G.C.E. Advanced Level
Combined Mathematics
PrActice questions
with Answers
(Prepared to suit the syllabus implemented from 2017)
Department of Mathematics
National Institute of Education
Maharagama
Sri Lanka
i
Combined Mathematics
G.C.E. Advanced Level
National Institute of Education
First Print - 2018
ISBN :
Department of Mathematics
Faculty of Science and Technology
National Institute of Education
Web Site:
Email:
www.nie.lk
info@nie.lk
Printed by
:
Press,
National Institute of Education
ii
Message from the Director General
The Department of Mathematics of the National Institute of Education (NIE) has adopted a
variety of methodologies with a view to promoting mathematics education. This book entitled
"Practice questions with answers" is a result of such an exercise.
Since the General Certificate of Education (Advanced Level) examination is a highly
competitive examination, preparing the students is also a highly competitive task for the
teachers, who teach Combined Mathematics for grade 12 and 13. To overcome the difficulty
of this task teachers need supportive materials which are developed according to a proper
standardand. But such type of materials are quite rare. It is not a secret that most of the
instruments available in the market are composed of questions that lack validity and quality.
The Department of Mathematics of the NIE has prepared this Practice questions with answers
to rectify this situation and facilitate students to prepare well for the examination. This collection
comprises of questions prepared according to the syllabus implemented from 2017. Inclusion
of answers along with questions undoubtedly, makes it easier to use for the teachers.
I request teachers and students to make the evaluation process in Combined Mathematics
a success by having access to this book.
I wish to extend my gratitude to the donors of the Austraian Aid programe for assistance given
to make this book available to you and also to the staff of the Department of Mathematics and
external scholars who provided academic contribution to make this venture a success.
Dr. (Mrs.) T. A. R. J. Gunasekera
Director General
National Institute of Education
iii
Preface
Among the G.C.E. (A/L) subject areas there is a special place belonging to the area of mathematics
Most of the students who complete G.C.E.(O/L) with high grading wishes to continue their
education in maths stream. The past evidence shows that most of the creative inventors came
from the field of mathematics or related field of mathematics.
The syllabus prepared for G.C.E.(A/L) maths stream is with the intention of producing experts
in the field of Mathematics, Science and Technology.
The revised new syllabus for Mathematics and Combined Mathematics was introduced from the
year 2017.To make the learning of students easy, a book named ‘practice question with answers’
was prepared by “The Department of Mathematics of the National Institute of Education”.
The way of questions provided in this book will help students to make them feel more comfortable
and also help them to prepare for their G.C.E.(A/L) examination by self-measuring their
achievement level. For the above reason It is expected to develop skills and ability on writing
necessary steps while answering the question in the public examinations by gaining experience
by practicing questions on this book.
I kindly request you to send feedback to us about the benefits you gained by using this practice
questions at your schools. It will be useful to us to edit and publish future editions of this book.
I earnestly thank our Director General for granting permission to prepare such a book and also
thank the resource people contributed in the making of this book. Again, I request you to use
this book in a proper and prospective way and send your valuable positive criticism and comments
about this book it will motivate us to produce more books.
Mr. K. R. Pathmasiri
Director
Department of Mathematics.
iv
Curriculum Committee
Approval:
Academic Affairs Board
National Institute of Education
Guidence:
Dr.(Mrs).T. A. R. J. Gunesekara
Director General
National Institute of Education
Supervision :
Mr. K. R. Pathmasiri
Director, Department of Mathematics
National Institute of Education
Subject Coordination:
Mr. S. Rajendram
Project Leader (Grade 12-13 Mathematics)
Department of Mathematics
National Institute of Education
Curriculum Committee:
Mr. G.P.H. Jagath Kumara
Senior Lecturer
National Institute of Education
Ms. M. Nilmini P. Peiris
Senior Lecturer
National Institute of Education
Mr. S. Rajendram
Senior Lectuer,
National Institute of Education
Mr. C. Sutheson
Assistant Lecturer
National Institute of Education
Mr. P. Vijaikumar
Assistant Lecturer
National Institute of Education
Miss. K.K.Vajeema S. Kankanamge Assistant Lecturer
National Institute of Education
v
Panel of Writters :
Mr. K.Ganeshlingan
Rtd Chief Project Officer,
National Institute of Education
Mr. V.Rajaratnam
Rtd Teacher
Mr. T.Sithamparanathan
Rtd Teacher
Mr. N.R.Sahabandu
Rtd Teacher
Mr. G.H.Asoka
Teacher Service, Rahula Vidyalaya, Matara.
Mr. H.D.C.S. Fernando
Teacher Service, Vivekananda College, Colmbo 13.
Mr. S.G. Doluweera
Teacher Service, Wesley College, Colmbo 09.
Type Setting:
Miss. Kamalaverny Kandiah
Press,
National Institute of Education.
Cover Design:
Mr. E. L. A. K. Liyanage
Press,
National Institute of Education.
Mrs. A. D. Anusha Tharangani
Press,
National Institute of Education.
Supporting Staff:
Mr. S. Hettiarachchi,
Department of Mathematics,
National Institute of Education.
Mrs. K. N. Senani,
Department of Mathematics,
National Institute of Education.
Mr. R.M. Rupasinghe,
Department of Mathematics,
National Institute of Education.
vi
Introduction
This book is prepared with the intend of giving more practice and revision in order to get ready
for the G.C.E.(A/L) final examination from the year 2019 onwards. After completing learning
of the syllabus, students can test their knowledge by practicing the questions in this book.
The subject teachers and students should note that this is not a book consisting of model questions
but rather, a collection of practice questions.
After practicing the given questions, students can check and compare their solutions with the
solutions provided, although the solutions obtained by the students need not to be exactly the
same as the solutions given in the book .The solutions provided can be considered as guidelines
for the student to learn how to obtain the answer. Also note that the solutions given here are to
check and follow steps needed to present the proper solution in proper ways.
Although this “Practice Questions with Answers” book is prepared to facilitate the students
siting for the G.C.E.(A/L) examination from 2019 onwards, under the revised syllabus implemented
from 2017, students who study Mathematics or Higher mathematics can also use relevant parts
of this book.
We have planned to publish the “Statics – I”, “Statics – II” and “Unit wise Practice Questions
Book I and II”, followed by the publication of this book, “Practice Questions with answers”.
Please, do not hesitate to point out the short comings and weaknesses of this book. Your comments
will help us to improve the quality of the book. Moreover, we want to stress out that we will
highly appreciate your valuable comments.
Thank You.
Mr. S. Rajendram
Project Leader
Grade 12-13 Mathematics.
vii
Content
Page
Message from the Director General
iii
Preface
iv
Curriculum Committee
v - vi
Introduction
vii
Combined Mathematics I
Combined Mathematics II
Part A
1- 5
Part B
6 - 12
Part A
13 - 19
Part B
20 - 28
Solutions for Practice Questions
29 - 145
viii
Combined Mathematics
Practice Questions (With Answers)
Combined Mathematics I
Part A
1  
1

2  x 2  2    x    14  0
x  
x

1.
Solve :
2.
Solve :
3.
Show that
3x  1  2  x  2 x  1
 
log 9 xy 2 
1
log 3 x  log 3 y
2

log c a 
 Hint : log b a 

log c b 

Hence Solve the Simultaneous equations
log 9 ( xy 2 ) 
1
2
log 3 x.log 3 y  3
4.
Let f ( x)  3 x 3  Ax 2  4 x  B @ where A, B are constants. Given that (3 x  2), is a
factor of f ( x ) and when f ( x ) is divided by ( x  1) the reminder is 2
(i) Find the values of A and B
(ii) Express f ( x ) as a product of lincer factors
5.
Let f ( x)  x4  hx3  gx 2  16 x  12 ; where h and g are constants. Given that ( x  1) is
a factor of f ( x ) and when f ( x ) is divided by ( x  1) the reminder is -24
(i)
Find the values of h and g
(ii) Show that ( x  2) factor of f ( x ) and find the remaining liner factors.
6.
The roots of the equation ax 2  bx  c  0 are  and  . find the roots of the equation
x2
7.
1 b2

in terms of  and 
x ac
If the equations x 2  bx  ca  0, and x 2  cx  ab  0 have a common root and a, b, c
are all different, prove that their other roots will satisfy the equation x 2  cx  bc  0
1
Combined Mathematics
8.
Practice Questions (With Answers)
If g ( x)  ax 2  2 x  (3a  2) , find the set of values of a for which g ( x ) is positive for
all real values of x
Sketch the graph of y  g ( x ) when a 
1
3
12
 x 1
x 3
9.
Find the solution set of inequality
10.
Solve 1  2 x  x  2  2
11.
How many ways can four boys and four girls sit in a row?
Find the number of ways, if
(i)
Two particular girls do not sit together
(ii) No girls sit next to each other.
10
12.
 2 2k 
For what values of k does the coefficient of x in the expansion of  x 

x 

2
the coefficient of
13.
equal
1
x
Find the coefficient of x 2 and x 3 in terms of k and n, in the expansion of (1  2 x  kx 2 ) n
where n is a positive integer. If the coefficient of x 2 and x 3 are 30 and 0 respectively..
Find the values of k and n.
14.
Let Z  1  i 3 be a complex number
(i)
Find Z
(ii)
Express Z 2 in the form of a  ib where a, b  
Find the values of the real number p such that Z 2  pz is real.
(iii)
and Arg ( Z )
2
(iv) Find the value of a real number q such that Arg ( z  qz ) 
15.
5
6
Let Z1  1, Z 2  cos   i sin  (0     ) be two complex numbers. Represent the
complex numbers Z1 and Z 2 by the points A and B respectively in the Argand diagram.
Find the points Cand D to represent the complex numbers Z 1  Z 2 and Z2  Z1
respectively. Using your diagram, find
(i)
Z1  Z 2
and Arg ( Z1  Z 2 )
Deduct that Z 1  Z 2
2
(ii)
2
Z 2  Z1 and Arg ( Z 2  Z1 )
+ Z 2  Z1 is independent of 
2
Combined Mathematics
16.
Practice Questions (With Answers)
sin x  sin a
xa
(a)
Find lim
xa
(b)
If sin y  x sin( y  a )
dy sin 2 ( y  a )

Show that
vdf; fhl;Lf.
dx
sin a
17.
tan x  sin x
x3
(a)
Find lim
x0
(b)
If y  x n lnx , find the value of n
d2 y
dy
2
Such that 2  2  3x for all values of x
dx
dx
18.
y  t  lnt (t  0)
If x  t  lnt and
Find
(i)
dy
dx
(ii)
d2y
interms of t
dx 2
d2y
8( x  y )

Also show that
2
dx
( x  y  2)3
19.
1
1
Simplify 1  x 2  (1  x ) 2
1
 (1  x
Hence find
2
0
20.
Use the substitution x  2(1  cos 2  )
3
to evaluate 
2
21.
x
)(1  x)2
x 2
dx
4 x
Using the method of integration by parts,
find
e
4x
.cos 3 x.dx
3
Combined Mathematics
22.
Practice Questions (With Answers)
The line 3 x  2 y  24 meets the y axis at A and x axis at B. The perpendicular bisector
of AB meets the line parallel to x axis through (0, -1) at C. Find the area of the triangle
ABC.
23.
5 5
The equation of a side of a square is x  2 y  0 and its diagonal intersect at  , 
2 2
find the equations of the remaining sides of the square.
24.
ABC is a triangle and AB=AC. A  (0, 8). The equations of the medians through B and
C are x  3 y  14 and 3 x  y  2 respectively. Find the equations of the sides of the
triangle ABC
25.
A straight line x cos   y sin   p  0 intersect in circle x 2  y 2  a 2  0 at A and B.
Find the equation of the circle AB as a diameter.
26.
S is the circle S  x2  y2  4x  2y  4  0 and P is the point P  (4, 2)
(i)
Show that the point P lies outside S.
27.
(ii)
Find the length of the tangents from P to S
(iii)
Find the equations of the tangents from P to S
Find the general equation of all circles which make an intercept 3 units on the x axis and
touch the y axis.
Show that their centers lie on the curve whose equation is 4 x 2  4 y 2  9
28.
Solve : cos 6  cos 4  cos 2  1  0 > where 0    
29.

1  1 
1  1 
Show that 2 tan    tan   
3
7 4
30.
In a triangle ABC, with the usual notation,
Prove that (b  c  a)(cot
B
C
A
 cot )  2a cot
2
2
2
4
Combined Mathematics
31.
Practice Questions (With Answers)
State De Movier’s theorem by using De Movier’s theorem find the modulas and argugment
of (1  3i ) 7
32.
Using De Movier’s theorem prove that
(i)
cos 3  4 cos3   cos 
(ii)
33.
sin 3  3sin   4sin 3 
Parametric equation of a curve is given by x  t (1  t ) 2 , y  t 2 (1  t ) where t is a real
2
2
parameter. Show that gradient of thetangent at the point  t (1  t ) , t (1  t )  is given
by
t (2  3t )
1
where t  1,
also show that equation of tangent drawn to the
(1  t )(1  3t )
3
curve at the point corresponding to t 
34.
1
is 4 x  4 y  1  0 .
2
Find the area bounded by the curve y  x ( x  3) and the x axis.
35.
(i)
Find the area shaded by the region.
(ii)
Find the volume of the solid form by rotating the shaded region through four
rectangles about x axis.
5
Combined Mathematics
Practice Questions (With Answers)
Part B
1.
(a)
The root of the equation x 2  px  q  0 are  and  .
(i)
(ii)
Given that the roots differ by 2 3 and the sum of the reciprocal of the roots
is 4, find the possible values of p and q.
find an equation whose roots are  
2

and  
2

expressing the
coefficients in terms of p and q.
(b)
x 2  3x  4
Find the possible values of k if
can take all values when x is real.
5x  k
Draw the graph of y 
2.
(a)
x 2  3x  4
, when k  5
5x  k
The roots of the quadratic equation f ( x)   2 ( x2  x)  2 x  3  0 > (  0)
are  and  . If 1 , 2 are the values of  . For which  and  are connected
12
2 2
  4
 
by the relation
  3 , find the equation. Whose roots are 2 and 1 .
Find the greatest integer  such that the quadratic function f ( x )  2 x for all
values of x .
2n
3.
(b)
Prove by mathematical induction that
(a)
Express
 (1)
r 1
r 1
2n
1
1
 
r r  n 1 r
2r  3
in partial fractions.
r (r  1)
2
Write the
r terms
th
U r of the series
Find Vr such that U r  Vr  Vr 1


Hence find
(b)
U
n 1
n
. Is the series
U
n 1
r
convergent. Justify your answer..
Sketch the graphs of y  2 x  1 and y  x  1  1 in same diagram.
Hence solve 2 x  1  x  1  1 .
6
3
5 1 7 1
9 1
 
  
   ...
1.2  3  2.3  3  3.4  3 
Combined Mathematics
4.
(a)
(b)
Practice Questions (With Answers)
Six boys and six girls sit in a row at random. Find the number of different ways that,
(i)
The six girls sit together
(ii)
The boys and girls sit alternatively.
Four digit numbers are formed by choosing digits from 0, 2, 3, 5, 7, 8
How many numbers can be formed If
(i)
Digits can be repeated in a number
(ii)
One digit can be used once only in a number
incase (ii) how many numbers are greater than 5000 and divisible by 2.
(c)
State and prove the binominal theorem for positive integral index.
Write down the binominal expansion of (1  x ) n and ( x  1) n ;; where n is a
positive integer by considering the first derivatives of both expansion, show that
(i)
(ii)
5.
(a)
1( n  1) nC12  2( n  2) nC2 2  ...  r.( n  r ). nCr 2  ...  (n  1).1. nCn 12
 n 2 .2 n  2 Cn  2
n
n 1
r 1
r 0
 r  nCr   (n  1)  nCr  n 2  22 n2
Find the three roots of Z 3  1
Given that is one of the complex roots of Z 3  1 , show that 1     2  0
Hence show that
(i)

1

 1 
(ii)
2
 
2 1
3k
(iii)
(b)
Get
3k
 2 
  

 2
  2, k is odd


  1 
  1
= + 2> k is even
u  2i and
1
3
v   i
be two complex numbers. Write u > v > uv >
2
2
u
in the form r (cos   i sin  ) where
      
v
In an argand diagram the points A, B and C represent the complex numbers u > uv
and
u
respectively. Show that ABC is an equilateral triangle.
v
7
Combined Mathematics
6.
(a)
 1 i 
Express 

 1 i 
integer.
Practice Questions (With Answers)
4 n 1
in the form of p  iq where p, q  ; and n is a positive
Show that the cube root of 1 is 1,  ,  2
  cos
Where
2
2
 i sin
3
3
Hence solve the equation  x  2   1
3
Also show that
(b)
(i)
 2  5  2 
(ii)
 p  q  p  q   p 2  q   p 3  q 3
(iii)
 a  b  c 2 


2 
 b  c  a 
2
6
 729
The point P ( x, y ) denotes the complex number Z  x  iy Argand diagram,
where x, y  
Given that Z  3  3i  2 find the locus of P and sketch it in the argand diagram
0  A rg  Z  3  3i  
Further if

3
shade the region which satisfies both
conditions in the Argand diagram. Also find the greatest value of Z in their region.
7.
(a)
(b)
Find
cos 4 x  cos 2 x
x 0
x2
(i)
lim
(ii)
lim
x 0
tan 2 x  2sin x
x3
1
1
Given that y  sin
x2  1

, Z  sec1 x x  2

Show that
(c)
(i)
cos y.
(ii)
dy

dz
dy
  cos ec 2 z
dz
x2
x
2

1 x  2
2

0
A wire of length l is bent in the shape of isosceles triangle. Show that the maximum
area included in the triangle is equilateral and find the maximum area.
8
Combined Mathematics
8.
(a)
Practice Questions (With Answers)
d
 f ( x)  2 cos 2 x
dx
Using the principle of mathematical induction prove that
If f ( x)  sin 2 x prove from first principle that
dn
 n

sin 2 x   2n sin 
 2x
n 
dx
 2

(b)
Let f ( x )  1 
1
Where x  0, 2
x  2x
2
Find the turning points of the graph of f ( x ) only by using first derivatives sketch
the graph of y  f ( x ) indicating the asymptotes and maxima or minima (if any)
Hence, sketch the graph of
9.
(a)
(i)
y  f ( x)
(ii)
y
1
f ( x)
1

Express 1  x
2
 x
2

 1 in partial fractions.
dx
Hence, find
(b)
 1  x  x
2
2

1
Given that sin x  cos x  t express sin 2x in terms of t.

4
Using the above substitution, evaluate
sin x  cos x
 9  16 sin 2 x dx
0

2
(c)
I 
0

2
cos x
sin xdx
dx, J  
a cos x  b sin x
a cos x  b sin x
0
(i)
Find aI  bJ
(ii)
By obtaining another linear combination in I and J , hence find the values of
I and J.
9
Combined Mathematics
a
10.
(a)
Prove that
Practice Questions (With Answers)
a
f ( x )dx   f (a  x)dx

0
0

x sin x
2
dx 
Show that 
1  cos 2 x
4
0
(b)
Using the method of integration by parts,
x.e x
dx
Find 
(1  x)2
11.
(c)
Find the area bounded by the curve y  x (2  x ) and the straight line y  x
(a)
A rectangle ABCD lies completely in the first quadrant. The equation of AD is
x  y  4  0 and the equation of AC is 3 x  y  8  0 and length of AB is 2 2
(i) Find the equation of AB
(b)
(ii)
Find the coordinates of B
(iii)
If BD is parallel to x  3 y  7  0 find the circle equations of BC and CD.
Show that the general equation of the circle S  0 passing through the points
(2> 0) and (0> -1) is
   4) 
S  x2  y 2  
 x     1 y    0
 2 
Where  is a parameter..
(i)
Hence find the equation of the circle S1  0 which passes through the points
(1> -1)> (2> 0) and (0> -1)
(ii)
S1  0 bisects the circumference of the circle S 2  0 of the given system
S  0 find the equation of S 2  0
(iii)
Two circles of the above system S  0 bisects orthogonally each other. Show
that
12  4 where, 1 and 2 are the corresponding parameters of the
circles.
10
Combined Mathematics
12.
(a)
Practice Questions (With Answers)
In a triangle ABC, the equation of internal bisector of C is x  4 y  10  0 and
the equation of the median through B is 6 x  10 y  59  0
The coordinates ofA is (3> -1) find
(b)
(i)
the coordinates of B and C
(ii)
the equations of the sides of the triangle ABC
(iii)
the equation of the perpendicular to AC through B
A circle S3  0 passes through the points of intersection of the circles
S1  3 x 2  3 y 2  6 x  1  0 > S 2  x 2  y 2  2 x  4 y  1  0 and also passes
through the centre of S1  0
Find the equation of S3  0 and verifies that S3  0 and S 2  0 intersect each other
orthogonally.
Find also the equation of the tangent to the circle S3  0 at the centre of S1  0
13.
(a)
(b)
Find the general solutions of the equations.
(i)
(2sin x  cos x)(1  cos x )  sin 2 x
(ii)
2 tan x  sec 2 x  2 tan 2 x
Prove that 2 cos 2   2 cos 2 2  cos 2  cos 4 and deduct that
cos360  cos720 
1
2
Hence, find the values of cos 360 and cos 720
(c)
14.
State and prove the since rule for a triangle ABC, with the usual notation.
In the usual notation for a triangle ABC,
a2  b2
b2  c2
c2  a2


0
cos A  cos B cos B  cos C cos C  cos A
(i)
show that
(ii)
if A  450 and B  750 , show that a  2c  2b
(a)
Solve the equation
(i)
(ii)
2(cos x  cos 2 x)  sin 2 x(1  cos x)  2 sin x
where   x  
 1 
1  1 
1  3 
1  1 
tan 1 
  tan 
  tan    tan  
 x 1 
 x 1 
5
3
where (2  x  4)
11
Combined Mathematics
(b)
Practice Questions (With Answers)
If (1  m )sin(   )  (1  m) cos(   ) , prove that




tan      m cot    
4

4

(c)
state and prove the cosine rule for a triangle ABC, with the usual notation.
(i) In a triangle ABC, AH is perpendicular to BC and AH = p show that
b  c 
(ii)
15.
(a)
2
 a 2  2ap cot
A
2
If a 4  b 4  c 4  2c 2 (a 2  b 2 ) Prove that C  450 or 1350
 3 1
Let A  
 be a 2  2 matrix.
 1 2 
Show that
A2  5 A  7 I  0 @ I is the identity matrix of order 2.
Hence Find A-1
Also furthe matrix B of order 2 such that BA = C
 9 4 
Where C  
 Find B
 6 16 
(b)
x, y are connected by equations x  y  a, x  y  b . Write down the equation
in the form AX = B, where A, X, B are matrices.
Find A -1
Hence find x, y interms of a , b .
1
2  p
It is given that A    B , without finding  A2  only by using matrices find
q
p, q interms of a and b.
12
Combined Mathematics
Practice Questions (With Answers)
Combined Mathematics II
Part A
1.
A train runs between two stations A and B which are 10km a part. It starts from A with an
initial velocity u and uniform accetertaion 1ms-2 . It moves for 40 seconds and reaches the
speed 60 ms-1 and maintains the speed for T Seconds. It comes to rest at B with uniform
1 2
retardation ms .
2
(i) Draw the velocity - time graph for the motion of the train.
(ii)
2.
From the graph find u and T
A particle A is projected vertically upwards with velocity u when A reaches its highest
point, another particle B is projected vertically upwards with velocity 2u from the same
point.
3.
(i)
Draw the velocity - time graph for A and B in the same diagram.
(ii)
Find the time taken for the particles to meet after B is projected.
A ship A is travelling due east at 2u kmh-1 and a second ship B is travelling S 300 E at
u kmh-1.. At midday the first ship is d km due south of the second
Find
4.
(i)
The velocity of A relative to B.
(ii)
The least diatance between the two ships and the time taken.
A particle a of mass m rest on a smooth horizonal
table and is connected by a light unextensibel string
A(m)
passing over a smooth fixed pulley at the edge of
the table and under a smooth light pulley C to a
fixed point on the ceiling as shown in the diagram.
The pulley C carries a particle of mass M. If the
system is released from rest. find the acceleraton of
C
C and the tension in the string.
5.
At time t the position vector of a particle is r>
M
r  a cos nt i  b sin ntj where
i and j are unit vectors along the Ox, Oy
a , b , ( a  b ) and n are constants and
axes respectively. Find v > the velocity vector and a the acceleration vector and hence
find the times at which the velocity is perpendicular to acceleration.
Also show that

v .v .  n 2 a 2  b 2  r . r
13

Combined Mathematics
6.
Practice Questions (With Answers)
A car of mass 1200 kg moves along a strainght horizontal road with a constant speed of
24 kmh-1 The resistance of motion to the car has magnitude 600N.
(i)
Find, in kW the rate at which the engine of the car.
(ii)
The car now moves up a hill inclined at  to the horizontal, here sin  
1
The
24
resistance to movion from non - gravitationnal forces remains of magutude 600N
The engine of the car now works at the rate of 30 kW .
Fine the accelaraton of the car when its speed is 20 ms-1
7.
Show that the velocity of water in a pipe of cross section 100 cm2 which delivers 0.1m3 is
10 ms-1.
Calculate the power of an engine which raise the water in this pipe to a height of 12m and
then delivers at this hight at 10 ms-1 (neglect friction)
8.
A gun of mass M is mounted on a smooth railway, and is fired in the direction of the track.
It fires a shell of mass m, with velocity v relative to the gun. If the angle of elevation of the

1  M  m
tan  
gun is  > prove that the unitial direction of the motion of the shell is tan 
 M

to the horizontal.
9.
Three particles A, B, C of masses m, 2m, 3m respectively lie at rest in that order in a
straight line on a horizontal table. The distance between consecutive particles is a. A slack
light in elastic string of lenght 2a connects A and B. An exaclty similar string connects B
and C. If A is prjected in the direction CBA with speed v find the speed with which C
begins to move after the two springs become tant that the ratio of the unipulsire tensions in
BC and AB when C is jerked into motion is 3:1. Find also the total loss of kinetic energy
when C has started to move
10.
Two small uniform smooth spheres A and B of equal size and of masses m, 4m and 4m
respectively are moving directly towardseach other with speeds 2u and 6u respectively.
The coefficient of restitution between the spheres is
Fine,
(i)
The speed of B Unimediatley after collion.
(ii)
The momentum transfered from one to other.
14
1
.
2
Combined Mathematics
11.
Practice Questions (With Answers)
Two partcles A and B move on a smooth horizontal table. The mass of A is m, and the mass
of B is 4m. Initially A is moving with speed u when it collides directly with B, which is at
rest on the table. As a result of the collision, the direction of motion of A is reversed. The
coefficient of restitution between the particles is e. Find expressions for the speed of A and
the speed of B unimediately after colision. In the susequent motion, B strikes a smooth
vertical wall and rebound. The wall is perpendicular to the direction of motion of B. The
coefficient of restitution between B and the wall is
collision between A and B, show that
12.
4
. Given that there is a second
5
1
9
e .
4
16
A vertical cliff is 73.5m high. Two stones A and B are projected simultaneously. Stone
A is projected horizontally from the top of the cliff with speed 28ms-1. Stone B is
projected fromthe botton of the cliff with speed 35ms-1 at an angle  above the
horizontal. The stones move freely under gravity in the same vertical plane and collide in
mid - air.
(i)
(ii)
13
4
.
5
Find the time which elapses between the mistant when the stones are projected and
the mistant when they collide. (g = 9.8ms-2)
Prove that cos  
.A projectile is fired with unitial speed
2ag to hit a target at a horizontal distance a
from the point of projection and at a vertical distance
a
above at. Find the two possible
2
angles of projection and the ratio of the time of flight along the two paths.
14.
An elastic string AB of natural length a and modulus of elasticity 2mg has one end A fixed.
A particle of mass m is attached to the end B and performs horizontal circles with angular
velocity
3g
. Find the extension in the string and cosine of the angle between the string
4a
and the vertical.
15
Combined Mathematics
15.
Practice Questions (With Answers)
A small bead of mass 2kg is threaded on to a smooth circular wire of radions 0.6m,
which is fixed in a vertical plane. If the bead is slightly disturbed from rest at the highest
point of the wire, find its speed when it reaches the lowest point. Find also the height above
the centre, of the print at which the reaction between the bead and the wire becomes zero.
(g = 10ms-2)
16.
A particel is moving in a straight line with Simple Harmonic Motion. Its velocity has the
values 1.2m and 0.9m when its distances from the centre of ocillation are 0.9ms-1 and
1.2ms-1 respectively.
Find the amplitude and period of the motion.
17.
A particle of mass m is attached to the mid point of an elastic string of natural length a and
modulus 2mg. The ends of the string are fixed to two points in a vertical line at a distance
of 2a. positon of equilibriun both parts of the string are intension. If the partcle is given
small vertical displacement and it performs simple hrmonic oscillation, find the period.
18.
 
ABC is an equilaberal triangle of side 2a. Forces p, 2 p and 3 p act along AB, BC

and CA respectively A.
C
Fine
(i)
The magnitude and resultant of the system of
3p
2p
forces.
(ii)
The distance from A of the point where its
line of action ents BA produced.
19.
A
p
B
In the rectangle ABCD, AB = 4a, and BC = 3a. Forces 2 p, 4 p, 6 p, 7 p and 5 p act
   

along AB, BC , CD, DA and AC trespectively. Show that the system reduces to a couple.

Find the magnitude and sense of the couple. If the force acting along BC is removed
find the magnitude, direction and the line of action of the resultant of the new system.
16
Combined Mathematics
20.
Practice Questions (With Answers)
A uniform rod AB of length 2a and weight W is smothly pivoted at A to a fixed point. It
1  3 
is held in equilibrium at an angle tan   to the downward vertical by a force of mag4
nitude P applied at B.
21.
(i)
using triangle of forces Find P, if the force P is horizontal.
(ii)
What is the least possible value of P and its direction.
A sphere of redus 9cm and weight W rests on a smooth inclined plane (angle 300). It
is attached by a string fixed to a point on its surface to a point on the plane 12cm from the
point of contact and on the same line of greatest slope. Mark the forces acting on the
sphere.
Draw the triangle of forces for the equlibrium of the sphere and hence find.
22.
(i)
the tension in the string
(ii)
Raction on the sphere by the plane
Three uniform rods AB, BC and CA of equal length a and weight W are feely jointed
together to form a triangle ABC. The framewore rests in a vertical plane on smooth
supports at A and C. so that AC is horizontal and B is above AC . A mass of weight W is
attached to a point on D on AB where AD 
a
.
3
Find the reaction between the rods AB and BC.
23.
A framewore consists of four light rods as
D
shown in the diagram.
A
AB = BC = CA = 2a, and AD = a
It is smoothly hinged to a vertical wall at band
D with BC horizontal, and carries a weight
W at C. Using bow’s notation draw a stress
diagram and find the stress in each rod.
Distaquish tension and thrust.
B
C
W
17
Combined Mathematics
24.
Practice Questions (With Answers)
A force P acting parallel to and up a rough plane of unclinaton  is just safficient to
prevent a body of mass m. From sliding down the plane. A force 3P acting parallel to and
up the same plane causes the same mass to be on the point of moving up the plane. If 
is the coefficient of friction show that 2   tan  .
T
25.
A uniform rod AB of weight W is in equilibrium in a
vertical plane as shown in the diagram. A vertical
A
string is attached to A
Fine
(i)
T in terms of W
(ii)
For equilibrium find the minimum value
Rough plane
B
of  > the coefficient of friction at B
300
300
W
26.
A uniform lamina OABCD consists of a rectangle
D
C
OACD and a right - angled triangle ABC as shown
in the diagram. OA = 2a, OD = a, AB = a .
(i)
a
Find the centre of gracityof the limina from
O
OB and OD.
(ii)
27.
2a
A
a
B
If it is feely suspended from O, find the angle OAB makes with the horizontal.
2
5
3
A and B are two events such that P  B   , P  A  B  
and P  A | B   .
3
8
4
Find P  B  , P  A  B  , P  A  and P  A  B  .
28.
(a)
Events A and B are such that P  A  0.3, P  B   0.4 and A and B are
independent. Find
(b)
(i)
P  A  B
(ii)
P  A  B
If 20% of the bulbs produced by a machine is defective determine the probability
that out of 4 balls chsen at raudom 3 will be defective.
18
Combined Mathematics
29.
Practice Questions (With Answers)
The marks obtained by 9 students in an examination given below.
7, 11, 5, 8, 13, 12, 11, 9, 14
Fine
30.
(i)
Mean
(ii)
Median
(iii)
Standard deviation and
(iv)
Coefficient of skewness
The age in years of the sesidents in a hotel is given below in the stem and leaf diagram.
0
2
(01)
1
1
5
7
9
(04)
2
1
3
8
9
(04)
3
2
3
3
5
6
6 7
4
0
5
7
7
8
9
5
8
9
9
9
9
(11)
(06)
(01)
2/3 Means 23 Years
(i)
Write down the minimum value, maximum value end mode of the age of residen.
(ii)
Find the values of Q1 , Q3 and median.
(iii)
Outlines are given by Q1  1.5(Q3  Q1 ) and Q3  1.5(Q3  Q1 )
check whether there is any outolines.
19
Combined Mathematics
Practice Questions (With Answers)
Part B
(01) (a)
A particle P starts from rest, moves with uniform acceleration a in a straight line.
After t seconds another particle Q. Starts from the same point with initial velocity
3a
. Both particles move in the same direction and
2
attam the same maximum velocity at the same time. Immediately they decelerate
u and uniform acceleration
with uniform deceleration a and 2a respectively and come to rest.
Draw the velocity - timegraphs of P and Q in the same diagram.
Hence
(i) Show that the maximum speed is 3at  2u .
(b)
(ii)
Show that the time difference in their hourney is
(iii)
Find the distance travelled by each particle.
5t u

2 a
Two straight roads OA and OB meet at an acute angle  . A car P moves along
OA towards O with uniform speed u, while a second car Q moves along OB, away
from O with uniform speed V. At t = 0 , the car P is at a distance a from O and the
car Q is at O. Find the relative velocity of P, relative to Q
(i)
Show that the shortest distance between the cars is
av sin 
u  v 2  2uv cos 
2
and find the time taken to reach shortest distance.
(ii)
(02) (a)
Show that the ratio of the distances from O when they are at shortest
distance is v  u cos  : u  v cos 
A car weight W has maximum power H. In all circumstances there is a constant
1  1 
resistance R due to friction. When the cat is moving up a slope of sin   its
n
maximum speed is v and when it is moving down the same slope its maximum
speed is 2v.
Find R in terms of W and n.
The maximum speed of the car on the level road is u.
Find the maximum acceleration of the car when it is moving with speed
given slop.
20
u
up the
2
Combined Mathematics
(b)
Practice Questions (With Answers)
Two particles A and B are free to move in the plane of the unit vectors i and j


which are perpendicular to each other. The velocity of A is 3 i  2qj ms 1 and
the velocity of B is v i  7 j m s  1 where v is a constant. Determine the

velocity of B relative to A and fine the vector AB at time t seconds given that,

when t = O, AB  56 i  8 j m




Find also the value of v such that the particles collide.


Show that, when v = 3> AB at time t is given by AB   6t  56  i  8(1  t ) j
and hence find t where A and B are closest together.
By evaluating a suitable scalar product show that, for your value of t and with v = 3

AB is perpendicular to the velocity of B relative to A.
(03) (a)
Two particles A and B of mass m and
2m are connected by a hight inextesible
B
A
m
2m
string passing under a smooth movable
pulley of mass M. A and B rest on rough
horizontal tables, as shown in the
diagram, the coefficients of friction are
 and   respectively. The system is
M
released from rest.
(i) Show that the tension in the string is
2 Mmg  2      
 3M  8m 
(ii)
Given that   2   > show that for the motion to take place.

  2
(b)

M  8m
2M
A
One end of a light uniextensible string ABCD in
a
which AB = BC = a is attached to a fixed point
A. Asomooth narrow tube CD is fixed below A so
B (m)
b
that ACD is a vertical line and AC = b. The end D
of the string is threeaded through the tube and
attached to a body of mass km which cannot pass
through the tube. A particle of mass m is fastened to
the string at B, and rotates about the line AC with
constanta angular velocity w in a horizontal circle,
21
a
c
D
O
km
Combined Mathematics
Practice Questions (With Answers)
tube at D. find the tensions in the two parts of the string and the vertical force
exerted at D by the tube on the body, and show that w2 ab  2 g (a  kb) Given
that the greatest tension the string can sustain without breaking is  mg , show that
the motion is possible only if    k  b  2a .
(04) (a)
Three particle A, B, C of equal mass m lie at rest in that order in a straight on a
smooth horizontal table. such that AB = BC = d. A is projected towards B with
speed u and at the same time B is projectd towards C with the same speed
u along the table. The coefficient restitution between any two particles is e>
(b)
(i)
The time taken for A to collide with B
(ii)
Find the distace travelled by A upto the above collision.
(iii)
show that there is another collision between B and C.
A particle P of mass m moves in a vertical circle along the smooth inner surface of a
fixed. hollw sphere of internal radius a and centre O, the plane of the circle passing
through O. The particle is projected from the lowest point of the sphere with a
horizontal velocity u. Where u 2  2ag . When OP makes an angle  with the
upward vertical, the velocity of the particle is v and the normal reaction between
the particle and the sphere is R. Find expressions for V and R in terms of m, a,
u,  and g show thati if
u 2  5ag the partcle leaves the shpere befor it
reaches the hishest point of the sphere and find cos  in terms of u, a and g when
it leaves the sphere. If the particle leaves the sphere at a point A and its trayctory
meets the sphere again at a point B such that AB is a diameter of the shpere, show
that OA makes an angle of 450 with the vertical, and find the requiste value of u.
(05) (a)
A particle is projected at an angle  to the horizontal from a point at height h from
horizontal ground. The particle reaches the ground at a point of horizontal distance
2h from the point of projection.
find the speed of projection in terms of g,  and h Given that the direction of
motion of particle with horizontal when it reaches the group is  , show that t
tan   1  tan 
22
Combined Mathematics
(b)
Practice Questions (With Answers)
On a smooth inclime place of angle  there is placed a smooth wedge of mass M
and angle  , in such a way that the upper face of the wedge is horizontal; on this
horizontal face is placed a particle of mass M. The system is released from rest.
Find the acceleraton of the wedge. Show that the reaction between the wedge and
the plane is
(06)
M ( M  m) g cos 
M  m sin 2 
Two points A and B on a smooth horizontal table are at a distance 8l apart. A
particle of mass m between A and B is attached to A by means of a light elastic
string of modulus  and natural length 2l and to B by means of a light elastic string
of modulus 4  and natural length 3l. If M is the midpoint of AB, O is the point
2l
11
If the particle is held at M and then released, show that it will move with simple
harmonic motion, and find the period of motion.
between M and B at which the particle would rest in equilibrium, prove that OM 
Find the velocity of the particle when it is at a point C distant
3l
from M, and is
11
moving towards B.
(07)
Particle of mass m is attached to one end of a light elastic string of natural length 6a
and modulus of elasticity 3mg. The other end of the string is fixed to a point O on a
smooth plane inclined at an angle 300 to the horizontal. The string lies along a line
of greatest slope of the plane and the particle rests in equilibrium at a point C on the
plane. calculate the distance OC.
The Particle is now pulled a further distance 2a down the line of greatest slope
through C and released from rest. At time t later, the displacement of the particle
from C is x is down the plane using the conservation of energy equation show that,
x 
x Satisfies the differential equation. 
gx
 0 , untill the string becomes
2a
slack.
g
2
Given that x  A cos t  B sin t , where  
, is the solution of above
2a
differential equation find A and B in terms of a.
Hence find the time at which the string slakckens and determine the speed of the
partical at this time.
23
Combined Mathematics
(08) (a)
Practice Questions (With Answers)
ABC is an equlaberal triangle of side 2a. The moments of a system of foces acting
M
and 2M
2
respectively in the same sense. Prove that the magnitude of the resultant of the
in the plane of the triangle ABC, about A, B, and C are
M,
7 M
and find tis direction with AB.
12 a
If the line of action of the resultant cut AB at D, find AD.
system is
(b)
A heavy uniform sphere of radius a has a light inextensible string attached to a point
on its surface. The other end of the string is fixed to a point on a rough vertical wall.
The sphere rests in equilibrium touching the wall at a point distant h below the
fixed point. If the point of the sphere in contact with the wall is about to slip down
wards and the coefficient of friction between the sphere and the wall is  , find the
inclination of the string to the vertica.
If  
is
(09) (a)
(b)
h
and the weight of the sphere is W , show that the tension in the string
2a
W
1 2
2
ABCDEF is a regular hexagon with sides of length 2a. Forces P, P, Q, P 3 N
  

act along AB , DA, CE and AE respectively..
(i)
Show that the system cannot reduce to a couple.
(ii)
Find the resultant of the system when Q  3 P
(iii)
If the line of action of the resultant cuts AB at G find AG
Two equal uniform rods AB and BC, each of weight W are freely joined at B. The
system is suspended frely from A and horizontal force P is applied at the lowest
point C. If, in the equilibrium position, the inclination of AB to the down ward
vertical is 300 find the corresponding inclination of BC and show that P 
Determine the resultant action at B.
24
W 3
2
Combined Mathematics
(10) (a)
Practice Questions (With Answers)
Two uniform beams AB and AC, equal in length and of weights 3W and W
respectively, are smoothly jointed at A; the system rests in a vertical plane with the
ends B and C in contact with a rough horizontal plane, the coefficient of limiting
friction at B and C being the same and equal to  .
If R and S are the normal reactions of the plane on AB and AC respectively and
ˆ  2
angle BAC
(i)
5
3
w, S  w
2
2
Stating at which point B or C the friction first becomes liniting as  is in
R
creased from zero.
(ii)
Prove also that, tan  
3
the reaction of one beam on the other makes
2
an angle of tan 1 (3 ) with the vertical.
(b)
In the framework shown, BC = 6a The framework is hinged at A and is kept with
BC horizontal by a force at B which acts downwards, perpendicular to BD Weights
60N and 40N
hang from C and D respectively find the magnitude and direction of the force a hinge
A draw a stress diagram using bow’s notation.
Hence, find the force in each rod distangnish - between thrust and tesions.
B
C

600
30
0
30
0
60N
(i)
Force at B.
(ii)
Magnitude and direction of the
force at the hinge.
D
(iii)
40N
By using Bow’s notaion draw
stress diagram and find forces
in each rods also verify that
tension and thrust.
A
(11) (a)
i , j are the unit vectors along Ox and Oy respectively forces F1  3 i  4 j >
F2   i  6 j > F3  3 i  3 j act at the points whose position vectors are
r1  2 i  3 j , r2  6 i  j , r3  3 i  2 j respectively find the resultant force R
and the cartesian equation of line of action of the resultant. If a fourth force F4 ,
acting at the origin, and a couple G in the plane are added to the system to be in
equlibrium find F4 and G .
25
Combined Mathematics
(b)
Practice Questions (With Answers)
 

In a triangle ABC, forces  BC ,  CA and  AB act along BC, CA and AB
respectively. show that system of forces reduces to a couple if and only if      .
(c)
The least force move a mass of M kg up a plane of inclination  is P. Show that
P  Mg sin(   ) where  is the angle of friction between the particle and the
plane.
Show that the least force acting parallel to the plane which will move the mass up the
slope is P sec 
(12) (a)
A uniform circular lamina of radius a and weight W rests with its plane vertical on
two fixed rough planes each incline at an angle  to the horizontal, their line of
intersection being perpendicular to the plane of the lamina. If the coefficient of
friction at each contact is  , prove that the least couple required to rotate the
Wa
lamina in the plane about its centre is of moment 1   2 cos 


(b)
A solid consists of a uniform right circular cone of density  , radius r,, and height 4r,
mounted on a uniform hemisphere of density  and radius r, , so that the plane
faces cokncide. Show that the distance to the centre of mass of the whole solid from
the common plance face is
r 16   3 
8  2    
If    and the solid is suspended frecly by a string attached to a point on the rim
of the common plane face, find the inclination of the axis of the cone to the vertical.
(13)
Show that the centre of mass of a uniform solid hemisphere of radus a is at a
3a
from the centre.
8
A bowl is made by removing a hemisphere of radius a from a solid hemisphere of
distance
radius 2a both have the same centre O.
Find the distance of centre of mass of the bowl from O.
(i)
The bowl is suspended from a point on the outer rim of the bow.. Show that
the plane surface makes an angle  with the horizontal where
 112 

 45 
  tan 1 
26
Combined Mathematics
(ii)
Practice Questions (With Answers)
If thes bowl rests in equilibrium with its curved surface in contact with a plane
inclimed to the horizontal at an angle  and sufficiently rough to prevent
sliding, find the maxium value of  .
(14) (a)
Each sunday a fisherman visits on the three possible locations near his home; he
goes to the sea with probability
1
1
@ to a river with probability @ or to a lake
2
4
1
. If he goes to the sea there is an 80% chnce that he will catch
4
fish; corresponding fiqures for the river and the lake are 40%, 60% respectively.
with probability
(i)
Find the probability that, on a given sunday, he catches fish.
(ii)
Find the probability that he catches fish on at least two of three consecutive
sundays.
(iii)
If, on a particular sunday, he comes home without catching anything, where is
it most likely that he has been.
(iv) His friend, who goes fishing every sunday, chooses among the three locations
with equal probabilities. Find the probability that the two fishermen will meet
at least once in the next two weekends.
(b)
The table below gives the wages paid per hour and the number of employees of a
factory.
Wages/ hour
(In rupees)
Number of
employees
900 - 800
800 - 700
700 - 600
600 - 500
500 - 400
400 - 300
300 - 200
200 - 100
14
44
96
175
381
527
615
660
Calculate
(i)
the mean wage
(ii)
Standard diviation
(iii)
Median
(iv)
Ciefficient of skewnees
and draw the shape of the distribution.
27
Combined Mathematics
(15) (a)
Practice Questions (With Answers)
3
1
of a sports club are adults, and are chidre, Three quarters of the adults, and
4
4
three fifth of the children, are male, Half the adult males, and on third of the adult
females, use the swimming pool at the club; the corresponding proportion for children of either sex is four fifths.
(i)
Find the probability that a memnber of a club uses the swimming pool.
(ii)
Find the probability that a member of the club who uses the swimming pool is
a female.
(iii)
Find the probability that a male user of the swimming pool is child.
(iv) Find the probability that a member of the club who does not use the swimming poo is either female or an adult.
(b)
A population consists of n1 males and n2 females. The mean height of the males
and females are 1 and 2 respectively and the vaniance of the heights are  12
and  2 2 respectively..
Show that mean height of the whole population is 1 w1  2 w2 and the variance
is w1 12  w2 22  w1 w2  1   2 
Where w1 
2
n1
n2
and w2 
n1  n2
n1  n2
The mean and standard deviation of a test for a group of 20 pupils were calculated
as 40 and 5 respectively. But while calculating them a mark of 15 was misread as
50. Find the correct mean and standard deviation. The mean and standard deviation
of another group of 30. pupils for the same lest is 40.25 and 8 respectively.
Calculate the mean and standard deviation of the combined group of 50 pupils.
28
Combined Mathematics
Practice Questions (With Answers)
Solutions
for Practice Questions
29
Combined Mathematics
Practice Questions (With Answers)
30
Combined Mathematics
Practice Questions (With Answers)
Combined Mathematics I
Part A
1.
1  
1

2  x 2  2    x    14  0
x  
x

Let y  x 
1
x
1
 y2  2
2
x
2
2( y  2)  y  14  0
x2 
2 y 2  y  10  0
(2 y  5)( y  2)  0
y
5
or y  2
2
1
 2
x
x2  2 x  1  0
x
2  8
2
x  1  2
x
2.
1 5

x 2
2x2  5x  2  0
x
x
5  41
4
3x  1  2  x  2 x  1
x
1
1
and x  2 and x 
3
2
1
x2
2
Squaring both sides
(3 x  1)  (2  x)  2 (3x  1)(2  x)  2 x  1
2  (3 x  1)(2  x)
4  (3x  1)(2  x)
3x 2  5 x  2  0
(3 x  2)( x  1)  0
31
Combined Mathematics
x
Practice Questions (With Answers)
2
or 1
3
When x  1 >
2
When x  >
3
L.H.S
=
R.H.S
L.H.S
= 1 1
= R.H.S
R.H.S
=
L.H.S
=
R.H.S
x
Hence,
3.
4  1  2 1  1
3
2
1

3
3
1
3
= L.H.S
2
or 1
3
 
log 9 xy 2  log 9 x  log 9 y 2

log 3 x log 3 y 2

log 3 9 log 3 9
log 3 x 2 log 3 y

2
2
1
 log 3 x  log 3 y
2

Let log 3 x  a and
log 9 ( xy 2 ) 
log 3 y  b
1
2
1
1
ab 
2
2
log 3 x.log 3 y   3
a  2b  1
(1)
ab  3
(2)
From (1) and (2)
b(1  2b)  3
2b 2  b  3  0
(2b  3)(b  1)  0
If b 
3
>
2 a  2


 or
y  3 3 
x
1
9
If b  1 > a  3
x  27 

1 
y 
3 
32
Combined Mathematics
4.
Practice Questions (With Answers)
f ( x)  3 x 3  Ax 2  4 x  B
8 4A 8
 2
f     
 B0
9 9 3
 3
4 A  9 B  16
f ( 1)  3  A  4  B  2
(1)
A B 1
(2)
From (1) and (2) A = 5, B = 4
3x3  5 x 2  4 x  4  (3x  2)( x 2  x  2)
 (3 x  2)( x  2)( x  1)
5.
f ( x)  x 4  hx3  gx 2  16 x  12
f (1)  1  h  g  16  12  0
h g 5
f (1)  1  h  g  16  12  24
(1)
h g 3
(2)
From (1) and (2)
h = 4, g =  1
f ( x )  x 4  4 x 3  x 2  16 x  12
f (2)  16  32  4  32  12  0
( x  2) is a factor of f ( x )
f ( 1)  1  4  1  16  12  0
( x  1) is a factor of f ( x )
x 4  4 x 3  x 2  16 x  12  ( x  1)( x 3  3 x 2  4 x  12)
 ( x  1)( x  2)( x 2  5 x  6)
 ( x  1)( x  2)( x  2)( x  3)
6.
ax 2  bx  c  0
b
a
     ,  
c
a
1 b2

x ac
 b2

x2    2  x  1  0
 ac

x2
33
Combined Mathematics
Practice Questions (With Answers)
     2

x 
 2  x 1  0
 



2
  
x2     x  1  0
  


x


x
7.


 x    0




or


Let  be the common root of the equations x 2  bx  ca  0, x 2  cx  ab  0
(1)
Then
 2  b  ca  0
 2  c  ab  0
(1) - (2) gives

(2)
a (b  c)
a
(b  c)
If  ,  are the roots of the Equations rkd;ghL (1)>
  ca and   a Implies that   c
If  ,  are the roots of the Equations rkd;ghL (2)>
  ab and   a Implies that   b
    b vd;gjhy; a  c  b
The equation whose roots are  and  is
( x   )( x   )  0
x2  (   ) x    0
x2  (b  c) x  bc  0
x2  ax  bc  0
8.
g ( x)  ax 2  2 x  (3a  2)
g ( x ) to be positive for all real values of x
a  0 and   0
4  4a (3a  2)  0
a  0 and
3a 2  2a  1  0
(3a  1)(a  1)  0
1
a  1 or a  3
34
Combined Mathematics
Practice Questions (With Answers)
a 
a  0 vd;gjhy;>
1
3
1

Solution:  x : x  , x  
3

1
When a  >
3
g ( x) 
1 2
x  2x  3
3
1
  x 2  6 x  9 
3
1
 ( x  3) 2
3
y
(0,3)
x
O
09.
(3,0)
12
 x 1
x 3
12
 ( x  1)  0
x 3
( x 2  2 x  15)

0
x3
( x  5)( x  3)

0
x 3
( x  5)( x  3)( x  3)  0( x  3)
-3
0
3
5
3  x  3 or x  5
35
Combined Mathematics
10.
Practice Questions (With Answers)
1 2x  x  2  2
When x  2 > 1  2 x  ( x  2)  2
x  1 no solution
When 2  x 
(1)
1
> 1  2 x  ( x  2)  2
2
3 x  1  2
x  1
Solution is 1  x 
When x 
1
>
2
1
2
(2)
(1  2 x )  ( x  2)  2
1  2 x  x  2  2
x5
1
 x5
2
(3)
1  x  5
Hence, solution is   x : x  R, 1  x  5
11.
8 children can sit ins 8! ways
8! = 40 320
(i)
The number of ways that the two particular girls can sit together is 2  7!
Hence the number of ways the two particular girls do not si together is
8! 2  7!
 7!(8  2)
 7! 6  30240
(ii)
4 boys can be arranged in 4! ways  4!
*
*
*
*
 B1  B2  B3  B4 
4 girls can be seated in 5  4  3  2  5!
Hence number of ways
 4! 5!  2880
36
Combined Mathematics
Practice Questions (With Answers)
10
12.
 2 2k 
x 

x 

Tr 1 
10

Cr  x 2 
10  r
10
 2k 


 x 
r
C r (  2 k ) r x 20  3 r
Coefficient of x 2 :
Coefficient of x 2 :
20  3r  2
r6
10
C6 ( 2 k ) 6
Coefficient of x 1 :; 20  3r  1
r7
Coefficient of x 1 ; is
10
C7 (2k )7
C6 ( 2 k ) 6  10C7 ( 2 k ) 7
10
10!
10!
( 2 k ) 6 
( 2 k ) 7
6! 4!
7! 3!
7
k
8
13.
(1  2 x  kx 2 ) n
 1  x(2  kx)
n
 1  nC1 x(2  kx)  nC2 x 2 (2  kx)2  nC3 x3 (2  kx)3  ...
Coefficient of x 2 :
k . n C 1  4. n C 2
 nk  2 n ( n  1)
Coefficient of x 3 :
4 k . n C 2  8. n C 3
 2n( n  1) k 
nk  2 n( n  1)  30
2 n ( n  1) k 
From (2)
(1)
4 n ( n  1)( n  2)
0
3
k
4n( n  1)( n  2)
3
(2)
2( n  2)
0
3
37
Combined Mathematics
Practice Questions (With Answers)
 2 n ( n  2)
 2n( n  1)  30
3
Substituting in (1)>
2n 2  n  45  0
(2n  9)(n  5)  0
n  5 > since n is positive in teger
n  5 and k  2
14.
Z  1  i 3
 1
3
2
2 

 2    i
 i sin
  2  cos

2 
3
3 

 2
2
Z  2, Arg ( Z ) 
3

Z 2  1  i 3


2
 2  i 2 3


3
Z 2  pz   2  i 2 3  p  1  i 3
 (2  p )  i

Z 2  pz real,
3p  2

3 p  2 3  0; p  2
Z 2  qz  (  2  q )  i

3q  2 3

3( q  2)
5
1
 tan

 ( q  2)
6
3
q4
15.
OA  z  1
OB  cos   i sin   1
D
OACB a parallelogram
Point C, represents Z1  Z 2
Since OA  OB > OACB is rhombus
OD  AB , OD is parallel to AB
Point D, represents Z2  Z 1
Z1  Z 2  OC  2 cos
Arg ( Z1  Z 2 ) 


2
2
38
B
C

2
O

2
A
(1, 0)
Combined Mathematics
Practice Questions (With Answers)
AB  Z 2  Z 1  2sin
Arg ( Z 2  Z 1 ) 

2
2

Z1  Z 2  Z 2  Z 1


2
2
2
2
2
 


  2 cos    2sin   4
2 
2

16.
(a)
lim
xa
sin x  sin a
xa
 x  a)   x  a) 
2 cos 
 sin 

2   2 

 lim
xa
 x  a) 
2

 2 
 x  a) 
sin 

 x  a) 
 2 
 lim cos 


xa
 x  a) 
 2 


 2 
 cos a
(b)
sin y  x. sin( y  a )
Differentiating w.r.t x
cos y.
dy
dy
 x. cos( y  a ).  sin( y  a )
dx
dx
From (1)
cos y.
(1)
x
sin y
sin( y  a )
dy
sin y
dy

. cos( y  a).  sin( y  a)
dx sin( y  a)
dx

sin y. cos( y  a)  dy
cos y  sin( y  a)  dx  sin( y  a )


sin a
dy
.  sin( y  a )
sin( y  a) dx
dy sin 2 ( y  a)

dx
sin a
39
Combined Mathematics
17.
lim
(a)
x0
Practice Questions (With Answers)
tan x  sin x
x3
sin x(1  cos x)
x 0
cos x.x 3
sin x 1  cos x
1
 lim


2
x 0
x
x
cos x
x
2 sin 2
sin x
2  1  1 1  1  1
 lim

2
x 0
x
2
2
 x  cos x
4 
2
 lim
y  x n .lnx
(b)
dy
1
 x n .  lnx.n.x n 1
dx
x
dy
x.  x n  n.lnx.x n
dx
dy
x.  x n  ny
dx
d 2 y dy
dy
x. 2 
 n.x n 1  n.
dx
dx
dx
2
d y
dy
x. 2  (1  n)
 n.x n 1
dx
dx
Hence n  3
18.
x  t  lm t
y  t  lm t
dx
1
 1
dt
t
dy
1
 1
dt
t

t 1
t
dy
dy dt t  1


dx dx t  1
dt

t 1
t
(1)
40
Combined Mathematics
Practice Questions (With Answers)
d2y
d  dy 



2
dx  dx 
dx

d  dy  dt
 
dt  dx  dx
d  t 1 
t


dt  t  1  t  1
2t

(t  1) 3

Since t 
x y
2
d2y
x y
8( x  y )


3
2
dx
( x  y  2) 3
 x y 
 1

 2

19.
1
1

2
1 x
(1  x ) 2

1  x 2  1  x 2 
1  x 2 1  x 2

2x
2
1  x 1  x 

1
1

x
1
1
1
dx

dx

dx

2
 1  x 2 
2  0 1  x 2
1  x 2 1  x 

0
2


1
1
1 
  tan 1 x 
2
1  x  0


1  1 1 
 tan 1    0  1

2 
2


1  1 1 
tan 1  
2 
2

1  1   1
  
2  4 2  8 4
41
Combined Mathematics
20.
Practice Questions (With Answers)
x  2(1  cos 2  )

x2

x3

2

4
dx
 4 cos  .sin 
d
3
x2
dx
4 x

2


4


2 cos 2 
.( 4 cos  .sin  .d )
2sin 2 
2


4
cos 
.(4 cos  .sin  )d

 sin 
2


2
2


  4 cos 2  d  2  1  cos 2  d
4
4

sin 2  2

 2  
2  

4
 
   1 
 2   0      
  4 2 
 2

21.

2
1
I   e 4 x . cos 3 xdx,
I   e4 x .cos3xdx  e4 x .
J   e 4 x . sin 3 x.dx
sin3x
sin3x

 4e4 x dx
3
3
3I  4 J  e 4 x . sin 3 x
(1)
  cos3x    cos3x 
4x
J   e4 x .sin 3xdx  e4 x .
  
  4e dx
 3   3 
4 I  3 J  e 4 x . cos 3 x
From (1) and (2)
(2)
I
1 4x
e (3 sin 3 x  4 cos 3 x)
25
42
Combined Mathematics
Practice Questions (With Answers)
y
22.
A  (0,12),
B  (8,0)
M  ( 4,6)
A
2
Equation of MC is y  6  ( x  4)
3
3 y  2 x  10  0
At C> y  1,
C  (
M
x
13
2
O
13
,1)
2
B
x
C
Distance AB  82  122  208
2
2
Distance MC  (4  13)  (6  1) 
441
 49
4
1
637 364

Area of the tri angle   208 
2
4
4
= 91 square units.
23.
Equation of AB:
C
D
x  2y  0
5 5
P , 
2 2
PN is perpendicular to AB and
P
5
5
5
2
PN 

2
5
Since DC is parallel to AB
Equation of DC is x  2 y  k  0 .
A
and the pependicular distance from P to CD is
5
5 k
5
2

2
5
2k  5  5, k  5 or 0
Hence equation of CD is x  2 y  5  0
43
5
2
N
B
Combined Mathematics
Practice Questions (With Answers)
BC and AD are perpendicular to x  2 y  0
Equation of BC and AD are of the form 2 x  y  d  0
and the perpendicular distance from P ;
5
2
5 5
2   d
5
2 2

2
5
d  10,5
Hence the equations are 2 x  y  5  0, 2 x  y  10  0
24.
A
It is given that AB = AC. So AD is pependicular to BC.
All three medians meet at a point G.
A  (0,8)
BE : x  3 y  14
E
F
CF : 3 x  y  2
G
G  ( 2,4), D  ( x0 , y0 )
AG : GC  2 : 1
2 x0  0
2 y0  8
 2,
4
2 1
2 1
B
D  ( x0 , y0 )  (3,2)
Equation of BC is
y2
1
( x  3)
2
2 y  x 1  0
BC : 2 y  x  1  0 

BE : 3 y  x  14  0
B  (5,3)
BC : 2 y  x  1  0 

CF : 3 x  y  2  0
C  (1,1)
Equation of AB is
y  3  1( x  5)
y  x 8  0
Equation of AC is
y  1  7 ( x  1)
y  7x  8  0
44
D
C
Combined Mathematics
25.
Practice Questions (With Answers)
A
S  x2  y2  a2  0

l  x cos   y sin   p  0
Any circle passing through the points A and B can be
written
x
2

 y  a   x cos   y sin   p   0
2
2
B
  cos  sin  
,

center is  
2
2 

AB is the diameta of the circle
  cos   sin  
,

, lies on x cos   y sin   p  0
2
2 


 cos 
2
. cos  
 sin 
2
. sin   p  0
  2p
Equation of the requird circle is
x
26.
2

 y 2  a 2  2 p  x cos   y sin   p   0
C  (2,1)
Centre
P  (4,2)
Radius
4  22  2  12
CP 
(1)
 4 1 4  1
 5
CP  1, P lies outside S
(11) PT  CP 2  12  5  1  2
Suppose that the equation of tangent is y  mx  c
It passes through P(4, 2)
2  4m  c
T
y  mx  (2  4m)
y  mx  (2  4m  0
CT  1
1  2 m  2  4m
1  m2
C
1
2m  1  1  m 2
2m  12  m 2  1
45
P
Combined Mathematics
Practice Questions (With Answers)
4
m  0 or 3
If m  0 >
C=2
If m 
4
>
3
10
C
3
Equatons of the tangentsare y  2, and 3 y  4 x  10  0
27.
Equation of the circle is x 2  y 2  2 gx  2 fy  c  0
Centre (  g , f ) and radius is equal to
g2  f 2  c
Perpendicular distance from C to y axis is equal to the radius of the circle.
Equation of y axis is x = 0
g
 g2  f 2  c
1
g2  g2  f 2  c
c f2
Equation: x2  y 2  2gx  2 fy  f 2  0
AC 2  AM 2  MC 2
C
2
3
g2  f 2  f 2     f 2
2
9
g2  f 2 
4
Therefore, the general equation is

9
x 2  y 2  2 f 2   x  2 fy  f 2  0
4


Centre  

9
f 2  , f
4
9
x0   f 2  ,
4




y0   f
9
4
2
2
4 x0  4 y 0  9
2
2
x0  y 0 
Locus of ( x0 , y0 ) is 4 x 2  4 y 2  9
46
A
M
Combined Mathematics
28.
Practice Questions (With Answers)
(o     )
cos 6  cos 4  cos 2  1  0
2 cos 5 .cos   2 cos 2   0
2 cos  (cos 5  cos  )  0
4 cos  .cos 3 .cos 2  0
cos   0
cos 3  0
  2n 


2

3  2n 
2

;
cos 2  0

2
  5
,
6 2
,
2  2n 
6
;
    3 5 
  , , , , 
6 4 2
29.
4
6 
1
tan 1    A
3
1
tan A  ,
3
tan 2 A 
0  2A 
2 tan A
3

2
1  tan A 4

4
2 A  tan 1
,
3
4
1  1 
Let tan    B
7
tan B 
1
and
7
0 B

4
1
1
2 tan 1    tan 1  
3
7
 2 A  B and
tan(2 A  B ) 
2A  B 
0  2A  B 
tan 2 A  tan B
1  tan 2 A.tan B

2
3 1

4
7 1

3 1
1 
4 7

4
1
1 
2 tan 1    tan 1   
3
7 4
47

 3
4
,
4

2
;n
Combined Mathematics
30.
Practice Questions (With Answers)
a
b
c
bca



sin A sin B sin C sin B  sin C  sin A
a
bca

sin A sin B  sin C  sin A
a
bca

A
A
A
A
 BC 
 BC 
2 sin cos
2sin 
 cos 
 2 sin cos


2
2
2
2
 2 
 2 
a
bca

A
A
 BC 
sin
cos 
 sin

2
2
 2 
a
bca

A
 BC 
 BC 
sin
cos 
 cos 


2
 2 
 2 
a
bca

A
B
C
sin
2sin .sin
2
2
2
a
A
bca
A
.cos 
.cos
A
B
C
2
2
sin
2sin .sin
2
2
2
 BC 
sin 

A
 2 
2a cot  (b  c  a)
B
C
2
sin .sin
2
2
B
C
B
C
sin .cos  cos .sin
A
2
2
2
2
2a cot  (b  c  a)
B
C
2
sin .sin
2
2
A
C
B

2a cot  (b  c  a)  cot  cot 
2
2
2

48
Combined Mathematics
31.
Practice Questions (With Answers)
Let   r (cos   i sin  ) for all n    >  n  r n (cos n  i sin n ) .
  1  3i
1
3 
   
i 
2 2 



   cos  i sin 
3
3

Demovier’s theorem,
7
7 

 7  27  cos
 i sin

3
3 




 128  cos  i sin 
3
3

7  128
Arg ( )7 
32.
Let

3
  r (cos   i sin  ) for all n    >  n  r n (cos n  i sin n ) .
If   cos   i sin  then
3  (cos   i sin  )3  (cos 3  i sin 3 )
cos3   3cos 2  (i sin  )  3cos  (i sin  ) 2  (2 sin  )3  cos 3  i sin 3
(cos3   3cos  sin 2  )  i (3cos 2  sin   sin 3  )  cos 3  i sin 3
Equating the real part
cos 3   3cos  sin 2   cos 3
cos3   3cos  (1  cos 2  )  cos 3
cos 3   3cos   3cos3   cos 3
 cos 3  4 cos3   3cos 
49
Combined Mathematics
Practice Questions (With Answers)
By equating the imaginary part
3cos 2  sin   sin 3   sin 3
3(1  sin 2  ) sin   sin 3   sin 3
sin 3   3sin   3sin 3   sin 3
 sin 3  3sin   4sin 3 
33.
y  t 2 (1  t )
dy 2
 t ( 1)  (1  t )2t  t 2  2t  3t 2
dx
 3t 2  2t
 t (2  3t )
x  t (1  t ) 2
dx
 t.2(1  t )( 1)  (1  t ) 2 .1
dy
 2t  2t 2  t 2  1  2t
 3t 2  4t  1
 (1  t )(1  3t )
dx
t (2  3t )

dy (1  t )(1  3t )
t T
dx
t (2  3t )
t T 
dy
(1  t )(1  3t )
1
dx
 4  1
dy  1
4
1 1 1
y  1   
4 2 8
2
1 1 1
x  1   
2 2 8
50
Combined Mathematics
Practice Questions (With Answers)
1
8  1
1
x
8
y
8 y 1
 1
8x 1
8 y  1  8 x  1
8 y  8x  2  0
4 y  4x 1  0
34.
y  x2  3x
y  x 2  3x 
9 9

4 4
2
3 9

y x  
2 4

3
 (x
2
 3x)dx
0
3
 x3 3x 2 
  
2  0
3

27 27

3
2

27
6

27
square units
6
51
Combined Mathematics
Practice Questions (With Answers)
(35)
3
R   e x dx
1
  e x 
3
1
 e3  e1 = e(e 2  1)
3
V    y 2 dx
1
3
   (e x ) 2 dx
1
3
 e2 x 
   e dx =   
 2 1
1
3
2x
 e6 e 2 
=  
2 2

V

2
e 2 (e 4  1)
 e2
2
(e 4  1)
52
Combined Mathematics
Practice Questions (With Answers)
Part B
1.
(a)
x 2  px  q  0
     p,
(i)
  q
1
  2 3

1
4
 

4

 p  4q
   
      4
2
2
12  p 2  4q
p 2  p  12  0
( p  4)( p  3)  0
p3 

3
q 
4
p  4 

q 1 
(ii)


2

2


  2 q  2




  2 q  2



(1)
x  px  q  0
2
Let
y
q2
x
x
q2
y
Replacing x by
q2
in equation (1),
y
we have
2
q2
q2

  p
q 0
 y 
 y 
 q  2
2
 p  q  2  y  qy 2  0
qy 2  p ( q  2) y  (q  2) 2  0
i.e The equation whose roots are  
qx 2  p ( q  2) x  (q  2) 2  0
53
2

, 
2

is
Combined Mathematics
(b)
Let
Practice Questions (With Answers)
x 2  3x  4
y
5x  k
x 2  (3  5 y) x  (ky  4)  0
  (3  5 y)2  4(ky  4)
 25 y 2  (4k  30) y  25
For real values of x,
0
i.e 25 y 2  (4k  30) y  25  0
For all values of y> 25 y 2  (4k  30) y  25 to be greaten than or equal to zero
(i)
Coefficent of y 2  25  0 and
(ii)
1  (4k  30) 2  4  25  25  0
(4k  30) 2  50 2  0
(4k  20)(4k  80)  0
(k  5)(k  20)  0
20  k  5
k=-5
f ( x) 
( x  4)( x  1)
5( x  1)
(1)
When x = 0 , f ( x )  
(2)
When y  0 > -4, 1
(3)
x = -1 is an assymptot.
(4)
4
5
 1
( x  4)  1  
x

f ( x)  
1

5 1  
x

  , x 
  as
f ( x ) 
f ( x ) 
 -  , x 
 -  as
(5)
x  4,
f ( x)  0
4  x  1,
f ( x)  0
 1  x  1,
f ( x)  0
x  1,
f ( x)  0
54
Combined Mathematics
Practice Questions (With Answers)
y
-4
-1
0
4
5
02.
f ( x)   2 x 2  ( 2  2 ) x  3  0
  
 
 2  2
2
3
2
   2   2      2 4
 


 


3
2
2
  2  2 
3

  2 2
2

4
 


3
3
2

2
 2

3
2
2
 6 2

4
3
3 4  12 3  12 2  18 2  12 2
3 4  12 3  18 2  0
55
1
x
Combined Mathematics
Practice Questions (With Answers)
 2   2  4  6   0
 2  4  6  0
1  2  4
12  6
12 22
Equaton whose roots are
>
2 1
2 2 
x 2   1  2  x  12  0
 2 1 

 13  23
x 
 12

2
  x   



1 2
0
 (1  2 ) 12  12  22
x 
12


2
  x   
1 2


0
 4 16  18 
x2  
x6  0
6


2
3 x  68 x  18  0
If f ( x )  2 x then
f ( x)  2 x  0 for all x  
 2 x2   2 x  3  0
x2  x 
x2  x 
3
2
0
1 1 3
 
0
4 4 2
2
1   12   2 

0
x  
2   4 2 

2
1

Since  x    0 for all x  
2

12   2
0
4 2
12   2  0
56
Combined Mathematics
Practice Questions (With Answers)
 2  12  0
   2 3    2 3   0
2 3    2 3
3.42    3.42
Greatest integer value of  is 3
2n
(b)
 (1)r 1
r 1
2n
1
1
 
r r  n 1 r
2
r 1 1

When n  1> L.H.S  (1)
r
r 1
 1
1 1

2 2
2
1 1


R.H.S

2
r 2 r
L.H.S
= R.H.S
The result is true for n  1
Assume that the result is true for n  p
2p
 (1)
r 1
r 1
i.e
1
2p
1
1
 
r r  p 1 r
1 1 1
1
1
1
1
  ... 


 ... 
2 3 4
2 p p 1 p  2
2p
2( p 1)
n  p  1 as,

r 1
(1) r 1
1
1 1 1
1
1
1
 1    ... 


r
2 3 4
2 p 2 p 1 2 p  2
 1
1
1 
1
1


 ... 


2 p  2 p 1 2 p  2
 p 1 p  2

1
1
1
1
1
1

 ... 



p2 p3
2 p 2 p 1 2 p  2 p 1

1
1
1
1
1

 ... 


p2 p3
2 p 2 p 1 2 p  2

2( p 1)
1
r  p2 r

The result is true for n  p  1
By the principle of mathematic induction, the result is true for all positive intergern.
57
Combined Mathematics
03. (a)
Practice Questions (With Answers)
2r  3 A
B
 
r (r  1) r r  1
A(r  1)  Br
r (r  1)
( A  B)r  A

r (r  1)

2r  3  ( A  B)r  A
A  3, B  1
2r  3
3
1
 =
r (r  1)
r
r 1
2r  3 1

r (r  1) 3r
Ur 
1  1
3
 
. r
 r r  1 3
1 1
1 1
  . r 1 
.
 Vr  Vr 1
r  1 3r 
r 3
Vr 
1
r.3r 1
U r  Vr  Vr 1
u1  v1  v2
u2  v2  v3
u3  v3  v4
.....................
un1  vn1  vn
un  vn  vn1
n
U
r 1
r
n
U
r 1
r
 V1  Vn 1
1
1 1
 
.
1 n  1 3n
n   Mf
1 1
. 0
n  1 3n
n
lim  U r  1
n   r 1

Hence the series is convergent and
58
U
r 1
n
1
Combined Mathematics
(b)
Practice Questions (With Answers)
1

2 x  1, x  2 
y  2x 1 = 

2 x  1, x  1 

2 
 x  2, x  1
y  x  1 +1 = 

x  1
  x,
y  x2
y  2 x  1
y  x2
x  2  2 x  1
x  2  2x 1
x3
y  2x 1
3 x  1
x
1
3
2x 1  x 1  1
2x 1  1 x  1
/
04.(a)(i)
solution
x  3 and
x
1
3
Consider the six girls as one group.
Now 7 can be arranged in a row in 7! ways.
6 girls can be arrange among them selves in 6! ways.
Hence the total number of ways that the six girls.
can sit together is 7! 6! ! ways.
 5040  720
 3628800
59
Combined Mathematics
(ii)
G
G
Practice Questions (With Answers)
G
G
G
G
(1)






G
G
G
G
G
G





(2)

Six girls can sit in 6! ways.
Six boys can be arranged as shown above in two ways.
In each way, boys can be arranged in 6! ways.
Hence the number of ways the boys and girls sit alternatively is 2  6! 6!
 2  720  720
 1036800
(a)
0> 2> 3> 5> 7> 8
*
*
*
*
(i)
 5 6  6  6  1080 numbers can be formed.
(ii)
One digit only once
 5  5  4  3  300 numbers can be formed.
Greater than 5000 and divisible by 2
* *
*
0
ends in zero,
2
ends in 2,
3  4  3 1  36
8
ends in 8,
2  4  3 1  24
3  4  3 1  36
Total  36  36  24  96
(c)
(1  x) n  nC0  nC1 x  nC2 x 2  ...  nCr x r  ...  nCn x n
( x  1)n  nC0 x n  nC1 x n 1  nC2 x n  2  ...   nCr x n  r  ...  nCn
differentiating w.r.t x
(1)
n(1  x )n 1  nC1  2. nC2 x  ...  r. nCr x r 1  ...  n. nCn x n 1
(2)
n( x  1) n 1  n. nC0 x n 1  (n  1) nC1 x n  2  ...  ( n  r ) nCr x n  r 1  1. nCn 1
Consider (1)  (2)
n 2 (1  x ) 2 n  2 

n

C1  ...  n. nCn x n 1 n. nC0 x n1  ...  Cn 1
60

Combined Mathematics
Practice Questions (With Answers)
Coefficient of x n  2
In R.H.S, coefficient x n  2 is
( n  1) n C12  2( n  2) n C22  ...  r ( n  r ) n Cr21  ...  ( n  1) n Cn2
In L.H.S, coefficient of x n  2 is n 2 . 2 n  2Cn  2
Hence the result.
In (3) put x  1 .
n 2 .2 2 n  2 

C1  2. nC 21  ...  n. nCn
n
n
n 1
r 1
r 0
 n. C
n 2 .22 n  2   r. nCr . ( n  r ). nCr
05.
( a)
Z3 1
 Z  1  Z 2  Z  1  0
Z  1  0 or Z 2  Z  1  0
Z 1
Z
1  3
2
1
3
1
3
or   i
Z  1 or   i
2
2
2
2
Let  be a comlex root of Z 3  1  0
Now
3 1  0
  1  2    1  0
  1 Therefore 1     2  0
(i) 1     2
1
1
 2
1 

1

1 

(ii)
1   2  
1
1

2

1 
2
 
2 1
61
n
0
 ...  1. nCn 1

Combined Mathematics
Practice Questions (With Answers)
3k
(iii)
 2 
  





1  
 1  
3k
3k
3k
 1
       
 
  1

 1
 3

  1
1  1 
3k
3k
 
k
 
 
3 k




  1 .2
3k
(b)
If
k odd,
If
k even,
 1
3k
 1
.2  2
3k
.2  2



u  2i  2  cos  i sin 
2
2

1
3
2
2 

v   i
 1 cos
 i sin

2
2
3
3 


7
7 
  2 
  2  

uv  2  cos  
 i sin
  i sin  
   2  cos

3 
6
6 
2
 2 3 



 5
 2  cos  
 6


 5  
  i sin  


 6 

u
  2 
  2  
 2  cos  
  i sin  

v
3 
3 
2
2


 
  
 2  cos     i sin    
 6
 6 

A
OA = OB = OC
It can be easily proved that
ˆ  ABC
ˆ  600
ˆ  ACB
BAC

6
B
Hence ABC is an equilateral triangle.
62

O
6
C
Combined Mathematics
06. (a)
 1 i 


 1 i 
Practice Questions (With Answers)
4 n 1
 1 i 1 i 



 1 i 1 i 
 2i 
 
2
4 n 1
4 n 1
 i 4 n 1   i 4  i  i
n
x3  1  0
 x  1  x 2  x  1  0
x  1,
x 2  x  1  0, x 
1  3
2
x  1,
1
3
x   i
,
2
2
1
3
x   i
2
2
x  1,
x  cos
x  1,
, 2
2
2
4
4
 i sin
, x  cos
 i sin
3
3
3
3
Also, 1     2  0,  3  1
 x  2
3
 1; y 3  1; y  1,  ,  2
x+2 = y
1
3
1
3
, x  2   i
x  2  1, x  2    i
2
2
2
2
x   1,
5
3
5
3
x   i
, x   i
2
2
2
2
 2  5  2    2  2  2
2 6
2
 3 
  3   36. 6  729
6
 p  q  p  q   p 2  q 
  p  q   p 2 3  pq 2  pq  q 2 
 ( p  q)  p 2  pq  q 2   p 3  q 3
  b  c  a 2   b  c 2  a 3
 a  b  c 2
a  b  c 2

b  c  a 2
63
6
Combined Mathematics
Practice Questions (With Answers)
(b)
B

C
Z  3  3i  2

4
Locus of P is a circle with centre (3, 3)
and radius 2.
Equation of the locus is ( x  3) 2  ( y  3)2  2 2
The treatest value of Z in the region is 3 2  2
07. (a)
cos 4 x  cos 2 x
lim
x 0
x2
1  2 sin 2 2 x  cos 2 x
x 0
x2
2
sin x
sin 2 2 x
 lim

2.
x 0
x2
x2
 lim
2
sin 2 2 x
 sin x 
 lim 
 24

2
x 0
 x 
2 x 
 1  8 1
 7
lim
x 0
tan 2 x  2sin x
x3
sin 2x
 2sin x
cos
2
x
 lim
x 0
x3
2sin x  cos x  cos 2x 
1
 lim


2
x 0
x 
x
 cos 2 x
3x
x
2sin
sin
2sin x
2  2 1
 lim

x 0
x
x
x
cos 2x
3
x
2sin x
sin
2sin x
3
2  
2 1
 lim

x 0
3x
x cos 2x
x
2
2
2
2
3 1
 2  2   1  3
2 2
64
3
A
Combined Mathematics
1
1
(b) (i) y  sin
sin y 
sin y 
cos y.
x2  1
Practice Questions (With Answers)
,
x  2
Z  sec 1 x
1
x  sec z
x2  1
1
sec z 2  1
1

 cot z
tan 2 z
dy
  cos ec 2 z
dz
Since x  2, 0  y 

2
, 0 z

2
dy
  1  cot 2 z
dz
1 dy
1 

1  2 .   1 

2
x  1 dz
 tan z 

1  sin 2 y .

x 2  2 dy
sec2 z
x2
.




x 2  1 dz
sec 2 z  1
x2 1
dy
x2
x2  1
 2
 2
dz
x 1
x 2
2
dy
x

dz
x2  2 x2  1

(ii)
(c)

x2
dy

dz
dy

dz

x
2

x2
x
2
0

 2 x2  1

 1 x2  2

0
l
 2 l

Area A    x  x    x 
2

2

l
l

   x  lx 
4
2

2
x
x
2
(l-2x)
65
Combined Mathematics
dA  l
 1
   x 
dx  2
 2

Practice Questions (With Answers)
2
1
l
 l  lx  (1)
2
4
l
lx 
4
2
ll
l 
 

x

lx





22
4 
 
2
l
lx 
4
2
l
3lx

2
 2
2
l
lx 
4
3l
l


x 

2
3
l 
2 lx 
4
l
l dA
x ,
 0 A increases.
4
3 dx
l
x ,
3
dA
 0 A decreases.
dx
Hence A has a maximum at x 
Area 
l
and the triangle is equitlateral triange.
3
1 l l
   sin 60
2 3 3
1 l l
3
3l 2
   

sq.units.
2 3 3 2
36
08. (a) (i)
f ( x)  sin 2 x
f ( x)  lim
h0
 lim
h0
sin  2 x  2h   sin 2 x
h
2 cos  2 x  h   sin h
h
sin h
h0
h
 2 cos 2 x.  1  2 cos 2 x
 lim 2 cos  2 x  2h 
66
Combined Mathematics
(ii)
Practice Questions (With Answers)
dn
 n

sin 2 x   2n sin 
 2x
n 
dx
 2

when n=1
L.H.S. 
d
(sin 2 x )  2 cos 2 x
dx


R.H.S.  2 sin   2 x   2 cos 2 x
2

The result is true for n=1.
Assume that the result is true for n  p
dp
 p

(sin 2 x)  2 p.sin 
 2x 
p
dx
 2

d p1
d 
 p

(sin 2 x)  2 p.sin 
 2 x 
p 1
dx 
dx
 2

 p

 2 p.cos 
 2 x   ( 2)
 2


 p

 2 p 1   cos 
 2 x 
 2


   p

 2 p 1.sin   
 2 x 

2  2



 2 p 1.sin ( p  1)  2 x 
2


Therefore the result is true for n  p  1
By the principle of mathematical induction the result is true for n all positive intergern.
(b)
f ( x)  1 
1
x( x  2)
f ( x) 
(2 x  2)
x 2 ( x  2) 2

2( x  1)
x 2 ( x  2) 2
67
Combined Mathematics
Practice Questions (With Answers)
When x  1 > f ( x )  0
x  0 and x  2 are asymptotes.
0
1
2
x0
f ( x )  0
f is increasing.
0  x 1
f ( x )  0
f is increasing.
ix2
f ( x )  0
f is decreasing.
x2
f ( x )  0
f is decreasing.
At x  1 > f has a maximum and f (1)  0 .
f ( x )  1 as x  
y  1 is an asymptot.
(i)
y  f ( x)
y
1
0
68
2
x
Combined Mathematics
(ii)
Practice Questions (With Answers)
y  f ( x)
y
0
f ( x)  1 

x
1
1
x( x  2)
( x  1) 2
x( x  2)
1
x ( x  2)

f ( x ) ( x  1) 2
f ( x )  1 as
x   ;
At x  0, 2>
1
0
f ( x)
1
 1 as
f ( x)
When x  0
1
is increasing.
f ( x)
0  x 1
1
is increasing.
f ( x)
1
Thereforce x  1
is decreasing.
f ( x)
f ( x)  0
1
At x  1 > y 
f ( x)
69
x  
Combined Mathematics
Practice Questions (With Answers)
Therefore is an asymptot of y 
1
f ( x)
Since 1  x  2, and x  2 >
f ( x ) decreases.
1
increases.
f ( x)
y
0
1
09.
(a)
1  x  x
2
2

1

2
1
x
A
B
Cx  D


1  x 1  x 1  x2
1  A(1  x 2 )(1  x)  B (1  x)(1  x 2 )  (Cx  D )(1  x)(1  x )
x  1, 1  4 B, B
1
4
x 1, 1  4 A, A
1
4
1
x  0, 1  A  B  D, D
2
Coefficient of x 3
0   A  B  C, C  0
1
1
1
1
1
dx
 1  x 1  x dx   4 1  x dx   4 1  x dx  2  1  x
2
2

1
1
1
ln 1  x  ln 1  x  tan 1 x  c
4
4
2
70
2
Combined Mathematics
(b)
Practice Questions (With Answers)
t  sin x  cos x
t 2   sin x  cos x   1  sin 2 x
2
sin 2 x  1  t 2

t  sin x  cos x
x:0 
dt
 cos x  sin x
dx
t : 1  0

4
0
dt
sin x  cos x
0 9  16 sin 2 x dx  1 9  16(1  t 2 )
4
0

dt
 (5  4t )(5  4t )
1
0
 A
B
 

(5  4t ) (5  4t )
1 
5 A  5B  1
4 A  4B  0
1
1
A ,B 
10
10
0
0

1
dt
1
dt
 

10 1 (5  4t ) 10 1 (5  4t )

1
1
ln 5  4t  ln 5  4t
40
40
0
1  5  4t 

ln
40  5  4t  1
1 
1
ln1  ln 

40 
9
1

ln 9
40

71

dt

Combined Mathematics
Practice Questions (With Answers)


2
(c)
2
cos x
sin xdx
I 
dx, J  
dx
a cos x  b sin x
a cos x  b sin x
0
0


a cos x  b sin x

dx   x 02 
0 a cos x  b sin x
2

b cos x  a sin x
bI  aJ   2
dx
0 a cos x  b sin x
aI  bJ   2

 ln a cos x  b sin x  02
 ln
b
a
1  a
b
 bln 
2 
a
a b  2
1  b
b
J 2
 aln 
2 
a
a b  2
I
2


  x  sin   x 
  x  sin x
x sin x
dx

dx

0 1  cos2 x 0 1  cos2   x 
0 1  cos2 x dx

10.
(a)


0
 sin xdx
1  cos x

2

x sin x
dx
2
0 1  cos x


x sin x
sin x
dx   
dx
2
2
1

cos
x
1

cos
x
0
0
2
Put u  cos x
du
  sin x
dx
x:0
u :1  1


x sin x

sin x
0 1  cos2 xdx  2 0 1  cos2 xdx


1
du
2  1 u
2
1


1
 du
2  1 u
2
1
72
Combined Mathematics



(b)

Practice Questions (With Answers)
1
 tan 1 u 
1
2

 tan 1 (1)  tan 1 (1) 
2
 
2
   





2  4  4   4
x.e x
 (1  x)2 dx
  x.e x .
u  x.e
1
1  x 
2
dx
dv
1

dx 1  x 2
x
v
x.e x
1
1  x 
1
1
 1  x  dx  x.e 1  x    1  x  .e
x
2
x
( x  1) dx
 x.e x
  e x .dx
1 x
 x.e x

 ex
1 x
ex

1 x

(c)
y  x (2  x )
y

  x2  2x

   x 2  2 x  1  1
1
y    x  1  1
2
0
x(2  x)  x
x(2  x)  x  0
x(1  x)  0
x  0,1
73
1
2
x
Combined Mathematics
1

Practice Questions (With Answers)
1

2
Area   2 x  x dx   x.dx
0
0
1


0

11.
(a)
1
 x 2 x3 
x  x dx    
 2 3 0
2

1
sq. units.
6
AD : x  y  4  0
D
AC : 3 x  y  8  0
A  (3,1)
Equaton of AB is
( y  x  4)   ( y  3 x  8)  0
C
(1  3 ) x  (1   ) y  (8  4)  0
Gradient of AB 
A
(3,1)
3  1
 1
Gradien of AD is  1
Gradient of AB is  1 
B
3  1
 1
 1
Equaton of AB is  x  y  2  0
Let B  ( x0 , y0 )
y0  1
1
x0  3
y0  1 x0  3

(=t , say)
1
1
 x0  3   y0  1
2
2

 2 2
t 2  t 2  8
2t 2  8
t 2  4
t  2
if t  2,
B  (5, 3)
t  2,
B  ( 1, 1)
74

2
Combined Mathematics
Practice Questions (With Answers)
Since B lies in the first quadrant
B  (5,3)
Equation of BC is y  x  k ( AD // BC )
53  k
k 8
Equation of BC is
yx8
BD, x  3 y  7  0
= x  3y  c  0
Equation of BD is
59c  0
c4
Equation of BD is
x  3y  4  0
BD : x  3y  4  0
AD : x  y  4  0
D  (2,2)
Equation of CD is x  y  k
22  k
k  0
Equation of CD is
(b)
x y 0
S  x 2  y 2  2 gx  2 fy  c  0
S: (2, 0), (0, 1)
4  0  4g  0  c  0
0 1 0  2 f  c  0
4  4g  c  0
1 2 f  c  0
g
(c  4)
,
4
f 
c 1
2
2(c  4)
2(c  1)
x
yc 0
4
2
2 x 2  2 y 2  (c  4) x  2(c  1) y  2c  0
S  x2  y2 
75
Combined Mathematics
Practice Questions (With Answers)
The general equation of the circle is
4
S  x2  y2  
 x     1 y    0 MFk;.
 2 
Since the circle passes through (1, -1)
4
11 
     1    0
 2 
  2
(i)
Equation of S1 is
1 1
C1   , 
2 2
centre
(ii)
x2  y 2  x  y  2  0
 4
2
2
S1  0 bisects S2  x  y  
 x     1 y    0
 2 
Common chord of S1  0 and S2  0 is S1  S 2  0
4 
 1 x     2  y     2   0

 2

   2 x  2    2 y  2    2  0
   4     1 
,
 of S 2  0
Common chord passes through the centre 
2
 4

  1
  2    2 
  2    2  0
 4 
 2 
   2 
 4
    2  0
  0 or   2
when   2 > S 2  S1
when   0 > S 2  x 2  y 2  2 x  y  0
76
Combined Mathematics
(iii)
Practice Questions (With Answers)
 4
x2  y2   1
 x   1  1 y  1  0
 2 
 4
x2  y 2   2
 x   2  1 y  2  0
 2 
Orthogonal to each other
 1  4 1  1 
,
centre C1  

2 
 4
   4 2  1 
C2   2
,

2 
 4
   4   2  4 
   1   2  1 
2 1
 2 1




  1  2
 4  4 
 2  2 
12  4
A (3, -1)
12.
(a)
x  4 y  10  0
Equation of CP is
Equation of BQ is 6 x  10 y  59  0
P
C lies on x  4 y  10  0
 t  10 
C   t,
 , A  (3, 1)
4 

B
t 3 t 6
Q
,

8 
 2
Since Q lies on BQ, 6 x  10 y  59  0
t 3
t6
6
  10 
  59  0
 2 
 8 
t 10
C  (10,5)
Gradient of AC is
6
7
Gradient of CP is
1
4
77
Q


C
Combined Mathematics
Practice Questions (With Answers)
Let Gradient of BC be m.
6 1
1

4  7 4
m
6 1
1
1 
4
7 4
m
4m  1 1

4m
2
4m  1
1

4m
2
m
6
2
or 
7
9
2
Equation of BC is y  5   ( x  10)
9
2 x  9 y  65  0
Equation of AC is y  1 
6
( x  3)
7
6 x  7 y  25  0
BC : 2 x  9 y  65  0
BQ : 6 x  10 y  59  0
 7 
B    ,8 
 2 
Equation of the line perpendicular to AC can be written in the form 7 x  6 y  c  0
 7 
Since this line passes through B    ,8 
 2 
 7
7   68  c  0
 2
c
47
2
Equation is 14 x  12 y  47  0
78
Combined Mathematics
(b)
Practice Questions (With Answers)
Equation of S3  0 is
 3x
2
 3 y 2  6 x  1    x 2  y 2  2 x  4 y  1  0
Centre of S1  0 is (1, 0)
S3  0 passes throug (1, 0)
 3  0  6  1   1  0  2  0  1  0
 1
S3  0
 3x
2
 3 y 2  6 x  1    x 2  y 2  2 x  4 y  1  0
4x2  4 y2  4x  4 y  0
x2  y2  x  y  0
S2  0
g 1
S3  0
g  
f  2
1
2
f
1
2
c 1
c  0
 1
 1
2 gg   2 ff   2  1     2  (2)      1  2  1
 2
 2
=
1
c  c =
1 0
2 gg   2 ff   c  c
S3  0 and S 2  0 intersect orthogonally..
centre of S1 is (1, 0)
Equation of the tangent at ( x1 , y1 ) to the
circle x 2  y 2  2 gx  2 fy  c  0
xx 1  yy1  g ( x  x1 )  f ( y  y1 )  c  0
Equation of the tangent at (1, 0) to the
circle x 2  y 2  x  y  0
1
1
x  1  y  0  ( x  1)  ( y  0)  0
2
2
x 1 y
x
 0
2
2
x  y 1  0
79
Combined Mathematics
Practice Questions (With Answers)
Equatio of AB is


 1  8  
7
y 8  
 


7
2
 3 

2 
9 2 
7
y 8 
  
13 
2
13 y  104  18 x  63
18 x  13 y  41  0
13. (a)(i) (2sin x  cos x)(1  cos x )  sin 2 x
 2sin x  cos x 1  cos x   1  cos 2 x   0
(1  cos x )  2sin x  cos x  1  cos x    0
(1  cos x )  2sin x  1  0
cos x  1  0
or
2sin x  1  0
1
2
cos x  1
sin x 
x  2n   , n  
x  n   1
n

(ii) 2 tan x  sec 2 x  2 tan 2 x
1  tan 2 x
4 tan x
2 tan x 

2
1  tan x 1  tan 2 x
2 tan x 1  tan 2 x   1  tan 2 x   4 tan x
2 tan x  2 tan 3 x  1  tan 2 x  4 tan x
2 tan 3 x  tan 2 x  2 tan x  1  0
tan 2 x  2 tan x  1  1  2 tan x  1  0
 tan
2
x  1  2 tan x  1  0
tan 2 x  1  0 @
tan x 
1
2

 1 
x  n     tan 1    n  
 2 

80
6
;n
Combined Mathematics
(b)
Practice Questions (With Answers)
2 cos 2   2 cos 2 2
 1  cos 2   1  cos 4 
 cos 2  cos 4
  360
2 cos 2 360  2 cos 2 720  cos 720  cos1440
2  cos 360  cos 720   cos 36  cos 72   cos 720  cos144 0
cos 360  cos 720 
cos 720  cos1440
2  cos 360  cos 720 
2sin1080 sin 360
4cos 540 cos180
2cos180 cos 540 1


4cos 540 cos180 2

cos 36  cos 72 
1
2
cos 36  x
1
2
2
2x  4x  2  1
x   2 x 2  1 
4 x2  2x  1  0
x
2  4  16
8
22 5
8
1 5

4

Since cos 360  0
cos360 
5 1
4
cos 72  cos 36 

5 1 1


4
2
1
2
5 1
4
81
Combined Mathematics
(c) (i)
Practice Questions (With Answers)
a
b
c


sin A sin B sin C
 k (say).
a2  b2
b2  c2
c2  a2


cos A  cos B cos B  cos C cos C  cos A



k 2 sin 2 A  sin 2 B
cos A  cos B

  k  sin
k 2 cos 2 B  cos 2 A
2
2
B  sin 2 C
cos B  cos C
  k  cos
2
2
  k  ain C  sin A
C  cos 2 B
2
2
2
cos C  cos A
  k  cos
2
2
A  cos 2 C
cos A  cos B
cos B  cos C
cos C  cos A
2
2
 k  cos B  cos A   k  cos C  cos B   k  cos A  cos C   0
2
(ii)
a
b
c


 t (say)
0
0
sin 45
sin 75
sin 60 0
a  2c  2b
 t sin 450  2t sin 600  2t sin 750
 t sin 450  2 sin 600  2sin 750 
 1
 3  1 
3
t
 2
 2 
 
2
 2 2  
 2
 1
3
3  1
t


0
2
2 
 2
a  2c  2b  0
a  2c  2b
14. (a) (i)
2(cos x  cos 2 x)  sin 2 x(1  2 cos x)  2sin x
2(cos x  cos 2 x)  2sin x cos x(1  2cos x)  2sin x  0
2(cos x  cos 2 x)  2sin x(cos x  2cos 2 x  1)  0
(cos x  cos 2 x)  sin x(cos x  cos 2 x)  0
(1  sin x)(cos x  cos 2 x)  0
82

Combined Mathematics
Practice Questions (With Answers)
sin x  1  0
cos x  cos 2 x  0
2 cos
sin x  1
x

cos
2
3x
x
.cos  0
2
2
x
0
2
or
x

 2n 
2
2
Solutions:
(ii)

cos
3x

 2n 
2
2

x  4n  
x
x  
x

   x   
 ,  ,
3 3
 1 
1  1 
1  3 
1  1 
tan 1 
  tan 
  tan    tan  
 x 1 
 x 1 
5
3
 1 
tan 1 
  A>
 x 1 
 1 
tan 1 
B
 x 1 
3
tan 1    C >
5
1
tan 1    D
3
A-B = C-D
tan( A  B )  tan(C  D )
tan A  tan B
tan C  tan D

1  tan A tan B 1  tan C tan D
1
1
3 1


x 1 x 1  5 3
1
3 1
1
1 
( x  1)( x  1)
5 3
2
4

2
x
18
2
x 9
x  3
Since 2  x  4 ,
3x
0
2
x3
83
3
 4n  1

3
, 
Combined Mathematics
(b)
Practice Questions (With Answers)
sin(   ) cos(   )

(1  m)
(1  m)
sin(   )  cos(   ) cos(   )  sin(   )

2
2m



  

m  sin(   )  sin   (   )     sin   (   )   sin(   ) 
2
   2











m  2 sin     cos      2 cos     sin    
4

4

4

4





m tan      tan    
4

4

  



tan      m.cot       
4


2  4




tan      m.cot    
4

4

A
(c)
(i)
cos A 
b2  c 2  a 2
2bc
b  c 
 a2
2
  b 2  2bc  c 2    b 2  c 2  2bc cos A 
 2bc(1  cos A)
 4bc.cos 2
B
A
2
(1)
1
1
Area of the triangle ABC  bc.sin A  a. p
2
2
bc.sin A  a. p
(2)
84
H
C
Combined Mathematics
Practice Questions (With Answers)
From (1) and (2)
b  c 
2
4ap
.cos 2
sin A
4ap

A
2 sin cos
2
A
 2ap cot
2

b  c 
(ii)
A
2
 a 2  4bc.cos2
2
A
2
A
2
.cos 2
 a 2  2ap.cot
A
2
A
2
a 4  b 4  c 4  2b 2 c 2  2c 2 a 2  2a 2b 2  2a 2b 2
a
2
 b2  c2

2
 2a 2b 2
a 2  b 2  c 2   ab 2
a 2  b 2  c 2  ab 2
1


2ab
2ab
2
cos c  
15.
(a)
1
,
2
c

4
or
3
4
 3 1
A

 1 2 
 3 1  3 1   8 5
A2  



 1 2   1 2   5 3 
Now A2  5 A  7 I
 8 5   15 5   7 0   0 0 
A2  




 5 3   5 10   0 7   0 0 
A2  5 A  7 I  0
7 I  5 A  A2
 A(5 I  A)
7 I  5 A  A2
 (5 I  A) A
1
I  A. (5I  A)
7
I
1
(5 I  A) A
7
85
Combined Mathematics
1
Hence A 
Practice Questions (With Answers)
1
(5 I  A)
7
1  5 0   3 1  



7  0 5   1 2  
 2 1 
1  2 1   7 7 
 


7 1 3   1 3 


7 7 

BA = C
(BA)A-1 = CA-1
B(AA-1) =CA-1
2
 9 4   7


B= CA-1  6 16   1

7
(b)
1 
7   2 3 


3  4 6 

7 
x  y  a ............... (1)
x  y  b ............... (2)
1

1
-1  x   a 
    
1  y  b
A
X
1
A
1
A -1 
= B
-1
 x
a
, X   , B   
1
 y
b
1
1
1  (1)   1
1 1
 
2  1
 1
 2
A -1  
 1

 2
1

1
1

1
1
2

1

2
86
Combined Mathematics
Practice Questions (With Answers)
AX  B
A1 AX  A1 B
X  A1 B
 1
 x  2
   
 y  1

 2
x 
1
2  a 
 
1  b 

2
a b

2 2
a b
y 
2 2
A2 X  B
A1 A2 X  A1 B
 b 
 p  2 

 
 q   a 


 2
p
b
2
q
a
2
87
Combined Mathematics
Practice Questions (With Answers)
Combained mathematics II
Part A
01.
tan   1 
60  u
40
Velocity
u  20ms 1
60
1 60
tan   
2
t
t  120
u 

Distance bavelled 10000 m
40
1
1
 60  u   40  60  T   60  t  10000
2
2
T
t
Time
80  20  60T  30  120  10000
T  80 Seconds
02.
OA  t1 ,
AB  t2
tan   g 
u
u
, t1 
t1
g
tan   g 
v1
v
, t2  1
t2
g
tan   g 
2u  v2
, v  2u  gt
2
2
t2
Velocity
1
1
Displacement of A  ut1  v1t2
2
2
Displacement of B 
2u

v2
u
1
 2u  v2  t2
2
t1
v1
1
1
1
u.t1  v1 .t2  (2u  v2 )t2
2
2
2
u.t1  v1 .t2  (2u  v2 )t2
u.t1  (2u  v1  v2 )t2
u
 (2u  gt2  2u  gt2 )t2
g
u
t2 
4g
u.
88

A  t2 B
Time
Combined Mathematics
03.
Practice Questions (With Answers)
B
VA, E  2u
VB , E 
300
300
u
u
VA, B  VA, E  VE , B
V  

2u
B
300
2u
N
u
V

v2  LN 2   2u    u   2  2u  u.cos60
2
2
V  3u
u
v

sin  sin60
u
3u  2

sin 
3
1
sin  
2
0
  30
2u
M
B

d
M
A
Shortest distance  d cos   d cos30 
04.
600
L
 3u 2
Time taken 
u

3d
2
d sin 30
d

v
2 3u
x  2 y  constant
x  2 y  0

x  2 
y0
R
Let 
y  a . then 
x  2a
Mg
x
i.e acceleration AM , E  a
y
Am, E  2a
T
T
Applying F  ma
(1)
M , Mg  2T  Ma

, T  m(2 a )
m
a
Mg
4M  m
, T
(2)
2 Mmg
M  4m
89
Mg
Combined Mathematics
05.
Practice Questions (With Answers)
r  a cos nt i  b sin ntj
dr
 v  an sin nt i  bn cos ntj
dt
dv
 f  an2 cos nt i  bn 2 sin ntj  n 2  a cos nt i  b sin ntj 
dt
If v is perpendicular to f then v. f  0
a 2 n3 cos nt sin nt  b 2 n3 sin nt cos nt  0
1 2
b  a 2 n3 sin 2nt  0
2


t
k
; where k  0,1, 2, 3...
2n
V ..V  a 2 n 2 sin 2 nt  b 2 n 2 cos 2 nt
 n 2  a 2 sin 2 nt  b 2 cos 2 nt 
r .r  a 2 cos 2 nt  b 2 sin 2 nt
a 2  b 2  r .r  a 2 sin 2 nt  b 2 cos 2 nt

V ..V  n 2 a 2  b 2  r .r
06.

Constant velocity means that acceleration is zero,
Applying F  ma
 P  600  1200  0
P  600 N
Power  600 
PN

RN

1200g
20
 4000 Watts.
3
 4kW
Applying F  ma
Q
60 0



1200g
Q  600  1200  10sin   1200a
Q  600  1200  10 
1
 1200a
24
 1200a  1100 
90
Combined Mathematics
Practice Questions (With Answers)
Q  20  30  1000
1200a  1100   20  30  1000
a
07.
1 2
ms
3
Let the velocity of water be Vms 1
100
1
V 
100  100
10
V  10ms 1
= Workdone in one second
= Change in energy in one second
Power
=
1 2
mv  mgh
2
1
 0.1  1000   102   0.1  1000   10  12
2
 17000 Watts

 17kW
08.
Let VM , E  u
v
Vm , M  
V m , E  Vm , M  VM , E
v
w


= 
+ 
u
Have of conservation of Momentam
u 

(for m and M)
(M, m) 
Mu  m(v cos   u )  0
u
mv cos 
M m
91

M
Combined Mathematics
Practice Questions (With Answers)
N
Velocity triangle from
sine rule
w
v
u

sin(180   ) sin(    )
v
mv cos 

sin  ( M  m) sin(    )
( M  m)sin(    )  m sin  cos 
L
u
M
v


( M  m) sin  cos   cos  sin    m sin  cos 
M sin  cos   ( M  m) cos  sin 
09.
tan  
M m
tan 
M
3m
2m
m
A

B
u
C
3m
V1
I


V1
B I
Using I  mv for the system
 m(v1  u )  2m(v1  0)  0
 mv1  2mv1  mu
v1 
V2
V2

I2 B I1
A I2
Using I  mv for the system
0  m(v2  v1 )  2m(v2  v1 )  3m(v2  0)
v2 
u
6
Applying I  ( mv )
for A >
  I1  m(V2  V1 )
u u
 m  
6 3
I1 
mu
6
92
u
3

I1 C
V2
Combined Mathematics
Practice Questions (With Answers)
Applying C,   I2  3m(V2  V0 )
u

 m  0
6

I1 
3mu
6
I 2 : I1  3 :1

Loss of K.E
2
2
 1  u 2 1
1
u 1
u 
mu 2   m    2m    3m   
2
2
2
6
 6  
 2  6 

10.
5
mu 2
12
V1
V2
I

 


A B
A B
Using I  mv for the system
2u
6u

 

A B
 m(v1  2u )  4m(v2  6u )  0
mv1  4mv2  4m  6u  m  2u
v1  4v2  22u
(1)
Newton’s Law of restitution
1
 6u  2u 
2
v1  v2  4u
v1  v2 
v2 
18u
5
(2)
Applying I  ( mv )
B
 ,
 I1  4m(v2  6u )
 18u

 4m 
 6u 
 5

I
48mu
5
Hence, momentum transferred is
48mu
5
93
Combined Mathematics
A B
11.
u
Practice Questions (With Answers)
A B
I
v2
I
A B
v 1
m 4m
Using I  mv for the system
 m(v2  u )  4m(v1  0)  0
4mv1  mv2  mu
4v1  v2  u
(1)
v1  v2  eu
(2)
v1 
1  e  u
5
, v2 
v2  0 implies that ;
e
 4e  1 u
5
1
4
(3)
Let velocity of B after the impact with the wall be W.
W
A
v2 A
W  ev1 
4 1 e 

u
5 5 
For the second collision W  V2
4 1 e 
4e  1

u
u 
5  5 
5
4(1  e)  5(4e  1)
e
9
16
(4)
From (3) and (4)
12.
1
9
e
4
16
Let t be the time taken for the collision
S  ut 
1 2
at
2
For A>


28
73.5
x  28t
For B>



x  35cos  t
35cos   28
x
94
Combined Mathematics
cos  
Practice Questions (With Answers)
4
5
(1)
1 2
gt
2
For A>
h1  0 
For B>
h2  35sin  t 
1 2
gt
2
h1  h2  35sin  .t
3
73.5  35   t
5
t  3.5 seconds
13.
S  ut 
1 2
at
2
u  2ag
u


a  u cos  t
(1)
a
1
 u sin  t  gt 2
2
2
From (1)>
t=

a
u cos 
a
ga 2
 a tan   2
2
2u cos 2 
a
a
 a tan   1  tan 2 
2
4
2
tan   4 tan   3  0


 tan   3 tan   1  0
tan   3 or tan   1
S  ut 
1 2
at
2
For A>

 a  u cos 1 .t1
For B>

 a  u cos 2 .t2
t1 cos  2
1
2
1




t2 cos 1
10 1
5
t1 : t2  1: 5
95
a
2
a
Combined Mathematics
14.
Practice Questions (With Answers)
Let AB = l
Natural length  a >   2mg
T
A

2mg  l  a 
a
Applying F  ma
B
T
 3g 
> T sin   ml sin  .  
 4a 
T
B

mg
3mgl
4a
(1)
From (1) and (2)
2mg  l  a 
a

l
Extension is
3mgl
4a
8a
5
(2)
8a
3a
a 
5
5
F  ma
T cos   mg  m  0
T cos   mg
6mg
cos   mg
5
cos  
15.
5
6
Energy equation
Energy at A = Energy at B
1 2

 mv  mgh  constant 
2

O  mga 
1
mw2  mga
2
w2  4ag  4  10  0.6
w2  24
w  2 6ms 1
96
Combined Mathematics
Practice Questions (With Answers)
A
Applying F  ma

mg cos   R 
R  mg cos  
mv 2
a

O
mv 2
a
(1)
Energy equation
B
1 2
mv  mga cos 
2
O  mga 
(2)
v  2ag (1  cos  )
2
w
From (1) and (2)
R  mg  3cos   2 
2
When R  0 > cos  
3
2
3
Height is  0.6 cos   0.6 
 0.4m
16.

x   2 x

v2   2 a2  x2

x
When x  0.9 > v  1.2
x  1.2 > v  0.9

a
1.22   2 a 2  0.9 2
0.9 2   2
2
 1.22


(2)
1.2
a  0.9
 2
2
0.9
a  1.22
2
(1)  (2)>
(1)
2

2

a 2 1.22  0.92  1.2 4  0.9 4
a 2  1.2 2  0.9 2
a  1.5m
Amplitude  1.5m
 2 1.52  0.9 2   1.2 2
2  1
 1
period T 
2

 2 Seconds
97

mg
Combined Mathematics
17.
Practice Questions (With Answers)
A
In equilibrium AC  d ,   mg
For equilibrium of the particle,
A
a
> T2  mg  T1  0
T1
2mg 
a
2mg 
a
 2a  d    mg 
d    0
a 
2
a 
2
mg
C 
T
a
2
a
2
a
 2a  d    1   d    0
a
2
a
2
2
B
B
2
a
a
 2a  d   d    1  0
a
2
2
5a
5a
3a
d
, AM 
, BM 
4
4
4
A
Applying F  ma
5a
4
T4  mg  T3  mx
2mg  3a
a
2mg  5a
a
 x    mg 
 x 


a 4
2
a 4
2
 mx
2 g  3a
a
2 g  5a
a
x g
 x    
x


a 4
2
a 4
2
2 g  3a
a 5a
a
x 
 x    g  
x

a 4
2 4
2
2g
a
a

x
 2 x  2   g  

x
4g
x
a

x   2 x
Time 
2

 2 4g 
  a 


 2
a
a

4g
g
98
T x
mg

T
3
3a
4
4
B
d
Combined Mathematics
Practice Questions (With Answers)
C
18.
3P
R
300
A
x
600
600
X  3P cos 60  2 P cos 60  P 
Y  3P sin 60  2 P sin 60 
2P
3P
2
3P
2
x
2
 3P   3 P 
2
R2  
  3P
  
2
2

 

Y
R  3P.N
2
tan  
B
P

R
Y
1

,
0
X
3   30
Taking moments about A,
Moment of the resultant about A = Moment of the system about A
R.x.sin 30  2 P.2a sin 60
P 3 x
1
3
 2 P  2a 
2
2
x  4a
19.
cos  
4
3
, sin  
5
5
D
X  2 P  6 P  5P cos 
 2P  6P  4P  0
6P
7P
Y  4 P  7 P  3P  0
A
 4 P  7 P  3P  0
99
5P

2P
C
4P
B
Combined Mathematics
Practice Questions (With Answers)
Moment about A,
G   4 P  4a  3P  3a  34 Pa
R  0, G  0
Hence the system reduces to a couple of moment = 34Pa 
If 4P is removed, the resultant will be 4P in the direction of CB, Parallel to CB
C
D
Moment of the resultant about A
= Moment of the system about A
4 P.x  18Pa
x
9a
2
A
The resultant meets BA lproduced at a distance
20.
The forces acting on the rod,
(i)
weight W
(ii) Horizontal force P
(iii) Reaction R at R
B
x
9a
from A.
2
R
A
Consider the triangle OAC
R 
 OA (R is represented by OA)
W 
 AC (W is represented by AC)
G
P 
 CO (P is represented by CO;)
tan  
3
4
C
R
W
P


OA AC CO
W
If AB  2a
AC  2a sin  
8a
5
CB  2 a sin  
6a
5
CO 
3a
5
P  W.
O
CO 3W

AC
8
100
B
P
Combined Mathematics
Practice Questions (With Answers)
For equilibrium of AB, W, P and
S meet at a point
S
A
P
G
L
S
P
W


M
B
D
W
Consider the triangle of force.
For least value P, P should be perpendicular to S
In the triangle ADB>
AG  GB, and
ADB  900
Implies that AG  GB  GD
and  

2
P  W sin   W sin
21.

2
R
For equilibrium of the sphere,
(i)
Weight W at G.
(ii) Reaction R at P
(iii) Tension T
P
300
three forces meet at G
12 4
tan  

9 3
4
sin  
5
T
G 
3
cos  
5
W
C
T

A
Sine rule for the triangle ABC>
T
W
R


sin 30 sin  sin  30   
T
W
R
W sin 30 5W

sin 
8
300
B
101
Combined Mathematics
R


Practice Questions (With Answers)
sin  30   
sin 
W sin 30 cos   cos 30sin  
sin 

W
3 4 3
8

Y
22.
For equilibrium of AB>
A  0
X
a
a
X .a sin 60  Y .a cos 60  W . cos 60  W . cos 60  0
2
3
W W 5W


2
3
6
For equilibrium of BC
3  Y 
(1)
A
C  0
a
 X .a sin 60  Y .a cos 60  W . cos 60  0
2
 3X  Y  
W
2
(2)
From (1) and (2) Y 
Reacton at B is
W
6
X 
X2 Y2

W 57
18
102
2W
3 3
R

D W

B

X
Y
W

W
C
S
Combined Mathematics
23.
Practice Questions (With Answers)
d
a
A
D
c
b
d
0
c
B
300
600
60
C

W
c
b
ab  W
ad  W tan 30 
W
Rod
3
BC
W
2W

cos 30
3
bd  bc  cd
bd 
AC
2W
3
2W

AB
W
3
2W
3
R2
F1 P
-
F1
3P
F2


W
W
24.
Thrust
3
AD
R1
Tension

For equilibrium,
For equilibrium,
, 3P  F  W sin   0
2
,
R2  W cos   0
, F1  P  W sin   0
, R  W cos   0
1
F1   R1
F2   R2
W sin   P  W cos 
From (1) and (2) P 
3 P  W sin   W cos 
W sin 
2
(1)
2   tan 
103
(2)
Combined Mathematics
25.
Practice Questions (With Answers)
T
A
For equilibrium of rod AB
300
G
R
F  T cos 60  W sin 30  0
(1)
300
R  T sin 60  W cos 30  0
B
Moment about B is equal to zero.
F
300
T .2a cos 60  Wa cos 60  0
T
W
W
2
W
4
W 3
R  W cos 30  T sin 60 
4
F  W sin 30  T cos 60 
1
F
,  min
 ,  
R
3
26.
1
3
Area of rectangle OACD  2a 2
Area of triangle ABC 
1 2
a
2
Let the mass of the rectangle OACD is 12m
Then the mass of the triangle ABC is 3m
Let G   x , y 
Taking moments about OB,
15my  12m 
D
G1(12m)
a
 ma
2
7a
O
15
Taking moments about OD >
15mx  12m  a  m  2a  m  2a  m  3a
y
x
C(m)
19a
15
Angle, OA makes with horizontal is  .
104
G  x , y 
A(m)
B(m)
Combined Mathematics
tan   cot  
Practice Questions (With Answers)
x 19a

y
7
 19 
tan   tan 1  
 7 
27.
P  B  
 A 3
2
5
, P  AUB   , P   
3
8
B 4
P ( B )  1  P ( B )  1 
2 1

3 3
 A  P  A  B
P  
P( B)
B
3 1 1
 A
P  A  B   P   .P ( B )   
4 3 4
B
P  A  B   P ( A)  P ( B )  P  A  B  ;  P  A  
P  A   B    P  ( A  B )   1  P  A  B   1 
28.
A and B are independent
P  A  B   P( A).P( B)  0.3  0.4  0.12
P  A  B   P( A)  P( B)  P  A  B 
 0.3  0.4  0.12
 0.58
P  A  B    P  A  B    1  P  A  B 


 1  0.58  0.42
20 1

100 5
P [one defective]

P [3defective out of 4]
1 4
 4C3   
5 5
3

16
625
105
5 1 1 13
  
8 3 4 24
1 3

4 4
Combined Mathematics
29.
Mean 
x
Practice Questions (With Answers)
7  11  5  8  13  12  11  9  14
9
90
 10
9
5
Median 
7
8
9
11
11
9 1
th score
2
= 5 th score = 11
9
  xi  x 
standard deviation  
i 1
n
25  9  4  1  1  1  4  9  16
9


Coefficient of skewness 

2
70
70

 2.78
9
3
3  mean - median 
standard deviation
3 10  11
2.78
 1.04
106
12
13
14
Combined Mathematics
30.
Practice Questions (With Answers)
0
2
(1)
1
1
5
7
9
(4)
2
1
3
8
9
(4)
3
2
3
3
56
4
0
5
7
78
5
8
6
9
7
9 9 9 9
(11)
(6)
(1)
27
2/3 means 23 years.
(i)
(ii)
Minimum value
02 yrs.
Maximum value
58 yrs.
Mode
39 yrs.
Q1 is the 
1
th
 27  1 score
4
= 7th score = 23 yrs.
Median
Q2 is the 
1
th
 27  1 score
2
= 14th score = 36 yrs.
Q3 is the 
3
 27  1 score
4
= 21th score = 40 yrs.
(iii)
Q1  1.5(Q3  Q1 )  23  1.5(40  23)
 23  25.5  2.5
Q3  1.5(Q3  Q1 )  40  1.5(40  23)
 40  25.5  65.5
107
Combined Mathematics
Practice Questions (With Answers)
Part B
01. (a)
v
Velocity

u


T
(i)
tan   a >
>
tan   2a >
tan  
v
,
T
tan  
3a v  u

2 T t
*
tan  
T
3a
2
v  aT
(1)
2(v  u )  3a (T  t )
(2)
From (1) and (2)
2  aT  u   3a(T  t )
(1) 

t2
t
3at  2u  aT
V  3at  2u
3u 

TP  Time taken for P is 2T  2  3t  
a 

TQ  Time taken Q is T  t   t2
2u v

a 2a
2u 3t u
 2t 
 
a
2 a
7t 3u
 
2 a
 2t 
108
Time
Combined Mathematics
(ii)
Practice Questions (With Answers)
 TP  TQ
Time difference
2u   7t 3u 

 2  3t 
  
a  2 a 


(iii)
5t u

2
a
Distance travelled by P is

1
.V .2 T  V T
2
  3at  2u  .

Distance travelled by Q is
 3at  2u 
 3at  2u 
a
2
a

1
1
 u  V T  t   V .t2
2
2

1
2u  1 

 3t u  
u   3at  2u    2t     3at  2u     
2
a  2

 2 a 

 2at  2u 
 3at  2u  
1
  3at  2u 
 3at  u 

2
a
2a

 3at  2u 
1 
 3at  u  2at  2u  

2a 
2

109
2



Combined Mathematics
Practice Questions (With Answers)
B
(b)

u
O
VP , E
A
P
V

u
VQ , E  

VP ,Q  VP , E  VE ,Q
u  v cos 

V
 u  


 u  v cos   v sin 
v sin 
V0
V0 2   u  v cos     v sin  
2
2
V0 2  u 2  v 2 cos 2   v 2 sin 2   2uv cos 
V0 2  u 2  v 2  2uv cos 
V0  u 2  v 2  2uv cos 
tan  
v sin 
u  v cos 
B
O

a

d
Shortest distance
VP,Q
d  a sin 

av sin 
u 2  v 2  2uv cos 
110
P
A
Combined Mathematics
t  Time taken is

Practice Questions (With Answers)
PM a cos 

V0
V0
t
t
a  u  v cos  
V0 2
a  u  v cos  
u  v 2  2uv cos 
2
Distance travelled by P is  ut
Distance travelled by Q is  vt
Ratio of the distances from O is

a  ut
Vt
a
a  u  v cos   u
V0 2
va  u  v cos  

v  u cos 
u  v cos 
V0 2
02.
At the maxium speed acceleration is zero.
s in  
1
n
T1  w s i n   R 
w
 0
g
R
T1  R  w sin 

H  ( w sin   R )v
w
(1)
R
2v
Applying F  ma
T 2  w s in   R 
T1
v
Applying F  ma
w
0
g
T2
T2  R  w sin 
H  ( R  w sin  )2v
(2)
111

w
Combined Mathematics
R
From (1) and (2)
Practice Questions (With Answers)
3w
n
Applying F  ma
T3  R 
T3  R 
u
w
 0
g
T3
R
3w
n
H  T3 .u 
3uw
n
u
2
Applying F  ma
T 4  R  w s in  
w
 a
g
H  T4 .
R
(a acceleraton)

4w wa

n
g
T4 
T4
w
u
2
u  4w wa  3wu



2 n
g 
n
a
(b)
2g
n
VA, E  (3 i  29 j )
VB , E  ( i  7 j )
VB , A  VB , E  VE , A
 V ( i  7 j )  (3 i  29 j )
VB , A  (v  3) i  (7v  29) j
(1)
B
At time t>


rA  a  3 i  29 j t


rB  b  v i  7 j t

AB  rB  rA
B0
b




  b  v i  7 j t    a  3 i  29 j t 

 


O
112
a

A0
A
Combined Mathematics
Practice Questions (With Answers)

AB   b  a    v  3 t i   7v  29  tj
 
When t  0 > AB  A0 B0  b  a   56 i  8 j 

AB   56 i  8 j    v  3 t i   7v  29  tj
  v  3 t  56  i   7v  29  t  8 j

When the particles collide AB  0
(m-J)
 v  3 t  56  0
(3)
 7v  29  t  8  0
(4)
From (3) and (4)
v4

AB   v  3 t  56  i   7v  29  t  8 j
When v  3

becomes AB   6t  56  i   8  8t  j

AB 
 6t  56 
2
  8  8t 

 100 t 2  8t  32

AB  10
t  4
2
2

 16

AB is minimum when t  4 and AB min  40 m
When v  3 and t  4

AB  32 i  24 j
VA, B  6 i  8 j

VA, B  AB  6 i  8 j  32 i  24 j



= -192 + 192
= 0

VA, B  AB  0

Hence, VA, B is perpendicular to AB .
113
(2)
Combined Mathematics
03. (a)
Practice Questions (With Answers)
 a1
Let AA, E  
a1
RA
F1  T
AB , E  
 a2
a1  a2
2
When the particles are moving
a2
T
T
m T
Then, AM , E 
T
F1   .mg , F2     2mg 
Applying F  ma
T   mg  ma1
A 
(1)
T
T

Applying F  ma
T     2mg   2ma2
B 
(2)
Mg
Applying F  ma
M 
Mg  2T  M
From (1)
a1 
 a1  a2 
2
T   mg
m
T  2  mg
2m
Substituting in (3)
a2 
From(2)
M  T   mg T  2  mg 


2 
m
2m

MT  Mg MT  Mg
Mg  2T 



2m
2
4m
2
Mg  2T 
M
M 
 Mg  Mg

T 2 

 Mg 


4m 2m 
2
2

T
2Mmg  2      
 3M  8m 
114
(3)
T
RB
F2
2mg
Combined Mathematics
(ii)
Practice Questions (With Answers)
Given that   2  
For the motion to take place a1  0
a1 
T
 g  0
m
T   mg
2Mmg  2      
 3M  8m 
2    


  mg
3M  8m
2M
   2 3M  8m

1

2m
   2 M  8m


2M

  2
(b)

2M
8m  M
Applying F  ma
B
T2 cos   T1 cos   mg  0
T2  T1  cos   mg

(1)
T2
a

b
Applying F  ma
T1  T2  sin   maw2 sin 
B

T1  T2   maw2
For equilibrium of D,
 T1  kmg  R  0
T1
D
R  T1  kmg
cos 
(3)
b
2a
From (1) T2  T1 
2mga
b
From (2) T2  T1  maw2
T1 

(2)
ma  2 2 g 
w 
>
2 
b 
T2 
ma  2 2 g 
w 
2 
b 
115
T2
a
T1
T1
km+R
T 1 mg
Combined Mathematics
From (3) R 
Practice Questions (With Answers)
ma  2 2 g 
w 
 kmg
2 
b 
R0
ma  2 2 g 
w 
 kmg
2 
b 
w2 ab  2 g  a  kb 
(4)
Greatest tension in the string is  mg
T1 , T2   mg
T2   mg
ma  2 2 g 
w 
  mg
2 
b 
w2 
2 g 2 g

a
b
2
From (4) w 
2 g 2kg

b
a
2 g 2kg
2 g 2 g

 w2 

b
a
a
b
2 g 2kg 2 g 2 g



b
a
a
b
1 k  1
  
b a a b
2  k

b
a
   k  b  2a
116
Combined Mathematics
04. (a)
A
u
Practice Questions (With Answers)
B
u
d
C
d
A u
A u
First collission (B and C)
Using I  mv for the system
 m(v2  u )  m(v1  0)  0
 mv1  mv2  mu
v1  v2  u
(1)
Newton’s experimental law,
v1  v2  eu
From (1) and (2) v1 
(2)
u
u
1  e  , v2  1  e 
2
2
Time taken for A to collide with B; (say t0)
t0 

d
d

u u  v2
d

u
d
u
1  e 
2
d
2d
 
u u 1  e 
u
d 3  e
u (1  e)
Distance travelled by A is  ut0

d 3  e
(1  e)
117
u
B C
 v2  v1
C
B
Combined Mathematics
Practice Questions (With Answers)
 v2
B
u  v 2
A B
 v4  v3
A B
u
A
d
Before conservation
After conservation
Second collission (A and B)
Using I  mv for the system
 m(v4  u )  m(v3  v2 )  0
(3)
v3  v4  u  v2
Newton’s experimental law,
v3  v4  e  u  v2 
From (3) and (4) v3 
(4)
v
u
1  e   2 1  e 
2
2
u
u
2
1  e   1  e 
2
4
u
  2  2e  1  2e  e 2 
4

v3 
Now
u
3  e 2 
4
v3  v1 
u
u
3  e 2   1  e 
4
2
u
1  2e  e 2 
4
u
2
 1  e 
4

v3  v1
Hence there will be another collision between A and B.
(b)
Law of conservation of energy
1
1
mu 2  0  mv 2  mga (1  cos )
2
2
(1)
v 2  u 2  2 ag (1  cos )
118
d
 v1
C
 v1
C
 v1
C
Combined Mathematics
Practice Questions (With Answers)
v

F  ma
R  mgcos 
mv
a
R
(2)
2
mg
m 2
u  2ag  3agcos 
a
When R  0 >    (say)
R
O  u 2  2 ag  3agcos
cos 
u 2  2ag
3ag
2ag  u 2  5ag (Given)
O  cos  1
Hence  is an acute angle.
hence the particle leaves the sphere before it reaches the highest point
when cos 
u 2  2ag
3ag
v0
When    > v  v0 (say)
R
From (1) v0  u  2ag 1  cos 
2
2

 u 2  2 ag  2 agcos
 3agcos  2 agcos
v0 2  agcos
S  ut 
(3)
1 2
at
2
2 a sin   v0 c os  .t0
(4)
2 a cos   v0 c os  .t0 
1 2
gt0
2
From (4) and (5)
2a sin 2  2a 2 g sin 2 
2a cos  
 2 .
cos 
v0 cos 2 
a 2 g sin 2 
a
.

2
2
v0 cos  cos 
119
(5)
mg
Combined Mathematics
Practice Questions (With Answers)
ag sin 2 
v0 
cos 
From (3) and (6)
2
(6)
tan 2   1
  450
u 2  2ag  3ag cos 450
 3

u2  
 2  ag
 2

05. (a)
Let the time of light be t vd;f.
S  ut 
(1)
1 2
at
2

 2h  u cos  .t
(2)
 h  u sin  .t 
1 2
gt
2
From (1)
Substituting in (2)
 h  u sin  .
2h
1
4h 2
 .g 2
u cos  2 u cos 2 
1  2 tan  
1  2 tan  
u 2 cos 2  
u 
2
2 gh
u cos 2 

2
2 gh
u cos 2 
h
2
2h
2 gh
1  2 tan 

2 gh 1  tan 2 
1  2 tan  

v  u  at
v1  u sin   gt

 u sin   g 
v1  u sin  
u
2h
u cos 
2 gh
u cos 
120

Combined Mathematics
Practice Questions (With Answers)
2 gh
 Velocity  u cos   u sin 
2 gh
 u sin 
u cos 
2 gh
 u sin 
tan   u cos 
u cos 
tan  
2 gh
 tan 
u cos 2 
2
 1  2 tan   tan 
tan   1  tan 
(b)
u cos 
R
F
m,E 

 m , M  
f
B
 m, E   m,M   M ,E

f+

S
m
m g 
S

F
Mg
Applying F  ma
(M, m) System
 > R sin   MF cos   m( F cos   f )

( M  m) g sin   MF  m( F  f cos  )
m for  F  ma
0  m ( F cos   f )
From (3) f  F cos 
Substituting in (2),
 M  m  g sin   MF  m( F  F cos2  )
 M  m sin 2   F   M  m  g sin 
F
(1)
 M  m  g sin  cos 
 M  m sin  
2
121
(3)
(2)
Combined Mathematics
f 
M  M  m  g cos 
 M  m sin  
2
From (1), R 
06.
Practice Questions (With Answers)
M  M  m  g cos 
 M  m sin  
2
M
A
T1 m T2

O
B
Let MO  d .
AM  MB  4l >
T1  T2
O
  d  2l 
2l

4  l  d 
3l
3  d  2l   8  l  d 
11d  2l
d
2l
2l
, OM 
11
11
M
A
T3 m T4
.

P
O
x
Let OP  x
F  ma
 T3  T4  mx
 
 4
2l

 4l   x   2l  

2l 
11

 3l
  24l
 4
 x 

2l  11
 3l


2l

 4l  11  x   3l   mx



 9l

11  x   mx


11
x
6ml
Hence, motion is S.H.M.
(i)
Centre of oscillation is x  O > (i.e) O

x
V2 
11
 A2  x 2  (A =amplitude)
6ml 
When x 
2l
>
11 v  0
Therefore A 
2l
11
122
M
B
Combined Mathematics
Practice Questions (With Answers)
Hence, when the particle comes to M  , where
2l
OM   > it instantance onsly comes to rest and
11
4l 40l

 3l
11 11
The string is always tant.
BM   4l 
The period of oscillation is
2  2 11 
 

 
6ml 
 2
V2 
2

11  2l 
2
   x 
6ml  11 

MC 
3l
>
11
When x 
OC 
3l 2l
l
 
11 11 11
l
> Let v  v0
11
V0 2 
2
2
11  2l   l  


  
 
6ml  11   11  
V0 2 
11
3l 2

6ml 11  11
V0 2 
V0 2 
l
22m
l
22m
123
6ml
11
Combined Mathematics
07.
Practice Questions (With Answers)
O
Let OC  d
For equilibrium of m
R
T  mg sin 300  0
2T  mg
2
T
C
300
3mg ( d  6a )
 mg
6a
mg
d  7a
Let CA  2a and CP  x
Energy at A
O
 3a 
1
 0  mg .2a.sin 300  .mg 
2
6a
Energy at P
2
A
a  x
1
1
 mx 2  mgx sin 300   3mg 
2
2
6a
Principle of conservation of energy
P
x
C
300
2
mg
1
mgx mg
2
 9a 2  mx 2 

a  x
4a
2
2
4a
Differentiating w.r.t. time t
 2mga.sin 300 
1
mgx mg
 
m2 xx

.2  a  x  x
2
2
4
g g
O  
x   a  x
2 2a
O

x
g
x0
2a
g 
 2
 

2a 

x  A cos t  B cos t
v
dx
 x   A sin  t  B sin  t
dt
t  0 > x  2a and x  0
(1)
2a  A
0  0  B
B0
x  2a cos t
(2)
124
Combined Mathematics
Practice Questions (With Answers)
The string becomes slack when x  a
Let x  a when t  t1
 a  2a cos t1
cos  t1  
 t1 
1
2
2
3
t1 
1 2
.
 3
t1 
2
3
2a
g
when  t1 
2
> 
3 x  2a sin t
x  2a
x  
2a
2
sin
g
3
3ag
>
2
3ag
2
Speed is
Y
08. (a) Suppose that the system reduces to
a couple G at A
and
C
600
A ,
M G
B ,
M
 Y .2a  G
2
C
X
G
, 2 M  X . 3a  Y .a  G
A
5M
600
D
x

600
M
G  M , Y  4a , X  4 3a
R
R 
X2 Y2 
M
a
tan  
M
a
1 25

16 48
Y
7
12

Y
5

X
3
125
x
B
Combined Mathematics
Practice Questions (With Answers)
Taking moment about A,
R. AD sin   M
 R sin   AD  M
Y . AD  M
AD 
M
 4a
Y
Forces acting on the sphere
O at W
C at T
B
Resultant of F and R is S

Now three forces T, W, S meet at a point M.
ˆ 
AB  h > OA  a > OAM
Where   tan 
T
C
OM  a tan   a 
F
a
a
tan  

h  OM h  a 
A

a 

 h  a 
  tan 1 
When  

a
h
1 
>   tan 

2a
 2a   a  
1


1

Consider the tringle AMB
T  MB 

W  BA 
S  AM 

M
R
w
  tan 1 
tan  
S
AMB is the triangle of forces.
T
W
S


sin(90   ) sin 90  (   )  sin 
T
W

cos  cos(   )
126
Combined Mathematics
Practice Questions (With Answers)
W cos 
cos(   )
W cos 
T
cos  cos   sin  sin 
W
T
cos   sin  tan 
W
T



2
1 
1 2
T
W 1 2
T
2
D
E
09. (a)
A
X  P  P cos 600  Q cos 300
Q
P Q 3
2
0
0
 Y  P 3  P sin 60  Q sin 30
X 
Y
F
P 3
P
QP 3
2
A
P
C
600 G
B
R
(i)
If the system reduces to couple
X  0, and Y  0
Then P  Q 3 and Q  0
But Q  0
The system cannot reduce to a couple.
(ii)
If Q  P 3
X   P, Y  P 3
/ R2  P2 

3P

2
2P
P 3
R  2P

tan   3 >   600
P
127
Combined Mathematics
(iii)
Practice Questions (With Answers)
Moments about A
Moment of the resultant about A = Moment of the forces about A
R. AG sin 600  Q.
3a
2
 R sin 60  AG  P
0
Y . AG 
3.
3a
2
3 3Pa
2
AG 
3 3Pa
1

2
3P
AG 
3a
2
(b)
A
300
Y
W
B
X
Y

W
Egulibrium of BC
B
Wa sin   P 2a cos   0
P
W
tan 
2
PX 0
X P
Y W  0
Y W
128
C
P
Combined Mathematics
Practice Questions (With Answers)
For equilibrium of AB,
A
Wa sin 300  Y .2 a sin 300  X .2 a cos 300  0
W
W  P 3  0
2
3W
2
Reaction at B is
P
R
X2 Y2
R
3W 2
W 2
4
tan  
R
7W
2
  600
tan  
2P
W
 3
Y
2

X
3
 2 

 3
  tan 1 
Y
10. (a) For equilibrium of AB and AC
R  S  4w  0
R  S  4w
B0
S .4a sin   w.3a sin   3w.a sin   0
S
3w
5w
> R
2
2

 F1  F2  0; F2  F2 (=F,, say)
For equilibrium of AB>
A0
F .2a cos   R.2a sin   3w.a sin   0
F  w tan 
5 w 3w

2
2
RS
1 1

R S
F F

R S
129
X
A
X

 
Y
R
B
3w
w
F1
F2
S
C
Combined Mathematics
For equilibrium ,
i.e
F
 >
R
Practice Questions (With Answers)
F

S
F F
 
R S

F
R


F
S
F
reaches  first and
S
Limiting occurs at C first.
When  increases
Now
F w tan   2 2 tan 


R
5w
5
F w tan   2 2 tan 


S
3w
3
Hence
F

S
2
tan 

3
tan  
3
2
For AB


FX 0
X  F  w tan 
Y  R  3w  0
Y

w
2
Y
X
tan  
 3
Y
x
  tan 1  3 
130
Combined Mathematics
(b)
(b)
300
P
Practice Questions (With Answers)
300
(e)
(a)
(c)

Rod
(d)
40N
(f)
X
(g)
60N
AB
100
-
BC
-
CD
120
DB
40
AD
80 3
60 3
-
f
Y
300
Y  220
Resultant at A
a
R
60N
c
30
300
n
600

b
X2 Y2
 40 3 
2
 2202
 20 133N
0
30
d
e

X  40 3
40N
0
Tension
Thrust
tan  
Y
X
tan  
220
tan  
g
131
40 3
11
2 3
Combined Mathematics
Practice Questions (With Answers)
11. (a)
4
(-3, 2)
3
(2, 3)
6
3
1
3
(6, 1)
0
Resultant R  F1  F2  F3
 (3i  4 j )  (i  6 j )  (3i  3 j )
R  i  7 j
X  1, Y  7
R  12  7 2  5 2 N
O  M  (9  6)  (36  1)  (8  9)  15  37  1  51
O Taking moment about = Algebraic sum of the moments of the forces about O
Y .x  X . y  M
7 x  y  51
Equation of line of action is 7 x  y  51  0
For equilibrium O
(b)
A
F4  i  7 j and G  51
1
=  BC.h1     2 BC  h1
2
 2.ABC
B
A
1
=  CA..h2    2   CA  h2
2


 CA
 AB
 2  .ABC
C
1
=  AB..h3    2   AB  h3
2
B
 2 .ABC

 BC
C
[h1 > h2 and h3 are the perpendicular distances from A, B and C to BC, CA, AB
respectively. ABC  Area of the triangle ABC]
132
Combined Mathematics
(i)
Practice Questions (With Answers)
Suppose that     
A=B=C 0
Since the moments about three points not in a straight line are constant and not equal
to zero, the system reduces to a couple.
(ii) Conversely assume that the system redues to a couple.
/ A=B=C
2 .ABC  2 ABC  2 .ABC
/   
(c)
For equilibrium of M,
T cos   F  Mg sin   0
R


R  T sin   Mg cos   0

At limiting,
T
F
F

R
Mg
T cos   Mg sin  sin 

Mg cos   T sin  cos 
T cos      Mg sin    
Mg sin    
T
cos    
(1)
For T to be minimum cos     should be maximum
cos      1
 
Tmini  Mg sin    
Hence to find the least force acting parallel to the place, put   0 in (1)
Hence the require force is 

Mg sin    
cos   
P
 P sec 
cos 
133
Combined Mathematics
Practice Questions (With Answers)
12 (a) For equilibrium mements about O
O
G
G  F1 .a  F2 .a  0
F2
G   F1  F2  a
S
O
R
At limiting equilibrium
F1   S > F2   R

G  a  R  S 
A
(1)
S .a tan   R.a tan   G  0
G  a tan  ( R  S )
w F1

A
(2)
( R  S ) cos   F2 sin   F1 sin   w  0
( R  S ) cos    sin  ( R  S )  w  0
From (1) and (2)
(3)
G cos 
G
  sin 
w0
a
a tan 

G cos   1
   w
a 

G
(b)
 aw
1   2  cos 
By symmetry centre of granity lies on OC.
A
C
G1 D
G2
B
Mass of the hemi sphere
Mass of the cone
O
M1 
2 3
r  >
3
DG1 
1
4
M 2   r 2  4r     r 3 
3
3
DG2 
1
 4r  r
4
134
3r
8
Combined Mathematics
Practice Questions (With Answers)
Total mass of the composite body is  M 1  M 2 
Let DG  x
D
 M1  M 2  x  M 2 .DG2  M1 .DG1
4 3 
4 3
2 3
3r
2 3
  r    r   x   r r   r  
3
3
3
8
3

2 3
4
3 

 r   2   x   r 4   

3
3
8 

x
If    >
x
r 16   3 
8   2  
13r
24
tan  
r 24

x 13
 24 

 13 
  tan 1 
135
Combined Mathematics
Practice Questions (With Answers)
M
13.
Vertical line
G

C
A
O

B
D
By symmetry centre of gravity lies on OM
Mass
Solid
M1 
Hemisphere CMD
Hemisphere ALB
M2 
O
2
3
  2a  
3
3
3a
OG1   2a 
8
4
2 3
a 
3
M1  M 2 
Bowl CD
Centre of gravity
OG2 
14 3
a 
3
3a
8
OG
 M1  M 2  OG  M1 .OG1  M 2 .OG2
14 3
16
3a 2
3a
 a  OG   a 3    a 3 
3
3
4 3
8
OG 
45a
56
tan  
2a 112

OG
45
 112 

 45 
At the point of toppling
a sin   OGSn   OG
  tan 1 
R
a sin   OG
sin  
sin  
45a
56a


45
56

 45 
  sin 1  
 56 
w
136
G
Combined Mathematics
Practice Questions (With Answers)
14. (a) Use the following notation.
S: he goes to sea.
R: he goes to the river.
L: he goes to the lake.
F: he catches fish.
P(S ) 
1
,
2
P (F | S) 
P( R) 
8
,
10
P(F | R)) 
1
2
1
4
S
R
1
4
L
(i)
1
,
4
P(L) 
4
,
10
1
4
P( F | L) 
6
10
8
10 F
2
10
F
4
F
10
6
10
F
6
F
10
4
10
F
1 8

2 10
1 2
P  S  F   
2 10
1 4
PR  F  
4 10
1 6
P  R  F   
4 10
1 6
PL  F   
4 10
1 4
P  L  F   
4 10
P S  F  
Using the law of total probability,
1 8 1 4 1 6
    
2 10 4 10 4 10
13
P( F ) 
20
P( F ) 
2
(ii)
P (Catches fish on 2 sundays)
7
 13 
 3C 2    
 20  20
 13 
P (Catches fish on all 3 sundays)  C3   
 20 
3
3
2
7
 13 
 3C 2  
 
20
 20 
The required probability is

137
2873
4000
 13 
 3C3   
 20 
3
Combined Mathematics
(iii)
P( F ) 
13
,
20
Practice Questions (With Answers)
P ( F ) 
7
20
1 2
P  S  F   2  10 2
P  S | F  


7
P  F 
7
20
1 6
P  R  F   4  10 3
P  R | F  


7
P  F 
7
20
1 4
P  L  F   4  10 2
P  L | F  


7
P  F 
7
20
Hence it is most likely that he has been to the river.
(iv)
On a given sunday,
P (Both go to sea) 
1 1 1
 
2 3 6
P (Both go to river) 
1 1 1
 
4 3 12
P (Both go to lake) 
1 1 1
 
4 3 12
P (Both meet) 
1 1
1 1
 

6 12 12 3
P (Both fail to meet on two sundays) 
2 2 4
 
3 3 9
Hence the probability that they meet at least once is  1 
138
4
5

9
9
Combined Mathematics
Practice Questions (With Answers)
14. (b)
f
x
d
x  450
100
4
fd
fd 2
56
224
800 - 900 14
700 - 800 30
850
750
3
90
270
600 - 700 52
500 - 600 79
650
2
104
208
550
1
79
79
400 - 500 206
300 - 400 146
450
0
0
0
350
-1
-146
146
200 - 300 88
100 - 200 45
250
-2
-176
352
150
-3
-135
405
-128
1684
660
(i) Mean
 128 
x  450  100 

 660 
x  469.39
(ii) stadard deviation
1684  128 
S  100


660  660 
2
S  158.55
(iii) Median class   400  500 
Median
 400 
100  660

 279 

206  2

 400  100 
51
206
 424.75
3
(iv) Coefficent of skewness
=3
 mean - median 
standard deviation
 469.39 - 424.75
158.55
 0.8446
139
Combined Mathematics
(v)
Practice Questions (With Answers)
Positively skewed curve.
Shape of the curve
Frequency
Wage
15. (a) Use the following notation.
A: Adult
C: Child
M: Male
F: Female
S: Using swimming pool.
3
P( A)  ,
4
P (C ) 
3
P( M | C )  ,
5
1
P( F | A)  ,
4
P S | A  M  
1
,
2
4
PS | C  M   ,
5
1
,
4
P( F | C ) 
1
P S | A  F   ,
3
P S | C  F  
140
4
5
2
5
Combined Mathematics
Practice Questions (With Answers)
1
2
M
3
4
A
3
4
1
4
(i)
F
3
5
C
2
5
M
F
1
3
2
3
4
5
1
5
4
5
1
5
S
S
S
S
S
S
S
 3 3 1 3 1 1 1 3 4 1 2 4
P( S )                    
 4 4 2  4 4 3  4 5 5   4 5 5 
P(S ) 
(ii)
1
4
1
2
S
9
1
3
2
87
 


32 16 25 25 160
PF | S  
PS  F 
P( S )
3 1 1 1 2 4
    
 4 4 3 4 5 5
87
160
1
2

 16 25
87
160
PF | S  
114
 0.262
435
141
Combined Mathematics
(iii)
P C | M  S  
Practice Questions (With Answers)
P M S C
P(M S)
1 3 4
 
4 5 5

3 3 1 1 3 4
    
4 4 2 4 5 5
3
 25
9
3

32 25
P C | M  S  
(iv)
3 25  32

 0.2999
25 321
P  A  F | S  
P  A  F   S 
P  S 
P  S   1 
87
73

160 160
Now  A  F   S    A  S    F  S 
Therefore  A  S     F  S     A  M  S     A  F  S     C  F  S  
Since all three events on R.H.S are mutnally exclusive.
Hence, P  A  F   S   P  A  M  S    A  F  S   P C  F  S 
 3 3 1  3 1 2 1 2 1
         
 4 4 2  4 4 3  4 5 5
P  A  F   S  
P  ( A  F ) | S  
9 1 1 341
 

32 8 50 800
P  A  F   S  
P  S 
341
 800
73
160

341
365
P  ( A  F ) | S   0.934
142
Combined Mathematics
Practice Questions (With Answers)
n1
15. (b)
x
i
1 
1
n1
n1
n2
 xi  n1 1 ,
x
i
1
n1
x
i
 12 
n1
x
1
n1
 n2 2
1
2
 21
2
i
 n1 12  n1 12
x
2
i
 n2 2 2  n2 2 2
1
n2
1
Mean of the population X 
n1 1  n2  2
n1  n2

n1
n2
1 
2
n1  n2
n1  n2
X  1 1  2 2
Where
1 
n1
n2
, 2 
n1  n2
n1  n2
n1  n2
Variancy the population S 2 
S2 
x
i 1
i
n1  n2
 X2
1  n1 2 n2 2 
2
 xi   xi   X
n1  n2  1
1

 n   n2  2 
1
 n1 12  n2 2 2  n1 12  n2  2 2    1 1
S 

n1  n2
 n1  n2 
2
2
 n   n2  2 
 n1  n2   2
n1 12
n2 2 2



n   n2 2 2    1 1

2  1 1
n1  n2 n1  n2  n1  n2 
 n1  n2 
143
2
Combined Mathematics
Practice Questions (With Answers)
n1 12
n2 2 2
1



n1  n2 n1  n2  n1  n2 2
 112  2 22 
n1n2
 n1  n2 
2
 12   2  212 
S 2  112  2 2 2  12  1  2 
X 
X
40 
X
 n1  n1  n2  12  n2  n1  n2   2 2 


2
2
2
 2n1 n2 1 2  n1 1  n2  2

2
n
20
 X  800
 X  800  50  15
Wrong value of
Correct value of
 765
765
 38.25
20
X 
Correct
x

 X2
x
25 
 40 2

2
2
i
n
2
i
20
Wrong
Correct
x
x
2
i
 500  1600  20  32500
2
i
 32500  2500  225
 30225
Correct
2 
30225
 38.252
20
 1511.25  38.252
 48.19
  48.19
  6.94
144
Combined Mathematics
Practice Questions (With Answers)
For the whole population  
20  38.25  30  40.25
20  30


765  1207.5
50
1972.5
50
  39.45
 2  1 12  2 22  12  1  2 
2
20
30
20  30
2
 6.94 2   82 
 40.25  30.25
50
50
50  50
2
3
6
  48.19   64 
4
5
5
25
481.9  960  24

25

2 
1465.9
25
 58.636
  58.636
  7.65
145