2 (A) Sketch the Graphs of and on a Single Diagram
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H1 JC1 Promotional Examinations 2006 – Marking Schemes
1The function f is defined by
(i)With the aid of a graphing calculator, sketch the graph of, labelling any asymptote, stationary point and axial intercept. [2]
Hence, state the range of f.[1]
(ii)Give a reason why f is a function.[1]
(iii)The domain of f is now restricted to , (where k is a constant) so as to produce f1.
What is the maximum value of kfor f1-1 to exist? Justify your answer.[2]
(i)
Range of f is
(ii)Every member of its domain maps onto a single image.
(Any vertical line cuts the graph only once).
(iii).
That is the largest value of k for which f1 is one-one.
2 (i) The diagram below shows the graph of .
Sketch the graph of, showing clearly the asymptotes.
[2]
(ii) Sketch the graphs of and on a single diagram. [1]
Hence solve the inequality . [3]
(i)
(ii)
= 0.571, = 2
Hence, or.
3Differentiate the following with respect to x,
(i) ,[2]
(ii) ,[3]
(iii) .[3]
(i)
(ii)
(iii)
Alternative method: Using Quotient Rule
4The functions g and h are defined by
, where ais a constant.
(i)Sketch the graph of g, and write down its range.[2]
(ii)Find g-1.[3]
(iii)Find the minimum value of a for hg-1 to exist, explaining briefly your choice.[2]
For your chosen value of a, find hg-1.[2]
(i) .
(ii) Let .
Hence,
(iii) For hg-1 to exist, .
Hence,
5Cylindrical tins, each 7D tall and measuring D in diameter, are used by a factory for packing golf balls for sale. The tins are then batched into cartons to be sold to wholesale dealers. Each carton contains 10 packed tins.
Assuming each golf ball to be spherical with a diameter of D, state M, the maximum number of golf balls that can be packed into each cylindrical tin. [1]
An attempt was made to determine the value of D. When the amount of water used to fill up a packed tin is transferred to a cylindrical measuring vessel measuring D tall and 6 units in diameter, the vessel becomes exactly one-third full.
Without using a graphic calculator, find the value of D.[5]
A flaw in one of the machines had resulted in the production of defective golf balls, each weighing 42 g, which is 3 g less than a golf ball without defects.
One carton weighs 3.129 kg. Assuming the weight of the tins and carton to be negligible, determine the number of defective golf balls present. [3]
M = .
Volume of water in vessel = ----- (1)
Volume occupied by empty space = ----- (2)
Equating (1) & (2):=
units
rejecting D = 0 and , hence D = 4 units.
Let B = number of defective balls, N = number of non-defective balls.
Each carton weights 3.129 kg ----- (3)
Each carton contains 10 packed tins----- (4)
Solving (3) & (4), N = 63,
B = 7
6Sketch the graph of , indicating clearly the co-ordinates of all turning points. [3]
State the line of symmetry for the above graph.[1]
On the same axes, sketch the graph of .
Hence determine the number of roots for the equation . [2]
Find the root(s) of the equation . [2]
Find the range of values of k for which the equation has no real roots.[2]
Line of symmetry: x = 0.
Number of roots: 2
Roots are -0.703 & 1.21.
is quadratic curve with maximum point .
Hence for to have no real roots, .
7 (i) Using a non-calculator method, solve[3]
(ii)Using a graphical method, solve the inequality[3]
(iii)The diagram below shows a circle with radius. Given thatAB is the diameter of the circle, and such that P is any other point on the circle, find the radius of the circle.
[5]
(i)
(ii)
(iii) Note that APB is a right-angled triangle.
Solving,
8(a) Find the following integrals:
(i) [2]
(ii) [2]
(iii) [3]
8(b) Find .[2]
Hence show that the exact value of is .[2]
8a(i)
(ii)
(iii)
8b
Hence,
9The figure above shows a square BCDE of sides with fixed length 2 units
inscribed in an isosceles triangle FGH in which FG = FH and angle HFG = 2.
I is the foot of he perpendicular from F to HG.
(i)Express FI and HI in terms of x, where x = tan. Hence show that
the area A of the triangle FGH is given by. [5]
Find the minimum value of A as x varies.[5]
(ii)The value of is increasing at a rate such that = 2 units/s, where t denotes time.
Find the rate at which the area A is changing when is 60o.[4]
(i) [M1]
FI = [A1]
HI = FI tan = [B1]
Hence [M1]
= [A1]
[M1]
(since is acute, tan > 0) [M1, A1]
.
ORAlternative Method
Tangent / /Hence, A is minimum.
(ii)Given: units/s,
[M1]
When = 60o, x = [B1]
[M1]
The area is increasing at a rate of units2/s when is 60o.
10(i) Sketch the graphs of xy = 2 and on the same diagram. [2]
(ii) Find the point(s) of intersection of the two graphs.[2]
(iii) Calculate the area of the finite region enclosed by the curve xy = 2 and the line.[3]
(iv) Find the equation of the normal to the curve xy = 2 at where a is a constant.[3]
Determine the negative value of a for which this normal is perpendicular to.[2]
Hence, find the area bounded by the normal, the x-axis and the line x = a.[4]
(i) (ii) Intersections at and.
(iii) Area enclosed
(Alternatively, GC may be used [M1])
(iv)
Equation of normal:
Reject value .
Hence, equation of normal:.
When y = 0, x = 31.5.
When .
Area enclosed =.
1