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SLOW LEARNERS MATERIAL FOR +2 MATHS
EASY TO GET PASS(ENGLISH MEDIUM)
IMPORTANT INSTRUCTIONS:
1 Students you should practice BOOK BACK one mark questions ,which will give 30
marks out of 40.
2 Writing formulas, diagrams and ending results should give step marks in Exam.
3 Should follow your class room Maths Teacher guidance.
4 Always concentrate: chapter 2-Ex(2.8- 7 to 14) with examples 2.50,2.51,2.52,Chapter
9- axioms of Group/Abelian Group and Truth tables and(cancellation and reversal)
laws, chapter 6-Trace the curve and Eulers theorem sums,Chapter 4-
Ex4.6(3),Ex4.5(2(ii) and all the applications problems,chapter 8- all the problems of
Ex 8.6 with all examples, and Text book one mark questions.
5. After exercising (written practice) this minimum material you may get FIRST class
marks.
S.no CHAPTERS POSSIBLE
MARKS TO GET
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1.APPLICATION OF MATRICES AND DETERMINANTS
(6 MARK QUESTIONS). = −.; verify the result () = () = 2Solution: = −
= − −− = − = −
(
) =
− − −−
=
−
−=
−
( ) = − −− − = − − = − ( ) = ( ) =
2. =
; Find the adj A,and verify the result ( ) = ( ) = 2
l i
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4. For = − − − − ; show that = − Solution: AA=A −
we have to prove = I
=−
− − − −
− − − = = I
5.Find the rank of
2751
5121
1513
Solution:
− −− − − = 1(2-35)+5(-4+25)-1(14-5)=-33+105-9
63 ≠ 0
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−
=0
=0
≠ −
=-1≠0 =
8) Find the rank of
7363
2142
3121
Solution:
− =0 − − −=0− − − =0
− − =0
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10) Find the rank of
1012
7436
3124
Solution:
=0 =0
=0 =0 ≠
= 5 ≠0
= 11) − −
Solution:
=0 =0
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Solution:
The matrix equation is
− − = = | | = − − = − + = −≠
Since A is non-singular, − exist.
=
− − − = | | = − − − = − −
the solution is = − = − − = −− = −
∴ =
,
=
−
14. Solve by matrix inversion method + = −, + = Solution:
The matrix equation is
=
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(
,
) =
−
,
,
∈ .
. − + + = , − + = , + + = Solution:
=
−
=
(
−−)
− (
− ) +
(
+3)
= −+ + = = − = −− − − + + = − + + = −≠
Since
=
,
≠ , The system is inconsistent (or) It has no solution.
+ +
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2.VECTOR ALGEBRA
10 MARK QUESTIONS
SOLUTION VECTOR EQUATION CARTESIAN EQUATION
1 point and 2 lines = + + − 1
− 1
− 1
1 1 12 2 2 = 0 1,3,2
To the lines +1
2=
+2−
1=
+33 and
−21
= +12
= +22
= + 3 + 2 , = 2 − + 3 = + 2 + 2 = + 3 + 2 + 2 − + 3
+ (+ 2 + 2 ) − 1 − 3 − 2
2
−1 3
1 2 2 = 0
8 + − 5 − 1 = 0
(2 1 3)
= 2 −− 3 ,= 3 + 2 4
2 + 1 + 3
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(−1,3,2) and perpendicular to the planes + 2 + 2 = 5
3 + + 2 = 8
= −+ 3 + 2 = + 2 + 2 = 3 + + 2 = − + 3 + 2 + + 2 + 2+ (3 + + 2 )
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−1,1,1 (1,−1,1) And perpendicular to plane + 2 + 2 = 5
= − + + , = −+ = + 2 + 2 = 1− − + + +
−+
+ (+ 2 + 2 )
+ 1 − 1 − 12 −2 01 2 2
= 0 2 + 2 − 3 + 3 = 0
Plane containing the line−22
=−2
3=
−1−
2 , point
1,1,−1
= − + − , = 2 + 2 + = 2 + 3 − 2 = 1− − + − +
2
+ 2
+
+ (2 + 3 − 2 ) + 1 − 1 + 13 1 2
2 3
−2
= 0 8 − 10 − 7 + 11 = 0
P i t 1 2 3
= − 2 + 3 , = − + 2 − = 2 + 3 + 4 + 1 + 2 3
2 4 4 0
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CEO,VILLUPURAM Page 11
PROVE THAT DIAGRAM , BY DEFINITION FROM DIAGRAM − =
+
= + = +
=
+
= + . = ( − )
= 1.1
(
− )
= − . = + . (
+
)
+ − = −
= + = + = + = +
× = × ( − ) = ( − )
× = 0 0 = ( − ) = ( − )
+ = − = + = + = +
= − . = ( + ) = 1.1 ( +)
= + . = + . ( − ) −
+
=
+ = +
=
+
= + = −
× = × ( +
)
= ( + ) × = − 0
0
= ( + ) = ( + )
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SOLUTION VECTOR EQUATION CARTESIAN EQUATION
3 POINTS = (1− − ) + + − 1
− 1
− 1
2 − 1 2 − 1 2 − 13 − 1 3 − 1 3 − 1 = 0 Points2,2,−1, 3,4,2 (7,0,6)
= 2 + 2 − , = 3 + 4 + 2 = 7 + 6 = 1− − (2 + 2 − )+
3
+ 4
+ 2
+
(7
+ 6
)
− 2 − 2 + 11 2 35 −2 7 = 0
5
+ 2
−3
= 17
Points 3 + 4 + 2 ,2 − 2 − , 7 +
= 1− − 3 + 4 + 2 + 2 − 2 − + (7 + ) −
3 − 4 − 2−1 −6 −34 −4 −1 = 0
6 + 13 − 28 − 1 4 = 0 Derive the equation of the
plane in intercept form
= (1− − )+ + + + = 1
(1,1, 1) = (, 0,0) (
2,
2,
2) = (0,
, 0)
(3 ,3, 3) = (0,0, ) − − 0 − 0− 0− 0 = 0 + + = 1 o
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=
1 1 1
2 0 1
2 1 1= 1
− = −5 − 3 − 4 Hence × × × = − [ ] 2.Show that the lines
−11
=+1−1 = 3 and −21 = −12 = −−11 . Intersects and find their point of
intersection.
Solution:
1 = 1,−1,02 = 2,1,−1, = 1,−1,3, = (1,2,−1)
The condition for intersecting is =2 − 1 = 0 2 − 1 = 1 2 −11 −1 31 2 −1 = 0
Take−1
1=
+1−1 = 3 = , Point ( + 1,− − 1, 3) Take.−
2
1 = −1
2 = −−1
1 = , Points ( + 2 , 2 + 1,− − 1) Since lines are intersecting + 1,− − 1, 3 = ( + 2 , 2 + 1,− − 1) + 1 = + 2 − − 1 = 2 + 1 = 0 = −1 The point of intersecting is (1,−1,0)
3 . .−1
3=
−11
=+1
0 and
−42
=0
=+1
3 intersect and hence find the
i t f i t ti
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Solution: × = −2 0 50 1 −3 = −5 − 6 − 2 × = 2 3 −1−5 −6 −2 = −12 + 9 + 3 (1)
.
= 6
. = −12 + 30 . = −9 . = −9 + 27 . − . = −12 + 9 + 3 (2) From (1) and (2)
×
×
=
.
−
.
4.Prove by vector method:Altitude of a triangle are concurrent. ⊥ ⇒ . = 0 . − = 0 . −. = 0 (1)⊥⇒ . = 0 . = 0
0 (2)
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= (+ ) ⟹−
+
= ±
+
= + ()−− . 3.Prove that + + − = +
Solution: + = +
⇒ =
=
+ = + + = + …………… (1) − = − …………… (2) (1) +(2)
+
+
−
=
+
4. Prove that
Solution: + = + = =
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6. Show that the points representing the complex numbers + ,− + , +
form a right angled triangle on the Argand diagram.
. Type equation here.
solution:
A ( , ) ,(−,) and (,) ; = | + – − + | = |
+
|
= + = . = | − + – + | = | − + | = + = = | + – +
= |
− −
|
= + = . + = + = =
= .It forms a right angled triangle.
7
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– =
−
=
−
.
– + . = ± ; ± , ± .9)
−
+
−+
=
+
Solution:
+ , − = .
=
+
=
.
– + − + −+ ( – + ) ( + + ) , – = − = − . – + . = ± , ± , ± .
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() ( + ) ( + ) ( + ) . . . ( + ) = + . () − + − + − + . . . + − = + − , ∈ .
solution :
i) Given
(
+
) (
+
) (
+
) . . . (
+
) =
+
,
| ( + ) ( + ) ( + ) . . . ( + ) = | + |, ( + )( + ) ( + ) . . . ( + )= A + i B , + + + . . . + = +
( + ) ( + ) ( + ) . . ( + ) = + . () { ( + ) ( + ) ( + ) . . . ( + ) }
= ( + ). + + + + + + . .+ + = ( + )
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∴
+
=
+
13.If cos2
1
x x
then prove that i) n x
xn
ncos2
1 (ii) ni
x
xn
nsin2
1 ; N n
Solution:
cos21
x
x
=
+
= − =( + ) =
+
----------(1)
= −-----------(2)(1)+(2) n
x
xn
ncos2
1
(1)- (2) ni x
xn
nsin2
1
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o
A
X+
<
+
− − −(
)
by the condition for the collinear + = + −− − () (), () | + | ≤ || + ||.
15. For any two complex numbers Z1 and Z2 then prove that
(i) 2121 Z Z Z Z
(ii) 2121 argargarg Z Z Z Z = ⟹ = ,arg() = = ⟹ = ,arg() = = = +
(i) = = (ii) arg () = +
= arg() + () 16.For any two complex numbers Z1 and Z2 then prove that
(i)1ZZ (ii) 1Z
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= - 4x=
−
= (+ ) = ((+ ) + (+ )) = ( cos (2k+1) + i sin (2k+1) )
where = ,,, .
(
+
)
( + ) Solution: + + = + − =+ –
= ( + ) + ( + ) 4.ANALYTICAL GEOMETRY
10 MARK QUESTIONS
1. Prove that the line 9125 y x touches the hyperbola 99 22 y x and find its point of contact
Given line is + = = +
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+
=
= , = = + = + = − + = + = :
− , = −, 3. Find the equation of the rectangular hyperbola which has for one of its asymptotes the line 052 y x
and passes through the points (6,0) and (-3,0).
+ − = − + = (+ −)(−+ ) =
.
+
−−+
=
(,) + = + = (1) (−,) −−+ =
= (2), =
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= −
+
−−+
=
−
5. A cable of a suspension bridge hangs in the form of a parabola when the load is uniformly distributed
horizontally. The distance between two towers if 1500ft, the points of support of the cable on the towers are 200ft
above the road way and the lowest point on the circle is 70ft above the roadway. Find the vertical distance to the
cable from a pole whose height is 122 ft.
= (,). = 130)
= × = ×
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= ⇒
=
×
=
= . (,) ∴ = × = × × ⇒
=
×
×
=
= = m7. The girder of a railway bridge is in the parabolic form with span 100ft. and the highest point on the arch is 10ft,above the bridge. Find the height of the bridge at 10ft, to the left or right from the midpoint of the bridge.
= −, (,−)
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the or bit. Find (i) the equation of the comet’s orbit (ii) how close does the comet nearer to the sun?( Take the orbit as
open rightward ).
=
, = (+ , ).
=
(
+
)
× = + + − = + − = + − = = − = = −
=
= = 9. On lighting a rocket cracker it gets projected in a parabolic path and reaches a maximum height of 4mts when
it is 6 mts away from the point of projection. Finally it reaches the ground 12 mts away from the starting point. Find
the angle of projection.
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10. Assume that water issuing from the end of a horizontal pipe, 7.5m above the ground, describes a parabolic
path. The vertex of the parabolic path is at the end of the pipe. At a position 2.5m below the line of the pipe, the flow
of water has curved outward 3m beyond the vertical line through the end of the pipe. How far beyond this vertical line
will the water strike the ground?
= − (,− ⋅)
=
−−.
= ⇒ = = − ,− ⋅ , = − × ×− ⋅ = × = ×
=
= 11 An arch is in the form of a semi-ellipse whose span is 48 feet wide. The height of the arch is 20 feet. How wide is
the arch at a height of 10 feet above the base?
= ⇒ = =
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= Equation of the ellipse
+ = + =
(,) , ∴ + = = − = − = − = ∴ = =
=
= ft13. The ceiling in a hallway 20ft wide is in the shape of a semi ellipse and
18ft high at the centre. Find the height of the ceiling 4 feet from either wall if the
height of the side walls is 12ft.′ = = =
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=
×
=
=
⋅
= + ⋅ = ⋅ ft14. A satellite is traveling around the earth in an elliptical orbit having the earth at a focus and of eccentricity 1/2.
The shortest distance that the satellite gets to the earth is 400 kms. Find the longest distance that the satellite
gets from the earth.
e=
1/2
= a-ae=400a(1-e)=400a(1-1/2)= 400
a= 2x400
a=800
Longest distance = ′ = +
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∴ +
=
=
∴ = = = = = ⇒ =
= ( − ) = − = × − =
+
=
⇒
+
=
17. A ladder of length 15m moves with its ends always touching the vertical wall and the horizontal floor.
Determine the equation of the locus of a point P on the ladder, which is 6m from the end of the ladder in contact
with the floor.
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∴ , is
+
=
, Which is an ellipse.
6.DIFFERENTIAL CALCULAS II
1) = + + ; = .u = log ( tan x + tan y + tan z )
=sec2
tan
+ tan
+ tan
2 = 2 2
+ + = 2 tan + + 2 = 2 2 + + = 2 tan + + 2 = 2 2 + + = 2 tan + + sin 2x
u= 2.
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∆
100 =1
2
100
= 12 −0.132.1 100 = −0.156 % The percentage error in the time of swing is a decrease of 0.156% .
4) Use differentials to find an approximate value for the given number 1.36
=
=
=
1
2
⇒
=1
2
−1
2
36 = 6 = 6, = ∆ = 0.1 = 12 −12 = 36, = ∆ = 0.1; 36 = 6 = 1336120.1 = 0.1
12= 0.008 36.1 = 6 + 0.008 = 6.0083
5) Use differentials to find an approximate value for .653
= = 3 = 13 ⇒ = 13 −23 64 = 4 ; = 64, = ∆ = 1 = 13−23
= 1364−231 = 1
2 1
6413
2 = 13 1 642
= =1
3
1
4
2
=1
3×
1
16=
1
48= 0.021
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= 42 − 2 + 42 = 42 − 42 + 82
= 4
2 + 4
2 = 4
(
+
2)
= + = 2−2+ 22 = −42 − 2 + 42 = −42 − 43 + 82 = 42 − 43 = 42 + 2 8) Find
r
w
and
w if
22log y xw where sin,cos r yr x
∂w∂r = ∂w∂x ∂x∂r + ∂w∂y ∂y∂r ∂w∂r = 1x2 + y2 2xcosθ+ 1x2 + y2 2y sinθ =
2
2
2
+
2
=
2
∂w∂θ = ∂w∂x ∂x∂θ + ∂w∂y ∂y∂θ ∂w∂θ = 1x2 + y2 2x(−r sin θ) + 1x2 + y2 2y (r cos θ) =
2r
r2(−xsinθ + ycosθ)
2r
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10 MARK QUESTIONS
Trace the curves
1. y=x3 2. y=x3+1 3. y2=2x3
1.Domain,Extent,Intercepts
And Origin
Domain (-α,α) Domain (-α,α) Domain;
≥ (or)
0,
α)
Horizontal extent(-α,α) Horizontal extent(-α,α) Horizontal extent0,α) Vertical extent (-α,α) Vertical extent (-α,α) Vertical extent (-α,α)x -intercept=0 x- intercept =-1 x -intercept =0 y -intercept 0 y- intercept =1 y- intercept =0
It passes through theorigin
It does not passthrough the origin
It passes through theorigin
2.SymmetryIt is symmetrical about
origin
There is no
symmetrical property
It is symmetrical about x
-axis
3.AsymptotesThe curve does notadmit asymptotes
The curve does notadmit asymptotes
The curve does notadmit asymptotes
4.MonotonicityThe curve is increasingin(−∞,∞) The curve isincreasing in(−∞,∞)
ofthe curve = 23 2 isincreasing.(ii)for the branch of
the curve = 23
2
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= sin−1 − + sin= − − = (,) ,degree n=1/2 + = 12 sin+ sin = 12 sin cos + cos = 12 sin
+
=
1
2tan
5) Using Euler ’s theorem prove that
If
y x
y xu
33
1tan Prove that .2sin u
y
u y
x
u x
= tan−1 3+3− tan
=
3+3−
=
(
,
) , degree n=2
+ = 2f (tan) + (tan) = 2 tan sec2 + sec2 = 2t an + = 2 sin 2 6) Verify Euler’s theorem for
1, y x f
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2 = 2−22+22 ⟹ 2 = 2 8) = 2 − 2 2 = 2 = 12 + 23 = −23 − 12 2
=
=
−2
3
−1
2
= −23 + 23 (1)2 = = 12 + 23 =
−23 + 23 From (1) and (2)2
=2
9) = sin3 cos4 2 = 2 = sin3 cos4 = 3 cos 3 cos4 = 4sin3 sin4
2
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=0.07
2= 0.01167
1.02
3+
1.02
4= 2 + 0.01167 = 2.0116
8. DIFFERENTIAL EQUATIONS
6 – MARK QUESTIONS
1. Solve
+
cot
= 2cos
= cot = 2 = = cot = log sin = sin
Solution is
= + = 2cos + = 2 +
= −22
+ 2. Solve + =
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=
+
(2 + 1)2 = 1
(2 + 1)2 (2 + 1)2 + (2 + 1)2 = + Do it your self :Solve
1.
+ 2
= sin
2. 1 + 2
+ 2 = 3. + = 4. (2 + 14 + 49) = −7 + 4
10 MARK QUESTIONS
1.Solve:( + )2 = 1 = 1( + )2 + =z+ 1 =
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: + 2
=1
2
− 3 + 22
3
=2
32 − 3 × 3 + 2 3 = 3 . ∶ = . + . = + 2 + 3 …………………………………………………………………….. (1)
= 2 = 0 0 = 2 + 22 + 32 0 = 2 + 4 + 8 ∴ + 2 = −4 …………………………………………………………. (2) = 0 = 0 0 = + + 1
+
=
−1 ………………………………………………………….. (3).
2
− 3
⇒ =
−3
(3)⇒ = 2 (1)⇒ = 22 − 3 + 3 3.: 2 − 1 = 2 − 22
: 2 1 = 0
= ±1
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=
⟹ = When = 0 A = 1,30,000 1,30,000 = 0 1,30,000 =
∴ = 1,30,000
= 30
A = 1,60,000
1,60,000 = 1,30,00030 1613 = 30 = 60 ,A = ? = 1,30,00060 = 1,30,000(30)2 = 1,30,000
16
13
2
≈ 1,97,600 The required population = 1976005.The number of bacteria in a yeast culture grows at a rate which is proportional to the number present. If the
population of a colony of yeast bacteria triples in 1 hour. Show that the number of bacteria at the end of five hours will
be5
3 times of the population at initial time.
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6..A drug is excreted in a patients urine. The urine is monitored continuously using a catheter. A patient is
administered 10 mg of drug at time t = 0 , which is excreted at a Rate of 21
3 t mg/h.
(i) What is the general equation for the amount of drug in the patient at time t > 0 ?
(ii) When will the patient be drug free?
(i) = −312 ⇒ −312 = −312
=
−3
12
+1
1
2+1
+
,
=
−2
3
2 +
When t=0, = 10 ⇒ = 10, = −232 + 10 (ii)When A = 0, 0 = 1 0− 232 ⇒ 1 0 = 232 ⇒ 5 = 32 3 = 52 = 25 ⇒ 2.9 Hence the patient will be drug free in 2.9 hours.
7.Radium disappears at a rate proportional to the amount present. If 5% of the original amount disappears in 50
years, how much will remain at the end of 100 years. [ Take A0 as the initial amount ].
Let A be amount of Radium at time T ∝ ⇒ = = = 0, = 0 0 = = 0
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.
9.A bank pays interest by continuous compounding, that is by treating the interest rate as the instantaneous rate of
change of principal. Suppose in an account interest accrues at 8% per year compounded continuously. Calculate thepercentage increase in such an account over one year. [ Take 0833.1
08e ].
Let A be the principal amount at time t ∝ ⇒ = ⇒ = ∴ = 0.08 = 0.08
=
0.08 ,
s 1−(0) (0) × 100 ⇒ 1( 0 − 1 × 100 0.08 − 1 × 100 0.08 − 1 × 100 = 1.0833− 1 × 100 = 0.0833 × 100 = 8.33%
= 8.3.3%
10.The sum of Rs. 1000 is compounded continuously, the nominal rate of interest being four percent per annum. In
how many years will the amount be twice the original principal ? ( 6931.02log e
)
Let A be principal amount at time t ∝ ⇒ = ⇒ = = 0.04 , = 0.04 = 10000.04 0.04 = 2
0.04 = log 2
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38.82 C temperature, after a further interval of 5 minutes.
12.For a postmortem report, a doctor requires to know approximately the time of death of the deceased. He records
the first temperature at 10.00 a.m. to be 93.4 ° F. After 2 hours he finds the temperature to be 91.4° F. If the room
temperature ( which is constant) is 72° F, estimate the time of death. (Assume normal temperature of a human body
to be 98.6° F).
1 9 .4 2 6 .6lo g 0 .0 4 2 6 2 .3 0 3 lo g 0 .0 0 9 4 5 2 .3 0 3
2 1 .4 2 1 .4e ea n d
Let T be temperature at time t,
∝ − 72 = 72℃ = − 72 ⇒ − 72 = = 0, = 93.4 ⇒ 93.4 − 72 = 0 ⇒ = 21.4 ∴ − 72 = 21.4
= 120,
= 91.4
⇒91.4
−72 = 21.4
120
⇒19.4 = 21.4
120
120
=19.4
21.4 ⇒ =1
120 log19.4
21.4 ⇒ = 1120 (−0.0426 × 2.303), 1 . = 1, = 98.6⇒ 98.6 − 72 = 21.41 = 26.6 1 = log 26.2621.4 1 = 1 log 26.621.4 = 0.0925×2.303−0.0426×2.303 × 120 266 = 4 26
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= 1, = − sec2 .
=
=
=
. = . + = − sec2 = sec2 = tan + 9.DISCRETE MATHEMATICS
1.State and prove cancellation law.
Let G be a group .
,
,
∈
() ∗ = ∗ ⇒ = (.. ) ∗ = ∗ ⇒ = (R.C.L) : () ∗ = ∗ ⇒ −1 ∗ ∗ = −1 ∗ ∗ ⇒ −1 ∗ ∗ = −1 ∗ ∗ ⇒ ∗
=
∗
⇒ = () ∗ = ∗ ⇒ ∗ ∗ −1 = ∗ ∗ −1 ⇒ ∗ ∗ −1 = ∗ ∗ −1 ⇒ ∗ = ∗
=
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( 3 ) construct the truth table for (pq)(~q)
P q ∨ ∼ ∨ ∧ ∼ T T T F FT F T T T
F T T F F
F F F T F
( 4) construct the truth table for ~ [(~ p)(~ )] P q ∼ ∼ ∼ ∧ ∼ ∼ ∼ ∧ ∼
T T F F F T
T F F T F T
F T T F F T
F F T T T F
( 5 ) construct the truth table for (pq)(~ r )
P q r
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T F T T T
T F F T F
F T T T T
F T F T F
F F T F F
F F F F F
( 7) construct the truth table for (p q ) [~ (pq )]
P q ∧ ∼ ∧ ∧ ∨ ∼ ∧ T T T F T
T FF T T
F T F T T
(8) construct the truth table for (pq ) r
p q r ∨ ∨ ∨ T T T T T
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T F T F T
T F F F FF T T F T
F T F F F
F F T F T
F F F F F
10. show that ~ (p q ) ≡ (~ p) ( ~q )P q ∨ ∼ ∨ T T T F
T F T F
F T T F
P q ∼ ∼ ∼ ∧ ∼ T T F F F
T F F T F
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12.show that [(~ q)p) ]q is a contradictions.
P q ∼ ∼ ∧ ∼ ∧ ∧ T T F F FT F T T F
F T F F F
F F T F F
It is a contradiction.
@ Use the truth table to establish which of the following statements are tautologies and
which are contradictions.
13: [(~ p)q] [ p (~q )]
P q ∼ ∼ ∼ ∨ ∧ ∼ ∼ ∨ ∨ ∧ ∼ T T F F T F T
T F F T F T T
F T T F T F T
F F T T T F T
It is a tautology
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P q ∨ ∼ ∨ ∨ ∨ ∼ ∨ T T T F T
T F T F T
F T T F T
F F F T T
It is a tautology
16. [ p (~q ) ] [ ( ~ p ) q ]
P q ∼ ∼ ∧ ∼ ∼ ∨ ∧ ∼ ∨ ∼ ∨ T T
FF
F T T
T F F T T F T
F T T F F T T
F F T T F T T
It is a tautology
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P q ∼ ∼ ∧ ∼ ∼ ∧ ∧ ∼ ∧ ∼ ∨ T T F F F F F
T F F T F T F
F T T F F F F
F F T T F F F
It is a contradiction
19. show that p q (~ p ) q
P q ⟶ T T T
T F F
F T T
F F T
P q ∼ ∼ ∨ T T F T
F
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T T T T T
T FF T F
F T T F F
F F T T T
p q ( p q ) ( q p )
21. show that p q[ (~ p ) q ) [ (~q ) p )
P q ⟷ T T TT F F
F T F
F F
T
P q ∼ ∼ ∼ ∨ ∼ ∨ ∼ ∨ ∧ ∼ ∨ T T F F T T T
T F F T F T F
F T T F T F F
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T F F T T
F T T F T
F F T T T
~ (pq ) (~p)(~q )
23. show that p q and q p are not equivalent.
P q ⟶
⟶
T T T T
T F F T
F T T F
F F T T
hence p
q and q
p are not equivalent
24. show that pq p q tautology.
P q ∧ ∨ ∧ ⟶ ∨ T T T T T
F T T
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[4] [4] [1] [5] [9] [3]
[5] [5] [4] [9] [3] [1]
[9] [9] [5] [3] [1] [4]
From the table,
(i)Closure axiom: all the elements in G again in G,The closure axiom is true.
(ii)Associative axiom: multiplication modulo 11 is always associative.
(iii)Identity axiom: the identity element is = [1] (iv)Inverse axiom:
element
1
3
4
5
9
inverse: 1
4
3
9
5
(v)commutative axiom: multiplication modulo 11 is always commutative.
G is a group.
2 .Prove that the set of four functions4321 ,,, f f f f on the set of non-zero complex numbers 0C defined by
011
,,4321
C z
z
z f and
z
z f z z f z z f forms an abelian group with respect to the
composition of functions.
Let = 1, 2, 3 , 4 • 1 2 3 4 1 1 2 3 4
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G is an abelian group.
3. 7 − 0,•7 . Let 7,−0,•7= 0 = 1, 2, 3, 4, 5, 6
•7 [1] [2] [3] [4] [5] [6][1] [1] [2] [3] [4] [5] [6]
[2] [2] [4] [6] [1] [3] [5]
[3] [3] [6] [2] [5] [1] [4]
[4] [4] [1] [5] [2] [6] [3]
[5] [5] [3] [1] [6] [4] [2]
[6] [6] [5] [4] [3] [2] [1]From the table,
(i)Closure axiom: all the elements of G is again in G,The closure axiom is true.
(ii)Associative axiom: multiplication modulo 7 is always associative.
(iii)Identity axiom: the identity element is = [1] (iv)Identity axiom:
Elements 1 2 3 4 5 6
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Closureaxiom
= , = ∈ ;
,
∈ − 0
= = 2 22 2 ∈ ∵ ≠ 0, ≠ 0 ⇒ 2 = 0 , ⇒ closure axiom is true
= 00 0
, = 00 0
∈
,
∈ − 0
= 00 0 00 0 = 00 0 ∈ ∵ ≠ 0, ≠ 0 ⇒ ≠ 0 , ⇒ closure axiom is true Associative
axiom Matrix multiplication is associative.
Matrix multiplication is associative
Identity
axiom
=
=
=
.
⇒ 2 22 2 = ⇒ 2 = ⇒ =
1
2∈ ∵ ≠ 0
= 12 1212
1
2
∈ .
=
0
0 0
=
=
0
0 0
.
⇒ 00 0 = 00 0 ⇒ = ⇒ = 1∈ ∵ ≠ 0 = 1 00 0
∈ .
Inverse
axiom
∈
∈
= 1
2
1
21
2
1
2
= 2 22 2 ⇒ 2 = 1
2⇒ = 1
4≠ 0 ∵ ≠ 0
=
∈ ∈ = 1 00 0 = 00 0
⇒ = 1 ⇒ = 1 ≠ 0 ∵ ≠ 0
=0
0 0 =
10
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• I A B C D E
I I A B C D E
A A B I E C D
B B I A D E C
C C D E I A B
D D E C B I A
E E C D A B I
(i)Closure axiom: all the elements of G is again in G,The closure axiom is
true.(ii)Associative axiom: matrix multiplication is always associative.
(iii)Identity axiom: the identity element is I
(iv)Identity axiom:
Element I A B C D E
Inverse I B A C D E
(,•) is a Group.
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CEO,VILLUPURAM Page 56
Show that *, Z is an infinite abelian group
where * is defined as .2* baba
Show that the set G of allpositive rational forms a groupunder the composition *
defined by a3
*ab
ba for all
., Gba
Show that the set G of allrational numbers except -1forms an abelian group withrespect to the operation *given by
abbaba * for
all ., Gba
Let G be the set of all rationalnumbers except 1 and * bedefined on G by
abbaba * for all
., Gba Show that ( G , * )is
an infinite abelian group
Closure
axiom
∴ ∗ ∈ . Closure axiom is true. , ∈ ∗ = 3 ∈ Closure axiom is true.
= − −1. ,∈ , and is arational number and ≠−1, ≠ −1. ∗ = + + is a rationalnumber, ∗ ≠ −1 andhence ∗ ∈ Closureaxiom is true.
= − 1 ,∈ , and is arational number and ≠−1, ≠ −1. ∗ = +− is a rationalnumber, ∗ ≠ −1 ∗ ∈ Closure axiom is true.
Associative
axiom
∗ ∗ = ∗ ∗ + + + 4 = + + + 4 ∈ ;
∗ ∗ = ∗ ∗ 9
=
9∈ ∗ ∗ = ∗ ∗ + + + + +
+
=
+
+ + + +
∗ ∗ = ∗ ∗ + + −−−+
=
+
+ −− + ;
Identity
axiom
= −2 ∈ = 3 ∈ = 0 ∈ = 0 ∈ Inverse axiom −1 = − − 4 ∈ −1 = 9 ∈
G is a Group.−1 = −
1 + ∈ ;
−1 = − − 1 ∈ ;
Commutative
axiom ∗ =
∗
+ + 2 = + + 2 (,∗) is an abelian group. ∗ =
∗
+ + = + + ;G is an abelian group. ∗ =
∗
+ − = + − ; G is an abelian group.
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CEO,VILLUPURAM Page 57
(10 MARK QUESTIONS)
CHAPTER 7-EX7.5- 1,2,3,4 and EXAMPLES 7.40,7.39.7.38,7.37
CHAPTER 3-EX 3.5- 5,4(ii),EX 3.4-5,6,10,8,EX 3.2-8(i),(v) EXAMPLES 3.11,3.22,3.23,3.24,3.25
CHAPTER 1-EX 1.5-2,3,EX1.4-10,9,EXAMPLES 1.8,1.22,1.19,1.24,1.26.1.28
Wish you all the Best