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G.C.E(A/L)-2019
COMBINED MATHEMATICS
PROBLEMS AND SOLUTIONS
Pure Mathematics
Integration
15.(a)Using the substitution x=2sin2+3 for 0
,evaluate
dx=I(say)
x-3=2sin2 ,5-x=5-(2sin2+3)=2(1-sin2)=2cos2,
=tan2,
=tan,
,
4sin2d=2(1-cos2)d,cos2=1-2si,
When x=3=2siθθθθ,when x=4=2si+3,1=2si,
si
then I=
=2[
=2(
=
(b)Using partial fractions ,find
where A the
constant of integration. Let f(t)=
Deduce that f(t)=
ln(t-2)- ln(t-1)+ln2 for t2. f(t)=loge(x-2)
=loge( t-2)-loge (t-1)-(loge1-loge2)=ln(t-2)-ln(t-1)+ln2. Using integration
by parts ,find .Hence find
=(x-k)ln(x-k)-x,=(t-2)ln(t-
2)-t-[(t-1)ln(t-1)-t]+ln2[t]=(t-2)ln(t-2)-(t-1)ln(t-1)+tln2
2
(c) Using the formula
=
,where a and b are constants,
show that
,Hence, find the value of
=J(say),
=
,
=
dx=
=
=
=
,2
,J=
Differentiation
14.(a)Let f(x)=
Show that f (x),the derivative of f(x),is given by
f(x)=
=9
f(x)=
=-18
for x3
Sketch the graph of y=f(x) indicating the asymptotes ,y-intercept and the turning
points given that , the last expression for the second derivative of f(x).
Find the x-coordinates of the point of inflection of the graph of y=f(x)
when x→3 ,f(x)→ so that x=3 is the asymptote. x→,f(x)→0
At x=0 ,f(0)=
so that y=
is the intercept where graph cut the y-axis.
f(x)=0,when x2-4x-1=0,(x-2)2-5=0,x-2=5,x=25,x=5+2 or x=-(5-2) are the
points where graph cut the x-axis.5=2.2360 and therefore 4.2360 ,-0.2360 are
those points numerically.
,when (x+3)=0 or (x-5)=0 that is when x=-3or x=+5
3
When x=-3,
=18
so that x=-3 is a
minima. When x=+5,
so that x=+5 is a
maxima. Therefore the graph has to be drawn according to the information
extracted on the given function and the rest of the answering for the query has to
be decided after it is plotted.f(-3)=-
=-0.8the minima,f(+5)=+
=+4.5 the maxima.
(b)A basin is there in the form of a frustum of a right circular cone with a bottom
.The slant length of the basin is 30cm and the radius of the upper circular
edge is twice the radius of the bottom .Let the radius of the bottom be r cm.
Show that the volume V cm3 of the basin is given by V=
for
0r30 .Find the value of r such that volume of the basin is maximum. Let
the height of the entire cone used to form the frustum is h c.m and the height of
the small cone that has been removed while forming the frustum is H c.m. Let the
half angle of the full cone be and hence the half angle of the removed cone
also.sin=
,tan=
,r0,
h-H==,h=2H,2H-
H=,h=2 ,V=
=
,0r30,
+2r}=
=0,r=c.m,
(600-
(200-,
0 therefore V the volume of the frustum is maximized at
r=10=102.449c.m=24.49c.m,
,,=3.14159
then by integration uv=
=
,
=-
,-
The formulas used in differential and integral calculus last is a new finding.
Quadratic equations
11.(a) Let pR the set of real numbers and 0p1.Show that 1 is not a root of
the equation Let and be the roots of this equation. Show
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that and are both real. Write down + and in terms of p, and show that
.Show also that the quadratic equation whose roots are
,
(x+
(x+
,(x+
x=
since p1,1-p0 the roots and are real. If roots are equal
to 1 then ,,
since p0,(p
,p=
C the set
of complex numbers that contradict the original assumption that pR the set of
real numbers so 1 cannot be the root of the given equation.
(x-)(x-)=+=
=0,+=-
=
,
=
,(x-
,
x+
,
(2
(x-
=
,x=
]
1,,,1-,(1-p)(1+p+
therefore both roots are
positive. For p=1,x=
0,since p0,1
(b)Let c and d be two non-zero real numbers and let f(x)=
given that (x-c) is a factor of f(x) and that, the remainder when f(x) is
divided by (x-d) is cd . Find the values of c and d, find the remainder when f(x) is
divided by (x+2.
f(c)= since (x-c) is a factor
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of f(x),
=0,d=0 or d=-1 since (x-d) must be a factor
of =g(x) so that g(d)=0.Now take c=-2 and d=-1 non zero values.
f(x)=
,
Trigonometry
17.(c) Solve 2
.Hence, show that
cos(
-
=
Let ,2+=
,(x+
,x=-
,tan=x,
tan=1+x=1+
,=
or tan=-1,tan=1+-1=0,=0,=-
cos(
,cos=1-2
=2co
,sin
=
(a)Write down sin(A+B) in terms of sin A, cosA, sin B and cosB, and obtain a
similar expression for sin(A-B).
Deduce that 2sinAcosB=sin(A+B)+sin(A-B) and 2cosAsinB=sin(A+B)-sin(A-
B).Hence, solve 2sin3cos2=sin7 for 0
.
sin(A+B)=sinAcosB+cosAsinB,sin(A-B)=sinAcosB-cosAsinB,sin(A+B)+sin(A-
B)=2sinAcosB,sin(A+B)-sin(A-B)=2cosAsinB,2sin3cos2=sin5+sin=sin7
sin=sin7-sin5=2cos6sin,sin(2cos6-1)=0,sin=0 or 2cos6-
1=0,cos6=
=cos
6=
,=
for 0
.
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(b)In a triangle ABC, the point D lies on AC such that BD=DC and AD=BC. Let
angle BAC= angle ACB =.Using sine rule for suitable triangles ,show that
2sincos=sin(+2).If :=3:2,using the last result in (a) above, show that =
.
for triangle ABD the application of sine rule
and AD=BC,
for triangle ABC the application of sine rule.
,sin2=2sincos,
,If
=3,=2,+2=3+2(2)=3+4=7,therefore =3(
with comparison of
the results under section (a) .Further when two sides of a triangle are equal that is
then an isosceles triangle the corresponding opposite angles to those sides are also
equal. When a side of a triangle is extended the formed exterior angle is equal to
the sum of two angles opposite to the interior angle. Figures has to be sketched by
the reader in each problem.
Complex Numbers
13.(b)Let z,Show that (i) Rez|z|,z=x+iy,x the real part,y the imaginary
part of the complex number z ,i= ,modulus of z=modz=
|z|=x=Real z,(ii)|
for .Let
,
|
=
=
=
=
for
Deduce that Re(
for
Re
Verify that Re(
and show that |
7
Re
=
Re(
Re(
,|
(c)Let =
Express 1+ in the form r(cos+isin);where r(0) and (-
are constants
to be determined .
Using De Moivre’s theorem,show that (1+
1+=1+
1+=1+
i)=
r the modulus and the argument of the complex number . De Moivre’ s theorem
states ,
,
(1+
Z=x+iy=
,i= This can be depicted on a Cartesian
plane taking real part on the x-axis and imaginary part on the y-axis, r on the
diagonal and theta, the angle measured from the positive side of the x-axis. (r,)
polar diagram is known as the Argand diagram in the name of Argand who
discovered it first, the graphical representation of a complex number where x
part is real and y part is pure imaginary.
Series
12.(b)Let
for r the set of positive
integers. Show that .
8
=
=6.
Hence, show that
r=1,
r=2,6
r=3,
………………….
r=n-2,
r=n-1,
r=n,
_____________________________
for n
Let
Deduce that
.Hence, show
that, the infinite series
r=1,
r=2,
r=3,
…………………..
r=n-2,
r=n-1,
r=n,
________________________
=
for n
9
=
,
so that series converges to finite ratio 5/144.
Coordinate Geometry
Straight line and Circle
16.Write down the coordinates of the point of intersection A of the straight
lines 12x-5y-7=0 and y=1 .
12x-5(1)-7=0 ,12x-12=0,x=1,A(1,1)
Let l be the bisector of the acute angle formed by these lines. Find the equation of
the straight line l.
The gradient of the straight line y=1 be since it subtend an angle
zero with the x axis.
The gradient of the straight line 12x-5y-7=0,y=
,
,Let the angle in
between these two straight lines be =.tan=tan
,tan=tan(
=
=
5tan
,6
(3tan
)(2tan
,3tan
,tan
the acute angle, tan
0 the obtuse angle ,therefore the
gradient of the angle bisector of the acute angle is tan
of the straight line l and it passes through the point A(1,1) &
it’s equation
,x=3+1,y=2+1 so that, the point
P(3+1,2+1) the parametric form of it’s Cartesian coordinates, where R the
set of Real numbers .
Let B(6,0).Show that, the equation of the circle with the points B and P as ends
of a diameter can be written as S+U=0,where S and
U-3x-2y+18 .
If we consider a point Q(x, y) which lies on the given circle then line QB is
perpendicular to the line QP and therefore the product of their gradients must be
equal to -1.tan9==
=
y(y-(2+1))=-(x-6)(x-(3+1)),
,
10
S
Deduce that S=0 is the equation of the circle with AB as a diameter. For
=0,PA(1,1),S0,
,
Show that U=0 is the equation of the straight line through B, perpendicular to l.
The gradient of l is equal to m=+
,let gradient of the straight line
perpendicular to l ,m
=
,B(6,0),
-3x+18=2y,2y+3x-18=0,that U-3x-2y+18=0
Find the coordinates of the fixed point which is distinct from B, and lying on the
circles with the equation S+U=0 for all R.
The fixed point must be (1,1)A which is a point independent of .
Find the value of such that the circle given by S=0 is orthogonal to the circle
given by S+U=0.
S
,g=-
,S+U=,
=
2
,when they are orthogonal they intersect perpendicularly to each other then
=
2g,-7.-
,49+21+1+2=24+36,13=26,=2
(x+ g,
the radius of the circle with center at (-g,-f).
Metrices
13.(a)Let A=
be
metrices such that A ,a, bR the set of real numbers. Show that a=2 and
b=1.In transpose of the matrix B the .
then
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2a-3=b,a-4=-2,a=4-2=2,a=b+1,2=b+1,b=2-1=1,2a-3=1,2a=4,a=2 so that
a=2,b=1.Show also that , Let
=
,p-q=1,-2p+2q=0,p=q,0=1,r-s=0,r=s,-2r+2s=1,
-2r+2r=0=1,so that, the system of equations is inconsistent and therefore inverse
of the matrix C does not exist.
Let P=
Write down and find the matrix Q such that 2P(Q+3I)=P-I,
where I is the identity matrix of order 2.
P=
,Let
=I=
,-k=0,k=0,-m=1,m=-1,-
=1,-=1,l=-2,-
,
,2PQ+6PI=P-I,2PQ+6P-P=-I
2,Q=-
(5I+
Q=-
=
Permutations and Combinations
12.(a) Let and be the two sets given by {A,B,C,D,E,1,2,3,4} and
{F,G,H,I,J,5,6,7,8} respectively. It is required to form a password consisting of
six elements taken from of which three are different letters and three are
different digits. In each of the following cases, find the number of different such
passwords that can be formed :
(i) all six elements are chosen only from ,
(ii)three elements are chosen from .
(i)={A,B,C,D,E,1,2,3,4}
{A,B,C,D,E,F,G,H,I,J,1,2,3,4,5,6,7,8}
,C denotes
combinations, factorial six is due to permuting each six character
combinations.
, five C three means number of ways of selecting three
letter combinations out of five letters, four C three means number of ways of
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selecting three digit combinations out of four digits.
(ii)228800=57600 twice of the answer in(i),n!=n(n-1)(n-2)…….3.2.1=factorial n
A,B,C,D,E,5,6,7,8}, are two sets from both that
leads to the so far given answer. But it must be a lesser estimate .If we form the
number of passwords from
that must be the answer .
Applied Mathematics
Statistics
17.(b) Let the mean and the standard deviation of the set of values
{
of the set of values { ,where is a constant.
=
=mean,=
=standard deviation, averaging sum of squares
is due to the negative and positive signs that can arise while taking the difference
in between the element values and the mean value.=
-2
=mean of the
square values –square of the mean value= the standard deviation
,=
=square of the constant into the standard deviation
Monthly salaries of 50 employees at a certain company are summarized in the
following table :
Monthly Salary average value of the range Number of Employees
(in thousand rupees)
5-15 10 9
15-25 20 11
25-35 30 14
13
35-45 40 10
45-55 50 6
Estimate the mean and the standard deviation of the monthly salaries of the 50
employees.=mean=
(=
,n=
where
=28.61000=28600 rupees.=standard
deviation=
=-707.96-707960000 rupee2
At the beginning of a year ,the monthly salary of each employee is increased
by p%. It is given that the mean of the new monthly salaries of the above 50
employees is 29172 rupees. Estimate the value of p and the standard deviation
of the new monthly salaries of the 50 employees.
28600+
,p=(29172-28600)/286=572/286=2
So that the salary is increased by 2% (two percent)=0.02=
Therefore the new standard deviation is =(0.02
Readers may not be happy of the squaring method calculation of the standard
deviation due to higher values that arise in enumeration & difficulties in squaring.
Probability
17.(a) Initially a box contains 3 balls identical in all aspects except for their color ,
each of which is either white or black .Now, one white ball identical to balls in the
box in all aspects except for its color, is added into the box and then one ball is
drawn at random from the box. Assuming that ,four possible initial compositions
of the balls in the box are equally likely, find the probability that
(i)the ball drawn is white, and
(ii)initially there were exactly 2 black balls in the box ,given that the ball drawn is
white
(i)p=1/41/2+1/41/2+1/41/2+1/41=1/8+1/8+1/8+1/4=3/8+2/8=5/8=0.6251
use tree diagram by sketching it and probability of drawing any ball is one fourth
& in the case of first three balls if drawn the probability them to be white is half
since they are either black or white then you have to use product of probabilities, in
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the case of last ball the probability it to be white is one. You have to add
probabilities to get the total but it cannot exceed one.
(ii)=2/4=1/2=0.51
when two balls are given as black when white is drawn there are two possible
chances to get white out of four.
The probability should always be positive or zero and less than or equal to unit
value in numerical estimates. To get an initial idea of probability think of tossing a
coin out of two chances one can be either head or tail so probability of getting any
side is half. If you add half with half the total become one. If you think of tossing
a gambling cube with six dotted numbers the probability of getting any number up
to six is one sixth since there is one chance only out of six possible chances if you
add all six one sixths together it gives one as the total. 0p1,where p the
probability .
Vectors
14.(a) Let OABC be a parallelogram and let D be the point on AC such that
AD:DC=2:1 .The position vectors of points A and B with respect to O are a and
b, respectively ,where 0.Express the vectors
and
in terms of a, b and .
Now, let
Show that 3|aa. b)-|b=0 and
find the value of , if |a|=|b| and A
.
a,
b ,0,
=
=b+a,
a-
b,
=
=
b ,If
⊥
then
(a-
b).(b+a)=0,a.b+a.a-
b.b-
b.a =0,
a.b
-
,a.b=b.a =|a||b|cos
|a,|a|=|b|,a.a=|a||a| cos0=|a,a=b,2+
,from the equation,32|a|2+2(a.b)-|b (+
=0,=-
or
where =-
,|a|=a,|b|=b,cos00=1,cos600=
vector has a
direction and magnitude .velocity, displacement &force are vectors. scalar has a
magnitude only .addition of two vectors gives the diagonal vector of a
parallelogram. two vectors are equal if direction and magnitude both are equal.dot
product is the scalar product since it produce a scalar.
A.C.Wimal Lalith De Alwis
E-Mail:dealwis_a@yahoo.com
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