“FAILURE IS NOT AN OPTION BUT IT IS A PREVAILING ATTITUDE”
ALGEBRA
1.The remainder when the expression is divided by is five times the remainder when the same expression is divided by , and 12 less than the remainder when the same expression is divided by . Find the values of and .
2.Solve the equation:
Thus,
3.Solve for :
,
(factorise by grouping)
Either Or So,
4.Find if form an A.P.
5.The first, second, third and th terms of a series are 4, -3, -16 and respectively. Find and the sum of terms of the series.
Eliminate ,
Eliminate ,
6.The coefficients of the 5th, 6th and 7th terms in the expansion of are in an A.P. Find .
Coefficient of , Coefficient of , and Coefficient of
Since these are in an A.P, then, we have
,
7.Given that the first three terms in the expansion in ascending powers of of are the same as the first three terms in the expansion of , find the value of and .
Thus
Equating coefficients, we get
,
, for thus
8.Prove by induction that is divisible by 7 for all .
Let
For , divisible by 7.
For , divisible by 7.
For ,
When
Thus
which is still divisible by 8. Thus if its true for , then is divisible by 7 for all .
9.Given that is a factor of , solve the equation .
If is a factor, then, , thus the quadratic factor is
By long division,
Thus,
The roots are
10.Define the locus of the following points;
i)ii)
ANALYSIS
1.Sketch the curve , clearly stating the asymptotes.
Intercepts: when so, , thus curve crosses the axes at origin.
Vertical asymptotes: , so vertical asymptotes at
To find the slant asymptotes: we perform long division:
We get , thus as , so the line is the slant asymptote.
For the turning points, we have ,
these occur when , ,
, ,
i.e curve has a local minimum at and a local maximum at and a negative point of inflexion at
2.Show that has turning points at intervals of in . Distinguish between the maximum and minimum and show that these values are in a geometrical progression with common ratio .
, for t.p ,
so , thus ,
or thus values of form a geometric progression with common ratio .
3.A curve is given parametrically by Show that the gradient at the point parameter is and that the equation of the tangent to the curve at this point is
,, thus,
,
Equation of the tangent is
as required.
4.Show that the particular solution to the equation , for , is .
, thus
, for
,
,
5.A sample of a radioactive radium losses mass at a rate which is proportional to the amount present. If is the mass after time years,
(i)Form a differential equation connecting and .
(ii)If initially the mass of radium is , deduce that .
(iii)Given that its initial mass halved in 6000 years, find the value of the constant , hence determine the number of years it takes for of radium to reduce to
, so when thus
or thus
When
so
When ,
6.Prove that (hint: Use the substitution .
=
== =
=
7.Evaluate:i)ii)
8.Evaluate:
i)
= Let :
=
=
=
= =
= =
ii) Let
ALTERNATIVELY:
iii)
9.Use:
i)find
=
= =
= =
ii), use
10.Evaluate:i), use
,
let
=
ii)
iii) use
11.Prove that
12.If prove that
13.Find the solution to the differential equation:
,
Thus,
TRIGONOMETRY
1.Prove that:i) ii)
2.Prove that.
From L.H.S
=
= = as the R.H.S
3.Prove that in any triangle ,
4.Prove that , and and hence solve the equation .
5.Given that in a triangle , where and are sides of the triangle , show that
6.Prove that: .
7.Prove that
R.H.S=x
= =
= = as L.H.S
ALT: From L.H.S.
=
=
= =
==
= as required.
8.Prove that , prove also that and .
9.Prove that if , then . Find all the angles for which satisfy the equation
VECTORS
1.The point divides the line joining and in the ratio . Find and .
2.Show that the lines , intersect, hence, find the position vector of their point of intersection. Find also the Cartesian equation of the plane formed by these two lines.
3a)Determine the equation of the plane through the points , and .
b)A line through the point and parallel to the vector meets the plane in (a) at point . Find:
i)the coordinates of .
ii)The angle between the line and the plane.
4.The points A, B, C and D have coordinates , , and respectively. Find the vector equation of the line PQ where P divides AB in the ratio and Q divides CD in the ratio .
,
,
.
, thus
5.Given that A, B and C are three collinear points whose position vectors are respectively, satisfy the equation , where are scalars, prove that .
6.A plane passes through the point and is perpendicular to the vector . The plane meets the plane in and the plane in . Find:
i)the equation of the plane.ii)the distance .
7.a)Show that the lines and do not intersect. Hence, find the shortest distance between them.
b)Find the equation of the line of intersection of the planes, and .
8a)Show that the vectors form a triangle and determine the area of this triangle.
GEOMETRY
1.Find the equations to the lines through the point (2, 3) which makes angles of 45o with the line .
The equation has gradient .
Its clear from the diagram that there are two possible lines PM, PN which makes an angle 45o with AB, and the tangents of these angles are respectively,
and , hence, if is the slope of PM, gives
,
Similarly, if is the slope of PN, then
The required lines that pass through (2, 3) have slopes 3 and , thus the equations are;
or
or .
2. is a square; is the point and is the point , being the diagonal. Find the equations of the lines and
make angles of with diagonal .
So, the gradient of , so if are the gradients of , then, , where and so since are perpendicular.
Equation of is; ,
Equation of is; ,
3.The line and the curve intersect at the origin and meet again at a point . If is the midpoint of find the locus of .
4.A circle with centre P and radius touches externally both the circles and . Prove that the coordinate of P is .
has centre = and
For has centre = and
If they touch externally, ,
If then
And
We get ……. (3)
And …… (4)
Eqn(3)-eqn(4), we get
thus
coordinate ,
5.Determine the equations of the tangents to the parabola from the point .
Using , comparing and , .
So, , passés through , thus,
the equations of the tangents are:
6.If the tangents at points P and Q on the parabola are perpendicular, find the locus of the mid point of PQ.
Let P and Q be the points, .
Gradient of tangent at P = ,Gradient of tangent at Q =
Hence since tangents are perpendicular, then, i.e
Let the coordinates of the mid point of PQ be
Thus; and
To eliminate and square the second equation, giving
since
So the locus of the midpoint of PQ has the equation,
MATH SEMINAR algebra
10(ii)
-
MATH SEMINAR algebra
10(i)
Let
Equation of a circle with centre
MATH SEMINAR analysis
Let be sin u
But
But x
2
2
2
2
2
2
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© Prepared by Ssekyewa Edward GHS 2012(0775944290) Email: 1