 2 = ( 7 − 4 ) + (1 − 5 )
2 2 M1 OE. Either using their centre and A or C or using A
and C and dividing by 2.
r =5 A1 FT FT on their (4, 5) if used.
Equation is ( x − 4 ) + ( y − 5 ) = 25
2 2 A1 OE. Allow 52 for 25.
5
8(b) 9−5 4 B1 FT FT for use of their centre.
Gradient of radius = =
7−4 3
3 B1 − 3 x 57
Equation of tangent is y − 9 = − (x − 7) or y = +
4 4 4
2
The first term of a progression is cos 1, where 0 < 1 < 12 π.
1
(a) For the case where the progression is geometric, the sum to infinity is . cos 1
(i) Show that the second term is cos 1 sin2 1. [3]
(ii) Find the sum of the first 12 terms when 1 = 13 π, giving your answer correct to 4 significant figures. [2]
(b) For the case where the progression is arithmetic, the first two terms are again cos 1 and cos 1 sin2 1 respectively.
Find the 85th term when 1 = 13 π. [4]
Mark scheme
9(a)(i) cos θ 1 B1
=
1 − r cos θ
1 − r = cos 2θ leading to r = 1 − cos 2θ M1 Eliminate fractions
r = sin 2θ leading to 2nd term = cos θ sin 2θ A1 AG
3
9(a)(ii) M1 Evidence of correct substitution, use of S formula
π 2 π
12
n
cos 1 − sin
3 0.5 1 − ( 0.75)
12
3 and attempt to evaluate
S12 = =
2 π 1 − 0.75
1 − sin
3
1.937 A1
2
9(b) [ d =] cosθ sin θ − cosθ
2 M1 Use of d = u − u
2 1
1 A1
−
8
1 1 M1 Use of a + 84d with a calculated value of d
[85th term =] + 84 × −
2 8
−10 A1
4
A
ka ka a
E
D
B C
The diagram shows a sector ABC which is part of a circle of radius a. The points D and E lie on AB and AC respectively and are such that AD = AE = ka, where k < 1. The line DE divides the sector into two regions which are equal in area.
(a) For the case where angle BAC = 16 π radians, find k correct to 4 significant figures. [5]
1
(b) For the general case in which angle BAC = 1 radians, where 0 < 1 < 12 π, it is given that > 1. sin 1
Find the set of possible values of k. [3]
Mark scheme
10(a) 1 π M1 Attempt to find the area of ΔADE .
( ka ) sin
2
Δ ADE =
2 6
1 2 2 A1 OE.
k a
4
1 2 π B1
Sector ABC = a
2 6
1 2 2 1 2π M1 OE. For 2 × ΔADE = sectorABC with at least one
2× k a = a
4 2 6 correct area.
π A1
k = = 0.7236
6
5
10(b) 1 1 M1 Condone omission of ‘2’ or ‘1/2’ on LHS for M1
( ka ) sin θ = a 2θ
2
2× only.
2 2
θ A1
k2 =
2 sin θ
1 1 A1 1
k2 > leading to <k < 1 OE. Accept k > or k > 0.707 (AWRT) or
2 2 2
1
0.707(AWRT) < k < 1 or k > OE
2
3
y
A x
O
−1 −3
The diagram shows the curve with equation y = 9 x 2 − 4x 2 . The curve crosses the x-axis at the point A.
(a) Find the x-coordinate of A. [2]
(b) Find the equation of the tangent to the curve at A. [4]
(c) Find the x-coordinate of the maximum point of the curve. [2]
(d) Find the area of the region bounded by the curve, the x-axis and the line x = 9. [4]
Additional Page
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
11(a) −1 −
3 −
3
9 x 2 − 4 x 2 = 0 leading to 9 x 2 ( x − 4 ) = 0 M1 OE. Set y to zero and attempt to solve.
x = 4 only A1 From use of a correct method.
2
11(b) dy 1 − 32 −
5
1 −
3
−
5
= 9 − x + 6x 2 B2, 1, 0 B2; all 3 terms correct: 9, − x 2 and 6x 2
dx 2 2
B1; 2 of the 3 terms correct
1 + 6 = 9 M1 Using their x = 4 in their differentiated expression
At x = 4 gradient = 9 − and attempt to find equation of the tangent.
16 32 8
9 A1 9x 9
Equation is y = (x − 4) or y = − OE
8 8 2
4
11(c) −
5
1 M1 dy
9x 2 − x + 6 = 0 Set their to zero and an attempt to solve.
dx
2
x = 12 A1 Condone ( ±)12 from use of a correct method.
2
11(d) 1 1 B2, 1, 0 1
−
1
−
−21
−
3
x 2
4 x 2 x 2 −4 x 2
,
9 x − 4 x 2
dx = 9 1 −
1
B2; all 3 terms correct: 9,
1
−
1
− 2 2
2 2
B1; 2 of the 3 terms correct
8 M1 Apply limits their 4 → 9 to an integrated
9 6 + − ( 4 + 4 ) expression with no consideration of other areas.
3
6 A1 Use of π scores A0
4
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Mathematics 9709/12 for February/March 2021 is out of 75 marks, with 1 hour 50 minutes to answer it.
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