show your working and give your answer correct to three ...show your working and give your answer...
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WJEC January 2014 C2 Solutions ©AHB
1. Use the Trapezium Rule with five ordinates to find an approximate value for the integral
∫ √
Show your working and give your answer correct to three decimal places.
When √
When √
√
When √
√
When √
√
When √
√
∫
{ }
∫ √
{ √
(√
√ √
)}
∫ √
{ }
{ }
∫ √
2 4 3.5 3 2.5
Formula booklet page 7
WJEC January 2014 C2 Solutions ©AHB
2. Find all the values of in the range satisfying
The angles and are the three angles of a triangle. Given that and that , find the values of and . Give each angle correct to the nearest degree.
EITHER OR
(
) (
)
, 246.42 , 311.81
Angles cannot be negative
Angles in a triangle add up to :
Substituting into
Let ,
WJEC January 2014 C2 Solutions ©AHB
3. The sum of the third and eighth terms of an arithmetic series is zero. The sum of the fifth, seventh and tenth terms of the series is 22. Find the first term and the common difference of the series.
The first term of another arithmetic series is 9 and the common difference is 2. The sum of the first terms of this arithmetic series is 3 times the sum of the first terms of the series. Find the value of .
The general formula for a term in an arithmetic series is .
Using this formula, the third term is equal to and the eighth term is equal to .
Fifth term
Substituting into ,
First term, . Common difference, .
when and .
[ ]
Substituting and into
[ ]
[ ]
[ ]
or
Formula booklet page 2
Formula booklet page 2
WJEC January 2014 C2 Solutions ©AHB
4. A geometric series has first term and common ratio . Prove that the sum of the first terms is given by
The fourth term of a geometric series is and the seventh term is 4.
(i) Find the common ratio of the series.
(ii) Find the sum to infinity of the series.
(
(i) The general term for in a geometric series is .
(ii) The sum to infinity of a geometric series is
| | .
Substituting and
into
:
( )
Formula booklet page 2
Formula booklet page 2
WJEC January 2014 C2 Solutions ©AHB
5. The diagram shows a sketch of the triangle with The point is on such that
(i) By applying the cosine rule in each of the triangles and , show
that
and find a similar expression for
(ii) Noting that and are angles on a straight line, use the expressions derived in part (i) to write down an equation satisfied by . Hence show that .
Find the area of triangle . Give your answer correct to two decimal places.
(i) Recall that the cosine rule is .
Using this rule to find ,
QED
To find
WJEC January 2014 C2 Solutions ©AHB
(ii) Angles on a straight line add up to .
(
)
(
)
(
) (
)
Using
(
) (
)
(
)
as cannot be negative. QED
Recall that the area of a triangle is equal to
.
Applying this rule to triangle :
From part (i),
(from part (ii))
(
)
(2dp)
This is k ‘
negative argument ’
Let
WJEC January 2014 C2 Solutions ©AHB
6. Find ∫(
) .
The diagram below shows a sketch of the curve with equation
.
The shaded region is bounded by the curve, the -axis and the lines . Find the area of the shaded region.
∫(
) ∫(
)
∫ ∫ ∫
∫ (
)
Integrate between and .
∫
∫ (
)
[
] [
]
[
] [
]
∫
WJEC January 2014 C2 Solutions ©AHB
7. Given that , show that
.
Solve the equation
.
Show your working and give your answer correct to three decimal places.
Solve the equation
Let
Taking the th root of both sides,
As and ,
QED
Taking log of both sides,
(
)
(3dp)
If ,
then
(
)
WJEC January 2014 C2 Solutions ©AHB
8. The circle has centre and equation
(i) Write down the coordinates of .
(ii) The point has coordinates and lies on . Find the equation of the tangent to at .
The line has equation . Show that and do not intersect. (i)
Completing the square:
(ii) The radius of the circle is perpendicular to the tangent at as shown. The gradient of the radius, i.e. the normal to the tangent, at point can be found by:
Substituting , and
into :
Equation of a tangent
WJEC January 2014 C2 Solutions ©AHB
Substituting into (from part :
( )
If the line intersects the circle , then the above quadratic equation will have at least one solution.
Looking at the discriminant, :
As the discriminant is negative, there are no real solutions to this quadratic equation, so and do not intersect. QED
√
If the discriminant is negative, then we
would be taking the square root of a
negative number in the quadratic formula,
which is impossible.
Looking at the graphs of and ,
it is also clear that they do not
intersect.
WJEC January 2014 C2 Solutions ©AHB
9.
The diagram shows two concentric circles with a common centre . The radius of the larger circle is 7cm and the radius of the smaller circle is 4cm. The points and lie on the larger circle and and cut the smaller circle at the points and respectively. The area of the shaded region is . Find the perimeter of .
Let angle .
Area of a sector
Area of shaded region
radians
Length of cm
Length of cm
Length of Length of cm
Perimeter of
cm
WJEC January 2014 C2 Solutions ©AHB
10. The th term of a number sequence is denoted by The th term of the sequence satisfies
for all positive integers . Given that ,
evaluate and ,
describe the behaviour of the sequence and hence, without carrying out any further calculation, write down the value of
which means the pattern,
will repeat indefinitely.
When is a multiple of 3,
When is 1 more than a multiple of 3,
When is 2 more than a multiple of 3,
For , is 2 more than a multiple of 3.
where is a positive integer.
,