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Bi yani ' s Thi nk Tank
Concept based notes
MathsClass XII
MS. VershaInformation Technology
Biyani Girls College, Jaipur

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Published by :
Think TanksBiyani Group of Colleges
Concept & Copyright :
Biyani Shikshan SamitiSector3, Vidhyadhar Nagar,Jaipur302 023 (Rajasthan)
Ph : 01412338371, 233859195 Fax : 01412338007Email : [email protected] :www.gurukpo.com; www.biyanicolleges.org
First Edition : 2009
While every effort is taken to avoid errors or omissions in this Publication, any mistake oromission that may have crept in is not intentional. It may be taken note of that neither thepublisher nor the author will be responsible for any damage or loss of any kind arising toanyone in any manner on account of such errors and omissions.
Leaser Type Setted by :Biyani College Printing Department

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Preface
Iam glad to present this book, especially designed to serve the needs of
the students. The book has been written keeping in mind the general weaknessin understanding the fundamental concepts of the topics. The book is selfexplanatory and adopts the Teach Yourself style. It is based on questionanswer pattern. The language of book is quite easy and understandable basedon scientific approach.
Any further improvement in the contents of the book by making corrections,omission and inclusion is keen to be achieved based on suggestions from thereaders for which the author shall be obliged.
I acknowledge special thanks to Mr. Rajeev Biyani, Chairman & Dr. Sanjay
Biyani,Director
(Acad.
) Biyani Group of Colleges, who are the backbones andmain concept provider and also have been constant source of motivationthroughout this Endeavour. They played an active role in coordinating the variousstages of this Endeavour and spearheaded the publishing work.
I look forward to receiving valuable suggestions from professors of variouseducational institutions, other faculty members and students for improvement ofthe quality of the book. The reader may feel free to send in their comments andsuggestions to the under mentioned address.
Author

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Unit 1
Chapter 1
Relations and functions
Q.1 Let A be set of firs t ten natural numbers. If R be a relation on A defined
by xRy x + 2y = 10 then
i. Express R and as set of ordered pairs1
R
ii. Find domain of R and1
R
iii. Find range of R and1
R
Ans. Here
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10 }
Relation R is defined as
xRy x + 2y = 10
10xy =
2
Now for x = 1, y = 10 1 9
= A2 2
Hence 1 is not related to any element of A.
Similarly use can observe that 3, 5, 7, 9 and 10 are also not related to anyelement of A.
Again use observe that
When 102x = 2,y = = 4 A 2R42
When 104
x = 4, y = = 3 A 4R32
When 6 106x = 6, y = = 2 A R22
When 8 108x = 8, y = = 1 A R12
Hence
i. R = {(2,4), (4,3), (6,2), (8,1) }

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and { }1 = (4,2),(3,4),(2,6),(1,8)R
ii. Domain of R = {2, 4, 6, 8}
Domain of { }1 = 4,3,2,1R
iii. Range of R = {4, 3, 2, 1}
Range of { }1 = 2,4,6,8R
Q.2 Prove that the relation R on the set N x N defined by (a, b) R (c, d) a+d =b+
c fo r all (a, b), (c, d) N x N is an equivalence relation . Ans. To prove that the given relation is an equivalence relation we have relation to
show that it is reflexive, symmetric and transitive.
1) Reflexive Let (a, b) be an arbitary element of N x N. Then,
( a, b)N x N a+ b = b+ a [by commutativity ofaddition on N]
(a, b) R (a, b)
Thus (a, b) R (a, b) for all (a, b)N x NHence the given relation R is reflexive relation on N x N.
2) Symmetric Let (a. b), (c, d) N x N, Such that (a, b) R (c, d)Since (a, b) R (c, d) a + d = b + c
c + b = d + a [by commutativity ofaddition on N]
(c, d) R (a, b)Thus (a, b) R (c, d) (c, d) R (a, b) for all (a, b), (c, d)N x N .So R is symmetric relation on N x N
3) Transitive Let (a, b), (c, d) and (e, f)N x N.Such that (a, b) R (c, d) and (c, d) R (e, f)
Since (a, b) R (c, d) a + d = b + c (1)And (c, d) R (e, f) c + f = d + e (2)Adding equation (1) & (2), we get
a + d + c + f = b + c + d + e a + f = b+ e (a, b) R (e, f)
Thus (a, b) R (c, d) and (c, d) R (e, f) (a, b) R (e, f) for all (a, b), (c d), (e, f)
N x N.
So, R is transitive relation on N x N

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Hence R being reflexive, symmetric and transitivite is an equivalence relationon N x N.
Hence proved.
Q. 3 On the set N of natural numbers a relation R is defined asa R b a24ab+3b2=0 v (a, b N) . Prove that R is reflexive but no t
symmetric not transitivity.
Ans. Given set is N = {1, 2, 3,..}
Relation defined on N is
a R b a24ab+3b2=0 v a, b N.1) Reflexivity Let a N
2 2a  4a.a + 3a
2 2 2= a  4a + 3a2 2
= 4a  4a
= 0
(a, a)R v a N. R is reflexive
2) Symmetry  Let a, b N such that (a, b) R2 2
2 2
4 3
4 3 0
( , )
( , ) ab b
b ba a
b a R
a b R a 0 + =
+
Hence (a, b) R but (b, a)R R is not symmetric relationEx. (3, 1) R because = 912+3 =1212 = 023 4 3 1 3(1) + 2
But (1, 3) R because = 2 24(1)(3) 3(3)(1) + 1 12 27 0= +
3) Transitivity Let a, b, c N such that(a, b) R and (b, c) R
2 2( , ) 4 3 0a b R a ab b + =
and (b, c) R b24bc+3c2Then it is not necessary true that a24ac+3c2=0

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Ex. (9, 3) because = 81 108 + 27 = 02 4(9)(3) 3(3)9 + 2
and (3, 1) R because = 912+3 = 023 4(3)(1) 3(1) + 2
but (9, 1) R because = 8136+32 24(9)(1) 3(1)9 + 0
R is not transitivity. Hence from (1), (2) & (3) it is clear that R is reflexive but notsymmetric and transitivity.
Q. 4 Let A = {1,2,3} then give examples of relations which are
1) Reflexive, symmetric and transitive
2) Symmetric and transitive but not reflexive
3) Reflexive and transitive bu t no t symmetric
4) Reflexive and symmetric but no t transitive
Ans. 1) is reflexive, symmetric and transitive1 {(1,1), (2, 2), (3,3)}R =
2) is symmetric and transitive but not reflexive2 {(1,1), (2,2)}R =
3) is reflexive and transitive but not
symmetric
3 {(1,1),(1,2),(1,3),(2,2),(2,3),(3,3)}R =
4) is reflexive and symmetric but
not transitive
4 {(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)}R =
Q. 5 If a, b, {1, 2, 3, 4}, then check whether the following is function or not
f = {(a, b):b=a+1} also find it s range.
Ans. Here f = { (1, 2), (2, 3), (3, 4)}. Here we observe that an element 4 of the given setis not related to any element of the given set. So f is not a function.
Q. 6 If3
1
x
x
+
(x) =f then find [ { ( )}]f f f x
Ans. f (x) =3
1
x
x
+
Now ( ){ } =f x ( ) 3
( ) 1
f x
f x
=
+
3 31
31
1
xx
x
x
+
+ +
=3 3 3 2 6 3
3 1 2 2 1
x x x x
x x x x
+ = = =
+ +
Again

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f [f {f(x)}] = f3
1
x
x
+
=
33
1
31
1
x
x
x
x
+
+ +
3 3 3 43 1 4
x x x xx x
+ += =+ +
=
[ { ( )}]f f f x x =
Q. 7 Prove that the function f: R R, f(x) = cos x is many one function?Ans. Given f: R R such that f (x) = cos x
many one function: 
Let a, b R such that f (a) = f (b)
cos cos
2 ,
a b
a n b n
=
= I
f is many one function
Into function Let y R (Co domain)
If it is possible let f (x) =y
1
cos
cos y
x y
x =
=
x will exist if 1 1y When then pre image of y does not exist in R (Domain)[ 1,1]y R
Hence f is not on to function
f is in to function
Hence f is many one in to function.
Q. 8 If f: R R, f (x) = 2x 3 and g: R R, g(x) = 32
x +
then prove that fog =
gof =R
I
Ans. Given functions :f R R , f (x) = 2x3
3: , ( )
2
xg R R g x
+ =
, so:gof R R

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(gof) (x) = g [f(x)] = g (2x3) =(2 3) 3 2
2 2x
x x= =
+
gof (x) = x .. (1)
Again
fog (x) = f [g(x)] =
x + 3 x + 3= 2  3
2 2
= x + 3
= f
 3 = x
fog (x) = x (2)
and : R R such that (R
I R
I x ) = x , vx R .(3)
from (1), (2) & (3) we get,
fog = gof =R
I
Hence proved
Q. 9 Let f: R R be defined by f (x) = 3x 7. Show that f is invertible and hence
find
1f
Ans. A function f is invertible if f is a bijection
1) Injectivity Let x, y R thenf(x) = f (y)
3 7 3 7x y
x y
= =
Thus f (x) = f(y) x=y for all x, y R. So, f is an injection2) Surjectivity Let y be an arbitary element of R, then f (x) = y
3 7
7
3
x y
yx
=+ =
Clearly7
3
yR
+
for all y R
Thus for all y R, there exists7
3R
yx
+= such that

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7
3
73 7
3
( )
( )y
y
f x y
f x f+
+ =
=
=
f is surjection. Hence :f R R is bijection. Consequently it is invertible
Let f(x) = y
1
7
3
7( )
3
3 7
y
yf y
x y
x
+
+ =
=
=
Therefore,1 :R Rf is given by
1 7( )3
xxf
+ =
Chapter 2
Binary Operation
Q. 1. Discuss the commutativity and associativity of the binary operation on Rdefined by
a * b =4
ab for all a, b R.
Ans. Commutati vi ty
a * b =4
ab and b*a =4
ba
We know that multiplication on R is commutative
4 4ab ba
= for all a, b R
for all a, b* *a b b a = RSo * is commutative on R.

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Associatively
Let a, b, c R then
(a * b) * c
( )4
4 1
abc
abc ==
6
.............................................(1)
and a * (b * c) = a *4
4 4 1
bc
bca
abc
= = 6
..(2)
From (1) & (2), we observe that a * (b*c) = (a*b) *c
Hence * is associative.
Q. 2. Let * be a binary operation on set Q {1} defined by a*b = a+ b a b , a, b Q  {1}
Find the identity element wi th respect to * on Q. Also prove that every
element of Q {1} is invertib le.Ans. Let the identity element e exist in Q {1} w.r.t * on Q  {1}, then
a *e = a = e*a for all a Q  {1} a*e = a for all a Q  {1} a+ e ae =a
e (1a) = 0 e = 0
Thus o, is the identity element for * on Q {1}. Let a be an arbitary element ofQ {1} and
let b be inverse of a, then.
a * b = 0 = b *a [0 is identity element]
a * b = 0 a + b ab = 0
b (1 a) =  a
b =1
a
a
{1}
1 0
a Q
a
Since , therefore b={1}Qa {1}1
aQ
a

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Thus every element of Q {1} is invertible and the inverse of an element a is
1
a
a
Q. 3 Let * be an associative binary operation on a set S and a be an invertibleelement of
S then1 1)( aa =
Ans. Let e be the identity element in S for the binary operation * on S, then
1 1
1 1
*
* *
* e a a
a a e a a
a a
= =
= =
a is inverse of 1a
a = 1 1)(a
Hence proved
Chapter 3
Inverse trigonometric functions
Q. 1 Find the princ ipal values of the followings : 
1)
1 1cos2
(2) ( )1 sec 2
3) (4)( )1cosec 1
1
3
1cot
Ans. (1) Let
1 1cos =2
cos = 1/2

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4) Let
1 1 = 3
cot
( )
1
3
cot = 
cot = cot
3
cot = cot  /3
2cot = cot
3
2 =
3
1 21cot =
33
Hence principal value of1 21cot
33is
.
Q. 2 Prove that1 11 1 1tan cos tan cos
4 2 4 2a a
b b
2ba
+ +
=
Ans. Let
1cos1
2
a
b
=
cos2a
b
=
LHS = ( ) ( )tan4 4
tan
+ +

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( )( )
( ) ( )
2 2
2
4
tan
4
=
= +
tan
+
tan  tan
1+tan 1tan
tan = 11tan 1+tan 4
+(1tan)=
1tan 1+tan
1+tan + 2 tan=
tan +
4
1tan4
(1+tan)
tan
1+tan
2+1+tan 2tan
2
2
2
2
2
2=
)
1+tan = 2
1tan
=
= = RHS2b
a
(1tan
2
1tan1+tan
2
cos
=
tanA+tanBtan(A + B)=
1tanAtanB
tanAtanBtan(AB)
1+tanAtanB
Hence proved.
Q. 3 If1 15 12sin sin 90
x x
+ =
then find the value of x
Ans.1 15 12sin sin 90
x x
+ =

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1 1
1 1
1 1
1 1
2
2
5 1sin = 90  sin
x x
5 12sin = cos
x x
sin x + cos x = 90
25 12cos 1 = cos
x x
25 121 =
x x
2
Squaring both sides, we get
2 2
2
2 2
2
2
25 1441 =
x x
x  25 144=x x
x  25 = 144
x = 169
x = 13
Since x =  13 does not satisfy given equation . So x = 13 is correct solution.
Q. 4 Prove that
21 1
2
1cos tan {sin(cot )}
2
xx
x
+ = +
Ans. We have
1 1
2
2
2
1 1
2 2
2 2
22
1sin(cot ( )) sin sin
1
1
1
1 1cos tan cos cos1 2
1 1
22
xx
x
xx x
x x
xx
=
+
=+
+ =
+ +
+ += =
++
Hence proved

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Q. 5 If1 1cos cos
x y
a b
+
= then prove that
2 22
2 2
2cos sin
x xy y
a ab b + =
Ans. Given1 1cos cos
x y
a b
+ =
{ }
2 21
2 2
1 1 1 2 2
2 2
2 2
2 2 2
2 2
2 2
2 2
x y x y= cos  1 1 =
a b a b
cos x +cos y = cos xy + 1 x 1 y
xy x y 1 1 = cos
ab a b
xy x ycos = 1 1ab a b
x y
a b
22 2 2 2
2 2 2 2
2xy x y x y cos +cos = 1  +
ab a b a b
2 22
2 2
2 22
2 2
x 2xy y cos + = 1cos
aba b
x 2xy y cos + = sin
aba b
Hence proved
Q. 6 Solve the following equation
1 1 1
2
1 1tan + tan = tan
1+ 2x 4x +1 x
2
Ans. Given
1 1
2
  1 1tan + tan = tan1+ 2x 4x +1 x
1 2

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( )( )
1 2
2
1 1 1
2
1 1+
21+ 2x 1+ 4xtan = tan1 1 x
1
1+ 2x 1+ 4x
x + ytan x + tan y = tan
1xy
4x +1+1+ 2x 2=
1+ 2x 4x +1 1 x
6x+2
1 2+ 4x + 8x + 2x  1( )
2
2=
x
2 ( )3x+1
2 ( )
( ) ( )
( )( )
22
3 2 2
3 2
2
2
2
= x4x + 3x
3x + x = 8x + 6x
3x  7x  6x = 0
x 3x  7x  6 = 0
x 3x  9x + 2x  6 = 0
x 3x x  3 + 2 x  3 = 0
x 3x + 2 x  3 = 0
2x = 0,3,
3
Practice Problems
1) Find principal values of the followings
i)1 1sin
2
ii) ( )1tan 3
2) Solve the following equation
( ) ( )
1 1 1 1x xsec  sec = sec b  sec aa b
3) If ( ) ( )1 1tan 3 tan 24
x x + = then find value of x.
4) Solve the following  ( ) ( ) (1 1 1tan 1 tan ( ) tan 1 tan 3 )1x x x + + + = x .

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Chapter 2
Determinant
Q. 1 Find determinant of A =
1 6 2
2 1 1
4 1 3
Ans. A = (1)1 1 2 1 2 1
 6  21 3 4 3 4 1
= 1 (31) 6 (64) 2 (24) = 4+60+4
A = 68
Q. 2 Find determinant of A =
1 2 1 3
2 1 2 3
3 1 2 1
1 1 0 2
Ans. A =
1 2 3 2 2 3 2 1 3 2 1 2
1 2 1  2 3 2 1 1 3 1 1  3 3 1 2
1 0 2 1 0 2 1 1 2 1 1 0
1
= 1{1(40) +2(2+1) +3(0+2) {2{2(40) +2(61) +3(02)}1{2(2+1)1(61)
+3(31)}3{2(0+2)1(02)2(31)}= 1{4+6+6}2{8+106}1{6512}3{4+2+8}= 1{16}2{12}1{11}3{14}= 39
Q. 3 Check whether the following matrix is s ingular or not
A =
1 0 2
1 2 3
2 4 6
Ans.A matrix is singular if A =0

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1 0 2
A = 1 2 3 = 1[12 12] 0[6 6]+ 2[4 4] = 0
2 4 6
A = 0
Hence A is singular matrix.
Q. 4 Find the mines and cofactors of elements of the
determinant
2 3 5
6 0 4
1 5 7
Ans. We have0 4
M = = 0  20 = 2011 5 7
6 4M = = 42  4 = 46
12 1 7
6 0M = = 30  0 = 30
13 1 5
3 5M = = 21  25 = 4
21 5 7
2 5M = = 14  5 = 19
22 1 7
2 3M = = 10 + 3 = 1323 1 5
3 5M = = 12  0 = 12
31 0 4
2 5M = = 8  30 = 22
32 6 4
2 3M = = 0 + 18 = 18
33 6 0
Co factors

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( )
( ) ( )
( )
( )
1+1 2A = 1 .M = (1) (20) = 20
11 11
1+2 3A = 1 .M = 1 (46) = 46
12 12
1+3 4A = 1 .M = (1) 30 = 30
13 132+1 3A = 1 .M = (1) (4) = 421 21
2+2 4A = (1) .M = (1) (19) = 19
22 22
2+3 5A = (1) .M = (1) (13) = 1323 23
3+1 4A = (1) .M = (1) (12) = 12
31 31
A3
3+2 5= (1) .M = (1) (22) = 22
2 32
3+3 6A = (1) .M = (1) 18 = 1833 33
Q. 5 If W is one of the imaginary cubs root of unity, find thevalue of
2
2
2
1 w w
w w 1
w 1 w
=
Ans. Given
2
2
2
1 w w
w w 1
w 1 w
=
givesC C +C + C1 1 2 3
2 2
2 2
2
1+ w + w w w
w + w +1 w 1
w + w +1 1 w
=
Since 21+ w + w = 0
So
2
2
0 w w
= 0 w 10 1 w
Now finding its determinant & expending along first column
= 0 3 2 2 4w 1  0 w w + 0 ww = 0

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Q. 6 Prove that ( )2 2a b ax + by
b c bx + c2
y = (b  ac) ax + 2bxy + cy
ax + by bx + cy 0
Ans. Let A =a b ax + byb c bx + cy
ax + by bx + cy 0
Appling , we get3 3 1
c  xc ycc2
A =
a b ax + by  ax  by
b c bx + cy  bx  cy
ax + by bx + cy 0  x(ax +by)  y(bx +cy)
A = 2 2
a b 0
b c 0ax + by bx + cy (ax + 2bxy + cy )
Now expending along we get3
c
( ) 2 2
b bx + cy  c(ax + by)  0[a(bx + cy)  a(ax + by)] + 2bx2
y + cy )[ac  b ]A = 0 (ax
b
(a
(
2 2
2 2
+ 2bx 2
2
y + cy )(ac  b )
ac)(ax + 2bxy + cy )
A =  x
A =
Hence proved
Q. 7 Find the area of tr iangle with vert ices at the points (3, 8), (4,2) and (5,1).
Ans. Let A (3, 8), B (4, 2), C (5,1) are three given vertices oftriangle. So the area ofABC is
given by
12
=3 8 1
4 2 1
5 1 1

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{ }
[ ]
[ ]
1
3 2+1 8{45}+1{410}2
133+89+1(6)
2
19+7262
75
2
=
=
=
=
Q. 8 If the points (a, 0), (0, b) and (1, 1) are coll inear, Prove that a+ b = ab
Ans. The given points are collinear soa 0 1
0 b 1 = 0
1 1 1
a[b 1]+1[0 b] = 0
a b  a  b = 0
a b = a + b
Hence proved.Q. 9 using determinant find the equation of the line joining the
points (1, 2) and (3, 6).Ans. Let P (x, y) be a point on line AB i.e. the points A (1, 2), B (3, 6)and P(x, y) are collinear so
ABP = 0
1 2 11
3 6 12
x y 1
= 0
0=
1 2 1
3 6 1
x y 1
1[6  y]2[3  x]+ 1[3y  6x] = 0
6
 y  6
+2 x +3 y  6 x =04x+ 2y = 0
y = 2 x
Which is the required equation of line.Q. 10 Find the adjoint of the matrix

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1 3 3
A = 1 4 3
1 3 4
Ans. First we have to find cofactors
1+111
1+2
12
1+3
13
2+1
21
2+2
22
2+3
23
3+1
31
3+2
32
3+3
33
c = (1) 7 = 7
c = (1) (1) = 1
c = (1) (1) = 1
c = (1) (3) = 3
c = (1) (1) =1
c = (1) (0) = 0
c = (1) (3) = 3
c = (1) (0) = 0
c = (1) (1) =1
T7 1 1
adjA = 3 1 0
3 0 0
7 3 3
adjA = 1 1 0
1 0 1
Q. 11 Find the inverse of matrix
2 1
A = 3 4 .
Ans. A = 8 + 3 = 11 0
So, A is non singular matrix and therefore it is invertible. Nowfinding cofactors
1+1
11
1+2
12
2+1
21
2+2
22
c = (1) 4 = 4
c = (1) 3 = 3
c = (1) (1) = 1
c = (1) 2 = 2
T
4 3adj A =
1 2
4 1adjA =
3 2
Hence 11
A =A
.adj(A)

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=
4 1 4/11 1/111=
3 2 3/11 2/1111
=
4/11 1/11
3/11 2/11
1
Q. 12 Find the inverse of
1 3 3
A = 1 4 3
1 3 4
and verify that 13A A = I
Ans. From question No. 10 we find that
adj A =
7 3 3
1 1 0
1 0 1
A = 1[16  9]  3[4  3] + 3[4  3] = 7  3  3 = 1 0
So A is invertibleHence
1
1
1A = adjA
A
7 3 3 7 3 31
A = 1 1 0 = 1 1 01
1 0 1 1 0 1
Now
1
1
1
1
3
7 3 3 1 3 3
A A = 1 1 0 1 4 3
1 0 1 1 3 4
7 3  3 2112  9 21 912
A A = 1+1 + 0 3 + 4 + 0 3 + 3 + 0
1+ 0+1 3 + 0+ 3 3 + 0+ 4
1 0 0
A A = 0 1 0
0 0 1
A A = I
Hence verified
Q. 13 Solve the following system of equations by using cramersrule.
x + 2 y = 3
4x+ 8y = 12

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Ans. We have
1
2
1 2D = = 8  8 = 0
4 8
3 2D = = 24  24 = 0
12 8
1 3D = = 12 12 = 0
4 12
Since D, and all are equal to zero so the given system of
equations has infinitely1D 2D
many solutions.Let y=k then from equations x+2y=3x+2k=3x= 32k
Hence, x=32k, y= k is the solution of the given system ofequations, where k is arbitrary
real number.
Q. 14 solve the following system of equations by cramers rulex 2y = 4 3x +5y = 7
Ans. We have
1
2
1 2D = = 5  6 = 1 0
3 54 2
D = = 20 14 = 67 5
1 4D = = 7 +12 = 5
3 7
So, by Cramers rule, we have
1
2
D 6x = = = 6
D 1
D 5
y = = = D 1 5
x= 6, y= 5 is required solution.
Q. 15 Solve the following system of equations2x+3y+4z=0x+ y+ z = 02xy+3z=0

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Ans. We have2 3 4
D = 1 1 1
2 1 3
= 2(3 +1) 3(3  2) + 4(1 2)=8  3  1 2
D = 7 0
So, the given system of equations has only the trivial solutionsi.e x=0, y=0, z=0
Q. 16 Solve the following homogeneous system of equationsx+y2z=0..............................................................(1)
2x+y3z=0...(2)5x+4y9z=0.(3)
Ans. We have,1 1 2
D = 2 1 3
5 4 9
= 1(9 +12) 1(18 +15)  2(8  5)
= 3 + 3  6
D = 0
So, the system of equations has infinitely many solutions.Consider eq. (1) & (2). Put z=k in equations(1) and (2), we get
x+y =2k2x+y=3k
Solving these equations by cramers rule

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1
2
1
2
1 1D = = 1 2 = 1
2 1
2k 1D = = 2k  3k = k
3k 1
1 2kD = = 3k  4k = k
2 3k
D kx = = = k
D 1
D y = =
D
k
= k
1
x=k, y=k and z=k gives the solution for each value of k.
Q. 17 Use matrix method to solve the following system ofequations
5x 7y =27x 5y =3
Ans. The given system of equations can be written as
5 7 x 2=
7 5
y 3
Or A x = B, where
=
5 7 x 2A = , X , B
7 5
y 3
So, the solution is given by . So the find we have tofind co factors
1X = A B
1A
1+1
11
1+2
12
2+1
21
2+2
22
T
1
1
C = (1) (5) = 5
C = (1) (7) = 7
C = (1) (7) = 7
C = (1) (5) = 5
5 7 5 7adj A = =
7 5 7 5
A = 25 + 49 = 24
5 71 1A = adj(A) =
7 5A 24
X = A B

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5 7 21X =
7 5 324
10 + 21 111 1X = =
14 +15 124 24
11/24X =
1/24
Hence11
x =24
and y=1
24
Q. 18 Show that the following system of equations is consistent
2x y + 3z = 53x + 2y z = 7
4x + 5y 5z =9Ans. The given of equation can be written as
2 1 3 x 5
3 2 1
y = 7
4 5 5 z 9
A X =B
Where
= =
2 1 3 x 5
A 3 2 1 , X
y ,B = 7
4 5 5 z 9
Now
=
2 1 3
A 3 2 1
4 5 5
= 2[10 + 5]+ 1[15 + 4]+ 3[15  8]
= 1011+ 21 = 0
So A is singular. So the given system of equation is eitherinconsistent on consistent with
infinitely many solutions according as (adj A) B 0 or (adj A)B= 0 respectively.

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1+111
1+212
1+313
2+121
2+222
2+323
3+131
3+232
3+333
c = (1) (10 + 5) = 5
c = (1) (15 + 4) = 11
c = (1) (15  8) = 7
c = (1) (5  15) = 10
c = (1) (10 12) = 22
c = (1) (10 + 4) = 14
c = (1) (1  6) = 5
c = (1) (2  9) = 11
c = (1) (4 + 3) = 7
5 11 7
adj A = 10 
5
11
7
5 10 5
22 14 = 11 22 11
5 11 7 7 14 75 10 5
adj A) (B) = 11 22 7
7 14 9
25 + 70  45 0
= 55  154 + 99 = 0
35  98 + 63 0
(adj A) (B) = 0
(
Thus AX=B has infinitely many solutions and the given systemof equation is consistent.
Unit III
Chapter 1
Continuity and Differentiability
1. Check the continuity of the function f(x) at the origin :
( ) ; 0
1; 0
xf x x
x
x
=
=

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Ans. We have to show that the given function is continuous at x= 0,so
LHL lim ( ) lim (0 )0 0
f x f hx h
=
lim ( )
0
lim lim 1( )0
1
f h
hh h
h hh h o
LHL
=
= =
=
=
RHL lim ( ) lim (0 )00
f x fh
hx
= ++
lim ( )0
lim lim 1( )0
1
f hh
h h
h hh h o
RHL
=
= =
=
=
Now f(0) = 1Since LHL RHL, so the function f(x) is not continuous at the
origin.
2. Test the continuity of the function at x= 0
sincos , 0
( )
2, 0
xx when x
f x x
when x
+ =
=
Ans. LHL lim ( ) lim (0 )0 0
f x f hx h
= lim ( )
0f h
h=
sin( )lim cos( )
( )0
hh
hh
= +
sin( )lim lim cos( )
0 0
hh
hh h= +
1 1 2
2LHL
= + =
=
RHL lim ( ) lim (0 )00
f x fh
hx
= ++

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( )lim cos( )
( )0
( )lim lim ( )
0 0
1 1 22
Sin hh
hh
Sin hCos h
hh h
RHL
= +
= +
= + = =
And f(0) = 2Since f(0) = LHL = RHLSo the given function f(x) is continuous.
3. Find the values of a and b for which the following
function is continuous at x = 1.
2 1
( ) 1
5 2 1
x a when x
f x b when x
x when x
+ >
= =
0. So f
(x) is continuous
on [1,2] and differentiable on (1,2). Thus both the conditions of
langranges mean value
Theorem is satisfied. Hence there exist some ( )1,2c such that :

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( )
( )
2
(2) (1)'( )
2 1
( ) log
1'( )
(2) log 2, (1) log 1 0
(2) (1)'( )
2 1
log 2 01
11
log 2
1log
2log 22 4
log 2 log log 42 2 2
1 log 2
log 1,22
(2) (1)'
2 1
e
f ff c
f x xe
f x
xf fe e
f ff x
ex
ex
x ee
Now e
e
c e
f ff c =
=
=
=
= =
=
=
=
=
= =
<
<
< >
+ < >
< < >
< < Hence after 2 second and before 5 second velocity will be negative
and acceleration will
be positive.
3. Find the points on curve 2 2 2 3 0x y x+ = where tangent is
i)Parallel to x axis
ii)Perpendicular to x axis
iii)Makes equal angle with both axes
Ans. Given curve is _________________(1)2 2 2 3x y x+ = 0
Differentiating with respect to x we get
2 2 2 0
1
1_______________(2)
dyx y
dx
dyx y
dxdy x
dx y
+ =
+ =
=
i)Tangent is parallel to x axis
tan 0 0
10
1
dy
dx
x
y
x
= =
=
=
Substituting x = 1 in equation (1)21 2 3
2 4
2
y
y
y
+ =
=
=
0
Hence required points are (1,2) and (1, 2).
ii)Tangent is perpendicular to x  axis

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tan90dy
dx =
=
o
So from equation (2)1 1
0
0
x
y
y
= = =
Substituting y = 0 in equation (1)
( )( )
2 0 2 3 0
2 2 3 0
3 1 0
1,3
x x
x x
x x
x
= + =
= =
= + =
= =
Hence required points are (1, 0) and (3, 0).iii) Making equal angle with both axes
tan 45 1dy
dx = =o
From equation (2)1
1
1
x
y
y x
= =
=
Substituting y = (1 x) in equation (1)
( )
( )
22 1 2 3 02 21 2 2 3 0
22 4 2 0
2 2 1 0
1 2, 1 1 2
x x x
x x x x
x x
x x
x y
= + + == + + =
= =
= =
= = = m 2
Hence required points are ( ) ( ),1 2 2 1 2, 2and+
4. Find tangent and normal of the curve2 2
12 2x ya b
+ = , on
( )cos , sina b .
Ans. Given curve2 2
12 2
x y
a b+ =
Differentiating w. r. t. x, we get

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( )cos , sin
220
2 2
2
2
a b
y dyx
dxa b
dy xb
dx ya
dy a
dx
= + =
=
=
2cos b
b 2sin a
( )cos , sin
cos
sin
1 sin
cosa b
b
a
a
bdy
dx
=
= =
Hence equation of tangent at ( )cos , sina b is
( )
( )
cossin cos
sin2 2sin sin cos cos
2 2cos sin cos sin
cos sin
by b x a
aay ab bx ab
bx ay ab
bx ay ab
= =
= = +
= + = +
+ =
Equation of normal at ( )cos , sina b
( )
( )( )
sinsin cos
cos2 2cos sin cos sin sin cos
2 2sin cos sin cos
2 2sec cos
ay b x a
b
by b ax a
ax by a b
ax by ec a b
= =
= = +
= =
=
5. A balloon, which always remain spherical has a variable
diameter (3
2 32
x + ) . Determine the rate of change of volume with
respect to x.
Ans. Diameter of balloon = ( )3
2 32 x +
( )3
2 34
Radius x = +
Volume of a sphere (V) =4 33
r=

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( )
( )
34 3
2 33 4
392 3
16
x
V x
= +
= +
Rate of change of volume ( )29 3 2 3 216
dv xdx
= +
( )227
2 38
dvx
dx
= +
6. A stone is dropped in to a quite lake and waves move in a circle
at a speed of 3.5 cm/sec. At the instant when the radius of the
circular wave is 7.5 cm, how fast is the enclosed area increasing?
Ans. Let r be the radius and A be the area of the circular wave at any time t
then
( )
( )
2 3.5 / sec.
2
2
2 3.5
7
7 7.5
252.5 /sec.
drA r and cm
dt
dA dr r
dt dt
dA dr r
dt dt
dAr
dt
dA rdt
dA
dt
cm
= =
=
=
=
=
=
=
7. Sand is pouring from a pipe at the rate of The
falling sand forms a cone on the ground in such a way that the
height of the cone is always one sixth of the radius of the base.
How fast is the height of the sand cone increasing when the
height is 4 cm?
312 /sec.cm
Ans. Let r be the radius, h be the height and v be the volume of the sand
cone at any time t.
Then.

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( )
( )
1 23
1 2363
312
236
212 36
1
23
1 1
2 484 3 4
V r h
V h h
V hdV dh
hdt dt
dhh
dt
dh
dt h
dh
dt h
=
=
= =
=
=
= =
=
Thus, the height of the sand cone is increasing at the rate of1
/sec.48
cm
8. Find all the points of local maxima and minima of the function
( ) 3 26 9 8f x x x x= +
Ans. Given y = ( ) 3 26 9 8f x x x x= + . Then 23 12dy
9x xdx
= +
For maxima and minima 0dy
dx=
( ) ( )( )( )
23 12 9 0
2 4 3 0
2 3 3 0
3 1 3 0
1 3 0
1,3
x x
x x
x x x
x x x
x x
x
+ =
+ =
+ =
=
=
= Now we have to check that whether these points are the points of
maxima or minima. So

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( )
26 12
2
26 1 12 6 0
2 1
d yx
dx
d y
dx x
=
= =
=
Hence the function has minimum value at x = 3. The minimum
value of the function at x = 3 is
( ) ( ) ( ) ( )( )
3 23 6 3 9 3
27
f x
f x
= + =
8
54 27+
( )
8
8f x
=
9. Find the maxima and minima of the function ( ) ( )2
1 xf x x e= .
Ans. Given ( )2
1 xy x e=
( ) ( )2
2 1 1dy x xx e x edx
= +
For maxima and minima 0dydx
=

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( ) ( )2
2 1 1 0
2
xx x e
x
+ =
22 2x x +
( )
1 0
2 1 0
20, 1 0
1
xe
xx e
xe x
x
+ =
=
=
=
Now ( ) ( ) ( )2 2
2 2 1 2 1 12
d y x x xe x e x e xdx
= + + + xe
( ) ( )2 2
2 4 1 12
d y xe x xdx
= + +
Now for x= 1
2' 2 4 0 0
21
22 0
21
d ye
dx x
d ye
dx x
= + +
=
=