Strategic Action Plan for Slow Learners in Mathematics · PDF file1 strategic action plan for slow learners mathematics class xii sl no topic areas identified marks 1

1

STRATEGIC ACTION PLAN FOR SLOW LEARNERS

MATHEMATICS

CLASS XII

SL NO

TOPIC AREAS IDENTIFIED MARKS

1. INVERSE TRIGONOMETRIC FNS

Applications 4

2. • MATRICES & DETERMINANTS

• PROPERTIES OF DETERMINANTS

SOLVING EQUATIONSUSING MATRIX METHOD

6 4

3 DIFFERENTIATION

Using Logarithms 4

4. APPLICATION OF DERIVATIVES

INCREASING AND DECREASING FNS

4

5. APPLICATION OF INTEGRALS

AREA BETWEEN TWO CURVES USING FORMULAE

OR AREA OF TRIANGLE

6

6. DIFFERENTIAL EQUATIONS

LINEAR DIFFERENTIAL EQN.

4

7

VECTORS DOT PRODUCT VECTOR PRODUCT

4

8. THREE DIMENSIONAL GEOMETRY SHORTEST DISTANCE BETWEEN TWO LINES

4

9

LINEAR PROGRAMMING GRAPH 6

10 PROBABILITY

BAYE’S THEOREM 4

TOTAL 50

2

Inverse Trigonometric Functions

Areas to be revised:

1. Principal value branch table. 2. Properties of Inverse Trigonometric functions.

Properties:

tan ifxy< 1

1. tan +tan tan if x > 0, y > 0, xy> 1

tan ifx < 0, y < 0, xy> 1

tan ifxy>-1

2.tan tan tan ifx > 0, y < 0, xy<-1

tan ifx < 0, y> 0, xy<-1

Problems

1. Prove that

Sol: L.H.S = tan tan tan

=tan

. -

tan tan tan tan 1

= tan tan

= tan

( xy> -1)

= tan tan 1 = R.H.S.

3

2. If ,then find the value of x.

Sol.: We have sin sin cos 1,

= sin sin cos sin

= sin cos = x=

3. Write the value of . √

Sol:tan 2 sin 2 x since cos √

= tan 2 sin tan 2 x √

tan √3 3

4. Prove that √

=2 tan tan

=tan tan 2 tan tan

=tan tan

=tan

tan

tan tan , 1

=tan

Let tan tan

sin √

sin√

tan =sin√

= R.H.S

4

5. Find the value of |x|<1 ,y>0, xy<1

Sol: tan sin cos

=tan [2tan 2 tan 2 tan sin

cos

=tan x 2 [ tan tan

=tan [ tan

6. Prove that

Sol. L.H.S tan tan tan tan

= tan tan

= tan tan

= tan tan

= tan x 1

=tan tan 1 7. Prove that

Sol. LHS =cot 7 cot 8 cot 18

=tan tan tan since cot tan

=tan tan since x 1

tan3

11 tan1

18 tan3

111

181 3

11 x 118

=tan tan cot 3 RHS

5

8. Solve

Sol.=tan 2 tan 3

=tan

tan

=> 1 6x2 +5x-1=0 => (6x-1)(x+1) = 0

x= or 1 sincex= -1 doesn’t satisfy the equation, x=1/6 is the only solution of the given equation.

9. Solve for x,

Sol. Given tan tan

=tan

= tan

= 1 1

= 2 4 3 √

10. If 1, then solve the following for x

Sol. Given tan 1 tan 1 tan

= tan tan

= tan tan

= 16 8 62 =4 31 8 0 8 4 1 0

8 &14

As 0 <x <1 ,x ≠ -8 therefore x=1/4

6

Matrices & Determinants

1. Let A= express A as a sum of two matrices such that one is symmetric

and other is skew symmetric.

Sol. A can be expressed as A= [A+A1] + ½ [A-A1]-------------(1)

Where A+A1 and A-A1 are symmetric and skew symmetric matrices respectively.

A+A1=3 2 5 4 1 30 6 7

3 4 0 2 1 65 3 7

6 6 56 2 95 9 14

A- A1=3 2 5 4 1 30 6 7

3 4 0 2 1 65 3 7

0 2 52 0 35 3 0

Putting the values of A+A1 and A - A1 in equation (1) we get

A= ½ 6 6 56 2 95 9 14

0 2 52 0 35 3 0

2.Using properties of determinants, prove that

Sol.LHS let∆a ab ac

ba b bcca cb c

takinga, b, c common from R1, R2 and R3 respectively

∆ bca b c

a b ca b c

Now taking a, b, c common from C1, C2 and C3 respectively

∆1 1 1

1 1 11 1 1

applying R1 R1 + R2

∆0 0 21 1 11 1 1

expanding along first row, we get

∆ 0 0 2 1 1 = 4

3. Using properties of determinants, show that

Sol. LHS = let ∆1

11

applying R1 R1+R2 + R3 we get

∆1 1 1

11

7

Taking 1 common from R1

∆ 11 1 1

11

applyingC2 C2– C1,C3 C3– C1

∆ 11 0 0

11

expanding along R1

∆ 1 1 1 1

= 1 1 1

= 1 1 1

4. Using properties of determinants, show that

Sol. LHS ∆1

11

Multiplying C1, C2 and C3 by a, b and c respectively , we get

∆1 1

11

taking a, b, c common from R1, R2 and R3 respectively

∆1

11

applying C1 C1+ C2 + C3

∆11 11 1

Taking common (1 from C1

∆ 111 11 1

Applying R2 R2 – R1 , R3 R3 – R1

∆ 110 1 00 0 1

expanding along C1 , we get

∆ 1 1 1 0 0 0 1 RHS

8

5. Prove that

Sol. let ∆2

22

applying C1 C1+C2+C3

∆ 222 22 2

Taking 2 common from C1 , we get

∆ 211 21 2

applying R2 R2-R1 , R3 R3-R1

∆ 210 00 0

expanding along C1 we get

∆ 2 1 0

= 2 = RHS

6. Using properties of determinants prove that

Sol. LHS let ∆ applying C1 C1+C2+C3

∆

Taking ( common from C1 , we get

∆111

applying R2 R2-R1 , R3 R3-R1 and expanding along C1 we get

∆ 1 0 RHS

7. Prove that

Sol.LHS= let ∆1 1 1

1 1 11 1 1

9

Taking a,b,c common from R1,R2 and R3 respectively

∆1 1 1 1

1 1 1 1

1 1 1 1applying R1 R1+R2+R3

∆1 1 1 1 1 1 1 1 1 1 1 1

1 1 1 1

1 1 1 1

taking 1 1 1 1 common from R1

∆ 1 1 1 11 1 1

1 1 1 1

1 1 1 1

applying C2 C2– C1 , C3 C3– C1

∆ 1 1 1 11 0 0

1 1 01 0 1

expanding along R1, we get

∆ 1 1 1 1 1 1 0

= 1 1 1 1 RHS

8. If a, b, c are real numbers and show that either a+b+c=0

ora=b=c

Sol. Let ∆ applying C1 C1+C2+C3

∆222

= 2 111

applying R2 R2-R1 , R3 R3-R1 and expanding along C1 and on simplification we get

∆ 2 given∆ 0 either 0 or 0 or 0 or 0

Either a+b+c=0 or a=b=c

10

11. Two schools decided to award some of their selected students for the values honesty, regularity and hardwork at the rate of Rs. X ,Rs. y and Rs.z respectively per student the first school allotted a total of Rs.15,000 for its 2,2and 1 students for the respective values, while the second school kept Rs.19000 for theses values for 3,1and 2 students respectively. If the sum of three awards per students is Rs.10,000 then find the values of x,yand z using matrices. Suggest one more value which should also be included for the awards. Sol. We can represent given information , by the system of equation 2x 2y z 15000 3x y 2z 19000 x y z 10,000 Rewriting the above equations in matrix form 2 2 13 1 21 1 1

150001900010000

AX B

Where A2 2 13 1 21 1 1

X B150001900010000

|A| 2 2 2 2 0, so A‐1 exists and have unique solutions

adjA1 1 31 1 1

2 0 4 A‐1

| |

1 1 31 1 1

2 0 4

X A‐1B1 1 31 1 1

2 0 4

150001900010000

200030005000

200030005000

Hence the award for honesty Rs 2000, award for regularity Rs 3000 and award for handwork Rs 5000 Value: Any one value like sincerity or helpfulness etc can be awarded. 12. There are 3 families A,B and C. The no. of men, women and children in these families are as under

Men Women Children

Family A 2 3 1

Family B 2 1 3

Family C 4 2 6

Daily expenses of men, women and children are Rs200 , Rs150 and Rs200 respectively only men and women earn and children do not. Using matrix multiplication, calculate the daily expenses of each family. what impact does more children in the family create on the society ? Sol. The No. of men, women and children in families A,B and C can be represented by 3 x 3 matrix as

11

Xfamily Afamily Bfamily C

2 3 12 1 34 2 6

and daily expenses of men, women and children can be

represented by 3 x 1 matrix as Y Men

womenchildren

200150200

Daily expense of each family is given

by the product XY

XY family Afamily Bfamily C

2 3 12 1 34 2 6

200150200

family Afamily Bfamily C

105011502300

Hence daily expense of

i Family A Rs 1050 ii Family B Rs 1150 iii Family C Rs 2300 VALUE: More children in the family will increase the expenses of family, which will affect the economy of society.

13. For the matrix A2 1 11 2 1

1 1 2show that A2‐5A 4I 0 hence find A‐1

Sol. Given A2 1 11 2 1

1 1 2

A2 AA 2 1 11 2 1

1 1 2

2 1 11 2 1

1 1 2

6 5 55 6 5

5 5 6

A2‐5A 4I6 5 55 6 5

5 5 6

10 5 55 10 5

5 5 10

4 0 00 4 00 0 4

0 0 00 0 00 0 0

0

A2‐5A 4I 0 Premultiplying by A‐1 both sides, we get A‐1A2‐5A‐1A 4A‐1I A‐10 A‐5I 4A‐1 0 4A‐1 5I‐A

A‐1 5I‐A

A‐1 5 0 00 5 00 0 5

2 1 11 2 1

1 1 2

3 1 11 3 11 1 3

Try These

1. If A 2 11 2 and I is the identity matrix of order 2 then show that

A2‐4A 3I 0 hence find A‐1Ans. 2

31

31

32

3

2. To raise money for an Orphanage, students of three schools A,B and C organized

an exhibition in their locality, where they sold paper bags , scrap book and pastel sheets made by them using recycled paper, at the rate of Rs 20, Rs 15 and Rs 5 per unit respectively . School A sold 25 paper bags, 12 scrap books and 34 pastel

12

sheets. School B sold 22 paper bags, 15 scrap books and 28 pastel sheets while school C sold 26 paper bags1 18 scrap books and 36 pastel sheets. Using matrices find the total amount raised by each school. By such exhibition, which values are inculcated in the students ? Ans: School A Rs 850 B Rs 805 C Rs 970 Values: helping the orphans, use of recycle paper.

3. Find non‐zero values of x satisfying the matrix equation 2 23 2 8 5

4 4 2 8 2410 6

4. Librarian Mr.Ajeet Kumar has purchased 10 dozen autobiography of great person, 8 dozen historical books, 10 dozen story books related to moral teaching the cost prices are Rs.80 , Rs.60 and Rs.40 respectively. Find the total amount of money that he invested for library using matrix algebra. Which type of books is more useful for students and why ? Ans: Rs.20160, autobiography of great person is more useful for students as it educate a lesson to them for being a great person.

5. If A1 0 20 2 12 0 3

prove that A3‐6A2 7A 2I 0

6. Using matrix, solve 3x‐2y 3z 8 , 2x y‐z 1 , 4x‐3y 2z 4 Ans: x 1 , y 2 , z 3

7. Find A‐1, if A1 2 51 1 12 3 1

hence solve the following system of linear equation

x 2y 5z 10 , x‐y‐z ‐2 , 2x 3y‐z ‐11 8. Solve using matrix,

2 3 3

10 , 1 1 1

10 , 3 1 2

13

9. Using properties of Determinants, show that

4

Sol: L.H.S Let ∆

Applying R1‐> R1+R2+R3

∆2 2 2

Taking 2 common from R1

∆ 2

Applying R2‐> R2‐R1 , R3‐> R3‐R1

∆ 00

13

Applying R1‐>R1+R2+R3

∆0

00

Expanding along R1, we get

2 0 0 0 4 =RHS

10. If x, y ,z are all different and ∆111

0, then show that 1 0.

Sol: Let ∆111

= 111

111+

111

(Taking common x,y,z from R1,R2 and R3

respectively)

1111

111

= 1 111

Applying R2‐>R2‐R1, R3‐>R3‐R1

1 100

Expanding along C1 and simplifying, we get

1

Since ∆ 0 and x,y,z are all different 0, 0, we get 1 0.

TRY THESE

I. Using properties of determinants, prove the following

1. 4 2 2

2 4 22 2 4

5 4 4

14

2. 2

3. 2 3 2 4 3 23 6 3 10 6 3

=

4.

5.

II. Using properties of determinants solve for x.

1. 3 8 3 3

3 3 8 33 3 3 8

0

[Ans: , ,

2. 2 2 3 3 44 2 9 3 168 2 27 3 64

0

[Ans: x=4]

3. 1 1 1

1 1 11 1 1

0

[Ans: x=‐1, 2]

4. 0, 0

[Ans: x= ]

5. Using properties of det. Prove that

i. 2

22

9

ii. 111

1

Solutions of Linear Equations using Matrices

1. Solve 2 7; 3 4 5 5; 2 3 12 using Matrix method. Sol: The given system of equations can be expressed in Matrix from A x =B, where

1 1 23 4 52 1 3

, ,715

12

| |1 1 23 4 52 1 3

7 19 22 4 0

| | 0 exists and given system has unique solution X=A‐1B

15

7 1 319 1 1111 1 7

,

| |14

7 1 319 1 1111 1 7

X=A‐1B=7 1 319 1 1111 1 7

715

12

84

12

213

=>213

2, 1, 3

2. If 2 3 53 2 41 1 2

, find A‐1, using A‐1 solve the system of equations 2 3

5 11, 3 2 4 5, 2 3.

Sol: | |2 3 53 2 41 1 2

=0‐6+5=‐1 0

A is a non‐singular Matrix , so A‐1 exists.

adj A= 0 1 22 9 231 5 13

, | |

0 1 22 9 231 5 13

0 1 22 9 231 5 13

The given system of equations can be expressed as

A x =B 2 3 53 2 41 1 2

1153

Where A= 2 3 53 2 41 1 2

, X= , B=11

53

AX=B => A‐1 B => X=0 1 22 9 231 5 13

1153

123

=123 => x=1, y=2, z=3

3. Determine the product 4 4 47 1 3

5 3 1

1 1 11 2 22 1 3

and using it solve the

equations. 2 1, 2 4, 2 3 0

Sol: Let A= 1 1 11 2 22 1 3

,4 4 47 1 3

5 3 1

CA= 4 4 47 1 3

5 3 1

1 1 11 2 22 1 3

= 8 0 00 8 00 0 8

8

=> A‐1= [

18

4 4 47 1 3

5 3 1

16

The given system of equations can be written inmatrix form as PX=B

Where P= 1 1 21 2 1

1 2 3, X= , B=

140

PX=B = >

But P=1 1 21 2 1

1 2 3=

18

4 7 54 1 34 3 1

X=4 7 5

4 1 34 3 1

140

328

16

=4

12

4, 1, 2

4. Solve 4, 1 , 2 by using Matrix method.

Sol: Rewriting the given equations in Matrix form, we get

2 3 104 6 56 9 20

1

1

1

412

AX=B

Where A= 2 3 104 6 56 9 20

, ,412

|150 + 330 + 720 = 1200 0 A is non‐singular so A‐1 exists and X= A‐1B.

adj A= 75 150 75

110 100 3072 0 24

, A‐1=

| |

75 150 75110 100 3072 0 24

X=A‐1B =>75 150 75

110 100 3072 0 24

412

600400240

=> = => x=2, y=3, z=5

17

5. The management committee of a residential colony decided to award some of its members (say x) for honesty, some(say y) for helping other and some others (say z) for supervising the workers to kepp the colony neat and clean. The sum of all the awardees is 12. Three times the sum of awardees for cooperation and supervision added to two times the number of awardees for honesty is 33. If the sum of the number of awardees for honesty and supervision is twice the number of awardees for helping others, using matrix method, find the number of awardees of each category. Apart from these values, namely, honesty, cooperation and supervision, suggest one more value which the management of the colony must include for awards. Sol: According to the question, the system of values is 12, 2 33 33, 2 0. The above system of equations can be written in matrix for AX=B as 1 1 12 3 31 2 1

=12330

where A= 1 1 12 3 31 2 1

, ,12330

|A|=9+1‐7=3 0, So A‐1 exists. AX=B = > A‐1B

adj A=9 3 01 0 17 3 1

, | |

9 3 01 0 17 3 1

X=9 3 01 0 17 3 1

12330

= 9

1215

=345=> x=3, y=4, z=5

Number of awards for honesty = 3

Number of awards for helping others= 4

Number of awards for supervising = 5

Value: The management can include cleanliness for awarding the members.

Or the management can also include the persons, who work in the field of health and hygiene.

Or any other relevant answer.

11. Given A2 2 44 2 4

2 1 5 B

1 1 02 3 40 1 2

find BA and use this to solve the

system of equations y 2z 7 , x‐y 3 , 2x 3y 4z 17

18

10. Sum of three numbers is 20. If we multiply the first by 2 and add the second number and subtract the third we get 23. If we multiply the first by 3 and add second and third to it we get 46. Find the numbers.

Ans: 13, 2, 5

11. If A‐13 1 115 6 55 2 2

and B1 2 21 3 0

0 2 1 then find AB ‐1

use AB ‐1 B‐1 A‐1

12. Express the matrix A4 2 13 5 71 2 1

as the sum of a symmetric and a skew

symmetric matrix.

13. Find a matrix X such that 2A B X 0, when A 1 23 4 , B 3 2

1 5

14. A trust has fund Rs.50,000 that is to be invested in two different types of bonds. The first bond pays 10%P.A interest which will be given to adult education and second bond pays 12% interest P.A which will be given to financial benefits of the trust using matrix multiplication, determine how to divide Rs.50,000 among two types of bonds, if the trust fund obtains an annual total interest of Rs.1800. what are the values reflected in the question.

15. An agriculture firm possesses 100 acre cultivated land that must be cultivated in two different mode of cultivations : organic and inorganic. The yield for organic and inorganic system of cultivation is 15 quintals/acre and 20 quintals/acre respectively .using matrix method determine how to divide 100 acre land among two modes of cultivation to obtain yields of 1600 quintals.

Which mode of cultivation do you prefer most and why ?

19

DIFFERENTIABILITY

LOGARITHMIC DIFFERENTIATION :

Rules of logarithmic function

=

/ =

=n

Change of base rule =

loge = 1, log1 = 0,

PRACTICE QUESTIONS:

1. Differentiate

Solution:

20

2. If find

3. If , ,

21

4. Differentiate with respect to x:

5. If ,

22

6. If

7. find

23

8. Differentiate (

PRACTICE QUESTIONS:

Find for the following :

1. If . 2.

3. 4.

5.

6. 7. 8.

9. 10. If prove

that

11. Differentiate with respect to x:

12. If ,

24

APPLICATIONS OF DERIVATIVES

INCREASING AND DECREASING FUNCTIONS:

1. Steps for working rule :

i) Find f’(x) in factor form.

ii) Solve f’(x) = 0 and find the roots.

iii) If there are ‘n’ roots ,then divide the real number line R into (n+1 ) disjoint open intervals . iv) Find the sign of f’(x) in each of the above intervals .

v) f(x) is increasing or decreasing in the intervals when f’(x) is positive or negative respectively

Tips and Techniques :‐

1. If the coefficient of the highest power is +ve then the rightmost interval in the Real Line is +ve & the other intervals from right to left get alternatively signed. The given function is increasing in +ve signed intervals and decreasing in the –ve signed intervals.

2. If the coefficient of the highest power is ‐ve then the rightmost interval in the Real Line is ‐ve & the other intervals from right to left get alternatively signed. The given function is increasing in +ve signed intervals and decreasing in the –ve signed intervals.

25

SAMPLE QUESTIONS AND SOLUTION:

1. Find the intervals in which the function f given by is

a) strictly increasing b) strictly decreasing

26

2. Find the intervals in which the function f given by ,, is strictly increasing or strictly decreasing.

SOLUTION:

3. Find the intervals in which the function f given by 2 3 36 7 is a) strictly increasing b) strictly decreasing

27

4. Find the intervals in which the function f given by is


.5. Find the intervals in which the function f given by 4 45 51 is


28

5. Show that log 1 , x> ‐1 is an increasing function throughout its domain.

29

PRACTICE QUESTIONS:

Find the intervals in which the following functions are increasing and decreasing. 1. 2 3 9 2 12 15 2. 2 3 12 2 18 15 3. 2 3 12 2 18 7 4. 3 15 2 120 3 5. 2 3 3 2 12 6 6. 2 3 9 2 12 1 7. 8. 2 2 9. 3 3 2 3 100 10. Prove that the function f given by

,

11. Prove that the function f given by ,

12. Show that the function is neither increasing nor decreasing on (0,1)

TANGENTS AND NORMALS :

1. Find the equation of the tangent to the curve , at

30

2. Find the equation of the tangent to the curve √ , which is parallel to the line


31

4. Find the points on the curve at which the slope of the tangent is equal to the y‐coordinate of the point.

32

5. Find the equations of the tangent and the normal to the curve

,

33

6. Find the equation of the normal at the point ( am2 ,am3) for the curve a y2= x3

7. Find the equation of the tangent and normal to the curve

√ ,

34

PRACTICE QUESTIONS:

1. Find the points at which the tangent to the curve Is parallel to the x axis .

2. Find the slope of the normal to the curve ,

3. Find the points on the curve at which the tangent has the equation

4. Prove that the tangents to the curve

, , Are at right angles.


6. Find the equation of the tangent and normal to the curve

7. Find the equations of the normals to the curve which is parallel to the line

8. Find the equation of the tangent to the

curve , .

9. Find the equation of the tangent to the curve , That are parallel to the line x+2y = 0

10. Find the equations of the normal at a point on the curve x2 = y which passes through The point (1 , 2). Also find the equation of the tangent

35

INTEGRATION & APPLICATION OF INTEGRALS

POINTS TO REMEMBER: A: Integration of standard functions

36

1. 2. 3. 4. 5. 6. 7.

8. ; where m –1

9. | |

10.

11.

12.

B: Integration by substitution /

| |

1. /

2. / Where

C: Integration using trigonometric identities

1.

2.

3.

4.

5. . .

6. . .

7. . .

37

Every quadratic polynomial can be expressed in one of the three forms , or by completing square method

Example: 3 7 5 3 [see that x coefficient is 1]

[Add and subtract square of (half the x coefficient)i.e. ]

It is in the form of where and √

373

4936

4936

53

376

1136

376

√116

A rational expression is called proper if degree of is smaller than degree of

If is proper and the polynomial can be expressed as product of linear/quadratic factors,

then it can be decomposed into small fractions called partial fractions.

• (All non‐repeated linear factors only)

• (Repeated but linear factors only )

• (involve non‐repeated quadratic factors)

D: Integration of special functions

1.

2.

3.

4. √

5. √

6.

7. √ √ √

8. √ √ √

E: Integration of quadratic equations

1.

Express Q. E as and use 1,2 and 3 formulae

2.

Find A, B Such that L.E= A. . +B, separate integrals and

proceed

3.

Express Q. E as and use 4,5 and 6 formulae

4.

Find A, B Such that L.E= A. . +B, separate integrals and

proceed 5. Express Q. E as and use 7,8 and 9 formulae

38

6. . . Find A, B Such that L.E= A. . +B, separate integrals and

proceed F: Integration of a rational expression If it is proper, decompose it into partial fractions and then integrate

2. If it is not proper, divide by Q(x)

And can be split into partial fractions as it is proper

G: Integration by parts 1. / ( learn ILATE rule) 2. /

H: Definite Integrals

1.

2.

3.

4.

5.

6.

7. I: Definite integral by limit Sum method

1. To find ; follow the steps mentioned below. 2. Write (constant) 3. Find

4. Substitute the in the formula ∑

5. Use values : , ∑ , ∑ , ∑ ∑

J: Application of Integrals

denote the area under the curve bounded by three lines , and y=0.

To find the area bounded by a curve and a curve/straight line, a. First find the points of intersection where the curve intersects the curve/line. b. Draw the rough sketch of the curve and curve/straight line c. Write the required area using definite integrals and then solve.

Questions for practice

1. Evaluate

2. Evaluate

3. Evaluate

4. Evaluate

39

5. Evaluate

6. Evaluate

7. Evaluate

8. Evaluate

9. Evaluate

10. Evaluate

11. Evaluate

12. Evaluate

13. Evaluate √ 14. Evaluate √

15. Evaluate

16. Evaluate

40

17. Evaluate

18. Evaluate

19. Evaluate

20. Evaluate

21. Evaluate | | | | | |

22. Evaluate

23. Evaluate

24. Evaluate ⁄

25. Evaluate

26. Evaluate √

27. Evaluate

28. Evaluate √

29. Evaluate

30. Evaluate

31. Evaluate

32. Evaluate

33. Evaluate by the method of limit of sums.

34. Evaluate by the method of limit of sums

35. Evaluate by the method of limit of sums 36. Using integration, find the area of the region bounded by the curves | |

, , , 37. Using the integration, find the area of the region bounded by the curve

and the line 38. Sketch the graph of | |and evaluate the area under the curve |

|above x‐axis and between to . 39. Using the integration, find the area of the region bounded by the curve and

40. Find the area of the circle which is interior to the parabola . 41. Using integration find the area of the triangle ABC, coordinates of whose vertices

are A(4, 1), B(6, 6) and C(8, 4). 42. Using the integration find the area of the triangular region whose sides have

equations , and 43. Using the integration find the area of the triangular region whose sides have

equations , and 44. Using the integration, find the area of the region enclosed between the two circles

and

41

HINTS/SOLUTIONS 1

Write , expand Ans: |

|2

Put sin x = t, continue as in problem 13

Ans: | |

3 Divide Nr and Dr by cos x, replace 1 by

Ans:

4 Use identities Ans:

5 Write

Ans:

6

Multiply and divide by , and put

Obtain the form , let

Obtain the form =√ √

7 Write √ use

formula D5

Ans:

8

√ Write Sol:

9 Evaluate Similar to previous problem given for practice




13 Evaluate √ Similar to previous problem given for practice

14 Evaluate √

Similar to previous problem given for practice

15 Evaluate Write = and

integrate

Ans: | | | |

16 Evaluate Write then

integrate

42

17 Evaluate (by parts)

Ans:

18 Evaluate

Take and use by parts formula Ans: √

19 Evaluate = ,

Ans:

20 Evaluate

(put

2x=t) =

Ans: +C

21 Evaluate

| | | | | |

For 4, | | ; and For 2, | | ; For 4, | |; For 4, | |

=

22 Evaluate

Use rule H4, add both integrals,.

Put cosx =t,

= 23 Evaluate Use rule H4, add both integrals,

Use rule H7, , Use rule H4 again add integrals … Ans .

24 Evaluate ⁄

Ans: =

25 Evaluate

where Sin x – Cos x = t, after substitution,it becomes

=. .

=

26 Evaluate

√

√ √

43

where Sin x – Cos x = t, after substitution, it becomes

√

√ Ans: 2 √

27 Evaluate Put x = Tan t, with that

Use rule H4,

28 Evaluate

√ Use rule H5,

√ √

√ √

Add

29 Evaluate Refer problem 27

30 Evaluate Use Rule H4, simplify to get

let

31 Evaluate

Given for practice use H4

32 Evaluate

Use rule H4, simplify to get

Apply H7,

Ans:

44

34. Evaluate by the method of limit of sums. Here , and

= ∑ ∑

∑ ∑ ∑ =

. ..

35. Evaluate by the method of limit of sums Here , and

= ∑ ∑

∑ ∑ =

..

36. Evaluate by the method of limit of sums Here , and

= ∑ ∑

∑ ∑ =

37. Using integration, find the area of the region bounded by the curves | | , ,

, Draw the lines y=x+2, y=‐x, x=‐3, x=3, y=0 Area of Shaded region:

= .

38. Using the integration, find the area of the region

bounded by the curve and the line

The curves and intersect at (‐

1, ), (2,1)

Area of Shaded region:

=

45

39. Sketch the graph of | |and evaluate the area under the curve | |above x‐axis and between to . Area of Shaded region:

= .

=9

40. Using the integration, find the area of the region bounded by the curve and

Points of intersection (0,0), (1,1)

Area of Shaded region:=

41. Find the area of the circle which is interior to the parabola .

Circle and parabola intersect at √ , , √ ,

Area of Shaded region:

= √√

√

√

42. Using integration find the area of the triangle ABC, coordinates of whose vertices are A(4, 1), B(6, 6) and C(8, 4).

Eq to AB : , Eq to BC : , Eq to AC

:

Required area =

=

=

=7 43. Using the integration find the area of the triangular region whose sides have

equations , and Given for practice do Same as above

44. Using the integration find the area of the triangular region whose sides have equations , and Given for practice do Same as above

46

45. Using the integration, find the area of the region enclosed between the two circles and

Given for practice 46. Find the particular solution of the differential equation

= 0 for x=1, y=1. Given for practice

‐‐‐ ALL THE BEST ‐‐‐

47

POINTS TO REMEMBER:

1. An equation involving derivative (derivatives) of the dependent variable with respect to independent variable (variables) is called a differential equation

2. Order of a differential equation is defined as the order of the highest order derivative of the dependent variable with respect to the independent variable involved in the given differential equation

3. The highest power (positive integral index) of the highest order derivative involved in the given differential equation is defined as the degree of the differential equation.

4. The curve y = φ (x) is called the solution curve (integral curve) of the given differential equation if the derivatives of y, satisfy the differential equation.

5. The solution which contains arbitrary constants and satisfy the given differential equation is called the general solution (primitive) of the differential equation

6. The solution of a differential equation independent from arbitrary constants is called a particular solution of the differential equation

7. To obtain differential equation when the general solution(Family of curves) is given (i). Identify the number of arbitrary constants involved in general solution, say ‘n’ (ii). Derivate the general solution for ‘n’ times let , , , …. (iii). Eliminate the arbitrary constants by using , , , …. (iv). The equation then obtained involve differentials in place of arbitrary constants,

and it is required differential equation. 8. To obtain general/particular solution when the differential equation is given

(i). (VARIABLE SEPARABLE)If the given differential equation can be expressed as

OR then it can be solved by separating the variables and

integrating both sides.

OR

(ii). (HOMOGENEOUS)If the given equation is expressed as ,, such that both

the functions and are homogeneous then; substitute and

. Then the differential equation can be solved by using method discussed in

type (i).

(iii). (LINEAR D E)If the given differential equation is in the form where

are functions of , OR where are functions of ;

then it is called linear differential equation(LDE). It can be solved by multiplying both sides of the LDE by integrating factor I.F. , and integrating both sides.

and write

.

DIFFERENTIAL EQUATIONS

48

Questions for Practice

49

45. Solve the differential equation: 1 0 46. Solve the differential equation 1 1 0, given that 0, 1 47. Find the particular solution of the differential equation 1 , given that

0 when 1 48. If is a solution of the differential equation

and 0 1, then

find the value of

49. Find the particular solution of the differential equation 1 1 0, given that 1 when 0

50. Find the general solution of the differential equation 2

51. Solve the differential equation , given that 1.

52. Solve the differential equation 1

53. Solve the differential equation 1 2 ; | | 1

54. Solve the differential equation

55. Solve the differential equation 3

56. Find the particular solution of the differential equation 3 = 0 for x=1, y=1.

57. Obtain the differential equation of all circles of radius ‘r’

58. Show that the differential equation 2 2 0 is a homogeneous. Find

the particular solution of this differential equation, given that x=0 when y=1.

HINTS/SOLUTIONS

1. given differential equation: 1 0

1 1 implies √ 0 integrating both sides

√

0implies 1√ √

0

1 1 log1 √1

2. Given differential equation 1 1 0, given that 0, 1 0implies General Solution is tan tan

0, 1implies C = Particular Solution: tan tan

3. Given differential equation 1 1 implies 1

General Solution: log 1 given that 0 when 1implies

Particular Solution: log 1

4. Given differential equation implies

log 1 log 2 log General Solution : 2 1 and 0 1 implies C = 4 , Particular solution: 2 1 4 the value of is

5. given differential equation 1 1 0,

0integrating we get General Solution log

50

given that 1 when 0 that implies log log √2 P.S. 2 1

6. Given differential equation 2 OR put

implies integrate both sides

: log| | 2√3 tan2

√3

7. Given differential equation

integrating factor is , Multiply both sides by x

integrating both sides G.S.

given that 1 C=0, PS.=

8. Given differential equation 1

integrating factor is , Multiply both sides by

integrating both sides GS .

9. Given differential equation 1 2

integrating factor is , Multiply both sides by 1

integrating both sides G.S. 1 log

10. Given differential equation (Homogeneous)

Put y = Vx… Ans G.S. =

11. Given differential equation 3 can be written as 3 (LinearDE)

Ans: General Solution 3

12. Find the particular solution of the differential equation 3 = 0 for

x=1, y=1. Given for practice

13. Obtain the differential equation of all circles of radius ‘r’ Equation to a circle with radius r is given by

………..(1) where a, b are arbitrary constants

Derivating (1) both sides with respect ‘x’ we get 2 2 . 0 ………….(2)

Derivating (2) both sides with respect ‘x’ we get 2 2 . 2 0 ………….(3)

So, ; . substituting these in eq 1, we get

. OR 1

14. Show that the differential equation 2 2 0 is a homogeneous. Find

the particular solution of this differential equation, given that x=0 when y=1. (Put x = Vy )

51

52

===================================================================

53

======================================================================

Hint: ̂ 3 ̂ 2 and 3 ̂ 3

: . 3 6 9 and 9√2

d = √

√

Hint: Convert in to Vector form and proceed.

The vector form of the lines are = ̂ ̂ 7 ̂ 6 ̂ )

= 3 ̂ 5 ̂ 7 ̂ 2 ̂ )

54

============================================================================

=============================================================================

===============================================================================

1. Find the distance between two parallel lines

Hint : 2 ̂ ̂ , ( 9 ̂ 14 ̂ 4

= √293 , 7 √

55

STRATEGIC ACTION PLAN FOR SLOW LEARNERS

LINEAR PROGRAMMING

SOME IMPORTANT RESULTS/CONCEPTS

** Solving linear programming problem using Corner Point Method. The method comprises of the following steps:

1. Find the feasible region of the linear programming problem and determine its corner points

2. Evaluate the objective function Z = ax + by at each corner point. Let M and m, respectively denote the largest and smallest values of these points.

3. (i) When the feasible region is bounded, M and m are the maximum and minimum values of Z.

(ii) In case, the feasible region is unbounded, we have:

4.(a) M is the maximum value of Z, if the open half plane determined by

ax + by >M has no point in common with the feasible region. Otherwise, Z has no maximum value.

(b) Similarly, m is the minimum value of Z, if the open half plane determined by ax + by < m has no point in common with the feasible region. Otherwise, Z has no minimum value.

56

SOLVED PROBLEMS:

1) A Shopkeeper sells only tables and chairs. He has only Rs 6,000 to invest and has a space for at most 20 items. A Table costs him Rs 400 and a chairs costs him Rs 250. He can sell a table at a profit of Rs 40 and a chair of Rs 30. Supposing he can sell whatever he buys, formulate the problem as a LPP and solve it graphically for maximum profit.

Sol: Let x tables and y chairs are bought Y

Then LPP is 25

To maximize Z = 40x + 30y 20

Subject to constraints, C(0, 20) B(3

20,

340

)

X+y ≤ 20

400x + 250y ≤ 6000 X 0 A(15, 0), =>8x +5y ≤ 120,

X ≥ 0, y ≥ 0

Possible points for maximum Z are A(15, 0), B(3

20,

340

), C(0, 20)

POINT Z = 40x + 30y VALUE A(15, 0) 600 600

B(3

20,

340

) 3

12003

800+

666.66[MAXIMUM VALUE]

C(0, 20) 0 + 600 600

Z is maximum for B(3

20,

340

) , ie., 6 tables and 13 chairs must be purchased and sold for a

maximum profit of Rs 666.6

57

1) One kind of cake requires 200 g of flour and 25 g of fat and another kind of cake requires 100 g of flour and 50 g of fat. Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat. Assuming there is no shortage of other ingredients used in making the cake.

Sol: Let x cakes of type 1 and y cakes of type 2 are made. Y

Then LPP is 50

To maximize Z = x + y

Subject to constraints, C(0,20)

X ≥ 0, y≥ 0 20 B(20,10)

200x + 100y ≤ 5000 => 2x + y ≤ 50 0 A(25,0) X And 25x + 50y ≤ 1000 =>x + 2y ≤ 40 25 40

POINT Z = x+ y VALUE A(25, 0) 25 + 0 25 B(20, 10) 20 + 10 30 maximum C(0, 20) 0 +20 20

Z is maximum at B (20, 10)

Hence, 20 cakes of type 1 and 10 cakes of type 2 must be made for maximum number of 30 cakes.

2) A Manufacturing company makes two models A and B of a product. Each piece of

model A requires 9 labour hours for fabricating and 1 labour hour for finishing. Each piece of model B requires 12 labour hours for fabricating and 3 labour hour for finishing. For Fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of Rs 8000 on each piece of model A and Rs 12000 on each piece of Model B. How many pieces of model A and Model B should be manufactured per week to realise a maximum profit ? What is the maximum profit per week ?

Sol: Suppose x is the number of pieces of model A and y is the number of pieces of Model B. Then Total profit (in Rs ) = 8000x + 12000y Y Maximize Z = 8000x + 12000y Subject to the constraints: 9x + 12y ≤ 180 C(0,10) B(12,6) => 3x + 4y ≤ 60 X + 3y ≤ 30 0 A(20,0) X X ≥ 0, y ≥ 0

58

CORNER POINT Z = 8000x + 12000y 0(0, 0) 0 A(20, 0) 160000 B(12, 6) 168000 C(0, 10) 120000 We find that maximum value of Z is 162000 at B(12,6). Hence, the company should produce 12 pieces of Model A and 6 pieces of Model B to realize maximum profit and maximum profit then will be Rs 168000.

3) A Dealer in rural area wishes to purchase a number of sewing machines. He has only Rs 5760 to invest and has space for atmost 20 items for storage.An electronic sewing machine cost him rs 360 and a manually operated sewing machine rs 240.He can sell an electronic sewing machine at a profit of Rs 22 and a manually operated sewing machine at a profit of Rs 18. Assuming that he can sell all the items that he can buy, how should he invest his money in order to maximize his profot? Make it as LPP and solve it graphically.

Sol: Let dealer purchased x electronic sewing machines and y manually operated sewing machines.

Our problem is to maximize, Z = 22x + 18y ‐ (i) Subject to constraints x + y ≤ 20 ‐ (ii) 360x + 240y ≤ 5760 or 3x + 2y ≤ 48 ‐ (iii) X ≥ o, y ≥ 0 ‐ (IV) On solving Equations we get x = 8 and y = 12

So, the point of intersection of the lines is B(8,12). Graphical representation of the lines is given below Y

C(0,20)

3X + 2Y = 48

B(8, 12)

X + Y = 20

A(16,0)

X

0

59

:‐ Feasible region is OABCA

The corner points of the feasible region are O(0,0), A(16,0), B(8,12), C(0,20).

The value of Z at these points is as follows

CORNER POINTS Z = 22x + 18y 0(0,0) Z = 22(0) + 18(0) = 0 A(16,0) Z = 22 x 16 + 0 = 352 B(8,12) Z = 22 x 8 + 18 x 12 = 392 C(0,20) Z = 22 x 0 + 18 x 20 =360

The maximum value of Z= Rs 392 at point B(8,12).

Hence, dealer should purchased 8 electronic and 12 manually operated sewing machines to get maximum profit.

60

PRACTICE PROBLEMS

1)An aeroplane can carry a maximum of 200 passengers.A profit of Rs 1000 is made on each executive class ticket and a profit of Rs 600 is made on each economy class ticket.The airlines reserves at least 20 seats for executive class.However, atleast 4 times as many passengers prefer to travel by economy class than by the executive class.Determine how many tickets of each type must be sold inorder to maximize the profit for the airline.What is the maximum profit?

2)There are two types of fertilizers,F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil condition a farmer finds that she needs atleast 14 kg of nitrogen and 14kg of phosphoric acid for her crop. If F1 cost Rs 6/kg and F2 cost Rs5/kg, determine how much of each type of fertilizer should be used, so that nutrient requirements are met at a minimum cost?

3)A manufacturer considers that men and women workers are equally efficient and so he pays them at the same rate. He has 30 and 17 units of workers (male and female) and capital respectively, which he uses to produces two types of goods A and B. To produce one unit of A, two workers and three units of capital are required while three and one unit of capital is required to produce one unit of B. If A and B are priced at Rs.100 and 120 per unit respectively, how should he uses his resources to maximize the total revenue? From the above as an LPP and solve graphically. Do you agree with this view of the manufacturer that men and women workers are equally efficient and so should be paid at the same rate.

4)A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1,400 calories. Two foods X and Y are available at a cost of Rs. 4 and Rs. 3 per unit respectively. One unit of the food X contains 200 units of vitamins, 1 unit of mineral and 40 calories, whereas one unit of food Y contains 100 units of vitamins, 2 units of minerals and 40 calories. Find what combination of X and Y should be used to have least cost? Also find the least cost.

61

PROBABILITY

BAYE’S THEOREM

If E1, E2 ,..., En are n non empty events which constitute a partition of sample space S, i.e.

E1, E2 ,..., En are pairwise disjoint and E1∪ E2∪ ... ∪ En = S andA is any event of nonzero probability, then

P(Ei|A) = Ej)P(A/ P(Ej)

(A/Ei) (Ei).P Pn

1j∑=

for any i = 1, 2, 3, ..., n

Problems with solutions

Q1.Bag 1 contains 3 red and 4 black balls and Bag 2 contains 4 red and 5 black balls. One Ball is transferred from Bag 1 to Bag 2 and then two balls are drawn at random (without replacement) from Bag 2. The balls so drawn are found to be both red in color. Find the probability that the transferred ball is red. ANS: ‐ Total No. of balls in 1st bag = 3+4= 7

And total No. of balls in 2nd bag = 4+5= 9

Let, E1: transferred ball is red E2: transferred ball is black. A: Getting both red from 2nd bag (after transfer)

P (E1) = 73 and P (E2) =

74

P (A/E1) =P (getting both red balls from 2nd bag, when transfer ball is red)

= 5C2/10C2=

4510

=92

P (A/E2) =P (getting both red balls from 2nd bag, when transfer ball is black)

= 4C2/10C2=

456

=152

Therefore, by Baye’s theorem

P (E1/A) = (A/E2) .P (E2) P(A/E1) (E1).P P (A/E1) (E1).P P

+

=95

152.

74

92.

73

92.

73

=+

62

Q2.Given three identical boxes 1,2and 3, each containing two coins. In box 1, both coins are gold coins, in box 2, both are silver coins and in box 3, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is probability tha5t the other coin in the box is also of gold?

Solution: let E1,E2,and E3, be the events that boxes 1,2and 3 are chosen respectively.

Then, P (E1) = P (E2) = P (E3) = 31

Also, let A be the event that ‘the coin drawn is of gold’

Thus, P (A/E1) = P (a gold coin from bag1) = 122=

P (A/E2) = P (a gold coin from bag2) = 0

P (A/E3) = P (a gold coin from bag3) = 21

Now the probability that the other coin in the box is of gold

= The probability that the gold coin is drawn from the box 1

= P (E1/A)

By Baye’s theorem, we know that

P (E1/A) = (A/E3) P (E3) P (A/E2) P (E2) P(A/E1) (E1)P P (A/E1) (E1)P P++

=32

21

310

311

31

131

=++ XXX

X

Q3.A man is known to speak truth 3 out of 4 times. He throws a die and report that it is a six. Find

the probability that it is actually a six.

Solution:P (E1) = Probability that six occurs = 61

P (E2) = Probability that six does not occurs = 65

P (A/E1) = Probability that the man reports that six occurs when six has actually occurred on the die.

63

Probability that the man speaks the truth = 43

P (A/E2) = Probability that the man reports that six occurs when six has not actually occurred on the die.

Probability that the man speaks the truth = 41

431 =−

Thus by Baye’s theorem, we get

P (E1/A) = Probability that the report of the man that six has occurred actually a six.

P (E1/A) = (A/E2) .P (E2) P(A/E1) (E1).P P (A/E1) (E1).P P

+=

83

824

81

41

65

43

61

43

61

==+

XXX

X

Q4.There are three coins. One is two headed coin (having head on both faces), another is a biased coin that comes up heads 75% of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

SOLUTION:

Let E1,E2 and E3 be the events that coins I ,II and III are chosen respectively.

Let A be the event of getting a head.

P(E1)=P(E2)=P(E3)=31

P(A/E1)=1, P(A/E2)=75%=43, P(A/E3)=

21

Required probability= p(E1/A) = E3)P(E3).P(A/ (A/E2) .P (E2) P(A/E1) (E1).P P

(A/E1) (E1).P P++

= 94

Q5. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers.

The probability of an accidents are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

SOLUTION:

Insured scooter drivers=2000 Car drivers = 4000 Truck drivers=6000 Total drivers= 12000

64

Let E1,E2 and E3 be the events that scooter driver , car driver and truck driver are selecting respectively Let A be the event of meeting with an accident

P(E1)= 61

120002000

= , P(E2)= 31

120004000

= ,P(E3)= 21

120006000

=

P(A/E1)=0.01 and P(A/E2)=0.03 and P(A/E3)=0.15

By Bayes theorem P(E1/A) = E3)P(E3).P(A/ (A/E2) .P (E2) P(A/E1) (E1).P P

(A/E1) (E1).P P++

=521

PROBABILITY DISTRIBUTION MEAN & VARIANCE OF RANDOM VARIABLE

The probability distribution of a random variable X is the system of numbers

X : x1 x2….... xn P(X): p1 p2….... pn

Where, pi > 0, ∑=

n

iPi

1

=1

Problem with Solutions

1. From a lot of 30 bulbs which include 6 defectives, a sample of 4 bulbs is drawn at random with replacement. Find the probability distribution of the number of defective bulbs.

Sol: Let X denotes the number of defective bulbs X= 0,1,2,3 or 4

Probability of getting a defective bulb=51

306=

Probability of getting a non defective bulb=54

511 =−

P(X=0)=P(no defective bulb)=p( all 4 good ones)= (54)4 =

625256

P(x=1)= 4c1 (51 ) (

54)3 =

625256

P(x=2)= 4c2 (51) 2 (

54)2 =

62596

P(x=3)= 4c3 (51) 3 (

54) =

62516

P(x=4)= (51) 4 =

6251

Probability distribution of X is X 0 1 2 3 4 P(X) 625

256

625256

62596

62516

6251

65

PRACTICE QUESTIONS

1. Suppose a girl throws a die’ if she gets a 5 or 6, she tosses a coin three times and notes the number of heads. If she gets 1,2,3 or 4 , she tosses a coin once and notes whether a head or tail is obtained. If she obtained exactly one head, what is the probability that she threw1,2,3 or 4 with the die?

2. In answering a question on amultiple choice test, a student either knows the answer or guesses. Let ¾ be the probability that he knows the answer and ¼ be the probability that he guesses. Assuming that a student who guesses the answer will be correct with probability ¼. What is the probability that a student knows the answer given that he answered it correctly?

3. Two numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the expectation of X.

4. Two cards are drawn simultaneously(or successively without replacement) from a well shuffled pack of 52 cards. Find the mean, variance and standard deviation of the number of kings.

TOPIC: RELATIONS, FUNCTIONS & INVERSE TRIGONOMETRIC FUNCTIONS

1. Show that f: R →R defined by f (x) = [ x ] is neither one‐one nor onto. 2. Find fοg for f(x) = ex ; g(x) = log x 3. Check for commutative property for the operation : * : R x R → R

defined by a * b = a + 3b

4. Find, if the binary operation, *, given by a * b = 2

ba + , in the set of real

numbers is associative. 5. Let S = {1,2,3}. Find whether the function f : S → S defined as f = {(1,3),

(3,2), (2,1)} has inverse. If yes, find f‐1.

6. For θ = ⎟⎠

⎞⎜⎝

⎛−

21sin 1 , find the value of θ

7. Find the Principal value of tan‐1 ( 3 ) 8. Evaluate: sin(cot‐1x) 9. Evaluate: cos(tan‐1

43)X

10. Express in simplest form: sin‐1[3x – 4x3] 11. Prove: [ ]22111 11coscoscos yxxyyx −−+=− −−−

12. Prove: 21

2tan2

1tan 21

21 π

=⎟⎠⎞

⎜⎝⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛ − −−

xx

xx

13. Show that the function f : R →R given by f(x) = 3x – 4 is a bijection.

14. Find fοg and gοf if f(x) = x2 + 2 and g(x) = 1 ‐ x−1

1

15. Let f:R→R defined by f(x) = 2x – 3 and g : R→R defined by g(x) = 2

3+x

Show that fοg = IR = gοf.

66

16. Let * be a binary operation on N, given by a * b = lcm (a,b) for a,b∈N. Find a) 2 * 4 b) 3 * 5 c) Is * associative.

17. Solve: )7(tan1tan11tan 111 −=⎥⎦

⎤⎢⎣⎡ −

+⎥⎦⎤

⎢⎣⎡

−+ −−−

xx

xx

18. Prove that: 2

0,24cos1cos1

cos1cos1tan 1 ππ<<+=

⎭⎬⎫

⎩⎨⎧

−−+−++− xx

xxxx

19. Let R0 denote the set of all non‐zero real numbers and let A = R0 x R0. If * is a binary operation on A defined by : (a,b) * (c,d) = (ac, bd) for all (a,b), (c,d) ∈A. a) Show that * is both commutative and associative on A. b) Find the identity element in A c) Find the invertible element in A.

20. Show that the function f: R →R given by f(x) = x3 + x is a bijection. Find the inverse.

67

MATRICES & DETERMINANTS

1. If A = ⎟⎟⎠

⎞⎜⎜⎝

⎛−100i

and B = ⎟⎟⎠

⎞⎜⎜⎝

⎛0

0i

i, show that AB ≠ BA

2. Find a matrix X, for which ⎥⎦

⎤⎢⎣

⎡1145X = ⎥

⎦

⎤⎢⎣

⎡ −3121

3. If A = ⎟⎟⎠

⎞⎜⎜⎝

⎛5432, prove that A – AT is a skew‐symmetric matrix.

4. If A = ⎥⎦

⎤⎢⎣

⎡−−

2423find ‘k’ for A2 = kA – 2 I

5. If A and B are symmetric matrices, show that AB is symmetric, if AB = BA. 6. Find the equation of the line joining (1, 2) and (3 , 6) using determinants.

7. For what value of ‘k’ the matrix ⎥⎦

⎤⎢⎣

⎡432khas no inverse.

8. For ⎥⎦

⎤⎢⎣

⎡dcba, find determinant {A(adj A)}

9. Evaluate ‘x’ if 1542=

xx

642

10. Vertices of a triangle ABC are A(1,3), B(0,0) and C(k,0). Find the value of ‘k’ such that the area of the triangle ABC is 2 square units.

11. Express the matrix A = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−−

−

133452516as a sum of symmetric and skew‐

symmetric matrices.

12. If A = ⎟⎟⎠

⎞⎜⎜⎝

⎛θθθθ

cossinsincos

ii

, then prove by principle of Mathematical induction

that An= ⎟⎟⎠

⎞⎜⎜⎝

⎛θθθθ

nninin

cossinsincos

13. If A = ⎟⎟⎠

⎞⎜⎜⎝

⎛2132, evaluate A3 – 4 A2 + A

14. If f(x) = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −

1000cossin0sincos

xxxx

, show that f(x) f(y) = f(x + y)

15. By using elementary transformations, find the inverse of A = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

113321210

68

Show that the matrix A = ⎟⎟⎠

⎞⎜⎜⎝

⎛− 21

13satisfies the equation A2 – 5A + 7I = 0.

Hence find A‐1.

16. Using properties of determinants, show that cbbaacacbacb

+++

= (a + b+ c) (a –

c)2

17. Show that cxxxbxxxaxxx

+++++++++

433221

= 0, where a, b, c are in A.P.

18. Prove that ))()((111

αγγββααβγαβγγβα −−−=

19. Without expanding prove that 3

388102445 x

xxyxxxyx

xxyx=

+++

20. If A = ⎟⎟⎠

⎞⎜⎜⎝

⎛5723and B = ⎟⎟

⎠

⎞⎜⎜⎝

⎛9876verify that (AB)‐1 = B‐1 A‐1

21. If x ≠y ≠z and ,0111

32

32

32

=+++

zzzyyyxxx

show that xyz = ‐1

22. If A = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

122212221, prove that A2 – 4A – 5I = 0. Hence find A‐1

23. Using matrix method, solve the system: x + y + z = 3 ; 2x – y + z = 2, x – 2y + 3z = 2

24. Using matrix method, solve the system: x + y – z = 1 ; 3x + y – 2z = 3 ; x – y – z = ‐1

25. Solve the system using matrices: 10332=+−

zyx; 10111

=++zyx

;

13213=+−

zyx

26. Given A = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

121232405and B‐1 =

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

431341331, compute (AB)‐1

69

27. If A = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −

310015102, prove that A‐1 = A2 – 6A + 11I

28. Show that x = 2 is a root of the equation 223332

16

+−−−−−

xxxx

x= 0 and solve it

completely.

70

INVERSE TRIGONOMETRIC FUNCTIONS

1. Evaluate: √3 2

2. Prove that : 3 sin‐1x = sin‐1(3x – 4x3)

3. Evaluate:

4. Find the Principal value of

5. Evaluate: sin(cot‐1x)

6. Express in simplest form:

7. Prove: 21

2tan2

1tan 21

21 π

=⎟⎠⎞

⎜⎝⎛−

+⎟⎟⎠

⎞⎜⎜⎝

⎛ − −−

xx

xx

8. Evaluate: √3 √3

9. Solve for x: 2 2

10. Evaluate:

11. Solve: )7(tan1tan11tan 111 −=⎥⎦

⎤⎢⎣⎡ −

+⎥⎦⎤

⎢⎣⎡

−+ −−−

xx

xx

12. Prove that: 2

0,24cos1cos1

cos1cos1tan 1 ππ<<+=

⎭⎬⎫

⎩⎨⎧

−−+−++− xx

xxxx

13. Solve for x:

14. Prove that :

15. Solve for x: tan‐1(x + 1) + tan‐1(x – 1) =

16. Prove that :

17. Solve for x :

18. Prove that : tan

19. Write in simplest form:

20. Prove that :

71

72

TOPIC : DIFFERENTIATION

Questions 01 to 10 carry 01 mark each Questions 11 to 22 carry 04 marks each Questions 23 to 29 carry 06 marks each

1. Find dxdy for y = log [x + 21 x+ ]

2. Find dxdy for y = tan‐1 ⎥⎦

⎤⎢⎣⎡− 212

xx

3. Find dxdy for y = xsec x

4. Find y’ for y = cos 21 x− 5. Find y’’ for y = x sinx

6. For y = log xx

cos1cos1

+− , show that

dxdy = cosec x

7. Find the interval at which f(x) = x3 + 3 x2 – 4 is increasing.

8. Show that f(x) = tan x – 4x is decreasing in 3π

− < x < 0

9. The cost function of a firm is given by C = 4x2 – x + 70. Find the marginal cost when x = 3.

10. The radius of a spherical bubble is increasing at the rate of 0.5 cm / sec. At what rate is the volume of the bubble increasing when its radius is 1 cm?

11. Find dxdy if x6 + y6 + 6x2 y2 = 16.

12. If y = .........+++ xxx , prove that dxdy =

121−y

13. If y = ex tan‐1x, then prove that : (1 + x2) 2

2

dxyd ‐ 2 (1 – x +x2)

dxdy + (1 – x2)y =

0 14. If y = (log x)2, then prove that x2 y’’ + x y’ = 2

15. If x = 2 cos t – cos 2t , y = 2 sint – sin 2t, find 2

2

dxyd at t =

2π

16. Differentiate: tan‐1⎥⎥⎦

⎤

⎢⎢⎣

⎡ −+xx 11 2

with respect to tan‐1x

17. Find the equations of tangents to the curve y = x3 + 2x + 6 which are perpendicular to the line x + 14 y + 4 = 0.

18. Using differentials find the approximate value of 4 82

73

19. Find the largest possible area of a right angled triangle whose hypotenuse is 5 cm long.

20. Find the local maximum and local minimum values, if any for f(x) = sinx +

cos x for 0 < x <2π

21. Find two positive numbers whose sum is 16 and sum of whose cubes is maximum.

22. Find the equation of the tangent to the curve x + 3y – 3 = 0 which is parallel to the line 4x – y – 5 = 0

23. Find all the points of local maxima and minima and the corresponding maximum and minimum values of the function: f(x) =

1052458

43 234 +−−− xxx

24. Find all the points of local maxima and minima and the corresponding

maximum and minimum values of the function: f(x) = sin x + 21 cos 2x

where 2

0 π<< x

25. Show that the rectangle of maximum perimeter which can be inscribed in a circle of radius ‘a’ is a square of side a2

26. A figure consists of a semi‐circle with a rectangle on its diameter. Given the perimeter of the figure, find its dimensions in order that the area may be maximum.

27. Find the volume of the largest cylinder that can be inscribed in a sphere of radius ‘r’ cm.

28. Show that the semi‐vertical angle of a right circular cone of given surface

area and maximum volume is ⎟⎠⎞

⎜⎝⎛−

31sin 1

29. Show that the volume of the greatest cylinder which can be inscribed in

a cone of height ‘h’ and semi‐vertical angle ‘α’ is απ 23 tan274 h

74

WEIGHTAGE TO DIFFERENT TOPICS

S.No. Name of the Topic Marks allotted

01 RELATIONS AND FUNCTIONS 10

02 ALGEBRA 13

03 CALCULUS 44

04 VECTORS & 3 – D GEOMETRY 17

05 LINEAR PROGRAMMING 06

06 PROBABILITY 10

TOTAL 100

75

INTEGRATION AND APPLICATION

1. Evaluate: ∫ + 3

2

1 xdxx

2. If ( )2

34

24 xedxbxex

ax +=+∫ , find the values of ‘a’ and ‘b’

3. Evaluate: ∫ xdxx 7cos4sin

4. Evaluate: dxx

x∫

sin

5. Evaluate: ∫ −−+ 235

xxdx

6. Evaluate: dxx

xxx∫ −

−+−1

123

7. Evaluate: dxxx

∫ +

1

021

2

8. Evaluate: dxx

x∫ +

2

02cos1

sinπ

9. Evaluate using properties of definite integrals: ∫ +

2

0 tancotcot

π

xxx

10. Evaluate using properties of definite integrals: ∫−

2

2

3 2cos

π

π

xdxx

11. Evaluate: ∫ + dxxx )1log(

12. Evaluate: ∫+− 23 xx

dx

13. Evaluate : dxxx

x∫ ++

+1

124

2

14. Evaluate : dxxxe x ]seclog[tan +∫

15. Evaluate : dxx∫ tan

16. Evaluate: ∫ xdx4cos

17. Evaluate: ∫ −− )2()1( 2 xxdx

18. Evaluate: dxx∫π

0

cos

19. Prove that: dxxafdxxfaa

∫∫ −=00

)()( . Use it to evaluate: dxxx∫ −2

0

2

76

20. Evaluate: dxxx

x∫ +

2

0 cossin

π

21. Evaluate as limit of a sum: dxxx )52(3

0

2 ++∫

22. Evaluate as limit of a sum: dxxx )(4

1

2 +∫

23. Evaluate: dxx

x∫ +

+1

021

)1log(

24. Evaluate: dxxx

x∫ ++

2

0 cossin1

π

25. Find the area of the region: {(x,y) : y2 ≤ 4x , 4x2 + 4y2 ≤ 9}

26. Find the area of the smaller region bounded by the ellipse 12

2

2

2

=+by

ax

and the line 1=+by

ax

27. Using integration find the area of the region given by:

{(x,y) : 0 ≤ y ≤ x2 , 0 ≤ y ≤x+1, 0≤ x ≤2}

28. Using integration, find the area of the triangular region whose vertices

are (1,0), (2,2) and (3,1)

29. Using integration, find the area bounded by the lines:

x + 2y = 2, y –x =1 and 2x + y = 7.

77

DIFFERENTIAL EQUATIONS

1. Solve: dxdy

+ 2y = 2x

2. Solve: dxdy

+ y cot x = sec x.

3. Solve: (1 +y2)dx + x dy = 0 given that y(1) = 1.

4. Solve: dxdy

= y2 tan2x given that y(0) = 2.

5. Solve : dxdy

+ 2

2

11

xy

−−

= 0

6. Solve: (x – 1)dxdy

= 2xy, given that y(2) = 1.

7. Solve: dxdy – y = x ex

8. Solve: (x – xy)dy = y dx.

9. Show that y = e‐x +ax +b is a solution of the differential equation: exy2 = 1

10. Show that y = A cosx – B sin x is a solution of the differential equation y2 + y = 0 11. Form the differential equation representing the family of curves y = e2x(A+Bx),where A and B are

constants. 12. Form the differential equation corresponding to y2 = a(b – x2) by eliminating ‘a’ and ‘b’. 13. Show that y = x sin3x is the solution of the differential equation: y2+9y – 6cos3x = 0 14. Solve: x2y dx – (x3 + y3)dy = 0. 15. Solve: y2 dx + (x2 – xy) dy = 0. 16. Solve: x y1 = y ‐ 22 yx +

17. Solve: dxdy

= (y/x) + tan(y/x)

18. x 21 y− dx + y 21 x− dy = 0. 19. Determine the order and degree of the equation.

032

2

22

=⎟⎟⎠

⎞⎜⎜⎝

⎛+⎟

⎠⎞

⎜⎝⎛

dtsds

dtds

, Order -2, Degree – 2

20. Solve (y + xy)dx + (x – xy2)dy = 0

78

VECTOR ALGEBRA

1. If is a unit vector and • = 15, find | | 2. Given = 3 ̂ ‐ ‐ 5 and coordinate of the terminal point are (0, 1, 3). Find the

coordinate of the initial point. 3. If , are any two vectors, give the geometrical interpretation of the relation

= 4. If the sum of the two unit vector is a unit vector, prove that the magnitude of their

difference is √3. 5. If | | = 2, = 5 and × = 8, find the value of • .

6. If , are any two unit vectors and θ is the angle between them, then show that

Sin = ½

7. If + = + , show that the points P, Q, R are collinear. 8. If = ̂ +2 ‐ and =3 + ‐ 5 find a unit vector in the direction of 9. If the position vectors of the points A and B are 2 ̂ +3 ‐ and 3 ̂ +2 + then find

the vector of magnitude 6 units in the direction of 10. If P(1, 5, 4) and Q(4, 1, ‐2), find the direction ratios and direction cosines of 11. If the angle between two vectors and of equal magnitude is 30° and their scalar

product is 2√3, find their magnitudes. 12. Find the value of λ so that the vectors 3 ̂ ‐ ‐ 5 and 2 ̂ +3 ‐λ are parallel 13. Find the value of λ so that the vectors 3 ̂ ‐ ‐ 5 and 2 ̂ +3 ‐λ are perpendicular 14. If =4 ̂ +2 ‐ and =5 ̂ +2 ‐ 3 find the angle between and 15. Find the value of λ when the scalar projection of =λ ̂ + +4 on =2 ̂ +6 + 3 is 4

units. 16. Show that the vectors =3 ̂ ‐ 2 + , = ̂ ‐ 3 + 5 and =2 ̂ + ‐ 4 form a

right angled triangle. 17. If , are any two unit vectors and θ is the angle between them, then show that

cos = ½

18. If , and are three vectors such that + + = 0 and | | = 3, = 5, | | = 7, find the angle between and .

19. Find the position vector of a point R which divided the line segment joining the points P and Q with position vectors ̂ 2 ̂ and ̂ respectively in the ration 2 : 1 i) internally ii) externally.

20. Show that the points A(2, 6, 3), B(1, 2, 7) and C(3, 10, ‐1) are collinear. 21. If the points (α, ‐1), (2, 1) and (4, 5) are collinear, find α by vector method. 22. The position vectors , , of three given points satisfy the relation 4 9 5

= 0. Prove that the three points are collinear. 23. If two vectors and are such that | | = 2, = 3 and • = 4, find 24. Three vertices of a triangle are A(0, ‐1, ‐2), B(3, 1, 4) and C(5, 7, 1). Show that it is a

right angled triangle. Also find the other two angles. 25. For the points A(1, 1, 1), B(2, 5, 0), C(3, 2, ‐3) and D(1, ‐6, ‐1) find the angle between

and . Interpret the answer.

79

26. If , and are three vectors such that + + = 0, prove that = = .

27. If , and are three vectors such that • = • , = and ≠ 0, then prove that =

28. Find the area of the parallelogram with diagonals 3 ̂ + ‐ 2 and ̂ ‐3 + 4 . 29. Define and prove that × = ( • tanθ where θ is the angle between and

30. The scalar product of the vector ̂ + + with the unit vector along the sum of

vectors 2 ̂ +4 ‐ 5 and λ ̂ +2 + 3 is equal to 1. Find the value of λ.

80

THREE DIMENSIONAL GEOMETRY

1. If the direction cosines of a line are‐ , , , then what are its direction ratios?

2. The direction cosines of two lines are: , , and , , . Find the angle between

them. 3. Find the angle between the lines whose direction ratios are < a, b, c > and < b – c , c – a , a –

b > 4. Using direction numbers show that the points A(‐2, 4, 7), B(3, ‐6, ‐8) and C(1, ‐2 , ‐2) are

collinear. 5. A line makes an angle of with each of X – axis and Y‐ axis. What angle does it make with Z

– axis. 6. A line in the XY‐ plane makes an angle of 30° with the x – axis. Find the direction cosines of

the line. 7. If a line makes angles α , β , γ with the coordinate axes, prove that sin2α + sin2β +sin2γ = 2. 8. Find the vector equation of the line passing through the point 2 ̂ ‐ + and parallel to

the line joining the points ̂ +4 + 5 and ̂ +2 +2 . Also find the Cartesian equation of the line.

9. The Cartesian equation of a line are 2x ‐3 = 3y + 1 = 5 – 6z. Find the direction ratio of the line and write down the vector equation of the line through (7, ‐5, 0) which is parallel to the given line.

10. If the equation of a line AB is √

. Find the direction cosines of a line

parallel to AB. 11. Show the lines x = ‐ y = 2z and x + 2 = 2y – 1 = ‐ z + 1 are perpendicular to each other. 12. Find the equation of the plane passing through the point (2, 4, 6) and making equal

intercepts on axes. 13. Find the equation of the plane passing through the point (‐1, 0, 7) and parallel to the

plane 3x – 5y + 4z = 11. 14. Find the distance between the planes •( ̂ + 2 + 3 ) + 7 = 0 and •( 2 + 4 + 6 ) + 7 =

0. 15. Find the distance of the point (2, 5, ‐3) from the XY – plane. 16. Find the Cartesian and vector equations of the planes through the intersection of the

planes 2x + 6y + 12 = 0 and 3x – y + 4z = 0 which are at a unit distance from the origin.

17. Find the equation of the plane passing through the line of intersection of the plane x – 2y + z =1 and 2x + y + z = 8 and parallel to the line with direction ratios <1 , 2, 1>. Also find the perpendicular distance of P(1 , 3 2) from this plane.

18. A straight line passes through the point (2, ‐1, ‐1). It is parallel to the plane 4x + y +z +2 = 0 and is perpendicular to the line . Find its equation.

19. Show that the points (1, ‐1, 1), (2, 3, 1), (1 , 2 , 3) and (0, ‐2, 3) are coplanar. Also find the equation of the plane containing them.

20. Prove that the equation of the plane making intercepts a, b and c on the coordinate axes is = 1.

21. Find the equation of the plane which is parallel to x – axis and has intercepts 5 and 7 on y – axis and z – axis respectively.

22. Find the distance of the point (1 , ‐2, 3) from the plane x – y + z = 5 measure along a line

parallel to

81

23. Find the distance of the point A(‐2, 3, ‐4) from the line measured

parallel to the plane 4x + 12y – 3z + 1 = 0. 24. Find the equation of the line passing through the point (2, 1, 3) and perpendicular to the

lines and .

25. Find the foot of the perpendicular drawn from the point P(1, 6, 3) on the line

. Also find its distance from P.

82

LINEAR PROGRAMMING

1. Solve the following linear programming problem graphically: Max Z = 60x + 15y subject to the constraints x + y ≤ 50, 3x + y ≤ 90, x , y ≥ 0.

2. A manufacturer produces two types of steel trunks. He has two machines A and B. The first type of trunk requires 3 hours on machine A and 3 hours on machine B. The second type of trunk requires 3 hours on machine A and 2 hours on machine B. Machine A and B can work almost for 18 hours and 15 hours per day respectively. He earns a profit of Rs.30 and Rs.25 per trunk of I & II type. How many trunks of each type must he make each day to make maximum profit?

3. A man has Rs.1500 for purchase of rice and wheat. A bag of rice and a bag of wheat cost of Rs.180 and Rs.120 respectively. He has a storage capacity of 10 bags only. He earns a profit of Rs.11 and Rs.9 per bag of rice and wheat respectively. Formulate an L.P.P. to maximize the profit and solve it.

4. A dealer wishes to purchase a number of fans and sewing machines. He has only Rs.57600 to invest and has space for at most 20 items. A fan costs him Rs.3600 and a sewing machine Rs.2400. His expectation is that he can sell a fan at a profit of Rs.220 and a sewing machine at a profit of Rs.180. Assuming that he can sell all the items that he can buy, how should he invest his money in order to maximize his profit? Translate this problem mathematically and then solve it.

5. A farmer has a supply of chemical fertilizer of type A which contains 10% nitrogen and 5% phosphoric acid, and type B contains 6% nitrogen and 10% phosphoric acid. After testing the soil condition of the field, it was found that at least 14 kg of nitrogen and 14 kg of phosphoric acid is required for good crop. The fertilizer of type A cost Rs.5 per kg and type B costs Rs.3 per kg. How many kgs of each type of the fertilizer should be used to meet the requirement of the minimum possible cost? Using LPP solve the above problem graphically.

6. A factory owner purchases two types of machines, A and B. for his factory. The requirement and limitations for the machines are as follows:

Machine Area occupied by the machine

Labour force for each machine

Daily output in units

A 1000 sq m 12 men 90 B 1200 sq m 8 men 40

He has an area of 9000 sq m available and 72 skilled men who can operate the machines. How many machines of each type should he buy to maximize the daily output?

7. An oil company requires 13000, 20000 and 15000 barrels of high grade, medium grade and low grade oil respectively. Refinery A produces 100, 300 and 200 barrels per day of high, medium and low grade oil respectively whereas refinery B produces 200, 400 and 100 barrels per day respectively. If A cots Rs.400 per day and B costs Rs.300 per day to operate, how many days should each should be run to minimize the cost of requirement.

8. An aero plane can carry a maximum of 200 passengers. A profit of Rs.400 is made on first class ticket and a profit of Rs.300 is made on each second class ticket. The airlines reserves at least 20 seats for first class. However at least four times as many passengers prefer to travel by second class then by first class. Determine how many tickets of each type must be sold to maximize profit for the airlines. Form an LPP and solve it graphically.

9. A housewife wishes to mix together two kinds of food F1 and F2 in such a way that the mixture contains at least 10 units of Vitamin A, 12 units of vitamin B and 8 units of vitamin C. The vitamin contents of one kg of food F1 and F2 are as follows:

Vitamin A Vitamin B Vitamin C Food F1 1 2 3 Food F2 2 2 1

One kg of F1 cost Rs.6 and one kg of food F2 cost Rs.10. Formulate the above problem as a LPP and use corner point method to find the least cost of the mixture which ill product the diet.

10. A diet for a sick person must contain at least 4000 units of vitamins, 50 units of minerals and 1400 units of calories. The food A and B are available at a cost of Rs.4 and Rs.3 per unit respectively. If one unit of A contains 200 units of vitamin, 1 unit of mineral and 40 units of calories, and one unit of food B contains 100 units of vitamin, 2 units of minerals and 40 units of calories, find what combination of foods should be used to have the least cost?

83

11. A dietician wishes to mix two types of foods in such a way that the vitamin content of the mixture contains at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C while food II contain 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs.5 per kg to product food I and Rs.7 per kg to product food II. Find the minimum cost of such a mixture. Formulate the above LPP mathematically and solve it.

12. A brick manufacturer has two depots A and B which stock of 30,000 and 20,000 bricks respectively. He receives orders from three builders P,Q and R for 15000, 20000 and 15000 bricks respectively. The cost of transporting 1000 bricks in the builders from the depots in Rupees is given below:

To From

Transportation cost per 1000 bricks (in Rs.) P Q R

A 40 20 20 B 20 60 40

How should the manufacturer fulfill the orders so as to keep the cost of transportation minimum? Formulate LPP.

13. A company has two factories located at P and Q and has three depots situated at A,B and C. The weekly requirement of the depots at A, B and C is respectively 5 , 5 and 4 units, while the production capacity of the factories P and Q are respectively 8 and 6 units. The cost of transportation per unit is given below:

To From

Transportation cost per 1000 bricks (in Rs.) A B C

A 160 100 150 B 100 120 100

How many units should be transported from each factory to each depot in order that the transportation cost is minimum?

Q.No. LPP Answer Q.No. LPP Answer01 (30 , 0) 08 Max Z = 400x + 300 y

x +y ≤ 200; x ≥ 20, y ≥ 4x (40 , 160)

02 Max z = 30 x + 25 y 3x + 3y ≤ 18, 3x + 2y ≤ 15

(3, 3) 09 Min Z = 6x + 10yx + 2y ≥ 10; 2x + 2y ≥ 12; 3x + y ≥ 8

(2,4)

03 Max z = 11 x + 9y 180x + 120 y ≤ 1500 x + y ≤ 10

(5 , 5) 10 Min Z = 4x + 3y200x + 100 y ≥ 4000; 2x + y ≥ 40 40x + 40 y ≥ 1400

(5 , 30)

04 Max Z = 220x + 180 y 3600 x + 2400 y ≤ 57600 x + y ≤ 20

(8, 12) 11 Min z = 5x + 7y2x + y ≥8 ; x + 2y ≥ 10

(2, 4)

05 Min Z = 5x + 3y

+ ≥ 14 ; + ≥ 14

(80, 100) 12 Min z = 40x ‐20y+1500 X + y ≤ 30; x ≤15, y ≤20. x+y≥15

06 Max z = 60x + 40 y 1000 x + 1200 y ≤ 9000 12x + 8y ≤ 72

94 ,

458

13 Min Z = 10(x – 7y + 190) x + y ≥4 ; x + y ≤8, x ≤5 ; y ≤5

(0,5)

07 Min z = 400x + 300 y 100x + 200 y ≥ 13000 300x + 400 y ≥ 20000 200x + 100 y ≥ 15000

1703 ,

1103

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