Q.1. Show that f : N N given by

Transcription

1 Q.1. Show that f : N N given by xx + 1, iiii xx iiii oooooo ff(xx) = is both one-one and onto. xx 1, iiii xx iiii eeeeeeee (CBSE-2012) Q.2 Consider the binary operations * :R R R and o : R R R defined as. a* b = a b and aob=a for all a, b R. Show that * is commutative but not associative, o is associative but not commutative (CBSE-2012) Q.3 Let AA = NN NN and * be the binary operation on AA defined by (aa, bb) (cc, dd) = (aa + cc, bb + dd). Show that * is commutative and associative. Find the identity element for * on A, if any. (CBSE-2010) Q.4 Consider ff:r + [ 5, )R given by ff(xx) = xx 2 + 6xx 5. Show that ff is invertible with ff 1 (yy) = yy+6 1. (CBSE-2015) Q.5 Using elementary transformations, find the inverse of the following matrix: (CBSE-2008) Q.6. A school wants to awards its students for the values of Honesty, Regularity and Hard work with a total cash award of Rs Three times the award money for Hard work added to that given for Honesty amounts to Rs The award money given for Honesty and Hard work together is double the one given for Regularity. Represent the above situation algebraically and find the award money for each value, using matrix method. Apart from these values, suggest one more value which the school must include for awards. (CBSE-2013)

2 Q.7 Two schools A and B wants to award their selected students on the values of sincerity, truthfulness and helpfulness. The school A wants to award Rs. xx each, Rs. yy each and Rs. zz each for three respective values to 3, 2 and 1 students respectively with a total award money of Rs. 00. School B wants to spend Rs to award its 4, 1 and 3 students on the respective values (by giving the same award money to the three values as before). If the total amount for one prize on each value is Rs. 1200, using matrices, find the award money for each value. Apart from these three values, suggest one more value which should be considered for award. (CBSE-2014) Q.8 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 books and postal 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 postal sheets. School B sold paper bags, 15 scrap books and 28 postal sheets. School A sold 26 paper bags, 18 scrap books and 36 postal sheets. Using matrices, find the total amount raised by each school. By such exhibition which value are generated in the students Q.9. If AA = 3 2 4, find AA 1. Using AA 1 solve the system of equations xx 3yy + 5zz = (CBSE-2015) 3xx + 2yy 4zz = 5 xx + yy 2zz = 3. (CBSE-2015) Q. 10 Using the properties of determinant, prove the following: 1 xx xx + 1 2xx xx(xx 1) xx(xx + 1) = 6xx 2 (1 xx 2 ) 3xx(1 xx) xx(xx 1)(xx 2) xx(xx + 1)(xx 1) (CBSE-2015)

3 Q. Show that the altitude of the right circular cone of maximum volume that can be inscribed in a sphere of radius RR iiii 4RR 3. AAAAAAAA show that the maximum volume of cone is 8 27 of the volume of the sphere. (CBSE-2014) Q.12 If the function f(x) = 2xx 3 9mmxx mm 2 xx + 1, where m > 0 attains its maximum and minimum at p and q respectively such that pp 2 = qq, then find the value of m. (CBSE-2015) Q.13 Show that the right circular cone of least curved surface and given volume has an altitude equal to 2 times the radius of the base. (CBSE-20) Q.14 Prove that the radius of the right circular cylinder of greatest curved surface area which can be inscribed in a given cone is half of that of the cone. (CBSE-2012) Q.15 An open box with a square base is to be made out of a given quantity of cardboard of area CC 2 square units. Show that the maximum Q16. Define a binary operation *on the set {0, 1, 2, 3, 4, 5} as aa + bb, aa + bb < 6 aa bb = aa + bb 6 aa + bb 6 volume of the box is CC3 cubic units. 6 3 (CBSE-2012) Show that zero is the identity for this operation and each element a 0 of the set is invertible with 6 a being the inverse of a. (CBSE-20) Q 17 (CBSE-2017)

4 Q18. A window is in the form of rectangle surmounted by a semicircular opening. The total perimeter of the window is 10 m. Find the dimensions of the window to admit maximum light through the whole opening. (CBSE- 2017) Q.19 Show that height of the cylinder of greatest volume which can be inscribed in a right circular cone of height h and semi vertical angle α is one-third that of the cone and the greatest volume of cylinder is tan 2 α. (CBSE -2010,2016)

5 Ans 1: :

6 Ans 2.

7 Ans 3:

8 Ans 4: Ans 5

9

10 Ans 6. x+y+z=6000 x+3z=000 x-2y+z= A= B= xx X= yy zz A -1 = 1/ So x=500,y=2000,z=3500 Ans 7 3x+2y+z=00 4x+y+3z=3100 X+y+z= A=

11 00 B= xx X= yy zz A -1 = 1/ So x=300,y=400,z=500 Ans 8 as same 7 Ans 9 First we have to find A -1 then find the solution using X = A -1 B. Ans 10. Take common x from R 2 & x(x-1) from R 3 1 xx xx + 1 X 2 (x-1) 2 (xx 1) (xx + 1) (1 xx) (xx 2) (xx + 1) Take (x+1) common from C 3 1 xx 1 X 2 (x-1)(x+1) 2 (xx 1) 1 (1 xx) (xx 2) 1 R 2 =R 2 -R 1 & R 3= R 3 -R 1 Expand C3 to get answer

12 Ans

13

14 Ans 12 f(x) = 2xx 3 9mmxx mm 2 xx + 1 f (x) = 6x 2-18mx+12m 2 for max and min value 6x 2-18mx+12m 2 =0 (x-2m)(x-m)=0 X=2m, x=m F (x) = 12x-18m F (x)at x=2m =6m>0 so min value at p=2m and q=m A/q p 2 =q m=2,0

15 Ans 13

16 Ans 14

17

18 Ans 16 *

19 From the above composition table it is clear that 0 is identity of the operation. And a -1 =6-a for all a. Ans 17. Ans 18 Let x and y be the length and breadth of the rectangular window.

20 Radius of the semicircular opening It is given that the perimeter of the window is 10 m. Area of the window (A) is given by, Thus, when Therefore, by second derivative test, the area is the maximum when length.

21 Hence, the required dimensions of the window to admit maximum light is given by Ans 19. The given right circular cone of fixed height (h) and semi-vertical angle (α) can be drawn as: Here, a cylinder of radius R and height H is inscribed in the cone. Then, GAO = α, OG = r, OA = h, OE = R, and CE = H. We have, r = h tan α Now, since ΔAOG is similar to ΔCEG, we have: Now, the volume (V) of the cylinder is given by,

22 And, for, we have: By second derivative test, the volume of the cylinder is the greatest when Thus, the height of the cylinder is one-third the height of the cone when the volume of the cylinder is the greatest. Now, the maximum volume of the cylinder can be obtained as: Hence, the given result is proved.

23 1. Evaluate: 2. Evaluate:. 3. Evaluate: dx +tttttttt dx tttttttt dddd (xx 44)(xx 55) ππ dddd 00 +cccccc xx 4. Evaluate: xxxxxxxxxx 5. Evaluate: llllll ssssss xx dddd 6. Evaluate: ππ 00 xx (xx +)(xx +44) dx 7. Evaluate:. xx xx( ) dddd 8. Find xx + xx 44 + dddd 9. Evaluate: (xx 44) (xx ) 33 eexx dddd cccccc xx 10. Find dddd cccccc xx (+cccccc xx). Evaluate: dddd cccccc 44 xx+ssiiii 44 xx 12. Evaluate: INTEGRATION ssssss 44 xx+ssssss xx cccccc xx+cccccc 44 xx dddd 13. Evaluate: cccccc xx + tttttt xx dddd 14. Evaluate: llllll(+xx) dddd 00 +xx Find (xx + ee + ) dddd as the limit of a sum 16. Evaluate : ππ xx dddd 00 aa cccccc xx + bb ssssss xx ππ Evaluate: ( llllll ssssss xx llllll ssssss xx) dddd 18. Evaluate : xx + llllll xx + llllll xx dddd xx ππ xx ssssss xx cccccc xx 00 ssssss 44 xx +cccccc 44 xx ππ (+ssssss xx) ππ +cccccc xx 00 +eessssss xx Evaluate: [ xx + xx + xx 33 ] dddd 20. Evaluate: 21. Evaluate. Evaluate ddxx dddd, dddd 23. Evaluate: cccccc ( xx + xx ) dddd

24 ππ 24. Evaluate: cccccc xx dddd ssssss xx ππ ssssss xx+cccccc xx 25. Evaluate: dddd 33 +ssssss xx ππ ssssss 26. Evaluate: xx dddd 00 ssssss xx+cccccc xx xx 27. Evaluate: dddd xx 44 +xx 28. Evaluate: ssssss66 xx+cccccc 66 xx dddd ssssss xx. cccccc xx xx+ssssss xx 29. Evaluate : ddxx +cccccc xx xx ssssss xx 30. Evaluate : ddxx cccccc xx APPLICATION OF INTEGRALS 31. Find the area of the region included between the parabolas y 2 =4ax and x 2 =4ay, where a>0 CBSE DELHI 2008, Find the area of the region {(x,y):x 2 y xx. CBSE 2013( AI) 33. Find the area of the smaller region bounded by the ellipse =1 aa bb xx and the straight line + yy =1 aa bb 34. Using integration, find the area of ABC, where vertices are A(2,0), B(4,5) and C(6,3). CBSE 2009, Using integration, find the area of the region enclosed between the circles xx + yy = 44 and (xx ) + yy = 44 CBSE DELHI Find the area of the region bounded by the curve x 2 = 4y and x=4y-2 CBSE(AI) Find the area of the region {(xx, yy): 00 yy (xx + ), 00 yy (xx + ), 00 xx } Sketch the graph of y= xx + 33 and evaluate xx + 33 dddd. NCERT 66 EXAMPLER 39. Using integration, find the area of the region bounded by the curve xx + yy =1. NCERT EXAMPLER 40. Find the area bounded by the curve y=ssssss xx between x=0 and x=2ππ DIFFERENTIAL EQUATIONS 41..Solve the following differential equation: ( ) x y dy x y dx + = 0. xx + yy

25 xx xx 42. Show that the differential equation eeyydddd + yy eeyy dddd = 00 is homogeneous.find the particular solution of the differential equation, given that x=0 when y= Solve the differential equation (tttttt yy xx)dddd = ( + yy )dddd 44. Find the particular solution of the differential equation dddd = xxxx dddd xx +yy,given that y=1 when x= What is the solution of the equation? dy tan y = tan y sin y 2 dx x x THREE DIMENSIONAL SPACE 46. Find the coordinates of the foot of perpendicular and the perpendicular distance of the point P(3,2,1) from the plane 2x y + z + 1 = 0. Find also, the image of the point in the plane 47. Find the coordinates of the point where the line through the points A(3,4,1) and B(5,1,6) crosses the plane determined by the points P(2,1,2), Q(3,1,0) and R(4,-2,1). 48. If lines xx = yy+ 33 =zz 44 xx 33 and =yy kk =zz intersect, then find the value of k and hence find the equation of the plane containing these lines. 49. Find the equation of the plane through the line of intersection of the planes 50. x +y + z =1 and 2x + 3y +4z = 5 which is perpendicular to the plane x y +z = 0.Also find the distance of the plane obtained above,from the origin LPP 51. A manufacturer produces of two types of steel trunks. He has two machines A and B. For completing, the first types of trunk requires 3 hours on machine A and 3 hours on machine B, whereas the second type of trunk requires 3hours on machine A and 2 hours on machine B. Machine A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of 30 and 25 per trunk of the first type and the second type respectively. How many trunks of each type must he make each day to make maximum profit? 52. A dealer wishes to purchase a number of fans and sewing machines. He has only to invest and has space for most 20 items. A fan costs him and a sewing machine His expectation is that he can sell a fan at a profit of.00 and a sewing machine at a profit of Assuming that he can sell all the items that he can buy, how should he invest his money in

26 order to maximize his profit? Translate this problem mathematically and solve it. 53. A manufacturer makes two types of machines, A and B for his factory. The requirements and limitations for the machines are as follows: Machine A Machine B Area occupie d by the machine 1000 sq.m 1200sq. m Labour force for each machine 12men 60 8men 40 Daily output in units 54. He has an area of 7600 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? 55. A dietician wishes to mix two types of food in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units per kg of vitamin A and 1 unit per kg of vitamin C while food II contains 1 unit per kg of vitamin A and 2 units per kg of vitamin C. It costs 5.00 per kg to purchase food I and 7.00 per kg to produce food II. Formulate the above linear programming problem to minimize the cost of such a mixture and solve it. 56. A factory makes tennis racket and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftsman s time and 24 hours of craftsman s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman s time. If the profit on a racket and on a bat is 20 and 10 respectively. Make it as an LPP to get maximum profit. 57. An oil company requires 12,000, 20,000, and 15,000 barrels of high grade, medium grade and low grade oil, respectively. Refinery A produces 100,300 and 200 barrels per day of high grade, medium grade and low grade oil respectively, while refinery B produces 200,400 and 100 barrels per day of high grade, medium grade and low grade oil respectively. If refinery A costs 400per day and refinery costs 300per day to operate, how many days should each be run to minimize costs while satisfying requirements. PROBABILITY 58. Suppose a girl throws a die. If she gets 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 threw 1, 2, 3 or 4 with die?

27 59. Two cards are drawn simultaneously for successively (without replacement) from a well shuffled pack of 52 cards. Find the mean and variance of the no. of aces. 60. An urn contains 3 white and 6 red balls. 4balls are drawn one by one with replacement from the urn. Find the probability distribution of the number of red balls drawn also find the mean and variance of the distribution. 61. Suppose 5% of men and 0.25% of women have grey hair. A grey haired person is selected at random. What is the probability that there are equal numbers of men and women? 62. From a lot of 10 bulbs which includes three defectives, a sample of 2 bulbs is drawn at random. Find the probability distribution of the defective bulbs.

28 Q.1 Evaluate: INTEGRATION dx +tttttttt Q.2.Evaluate:. Ans: dx tttttttt Q.3 Evaluate: dddd (xx 44)(xx 55) Ans: 6x+7 = A d/dx(x 2-9x+20) +B, 6x+7 =A(2x-9) +B Equating the coefficients of x and constant term, we obtain: 2A = 6 A = 3, 9A + B = 7 B = 34 6x + 7 = 3 (2x 9) + 3

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30 ππ 00 +cccccc xx Q.4 Evaluate: xxxxxxxxxx dddd Put cos x = t so that sin x dx = dt. When x = 0, t = 1 and when x = p, t = 1. Therefore, (by P 1 ) we get

31 ππ Q.5. Evaluate: llllll ssssss xx 00 dddd Q.6 Evaluate: xx (xx +)(xx +44) dx Comparing coefficients of y and constant terms on both sides, we get A + B = 1 and 4A + B = 0, which give

32 Q.7 Evaluate:. xx xx( ) dddd Ans: Q.8 Find xx + xx 44 + dddd Ans: tttttt xx + cc xx Q.9Evaluate: (xx 44) (xx ) 33 eexx dddd Ans: ee xx (xx ) + cc cccccc xx Q.10 Find dddd cccccc xx (+cccccc xx) Ans: llllll ssssss xx + tttttt xx xx + cc Q. Evaluate: dddd cccccc 44 xx+ssiiii 44 xx Ans: : tttttt tttttt xx tttttttt + Q.12 Evaluate: dddd ssssss 44 xx+ssssss xx cccccc xx+cccccc 44 Ans: : xx 33 tttttt tttttt xx + cc 33 tttttttt tttttt xx Q.13 Evaluate: cccccc xx + tttttt xx dddd Ans: tttttt + cc OR xx ssssss (ssssss xx cccccc xx) + cc Q.14 Evaluate: llllll(+xx) dddd Ans: ππ 00 +xx llllll 88 Q.15 Find (xx + ee + ) dddd as the limit of a sum. Ans: ee55 ee 33

33 Q.16 Evaluate : ππ xx dddd 00 aa cccccc xx + bb ssssss xx ππ 00 Q.17 Evaluate: ( llllll ssssss xx llllll ssssss xx) dddd Ans: ππ Ans: ππ llllllll Q.18 Evaluate : xx + llllll xx 33 + llllll xx dddd Ans: - xx xx llllll ( + ) + xx 33 cc 44 Q.19 Evaluate: [ xx + xx + xx 33 ] dddd Ans: ππ xx ssssss xx cccccc xx Q.20 Evaluate: dddd, 00 ssssss 44 xx +cccccc 44 xx Ans: ππ ππ (+ssssss xx) Q.21 Evaluate dddd ππ +cccccc xx Q. Evaluate ddxx 00 +eessssss xx Q.23 Evaluate: cccccc ( xx + xx ) dddd 00 ππ Q.24 Evaluate: cccccc xx dddd ssssss xx Ans:ππ Ans: ππ Ans: ππ 66 Ans: ππ llllll ssssss xx+cccccc xx Q.25 Evaluate: ssssss xx ππ dddd Ans: llllll ssssss Q.26 Evaluate: xx dddd 00 ssssss xx+cccccc xx ππ Ans: llllll + xx Q.27 Evaluate: dddd xx 44 +xx cc Q.28 Evaluate: ssssss66 xx+cccccc 66 xx ssssss xx. cccccc xx dddd xx+ssssss xx Q.29 Evaluate : ddxx +cccccc xx xx ssssss xx Q.30 Evaluate : ddxx cccccc xx Ans: xx tttttt + llllll xx xx+ Ans:tttttt xx cccccc xx cc Ans: xx tttttt xx + cc Ans: xx cccccc xx + cc

34 Q1.Find the area of the region included between the parabolas y 2 =4ax and x 2 =4ay, where a>0. B X O* C D X A Soln: Y 2 =4ax (i) X 2 =4ay (ii) Putting x= yy 4444 from (i) in (ii) we get y4-64 a 3 y=0 or, y(y 3-64a 3 )=0 or,y=0 or,y=4a Thus, points of intersection of two parabolas are O(0,0) and A (4a,4a). Draw AD OX.Then,point D is (4a,0) Required area = area OCABO = (area OBADO)-( area OCADO) 4444 = yy dddd = aaaa 00 = 3333aa 33 = aa 33 aa 33 for (y 2 =4ax)- dx- sq. units Hence the required area is aa xx dx sq. units. Q2. Find the area of the region {(x,y):x 2 y xx. yyyyyy (for x 2 =4ay)

35 B A Sol. Let us Consider the equations * * E C O D X 2 =y..(i) And y= xx..(ii) xx, wwwwwwwwww 00 Also y= xx, wwwwwwww xx < 00 Each of the given equations remain unchanged when x is replaced by x. So, each of the given curve is symmetrical about y-axis. Therefore, Required area= 2(area OEAO). Solving, X 2 =y and y=x we get,x=0or x=1 and y=o or 1 Thus, pt. of intersection are O(0,0) and A(1,1) AD OX and BC OX Required Area = 2(area OEAO) = 2(area ODAO) - (area ODAEO) =2[ yy dddd 00 =2[ xx dddd 00 =(2x1/6) =1/3 sq.units. for (line OA)- yyyyyy 00 - xx dddd 00 ] (for the curve OEA)]

36 Q3. Find the area of the smaller region bounded by the ellipse the straight line xx aa + yy bb =1 * xx aa + yy bb =1 and Soln. Both ellipse and the straight line passes through (a,0) and (0.b).We have to find the area between them. Required area =(Area between the ellipse xx yy + =1 and the x-axis from x=0 to aa bb x=a)- (Area between the straight line xx + yy =1 and the x-axis from x=0 to x=a) aa bb aa =[ bb aa aa xx dddd 00 aa - bb(aa xx) 00 aa dddd =[ ππππππ 44 aaaa ] sq.units Q4.Using integration, find the area of ABC, where vertices are A(2,0), B(4,5) and C(6,3). * B * * C A L M

37 Soln. The eqn. of the line AB is y= 55 (x-2)..(i) BL OX and CM OX is drawn. Area of triangle ABC BC is y= -x+9 (ii) AC is y= 33 (x-2) (iii) 44 =( Area of triangle ALB)+( Area of trapblmc)-( Area of triangle AMC) = yy AAAA dddd + yy BBBB dddd - yy AAAA dddd = (xx ) dddd = =7 sq.units (99 xx) dddd (xx ) dddd Q5. Using integration, find the area of the region enclosed between the circles xx + yy = 44 and (xx ) + yy = 44. A O D** C B Soln. The radius of the two circles are 2 units. Centre of the first circle is (0,0) and second circle is (2,0). On solving the equations we get solutions as x=1, y=± 33 Therefore, the poins of intersections are A(1, 33),B(1,- 33). Bothe the circles are symmetrical with respect to x-axis. Required area=2(area AOCA)

38 ssssss }] =2(area ODAO- area CADC) =2 yy dddd 00 for circle (ii)+ yyyyyy =2 44 (xx ) dddd 00 = (xx ) 44 (xx ) =2[{ 33 for circle(i) )+ 44 xx dddd + 44 ssssss + xx xx 44 xx ssssss xx + + ssssss }-{0+2 ssssss ( )} +{2ssssss ()- 33 =2[{ 33 + ππ -2 ππ 66 +{2 ππ - 33 ππ }] 66 =2( ) sq.units Q6.Find the area of the region bounded by the curve x 2 = 4y and x=4y-2 B A * * * X L M X Soln. on solving the equation we get x=-1,y=1/4 and x=2,y=1 Thus the point of intersections are A (,, and B (, ). 44 AL OX and BM OX is drawn. Required Area = (area ALMBA)-( area AOBMLA) = (XX+) 44 dddd - XX 44 dddd = (XX+) 44 + XX dddd 44

39 = xx33 44 xx + 33 = ( + 33 ) = Sq. units. Q7.Find the area of the region {(xx, yy): 00 yy (xx + ), 00 yy (xx + ), 00 xx }. ** C E A B O D F The equations are: y xx +2 (i) y x+1 x 2 0 x..(ii) (iii)..(iv) On solving (i) and(ii) we get x=0, y=1 or x=1,y=2. CD OX and EF OX is drawn Required Area : =(Area ODCBA)+(area CDFEC) = (xx + ) dddd 00 + (xx + ) dddd

40 = xx33 + xx + xx + xx =( 33 + ) +( 4-33 ) = 66 sq.units Q8.Sketch the graph of y= xx + 33 and evaluate xx + 33 dddd. (xx + 33), wwwwwwww xx < 33. Soln. The equations are y= xx + 33 = xx + 33, wwwwwwww xx 33 When xx < 33,(x=-4,y=1),(x=-5,y=2),(x=-6,y=3) When xx 33,(x=-1,y=2),(x=-2,y=1),(x=-3,y=0) y C D ** ** ***** *** *** x X -6B A Therefore Required area = Area of region ABC +Area of region OAD 33 = xx + 33 dddd = ( xx 33)dddd 66 = xx 00 + xx + 33 dddd (xx + 33)dddd xx

41 = ( + ) + 99 = =9 Sq.units. Q9. Using integration, find the area of the region bounded by the curve xx + yy =1. Soln. The equations are: xx + yy =, wwwwwwww xx > 00, yy > 00 xx + yy =, wwwwwwww xx < 00, yy > 00 xx + yy =1. xx yy =, wwwwwwww xx < 00, yy < 00 xx yy =, wwwwwwww xx > 00, yy < 00 Y B X C * O* A X * D Required Area: =4(Area of shaded region in the fires quadrant) =4 ( xx)dddd 00 =4 xx xx 00 =4 00 =2 Sq. units. Y, in first quadrant x+y=1, y=1-x. Q10.Find the area bounded by the curve y=ssssss xx between x=0 and x=2ππ. Soln.

42 A O * B D C* Required area : =Area OABO + Area BCDB ππ = ssssss xx dddd 00 ππ = ssssss xx dddd 00 + ssssss xx dddd ππ =[ cccccc xx] 00 ππ +[cccccc xx] ππ + ( ssssss xx) dddd ππ = cccccc ππ + cccccc 00 + cccccc cccccc ππ =-(-1)+1+1-(-1) =4 sq.units 1.Solve the following differential equation: ( ) x y dy x y dx + = 0. DIFFERENTIAL EQUATIONS 2 dy x y SOL = 3 3 dx x + y Let y=vx dy dx = v + dv x dx dv v v + x = dx 1+ v dv v x = dx 1+ v 3 3 v 3 3 dv v 1+ v 1 1 x = dx = dx dv + dv = x + c dx + v v x v log v 1 x y + v = log x + c + v = log x + c + = log x + c v 2y x

43 2. What is the solution of the equation? dy dx 1 + tan y = x 1 2 x tan y sin y SOL-The given differential equation can be rewritten as dy 1 1 cot y cosec y + cosecy = 2 dx x x On putting dv dx 1 v = x dv v = cos ecyand = cosecy cot dx 1 2 x dy y dx 1 1 Here P = and Q = x 2 x I.F = e 1 pdx dx 1 x = e = x Required solution is v = dx + c = c x. x x x x + (1 2cx )sin y = 0 3.Show that the differential equation eeyydddd + yy eeyy dddd = 00 is homogeneous.find the particular solution of the differential equation, given that x=0 when y=1. xx xx Sol- eeyydddd + yy xxeeyy dddd = 00 xx dddd = xx yy eeyy xx dddd eeyy xx xx ee yy dddd = yy ee yy dddd xx dddd dddd = ee yy yy xx eeyy..(1) xx dddd dddd = ff(xx yy ) xx

44 TTTTTTTTTTTTTTTTTT gggggggggg dddddddddddddddddddddddd eeeeeeeeeeeeeeee iiii hhhhhhhhhhhhhhhhhhhhhh. Let x=vy dddd dddd = vv + yy..(ii) dddd dddd From equation (i) and (ii) xx yy ee xx yy xx eeyy ee vv ee vv = vv + yy dddd dddd = vv + yy dddd dddd ee vv ee vv vv = yy dddd dddd dddd = yy eevv dddd dddd yy = eevv dddd log y=-2ee vv +C xx llllllll = eeyy + CC (iii) Putting x=0,y=1 llllll = ee 00 + CC 0=-2+C C=2 Form equation(iii) llllllll = eeyy + xx 4.Solve the differential equation (tttttt yy xx)dddd = ( + yy )dddd 5.Find the particular solution of the differential equation dddd y=1 when x=1. 6.What is the solution of the equation? = xxxx dddd xx +yy,given that dy dx 1 + tan y = x 1 2 x tan y sin y THREE DIMENSIONAL GEOMETRY

45 1.Find the coordinates of the foot of perpendicular and the perpendicular distance of the point P(3,2,1) from the plane 2x y + z + 1 = 0. Find also, the image of the point in the plane. SOLUTION: Direction ratios of the normal to the given plane 2x y + z +1 = 0 are (2,-1,1). Let M(x,y,z) be the foot of perpendicular from the point P to the given plane. Equation of PM is xx 33 = yy = zz = λ (say).. 1 Coordinates of the point M are ( 2λ + 3, λ + 2, λ + 1).(1) But, M lies on the plane 2x y + z +1 = 0 2(2λ + 3) ( λ + 2 ) + ( λ + 1) + 1 = 0 4λ +6 +λ -2 +λ +1 =0 6λ +6 =0 λ =-1 1 P(3,2,1) From equation (1) Coordinates of the point M are (-2+3, 1+2, -1+1) =(1,3,0) Perpendicular distance PM = (1 3) + (3 2) + (0 1) M Let,P / ( αα, ββ, γγ ) be the required image of P (3,2,1) = = 6 units 1 Mid point of PP / = Point M.. P / (αα, ββ, γγ)

46 33+αα, +ββ, +γγ =(1,3,0) 33+αα =1, +ββ = 3, +γγ =0 αα =2-3 =-1,ββ = 6-2 =4,γγ =0-1 =-1 Image of the point P(3,2,1) is P / (-1,4,-1) 1 2.Find the coordinates of the point where the line through the points A(3,4,1) and B(5,1,6) crosses the plane determined by the points P(2,1,2), Q(3,1,0),and R(4,-2,1) SOLUTION: Equation of the line is xx 33 =yy =zz 55 1m Equation of the plane is xx yy zz =0 1m ( xx )(0-6) +( yy )(-4+1) +( zz )(- 3-0) =0 (-6)( ) +(- 3)( yy ) +(- 3)( zz ) = 0-6x +12 3y + 3 3z + 6 = 0-6x - 3 y - 3z +21 = 0 2x + y + z 7 = 0..(i) 1m General point on given line is (2λ+ 3, 3λ+ 4,5λ+ 1) lies on line (i) 1m 2( 2λ + 3) + ( 3λ 4) + + (5 1) 4λ+ 6 3λ+ 4+ 5λ+ 1 7= 0 λ = 0 λ = 2 3 1m

47 The point of intersection is 2. ( ) +3, -3. ( )+4,5.( ) i.e ( 55 33, 6, ) 1m 3.If lines xx = yy+ 33 =zz xx 33 and 44 =yy kk =zz intersect, then find the value of k and hence find the equation of the plane containing these lines. SOLUTION: Any point on line xx is (2 λ +1, 3 λ -1, 4 λ +1) and any point on line xx 33 is (μμ +3, 2 μμ + k, μμ ) = yy+ 33 =zz = λ. (1) 44 =yy kk =zz =μμ.. (2) If lines (1) and (2) intersect,then for some values of λ and μμ 2 λ +1 = μμ +3, 3 λ -1 = 2 μμ + k and 4 λ +1 = μμ Solving these equations we get 2 λ +1 = (4 λ +1) +3 and μμ = 4 (- 33 ) + 1 = = λ = -3 λ = λ -1 = 2 μμ + k 3( - 33 ) -1 =2 (-5) +k k -10 = - k = 10 - = 99 The lines (1) and (2) are parallel to the vectors b 1 and b 2 given by

48 b 1 = 2iˆ 3ˆj 4kˆ + + and b 2 = iˆ+ 2 ˆj+ kˆ Also line (1) passes through the point (1, - 1, 1).So, the plane containing the two lines passes through the point whose position vector is a = iˆ ˆj+ kˆ and is perpendicular to the vector N given by- N = b 1 x b 2 = iˆ ˆj kˆ = (3-8)î +(4 1) ĵ + (4 3) ˆk = -5î +3 ĵ + ˆk So, the vector equation of the plane containing the given line is ( ˆ ˆ). ˆ r a N = 0 r. ( 5 iˆ + 3 ˆj+ kˆ ) r. ( 5 iˆ + 3 ˆj+ kˆ ) = - 6 = ( iˆ ˆj+ kˆ).( 5iˆ+ 3 ˆj+ kˆ) or, ( xiˆ yj ˆ zkˆ).( 5iˆ 3 ˆj kˆ) = -6 or,5x +2y +z +6 = 0 4.Find the equation of the plane through the line of intersection of the planes x +y + z =1 and 2x + 3y +4z = 5 which is perpendicular to the plane x y +z = 0.Also find the distance of the plane obtained above,from the origin. SOLUTION: Equation of the plane through the intersection of given two planes is x+y+z -1 +λ(2x +3y + 4z -5 ) =0 1m or, (1 +2 λ)x + (1 + 3 λ) y + (1 +4 λ) z λ = 0..(i) ½ m Plane (i) is perpendicular to the plane x y +z =0, so,1(1 + 2 λ) 1 (1 +3 λ) + 1(1+4 λ) =0 1 m 3λ = -1 λ = 33 Equation of the plane is ( 1 - ) x + (1-1) y + ( 1-44 ) z = ½ m ½ m

49 I.e x z +2 =0 Distance of the above plane from the origin = = unit 1m 1m LPP 1. A manufacturer produces of two types of steel trunks. He has two machines A and B. For completing, the first types of trunk requires 3 hours on machine A and 3 hours on machine B, whereas the second type of trunk requires 3hours on machine A and 2 hours on machine B. Machine A and B can work at most for 18 hours and 15 hours per day respectively. He earns a profit of 30 and 25 per trunk of the first type and the second type respectively. How many trunks of each type must he make each day to make maximum profit? Solution: The given information can be summarized in the following tabular form: Machines Time required to products Max.Machine hour available Nut Bolt A B Profit (in ) Let the manufacturer produce x packages of nut and y packages of bolts each day. Since machine A takes one hour to produce one package of nuts and 3 hours to produce one package of bolts. Therefore, xx + 33yy Similarly, 33xx + yy Let Z denote the total profit. Then ZZ =. 55xx + yy Clearly xx 00, yy 00 Thus, Maximize ZZ =. 55xx + yy Subject to, xx + 33yy 33xx + yy And, xx 00, yy 00 To solve the LPP graphically, we first convert the inequations into equations and draw the corresponding lines. Thus the shaded region OA 2 PB 1 represents the feasible region of the given LPP.

50 The values of the objective function are given in the following table: Point(x,y) Value of the objective function Z=2.5x+y O(0,0) Z= =0 A 2 (4,0) Z= =10 P(3,3) Z= =10.50 B 1 (0,4) Z= =4 Hence Z is maximum at x=3,y=3 and maximum value is Hence, maximum profit is A dealer wishes to purchase a number of fans and sewing machines. He has only to invest and has space for most 20 items. A fan costs him and a sewing machine His expectation is that he can sell a fan at a profit of.00 and a sewing machine at a profit of 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 solve it. Solution:-Suppose the dealer buys x fans and y sewing machines. Since the dealer has space for at most 20 items. Therefore, xx + yy A fan costs 360 and a sewing machine costs 240, but the dealer has only 5760 to invests. Therefore, xx + yy The profit on a fan is and on a sewing machine is 18. Let the total profit be Z Thus, mathematical formulation of the given problem is Maximize ZZ = xx + yy Subject to, xx + yy xx + yy And xx 00, yy 00 To solve this LPP graphically, we first convert the inequations into equations and draw the corresponding lines. The feasible region of the LPP is shaded in fig. The

51 corner point of the feasible region OA 2 PB 1 are O(0, 0),A 2 (16,0), P(8,12) and B 1 (0,20). These points have been obtained by solving the corresponding interesting lines, simultaneously. The values of the objective function are given in the following table: Point(x,y) Value of the objective function Z=x+18y O(0,0) Z= =0 A 2 (16,0) Z= =352 P(8,12) Z= =392 B 1 (0,20) Z= =360 Hence Z is maximum at x=8, y=12 and maximum value is 392 Hence, maximum profit is obtained when dealer will purchase 8fans and 12 sewing machines. 3. Solve the following LPP graphically: Minimize ZZ = 3333xx + yy Subject to, xx + yy 88 xx + 44yy 55xx + 88yy xx, yy 00 Solution: To solve this LPP graphically, we first convert the inequations into equations and draw the corresponding lines. The feasible region of the LPP is shaded in fig. The corner point of the feasible region ABC are A(0,8),B(0,3), C(20/3,4/3) and. These points have been obtained by solving the corresponding interesting lines, simultaneously.

52 4. A manufacturer makes two types of machines, A and B for his factory. The requirements and limitations for the machines are as follows: Area occupied by the machine Labour force for each machine Daily output in units Machine A 1000 sq.m 12men 60 Machine B 1200sq.m 8men 40 He has an area of 7600 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? Solution: Let x machines of type A and y machines of the type B are bought to maximize the daily output. Then, the LPP is Maximize ZZ = 6666xx yy Subject to, xx + yy xx + 88yy 7777 xx, yy 00 i.e 55xx + 66yy xx + yy xx, yy 00 To solve this LPP graphically, we first convert the inequations into equations and draw the corresponding lines. The feasible region of the LPP is shaded in fig. The corner point of the feasible region OABC are O (0, 0), A (6, 0), B (4, 3) and C (0, 38/5).

53 These points have been obtained by solving the corresponding interesting lines, simultaneously. The values of the objective function are given in the following table: Point(x,y) Value of the objective function Z=60x+40y O(0,0) Z= =0 A(6,0) Z= =360 B(4,3) Z= =360 C(0,38/5) Z= (38/5)=304 Hence Z is maximum at A (6, 0), B(4, 3) Hence he should buy either 6 machines of type A and no machines of type B Or 4 machines of type A and 3 machines of type B to maximize daily output. 5. A dietician wishes to mix two types of food in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units per kg of vitamin A and 1 unit per kg of vitamin C while food II contains 1 unit per kg of vitamin A and 2 units per kg of vitamin C. It costs 5.00 per kg to purchase food I and 7.00 per kg to produce food II. Formulate the above linear programming problem to minimize the cost of such a mixture and solve it. Solution: Let the dietician mix x kg of food I with y kg of food II. Then the mathematical model of the LPP is as follows Minimize ZZ = 55xx + 77yy Subject to, xx + yy 88 xx + yy xx, yy 00

54 To solve this LPP graphically, we first convert the inequations into equations and draw the corresponding lines. The feasible region of the LPP is unbounded region which is shaded in fig. The corner point of the feasible region ABC are A(10,0), B(2,4) and C(0,8). These points have been obtained by solving the corresponding interesting lines, simultaneously. The values of the objective function are given in the following table: Point(x,y) Value of the objective function Z=5x+7y A(10,0) Z= =50 B(2,4) Z= =38 C(0,8) Z= =32 Hence Z is minimum at C(0,8) 6. A factory makes tennis racket and cricket bats. A tennis racket takes 1.5 hours of machine time and 3 hours of craftsman s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftsman s time and 24 hours of craftsman s time. In a day, the factory has the availability of not more than 42 hours of machine time and 24 hours of craftsman s time. If the profit on a racket and on a bat is 20 and 10 respectively. Make it as an LPP to get maximum profit. Solution: Let the factory makes x tennis racket and y cricket bats. Then tennis racket takes 15 hours and cricket bats 3 hours of machine time. Then,. 55xx + 33yy 4444 And tennis racket takes 3 hours and cricket bats 1 hours of craftsman s time. Then, 33xx + yy Clearly x,y 0 Let Z be the maximum profit. Then the formulated LPP is given below: Maximize ZZ = xx + yy Subject to,. 55xx + 33yy 4444

55 33xx + yy xx, yy An oil company requires 12,000, 20,000, and 15,000 barrels of high grade, medium grade and low grade oil, respectively. Refinery A produces 100,300 and 200 barrels per day of high grade, medium grade and low grade oil respectively, while refinery B produces 200,400 and 100 barrels per day of high grade, medium grade and low grade oil respectively. If refinery A costs 400per day and refinery costs 300per day to operate, how many days should each be run to minimize costs while satisfying requirements. Solution: The given data can be put in the tabular form: Refinery High grade Medium grade Low grade Costs per day A B Minimum Requirement 12,000 20,000 15,000 Suppose refinery A and B should run for x days and y days respectively to minimize the total cost. Then mathematical form of the above LPP is Minimize ZZ = xx yy Subject to, xx + yy xx yy xx + yy xx, yy 00 To solve this LPP graphically, we first convert the inequations into equations and draw the corresponding lines. The feasible region of the LPP is shaded in fig. The corner point of the feasible region OABC are A 2 (120,0),P(60,30), B 3 (0,150).These points have been obtained by solving the corresponding interesting lines, simultaneously.

56 The values of the objective function are given in the following table: Point(x,y) Value of the objective function Z=400x+300y A 2 (120,0) Z= =48000 P(60,30) Z= =33000 B 3 (0,150) Z= =45000 Hence Z is minimum at P(60,30).The feasible region is unbounded. So, we find the half-plane represented by 400x+300y Hence, the machine A should run for 60 days and the machine B run for 30 days to minimize the cost while satisfying the constraints. Q1-Suppose a girl throws a die. If she gets 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 threw 1, 2, 3 or 4 with die? Sol: 4. Let E 1 be the events that she gets 5 or 6 and E 2 be the events that she gets 1, 2, 3 or Then P(E 1 )= 66 = 33 P(E 2 )= = 33 Let A be the event that she obtained exactly one head then P(A/E 1 ) = 33 88, P(A/E 2 ) = We have to find the probability that she threw 1,2,3 or 4 with the die given that she obtained exactly one head, i.e.

57 P(E 2 /A)= PP(EE)PP( AA EE ) PP(EE)PP AA EE +PP(EE)PP( AA EE ) = = = 33 = 88 Q.2 Two cards are drawn simultaneously for successively (without replacement) from a well shuffled pack of 52 cards. Find the mean and variance of the no. of aces. Sol: Let E 1 and E 2 be the events of drawing ace in first and second draw respectively. Let X be the random variable representing the no. of aces then the value of X may be 0,1 and 2. So P(X=0)= P(EE EE ) =P(EE ) P(EE /EE )= = P(X=1)= P(EE EE oooo EE EE )=P(EE ) P(E2/EE )+P(E1) P(EE /EE)= = = 3333 P(X=2)= P(E1 E2) =P(E1) P(E2/E1)= = So the probability distribution of random variable X is shown in the following table: X=x i P(X=x i )=p i x i 2 p i x i p i x i p i x i = p i x i 2 = 3333

58 Mean(µ)= p i x i = Variance= p i x i 2 -( p i x i ) 2 = = I.F = e 1 pdx dx 1 x = e = x Required solution is v = dx + c = c x. x x x x + (1 2cx )sin y = 0