WJEC May 2014 C2 Solutions
1.
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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.
Use your answer to part
integral
to deduce an approximate value for the
∫
1
2
1.5
3
2.5
When
When
Formula
booklet
page 7
When
When
When
{
∫
}
{
∫
}
{
}
{
}
∫
Using the first rule of logarithms,
∫
∫
∫
.
WJEC May 2014 C2 Solutions
2.
Find all the values of
The angle
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in the range
satisfying
satisfies
√
√
and
Given that
, find the value of .
Find all the values of
in the range
satisfying
Let
EITHER
,
OR
( )
(
, 300
)
, 221.41
√
√
( )
√
The only value of
.
between
√
( )
which satisfies both equations is
WJEC May 2014 C2 Solutions
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(
)
WJEC May 2014 C2 Solutions
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3. The diagram below shows a sketch of the triangle
,
with
,
Show that
Given that
cm, use the cosine rule to find the exact value of .
Using the sine rule:
From the question,
.
The formula for the cosine rule is
Substituting the values we know into this formula,
(
As
)
, we can substitute
into the formula.
(
(
)
)
as cannot be negative
WJEC May 2014 C2 Solutions
4.
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An arithmetic series has first term and common difference .
Prove that the sum of the first terms of the series is given by
[
]
The first term of an arithmetic series is 3 and the common difference is 2.
The sum of the first terms of the series is 360.
Write down an equation satisfied by . Hence find the value of .
The tenth term of another arithmetic series is seven times the third term.
The sum of the eighth and ninth terms of the series is 80. Find the first term
and common difference of this arithmetic series.
[
lots of
]
[
]
QED
and
[
Substituting the above values into
[
],
]
[
]
Formula
booklet
page 2
as
cannot be negative.
The general term of an arithmetic series is
.
,
Substituting,
When
,
40d=240
WJEC May 2014 C2 Solutions
©AHB
5. A geometric series has first term
and third terms of the series is
series is 8.
Prove that
and common ratio . The sum of the second
. The sum of the fifth and sixth terms of the
.
Find the sum to infinity of the series.
The general term of a geometric series is
Formula
booklet
page 2
√
Formula
booklet
page 2
√
| |
The sum to infinity of a geometric series is
Recall that
Substituting
. With
and
into
,
(
)(
(
)( )
(
)
,
(
)
(
.
))
WJEC May 2014 C2 Solutions
6.
Find ∫ (
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√ )
The diagram shows a sketch of the curve
and the line
. The line and the curve intersect at the points and .
(i) Find the coordinates of
and .
(ii) Find the area of the shaded region.
∫(
√ )
∫(
)
∫
∫(
√ )
(i) At the points of intersection,
The -coordinates of the points of intersection are
When
When
,
,
Points of intersection:
and
and
.
WJEC May 2014 C2 Solutions
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(ii) The area of the shaded region is equal to the total area under the curve
between
and
minus the area of trapezium
.
∫
[
]
(
)
(
(
)
(
(
)
)
Area of
)
̇
∫
can also be found by
WJEC May 2014 C2 Solutions
7.
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Solve the equation
Show your working and give your answer correct to three decimal places.
The positive numbers
and are such that
(i) Express as a power of .
(ii) Using your answer to (i), evaluate
Take
of both sides of the equation:
(
(
)
)
( )
(3dp)
(i)
(ii)
If
then
,
WJEC May 2014 C2 Solutions
8.
The circle
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has centre
and radius 15.
(i) Show that
and
and radius 5. The circle
has centre
touch, justifying your answer.
(ii) Given that the circles touch at the point
common tangent at .
, find the equation of the
Gareth, who has been asked by his teacher to investigate the properties of
another circle , claims that the equation of this circle is given by
Show that Gareth cannot possibly be correct.
(i) The distance between the centres of
and
found by using the following formula:
(points
and ) can be
√
√
√
√
Since the distance between and is 20, and the radii of and
up to 20
, the circles must touch at exactly one point.
(ii) The gradient of line
is equal to
The common tangent at
is perpendicular to line
,
Common tangent
At
From this,
,
, which is impossible.
add
WJEC May 2014 C2 Solutions
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9.
The diagram shows a circle with centre and radius
. The points
̂
are on the circle and
radians. The tangent to the circle at
intersects the line
produced at the point .
and
Find an expression in terms of for
(i) the area of sector
(ii) the length of ,
(iii) the area of triangle
,
.
Given that the area of the shaded region is 95.22
, find the value of .
(i) Area of sector
(ii) Using
(iii) Area of a triangle
Area of
found by
Remember to
change the
calculator to radians.
can also be
( )
Area of shaded region = Area of triangle
From part
(i) and
(iii),
Area of sector
and
as cannot be negative.
.