WJEC May 2014 C1 Solutions
1. The points
and
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have coordinates
and
respectively.
(i) Find the gradient of
(ii) Find the equation of
The line is perpendicular to
and intersects the -axis at the point
. The lines
and intersect at the point .
(i) Write down the equation of .
(ii) Show that
has coordinates
(iii) Find the length of
(c) The line
.
and the length of
is extended to the point
.
so that
is the mid-point of
.
(i) Find the coordinates of .
(ii) Write down the geometrical name for the quadrilateral
.
(i) Gradient of
(ii) Using the equation
,
(
)
Equation of
(i) As and
are perpendicular, the gradient of line is 2.
Using point
Using
,
(ii) At the point of intersection, both lines share a common point.
in
When
,
.
Point
QED
WJEC May 2014 C1 Solutions
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(iii) Length of a line
Length of
√
√
√
√
Length of
√
√
√
√
√
√
√
(i) To find the coordinates of a midpoint, Midpoint
(
)
and
Point
(ii)
and
so the
is a kite.
(
)
WJEC May 2014 C1 Solutions
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2. Simplify
√
√
(√
√
√
)
√
√
√
( √
)
√
( √
)
√
√
√
√
√
√
√
√
Multiply the
numerator and
denominator by
the conjugate of
the denominator.
√
√
√
(√
√
( √
√
)
√
√ )
√
√
√
√
√
√
√
√
√
√
√
√
√
√
√
WJEC May 2014 C1 Solutions
3. The curve
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has equation
.
The point has coordinates
equation of the normal to at .
The point
lies on
and lies on the curve . Find the
and is such that the tangent to
at
has equation
,
where is a constant. Find the coordinates of
Differentiating
with respect to ,
At the point
and the value of .
.
,
Gradient of tangent at
Gradient of normal at
Using
,
Comparing the equation of a line
,
.
This means that
When
At ,
,
to the given equation
WJEC May 2014 C1 Solutions
4.
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Write down the expansion of
including the term in .
in ascending powers of
Showing all your working, substitute an appropriate value for
expansion in part
to find an approximate for
.
up to and
in your
| |
Formula booklet
page 2
, with
Substituting
into
WJEC May 2014 C1 Solutions
5.
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Express
in the form
constants whose values are to be found.
Use your answer to part
[
to find the greatest value of
]
[
[
, where
]
]
, with
and are
.
WJEC May 2014 C1 Solutions
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6. Given that the quadratic equation
has no real roots, show that
Find the range of values of
with
satisfying this inequality.
and
is a quadratic in the form
.
No real roots, means that
Inequality signs change
when multiplying or
dividing by negative
numbers.
or
,
WJEC May 2014 C1 Solutions
7.
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Given that
find
Differentiate
√
from first principles.
with respect to
Where
is a small
increase in and
is
a small increase in .
Take
As
Let
from both sides, where
,
.
and
√
(
)
(
)
(
)
WJEC May 2014 C1 Solutions
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8. Solve the equation
.
Let
In order to solve this equation, we need to find a value of
We can do this by using trial and error.
for which
is a solution.
is a factor of the equation.
Because
Compare
is a factor of
terms in
,
and
Compare constant terms in
Compare
terms in
can be rewritten as:
:
and
:
and
:
Substituting
(
)
.
WJEC May 2014 C1 Solutions
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9. The diagram shows a sketch of the graph
The graph passes through
the points
and
and has a maximum point of
Sketch the following graphs, using a separate set of axes for each graph. In
each case, you should include the coordinates of the stationary point and
the coordinates of the points of intersection of the graph with the -axis.
(i)
(ii)
Hence write down one root of the equation
(i)
, this means that the graph will shift 4 units to the left. In
other words, the value of each point decreases by 4.
For those who are interested,
the actual equation of
is:
WJEC May 2014 C1 Solutions
(ii)
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, this means that the graph will flip upside down (about the
-axis) and stretch by a factor of 2 in the direction. In other words, the
value of each point is multiplied by
.
means the graph of
values will increase by 4). The point
a point on the graph of
.
is a root of the equation.
will move 4 units upwards (
will shift to
), which is also
The other root is
.
WJEC May 2014 C1 Solutions
10. The curve
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has equation
Show that
point.
has only one stationary point. Find the coordinates of this
Verify that this stationary point is a point of inflection.
Sketch the graph of
indicating the coordinates of the stationary paper.
At stationary points,
.
stat point when
There is only one solution of the equation
there is only one stationary point in curve .
, therefore
At
(
)
0
(
)
Direction of graph
At
, the graph is upward sloping, due to a positive value of
At
, the graph is stationary, since
At
, the graph is upward sloping, due to a positive value of
.
.
.
This means the stationary point is a point of inflection.
Note: When verifying whether a stationary point is a minimum or a maximum,
can be used.
However the same method cannot be used to verify points of inflection. You should not fall
into the trap of thinking that if
the stationary point is always a point of inflection.
Think of
which has a minimum point.
WJEC May 2014 C1 Solutions
Curve
is a positive cubic with a point of inflection at
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.