scert +2 maths em

8/18/2019 SCERT +2 maths EM

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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


43/57

= 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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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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(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

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