4 Marks Questions. Relation and Function (4 marks)

Transcription

1 4 Marks Questions Relation and Function (4 marks) Q.1 Q. Q.3 Q.4 Q.5 Q.6 Q.7 Inverse trigonometric functions Q.1 [CBSE 008]

2 Q. [CBSE007] Q.3 [CBSE 011] Q.4 [CBSE 014] Q.5 [CBSE 005 AI] Q.6 Q.7 [NCERT EXEM.] Q.8 [CBSE 010] Q.9 CBSE 013 Q.10 Prove that Matrices CBSE 01 F Q1. If A= , show that A-AT is a skew symmetric matrix where A T is the transpose of matrix A. ( CBSE 014)

3 Q. Let A= 1 5, B= 5 and C =. Find a matrix D such that CD AB = [CBSE 017] Q3.(i) Prove that the sum of two skew symmetric matrix is a skew symmetric matrix. (ii) Express the following matrix as a sum of symmetric and skew symmetric matrix [CBSE 006] Q4. Show that A= satisfies the equation x 6x + 17 = 0. Hence find A -1. [CBSE 007] 3 5 Q5. Let A= 4 1 3, express A as a sum of two matrices such that one is symmetric and the other is skew symmetric. [CBSE 008] Q6. Find the value of x, is [1 xx 1] 5 1 =0. [CBSE 006] 15 3 xx Q7. For the following matrices A and B, verify that (AB) =B A. 1 A= 4 B=[ 1 1] [CBSE AI 010] 3 Q8. Obtain the inverse of the following matrix of the matrix using elementary operation [CBSE AI 011] Q9. Show that the matrix A= 3 1 satisfies the equation A 4A +I=0, where I is x identity matrix and O is x zero matrix using this equation, find A -1. [CBSE AI 010] 0 1 Q10. If A= 1 3, then find the value of A 3A + I. [CBSE DELHI 007] Q11. Show that the diagonal elements of a skew symmetric matrix are all zero. [CBSE F 010] n 1 3 n 1 3 n 1 Q1. If A= then provwe that A n = 3 n 1 3 n 1 3 n 1, when n N [CBSE 015] n 1 3 n 1 3 n 1 Q13. Show that B AB is symmetric or skew symmetric according as A is symmetric or skew symmetric. [NCERT]

4 Q14. Find the matrix X so that X = 7. [CBSE 015] Continuity and Differentiability 1) For what value of k, the following function is continuous at xx = 0? 1 cccccc4xx, xx 0 ff(xx) = 8xx kk, xx = 0 ) Find the relationship between aa and bb so that the function ff defined by: ff(xx) = aaaa + 1, iiiiii 3 is continuous at xx = 3. [CBSE AI 011] bbbb + 3, iiii xx > 3 3) Show that the function ff(xx) = xx 3 xx RR, is continuous but not differentiable at xx = 3. [CBSE DELHI 013] 4) Show that the function ff(xx) = xx xx is continuous at xx = 0. 1 cccccc4xx, iiii xx < 0 xx 5) Let ff(xx) = aa, iiii xx = 0 xx, iiii xx > xx 4 Determine the value of aa so that ff(xx) is continuous 6) If yy xx = ee yy xx, then prove that = (1+llllllll) llllllll 7) If 1 xx + 1 yy = aa(xx yy), then show that. [CBSE AI 013] = 1 yy 1 xx. [CBSE F 009] 8) Find, if (xx + yy ) = xxxx 9) Find ddxx, yy = (ssssssss)xx + sin 1 xx [CBSE DELHI 009, 013] 10) If yy = xx xx, then prove that dd yy 1 ddxx yy ( ) yy = 0 [CBSE DELHI 014,16] xx Application of derivative 1. Show that the right circular cylinder of given surface and maximum volume is such that is heights is equal to the diameter of the base.. Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic cms, find the dimensions of the can which has the minimum surface area? [CBSE 014]

5 3. Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is of the volume of the sphere. [CBSE 014] 4. Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is. [CBSE ] 5. A ladder of 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of cccc/ss. how fast is its height on the wall decreasing when the foot of the ladder is 4 m away from the wall. 6. Find the intervals in which the function f given by ff(xx) = 4xx 3 6xx 7xx + 30 is (i) strictly increasing (ii) strictly decreasing. 7.. Find the intervals in which the function f given by ff(xx) = sin xx + cos xx, 0 xx ππ is (i) strictly increasing (ii) strictly decreasing. xx Find the equations of the tangent to the curve yy = at the point where it cuts the X axis. [CBSE 010] (xx )(xx 3)

6 Sol.1 Relation and Function

7 Sol.

8 Sol.3

9 Sol.4 Sol. 5

10 Q6.

11 Q 7. Inverse trigonometric Q1.

12 Sol. Sol.3

13 Sol.4 Sol.5 Sol.6

14 Sol.7

15 Sol.8

16 Sol.9 Sol.10 Matrices 1. Given, A= Therefore, AT = A-A T = = Also, (A-A T ) T = T= = = -(A-AT ) So, (A-A T ) is a skew symmetry matrix.

17 . Since A, B, C are all square matrices of order, and CD AB is well defined, D must be a square matrix of order. Let D= aa bb. Then CD- AB = 0 gives cc dd 5 bb 1 aa cc dd =0 aa + 5cc bb + 5dd Or, = 0 3aa + 8cc 3bb + 8dd aa + 5cc 3 bb + 5dd 0 Or, = 0 3aa + 8cc 43 3bb + 8dd 0 0 Comparing, a +5c -3 =0 3a+8c-43=0 b+5d=0 3b+8d=0 Solving the above equation we get a=-191, b=-110, c=77 and d=44. Therefore; D= (i). Let A and B be two skew symmetric matrices. Then, A = -A and B = -B Therefore, (A+B) = A + B = (-A)+(-B) =-(A+B). Hence, (A+B) is again a skew-symmetric (ii). Let, A = and A = P=(A+A )/ = ½ and P = ½ Hence, (A+A )/ is a symmetric matrix Now, Q=(A-A )/ = ½ and Q = -½ Hence, (A-A )/ is a skew symmetric mtrix

18 (P+Q) = ½ ½ = =A Hence, A= (A+A )/ + (A-A )/ =Symmetric matrix + skew-symmetric matrix. 4. We have, A= A =A.A= = A= = 1 and 17 I= = Hence matrix A satisfies the equation, x 6x 17=0. A -6A =-17I A -1 = ( 1/17 )(6I - A) =(1/17) =1/ A can be expressed as A=(1/) (A+A ) + ½(A-A )..(1) Where A+A and A-A are symmetric and skew symmetric matrices respectively Now A+A = =

19 A-A = = Putting the values in 1 we get A=1/ / / 0 1 5/ = 3 1 9/ / 5/ 9/ 7 5/ 3/ Q11. [1 xx 1] 5 1 = xx [1 + xx xx xx + ] =0 xx (16 + xx) + (1 + 10xx) + (4xx + xx )=0 Solving we get x=-14 and x=-. Q AB= 4 [ 1 1] =

20 1 4 3 (AB) = B A = [1 4 3] = Therefore (AB) =B A Q13. A= A=IA = A = A R1=R1-R = /3 1 0 A /3 = /3 1 0 A R=R-R1 R=R/ /3 1/ /3 = /3 1 0 A R1=R1+R

21 1 0 1/3 1/ = /9 1/9 0 A R=R/ /3 1/ /9 = /9 1/9 0 A 0 0 1/9 8/9 4/9 1 R3=R3-4R /3 1/ /9 = /9 1/9 0 A R3=9*R / /9 = /9 1/9 0 A R1=R1+1/3R / = 1 A R=R-/9*R3 3 4/3 3 Therefore A -1 = Q14. A = A.A = = A -4A +I = AA -4A =-I AA (A-1) -4A A -1 =-IA -1 A-4I=-A -1 A -1 = =0

22 Q15. A = = A -5A +6I= Q Let A be a skew symmetric matrix. Therefore the diagonal elements be are denoted by a ii a ii =- a ii a ii =0 a ii =0 (Proved) 3 n 1 3 n 1 3 n 1 Q17. P(n): A n = 3 n 1 3 n 1 3 n 1 3 n 1 3 n 1 3 n Now P(1):A 1 = P(1) is true. 3 k 1 3 k 1 3 k 1 A k = 3 k 1 3 k 1 3 k 1 3 k 1 3 k 1 3 k k 1 3 k 1 3 k 1 P(k+1) : A (k+1) = A.A k = k 1 3 k 1 3 k k 1 3 k 1 3 k 1

23 3 k 3 k 3 k = 3 k 3 k 3 k 3 k 3 k 3 k Therefore P(k+1) is true. Q18. Let A be Symmetric matrix. Then A = A (B AB) = B A (B ) =B A B=B AB Therefore B AB is symmetric matrix. Let A be skew symmetric matrix. Then A =-A. (B AB) = B A (B ) =B A B=B (-A)B= -B AB. Therefore B AB is skew symmetric matrix. Q19. X will be a X order matrix. X= xx yy aa bb Now, xx yy = 7 aa bb xx + 4yy xx + 5yy 3xx + 6yy 8 9 = 7 aa + 4bb aa + 5bb 3aa + 6bb 4 6. Equating the corresponding elements and solving, we get X= 1 0 Continuity and Differentiability Sol.1 Given function, ff(xx) = 1 cccccc4xx 8xx, xx 0 kk, xx = 0 At xx = 0, we have ff(0) = kk

24 1 cccccc4xx LHL= lim ff(xx) = lim xx 0 xx 0 8xx ssssss = lim xx xx 0 8xx = lim xx 0 ssssssxx xx 1 = 1 1 cccccc4xx RHL= lim ff(xx) = lim xx 0 + xx 0 + 8xx ssssss = lim xx xx 0 + 8xx = lim xx 0 + ssssssxx xx 1 = 1 SOL: Since ff(xx) is continuous at xx = 3 lim ff(xx) = lim xx 3 xx 3 +ff(xx) = ff(3)... (i) Now, lim ff(xx) = lim ff(3 h) xx 3 h 0 = limaa(3 h) + 1 h 0 SOL3: Here, ff(xx) = xx 3 (xx 3), xx < 3 ff(xx) = 0, xx = 3 (xx 3), xx > 3 lim xx 3 = lim h 0 3aa aah + 1 = 3aa (ii) +ff(xx) = limff(3 + h) h 0 = lim h 0 bb(3 + h) + 3 = lim h 0 3bb + bbh + 3 = 3bb (iii) From equation (i), (ii) and (iii), 3aa + 1 = 3bb + 3 Or aa bb =, which is the required relation. 3

25 For continuity: Now, lim ff(xx) = limff(3 h) = lim (3 3 h) = lim h xx 3 h 0 h 0 h 0 lim xx 3 + = 0... (i) ff(xx) = limff(3 + h) = lim(3 + h 3) = lim h 0 h 0 h 0 h = (ii) Also, ff(3) = 0... (iii) From equation (i), (ii) and (iii) Hence, ff(xx) is continuous at xx = 3. For Differentiability: lim ff(xx) = lim xx 3 +ff(xx) = ff(3) xx 3 RHD LHD ff(3+h) ff(3) (3+h 3) =lim = lim h 0 h h 0 h = lim ff(3 h) ff(3) h (3 h 3) = lim h 0 h h = lim = 1... (iv) h 0 h h = lim = 1... (v) h 0 h Equation (iv) and (v) RHD LHD at xx = 3. Therefore,ff(xx) = xx 3, xx RR is continuous but not differentiable at xx = 3. xx xx, iiii xx 0 SOL4: ff(xx) = xx xx = xx ( xx). iiii xx < 0 LHL= lim ff(xx) = lim xx 0 xx 0 3xx = 0 RHL= lim ff(xx) = lim xx 0 + xx 0 +xx = 0 ff(0) = 0 = xx iiii xx 0 3xx iiii xx < 0 lim xx 0 ff(xx) = lim xx 0 +ff(xx) = ff(0) = 0. So, ff(xx) is continuous at xx = 0. SOL 5 : For ff(xx) to be continuous at xx = 0, we must have lim xx 0 ff(xx) = lim +ff(xx) = ff(0) xx 0

26 lim ff(xx) = lim xx 0 +ff(xx) = aa... (i) Now, LHL= lim xx 0 = 4lim ssssssxx xx 0 xx xx 0 ff(xx) = lim xx 0 1 cccccc4xx xx =8... (ii) xx = lim ssssss xx = lim xx 0 xx ssssssxx xx 0 xx RHL= lim ff(xx) = lim = lim ( 16 + xx + 4) xx 0 + xx xx 44 xx xx 1111 = lim xx xx + 4 = 8... (iii) From (i), (ii) and (iii), we get aa = 8. SOL 6: given, yy xx = ee yy xx Taking log on both sides, we get log yy xx = llllllee yy xx xx = 1+llllllll Differentiating on both sides with respect to yy yy xx xx log yy = yy xx =1+llllllll 1 (1+llllllll) = llllllll (1 + llllllll) (1 + llllllll) = llllllll SOL 7: 1 xx + 1 yy = aa(xx yy) Putting xx = ssssssss αα = sin 1 xx and yy = ssssssss ββ = sin 1 yy cccccccc + cccccccc = aa(ssssssss ssssssss) cccccc αα+ββ αα ββ cccccc cccccc αα ββ αα ββ = aa = cot 1 aa αα ββ = cot 1 aa 1 (ssssssss) + 1 (ssssssss) = aa(ssssssss ssssssss) = aa.. cccccc αα+ββ αα ββ ssssss sin 1 xx sin 1 yy = cot 1 aa

27 Differentiating both sides wrt x. 1 1 xx 1 1 yy = 0 SOL 8: given (xx + yy ) = xxxx Diff w.r.t x, we get = 1 yy 1 xx (xx + yy ). xx + yy = xx + yy 4xx(xx + yy ) + 4yy(xx + yy ) = xx + yy = yy 4xx(xx + yy ) 4yy(xx + yy ) xx. SOL 9: Given yy = (ssssssss) xx + sin 1 xx yy = uu + vv, where uu = (ssssssss) xx and vv = sin 1 xx i.e = + Now, uu = (ssssssss) xx Taking log on both sides, we get log uu = xlog ssssssss Differentiating both sides wrt x. = (ssssssss)xx (xxxxxxxxxx + llllllllllllll) Also, vv = sin 1 xx Therefore, = (ssssssss)xx (xxxxxxxxxx + llllllllllllll) + SOL 10: Given, yy = xx xx Taking log on both sides, we get = 1 xx(1 xx) 1 xx(1 xx).

28 Differentiating both sides wrt x. llllllll = xx. llllllll 1 yy. = 1 + llllllll = yy(1 + llllllll) Again, differentiating both sides wrt x dd yy ddxx = yy. 1 + (1 + llllllll) xx dd yy ddxx = yy xx + 1 yy. dd yy ddxx 1 yy ( ) yy xx = 0 AAAAAAAAAAAAAAAAAAAAAA oooo dd Sol 1: Let r and h be the radius and height of the cylinder respectively. Then, the surface area (S) of the cylinder is given by, Let V be the volume of the cylinder. Then, By second derivative test, the volume is the maximum when.

29 Hence, the volume is the maximum when the height is twice the radius i.e., when the height is equal to the diameter. Sol: Let r and h be the radius and height of the cylinder respectively. Then, volume (V) of the cylinder is given by, Surface area (S) of the cylinder is given by, By second derivative test, the surface area is the minimum when the radius of the cylinder is. Hence, the required dimensions of the can which has the minimum surface area is given by radius = and height =

30 Sol 3 : Let r and h be the radius and height of the cone respectively inscribed in a sphere of radius R. Let V be the volume of the cone. Then, Height of the cone is given by, h = R + AB

31 By second derivative test, the volume of the cone is the maximum when

32 Sol 4 : Let θ be the semi-vertical angle of the cone. It is clear that Let r, h, and l be the radius, height, and the slant height of the cone respectively. The slant height of the cone is given as constant. Now, r = l sin θ and h = l cos θ The volume (V) of the cone is given by,

33 By second derivative test, the volume (V) is the maximum when. Hence, for a given slant height, the semi-vertical angle of the cone of the maximum volume is. Sol 5. : let the height of the wall at where the top of the ladder touches the wall be x and distance between the wall and foot of the ladder be y By the Pythagoras theorem AAAA = AAAA + BBBB Putting the value of x=4 5 = xx + yy..(1) 5 = 4 + yy Or yy = 5 4 or yy = 3mm differentiating equation (1) with respect to x, we get 0 = xx + yy Putting the value of xx aaaaaa yy, wwww gggggg Or xx = yy 4() = ( 3) Or = 8 cccc/ssssss 3 Hence the rate of the decreasing of its height = 8 cccc/ssssss 3 Sol 6. :

34 We have f (x) = 4x 3 6x 7x + 30 Or f (x) = 1x 1x 7 = 1(x x 6) = 1(x 3) (x + ) Therefore, f (x) = 0 gives x =, 3. The points x = and x = 3 divides the real line into Three disjoint intervals, namely, (, ), (, 3) and (3, ). In the intervals (, ) and (3, ), f (x) is positive while in the interval (, 3), f (x) is negative. Consequently, the function f is strictly increasing in the intervals (, ) and (3, ) while the function is strictly decreasing in the interval (, 3). However, f is neither increasing nor decreasing in R. Sol 7. :We have f(x) = sin x + cos x, orf (x) = cos x sin x Now f (x) = 0 gives sin x = cos x which gives that xx = ππ, 5ππ 4 4 as0 x π The points xx = ππ aaaaaa xx = 5ππ 4 4 = divide the interval [0, π] into three disjoint intervals, Namely 0, ππ, 4 ππ, 5ππ, 4 4 (5ππ, ππ] 4 f (x)>0 if x [0, ππ 4 ) (5ππ 4, ππ] or f is strictly increasing in the intervals[0, ππ 4 )and(5ππ 4, ππ] and if f (x) < 0 than f is strictly decreasing in( ππ 4, 5ππ 4 ) Sol 8. : Here yy = xx 7 (xx )(xx 3) Since the curve cut the x axis therefore y=0 so x=7(from given equation) Differentiating the given function we get Putting x=7 and y=0 we get = 1 0 so equation of tangent x-0 y-7=0 (xx )(xx 3) (xx 7)[(xx ) + (xx 3)] = [(xx )(xx 3)]

35 GROUP- 3 B (4 MARKS QUESTIONS ON INTEGRATION ONWARDS) ASSIGNMENT FOR WORKSHOP ON THE TOPIC INTEGRAL 1. Evaluate xx +11 (xx +33) 44. Evaluate xx ππ ππ [CBSE Delhi 011] [CBSE F 010] 3. Evaluate (cccccc aaaa ssssss bbbb) [ [CBSE F 013] 4. Evaluate, xx +11 [CBSE AI 01] xx Evaluate, 11 xx xx(11 ) [CBSE Delhi 01] 6. Evaluate, xx ssssss 11 xx [CBSE C 016] 7. Evaluate, ssssssss ssssssssss ssssssssss [CBSE AI 013] 8. Evaluate, ππ xx ssssss xx cccccc xx [CBSE AI 01, Delhi 017] 9. Evaluate (xx + 33) xx. [CBSE AI 016] aa 10. Evaluate (aa xx). [CBSE F 011] 11. Evaluate aa (aa+xx) ee xx. [CBSE Delhi 009] 55 44ee xx ee 1. Evaluate dx [CBSE Delhi 010] Evaluate, ssssssss+cccccccc ππ ππ 66 ssssssssss 14. Evaluate, xx xx Evaluate, ssssss66 xx+cccccc 66 xx ssssss xx cccccc xx 16. Evaluate, cccccccccc cccccccccc 17. Evaluate, cccccccc cccccccc xx (xxxxxxxxxx+cccccccc) [CBSE AI 011]. [CBSE F 011]. [CBSE Delhi 01] [CBSE F 009]. [CBSE F 013] 18. Evaluate, xx +44 [CBSE Delhi 014] xx APPLICATIONS OF INTEGRALS 1. Find the area between the curves y = x and y = x [CBSE AI 009]. Find the area enclosed between the parabola y = 4ax and the line y = mx. [CBSE Delhi 011] 3. Find the area of the region enclosed by the parabola x = y, the line y = x + and x- axis. [CBSE Delhi 013] 4. Using the method of integration find the area bounded by the curve x + y =1 ( CBSE AI 010) 5. Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A (, 0), B (4, 5) and C (6, 3) ( CBSE F 009)

36 6. Using the method of integration find the area of the region bounded by lines: x + y = 4, 3x y = 6 and x 3y + 5 = 0 (CBSE F 010) 7. Find the area of the region (CBSE AI 010) 8. Find the area of the circle x + y = 16 exterior to the parabola y = 6x. ( CBSE D 016) 9. The area bounded by the y-axis, y = cos x and y = sin x when (CBSE AI 013 ) 10. Find the area bounded by curves ( CBSE D 014) 11. Find the area of region bounded by the ellipse xx + xx = 11. ( CBSE F 015 ) DIFFERENTIAL EQUATIONS 1. Solve the following differential equation: dy x 1 + y = ; y (0) = 0 dx x + 1 x + 1 ( ) ( CBSE D 010 ).Solve the following differential equation: ( ) 3 3 x y dy x y dx + = 0. ( CBSE F 014 ) 3. Solve the following differential equation x y dx + xydx = y = ( ) 0, (1) 1 ( CBSE D 015 ) 4. Form the differential equation of the family of curves y = a e x + b e -x by eliminating a and b. ( CBSE AI 011 ) dy 5. Solve the following differential equation: x + y = x log x. ( CBSE D 013 ) dx dy 6. Solve the differential equation. + y cot x = 4x cosecx, y = 0 when dx ( CBSE AI 013 ) dy 1 7. Solve the differential equation (1 + x ) + y = tan x. ( CBSE D 009 ) dx π x =

37 8. Solve the following differential equation x dy - ydx = x + y dx. ( CBSE F 011 ) 9. cccccccc + yy = sin xx, gggggggggg tthaaaa yy = wwheeee xx = 0.( CBSE AI 014 ) dy What is the solution of the equation? + tan y = tan y sin y dx x x ( CBSE D 014 ) dy y 11. Solve the differential equation x = y x tan ( CBSE F 008 ) dx x x x 1. Find the particular solution of the differential equation ( 1+ e ) dy + ( 1+ y ) e dx = 0 When y =1; x = 0. ( CBSE F 009 ) Q1. ( CBSE D 015 ) VECTORS Q( CBSE D 011 ) Q.3. ( CBSE AI 010 ) Q.4( CBSE F 013 ) Q.5. ( CBSE AI 01 )

38 Q.6. ( CBSE B 017 ) Q.7. ( CBSE B 017 ) Q.8. ( CBSE B 017 ) Q.9. ( CBSE F 015 ) Q.10. ( CBSE D 013 ) Q.11. ( CBSE AI 010 ) Q.1. ( CBSE AI 011 ) Q.13. ( CBSE D 014 )

39 Q.14. ( CBSE B 017 ) Q.15. ( CBSE B 017 ) Q.16. ( CBSE D 009 ) Q.17. ( CBSE D 014 ) Q.18. ( CBSE F 010 ) THREE DIMENSIONAL GEOMETRY 1. Find the coordinates of the foot of perpendicular and the perpendicular distance of the point P(3,,1) from the plane x y + z + 1 = 0. Find also, the image of the point in the plane. ( CBSE F 014 ). 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(,1,), Q(3,1,0),and R(4,-,1) ( CBSE D 013 )

40 3. If lines xx 11 = yy =zz 11 and xx =yy kk =zz intersect, then find the value of k and hence find 11 the equation of the plane containing these lines. ( CBSE AI 013 ) 4. Find the equation of the plane through the line of intersection of the planes x +y + z =1 and x + 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. ( CBSE AI 014 ) LINEAR PROGRAMMING PROBLEMS 1. Solve the LPP graphically: Maximize ZZ = 60xx + 15yy, Subject to, +yy 50, 3xx + yy 90, xx, yy 0 ( CBSE B 017 ). Solve the LPP graphically: Maximize ZZ = xx + yy Subject to, xx + 3yy 15, 3xx 4yy 1, xx, yy 0 ( CBSE C 017 ) 3. Solve the LPP graphically: Minimize ZZ = 3xx + 5yy Subject to, xx + 3yy 3, xx + yy, xx, yy 0 ( CBSE A 017 ) 4. 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 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 Rs 30 and Rs 5 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? ( CBSE D 015 ) 5. A dealer wishes to purchase a number of fans and sewing machines. He has only Rs to invest and has space for most 0 items. A fan costs him Rs and a sewing machine Rs His expectation is that he can sell a fan at a profit of Rs.00 and a sewing machine at a profit of Rs 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. ( CBSE AI 013 ) 6. Solve the following LPP graphically: Minimize ZZ = 30xx + 0yy Subject to, xx + yy 8, xx + 4yy 1. 5xx + 8yy 0, xx, yy 0( CBSE G 017 ) 7. 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 1men 60 Machine B 100sq.m 8men 40 He has an area of 7600 sq.m available and 7 skilled men who can operate the machines. How many machines of each type should he buy to maximize the daily output? ( CBSE D 009 )

41 8. 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 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 units per kg of vitamin C. It costs Rs 5.00 per kg to purchase food I and Rs 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. ( CBSE AI 013 ) 9. 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 4 hours of craftsman s time. In a day, the factory has the availability of not more than 4 hours of machine time and 4 hours of craftsman s time. If the profit on a racket and on a bat is Rs 0 and Rs 10 respectively. Make it as an LPP to get maximum profit. ( CBSE D 011 ) 10. An oil company requires 1,000, 0,000, and 15,000 barrels of high grade, medium grade and low grade oil, respectively. Refinery A produces 100,300 and 00 barrels per day of high grade, medium grade and low grade oil respectively, while refinery B produces 00,400 and 100 barrels per day of high grade, medium grade and low grade oil respectively. If refinery A costs Rs 400per day and refinery costs Rs 300per day to operate, how many days should each be run to minimize costs while satisfying requirements. ( CBSE AI 008 ) PROBABILITY Q 1. Probability of solving specific problem independently by A and B are 1 and 1 3 respectively.if both try to solve the problem independently, find the probability that (a) the problem is solved (b) exactly one of them. Solve the problem. ( CBSE D 015 ) Q. An urn contains 5 red and 5 black balls. A ball is drawn at random; its colour is noted and is returned to the urn. Moreover, additional balls of the colour are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red? ( CBSE F 015 ) Q.3 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,, 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,, 3 or 4 with die? ( CBSE AI 011 ) Q.4 Two cards are drawn simultaneously for successively (without replacement) from a well shuffled pack of 5 cards. Find the mean and variance of the no. of aces. ( CBSE F 013 )

42 Q 5. A die is rolled 0 times. Getting a no. greater than 4 is a success. Find mean and variance of the no. of success. ( CBSE AI 008 )

43 GROUP- 3 B (4 MARKS QUESTIONS ON INTEGRATION ONWARDS) ANSWERS / SOLUTIONS ON INTEGRAL 1. Evaluate xx +11 (xx +33) Solution-Given integral can be written as- xx (xx +1)(xx +3) = xx xx (xx +1) (xx +3) xx = xx (xx +1) xx (xx +3) = 1 xx 1 (xx +1) (xx +3) =1 log (xx + 1) - 1 log (xx + 3) +c, where c is integral constant. = 1 log xx +1 xx +3 +c =log xx +1 xx +3 +c. 44. Evaluate xx Solution- Let I= xx 1 0 It can be seen that,(x-1) 0 wwheeee 0 xx 1 aaaaaa (xx 1) 0 wwheeee 1 xx 4, 4 I= xx = xx 1 + xx = (xx 1) + (xx 1) = xx xx 1 xx + xx = =4+1=5 Ans. ππ 3. Evaluate (cccccc aaaa ssssss bbbb) ππ Solution- ππ Let I= (cos aaaa sin bbbb) ππ ππ = (cccccc aaaa cccccccccccccccccccc + ssssss bbbb) ππ π = cccccc π π aaaa + cccccccccccccccccccc + ssssss bbbb dx π π = 1 ππ (1 + ccccccaaaa) cccccccccccccccccccc + 1 ππ (1 ccccccbbbb) π ππ ππ ππ 0 ππ 0 ππ ππ 0 ππ 1 = 1 (1 + ccccccaaaa) -. cccccccccccccccccccc +. (1 ccccccbbbb) 0 ππ ππ ππ = + ccccccaaaa ccccccbbbb [ssssssssss ccccccaaaa iiii eeeeeeee ffffffffffffffff aaaaaa ssssssssss iiii oooooo ffffffffffffffff. ] =[xx] ππ 0 + ssssssaaaa ππ 0 + [xx]0 ππ ssssssbbbb ππ aa 0 bb 0 ππ

44 =π + ssssssaaaa 0 + ππ ssssssbbbb aa bb =ππ Ans. [ssssssssss ssssssssss = 0] 4. Evaluate, xx +11 xx Solution- Let I= xx +1 xx 4 +1 Dividing numerator and denominator by xx, wwww gggggg I= 1+ 1 xx xx + 1 xx = = 1+ 1 xx xx + 1 xx xx xx 1 xx + Now Let xx 1 xx = tt Orr, xx = (1) Therefore II = [bbbb uuuuuuuuuu (1)] tt + = 1 tan 1 tt + cc wwheeeeee CC iiii cccccccccccccccc oooo iiiiiiiiiiiiiiiiiiiiii. = 1 tan 1 xx 1 xx + CC = 1 tan 1 xx 1 + CC wwheeeeee CC iiii aa cccccccccccccccc oooo iiiiiiiiiiiiiiiiiiiiii. xx 5. Evaluate, 11 xx xx(11 ) Solution: 1 xx 1 xx = xx(1 xx) xx xx = xx 1 xx xx

45 = 1 (xx xx xx ) = 1 xx +(xx ) xx xx = 1 xx 1 + (1) xx xx By partial fraction--- xx xx xx = xx xx(xx 1) = AA xx + BB xx 1 xx = AA(xx 1) + BBBB () Equating coefficient we get--- A+B=1 and A=- So A=,B=-3 From equation (1) we get--- = xx + log xx 3 log 1 xx + cc 4 6. Evaluate, xx ssssss 11 xx Solution: let II = xx sin 1 xx Integrating by parts we have--- = xx sin 1 xx 1 ( 1 xx ) 1 sin 1 xx xx xx xx = xx xx xx 1 xx(1 xx) = xx xx xx II = xx sin 1 xx 1 xx II = xx sin 1 xx 1 1 xx 1 1 xx = xx sin 1 xx 1 sin 1 xx + 1 xx 1 xx + 1 sin 1 xx + cc

46 = xx sin 1 xx 1 4 sin 1 xx xx 1 xx + cc = 1 4 (xx 1) sin 1 xx + xx 1 xx + cc 7. Evaluate, ssssssss ssssssssss ssssssssss Solution: wwww kkkkkkkk tthaaaa ssssssss ssssssss = 1 [cos(aa BB) cos (AA + BB)] SSSS, ssssssss ssssssxx cccccc3xx = [ssssssss. 1 {cos(xx 3xx) cos(xx + 3xx)}] = 1 {ssssssss cos( xx) ssssssss cos 5xx} = 1 {ssssssss cos xx ssssssss cos 5xx} = 1 ssssssxx 1 ssssssss cos 5xx = 1 4 ccccccxx 1 1 sin(xx + 5xx) + sin (xx 5xx) = ccccccxx 8 = ccccccxx 8 = ccccccxx 8 = 1 8 cccccc6xx {ssssss6xx + sin( 4x)} 1 4 cccccc6xx cccccc6xx 3 cccccc4xx 8. Evaluate, + cccccc4xx + cc 4 + cccccc4xx + cc ccccccxx + cc where C is a constant of integration. ππ xx ssssss xx cccccc xx Solution: let, II = Then,II = = ππ (ππ xx) sin x 0 1+cccccc xx =π ππ xx ssssssss 0 1+cccccc xx II =π ππ 0 ππ (ππ xx)sin(ππ xx) 0 1+cccccc (ππ xx) ππ 0 II + II = xx sin xx 1+cos xx ππ xx ssssssss 0 1+cccccc xx xx ssssssss 1+cccccc xx ππ xx ssssssss 0 1+cccccc xx (1) II [bbbb uuuuuuuuuu (1)]

47 II = ππ ππ xx ssssssss 0 1+cccccc xx Now let, cccccccc = tt wwheeee xx = ππ, tt = 1 aaaaaa SSSS, ssssssss = wwheeee xx = 0, tt = 1 OOOO, = ssssssss So we get- II = ππ tt tt = ππ. [ssssssssss 1 1+tt iiii eeeeeeee ffffffffffffffff] = π[tan 1 tt] 1 0 = π[tan 1 1 tan 1 0] = ππ[ ππ 4 0] = ππ 4 9. Evaluate (xx + 33) xx. Solution: LLLLLL II = (xx + 33) xx. HHHHHHHH, (xx + 3)cccccc bbbb wwwwwwwwwwwwww aaaa dd (xx + 3) = AA. ( xx ) + BB oooo, (xx + 3) = AA( 4 xx) + BB.. (1) Onn cccccccccccccccccc tthee cccccccccccccccccccccccc oof oooo xx aaaaaa cccccccccccccccc tttttttt, wwww gggggg AA = 1 aaaaaa 4AA + BB = 3 oooo, AA = 1 and BB = 1 SSSS ffffffff (1) wwww gggggg (xx + 3) = 1 ( 4 xx) + 1

48 TTheeee, II = 1 ( 4 xx) xx = 1 ( 4 xx) xx xx = 1 tt + 7 (4xx + xx + 4) [pppppp 3 4xx xx = tt] = 1 tt (xx + ) = 1 3 (3 4xx xx ) (xx + ) xx + 77 xx + ssssss 11 + CC wwwwwwwwww CC iiii cccccccccccccccc oooo IIIIIIIIIIrrrrrrrrrrrr. 77 SSSSSSSSSS aa xx = xx aa xx + aa. sin 1 xx aa aa aa 10. Evaluate (aa xx) (aa+xx). Solution: Let xx = ccccccθθ = aa( ssssssθθ) IIII xx = aa, ttheeee ccccccθθ = 1. So, θθ = 0 Or, θθ = 0. AAAAAAAA iiii xx = aa, ccccccθθ = 1. θθ = ππ θθ = ππ 0 Now II = ππ ππ aa aa cos θθ aa+aa cos θθ = ssssss θθ.aa ssssssθθ. 0 cccccc θθ ππ =aa ssssss θθ. 0.aa( sin θθ).

49 ππ =aa (1 ccccccθθ). 0 = aa[θθ ssssssθθ =aa( ππ 0) = ππππ ] 0 ππ SO, aa (aa xx) aa (aa+xx) = ππππ. 11. Evaluate ee xx 55 44ee xx ee Solution: Let ee xx = tt So, = = ee xx 55 44ee xx ee 1 (tt +4tt 5) 1` 3 (tt+) = tt. oooo, ee xx = 1 tt+ =sin + c where c is constant of integration. 3 Thus ee xx = 55 44ee xx ee ssssss 11 eexx c where c is constant of integration. 1. Evaluate dx Solution: dx= dx = dx = =..(i) Now, we take It is known that, So, by (i) applying (ii) then. (ii) dx= = +c, where c is constant of integration.

50 Evaluate, ssssssss+cccccccc ππ Solution: GGGGGGGGGG II = = ππ 3 ssssssss+cccccccc ππ 1 (ssssssss cccccccc) 6 ππ 66 ssssssssss Let,ssssssss cccccccc = tt so(cccccccc + ssssssss) = or, = ssssssss+cccccccc ππ 3 ssssssss+cccccccc ππ 1 (1 ssssssxx) 6 when xx = ππ 3, tt = ssssss ππ 3 cccccc ππ 3 = 3 1 =>II = tt 3 =[sin 1 tt] 1 =sin 1 ( 3 1 ) sin 1 ( 1 3 ) =sin 1 ( 3 1 ) + sin 1 ( 3 1 ) =sin 1 ( 3 1 ). 14.Evaluate, Solution: LLLLLL II = xx. xx(xx ). xx xx = [ssssssssss mmmmmmmmmmmmmmmmmmmmmm nnnnnnnnnnnnnnnnnn aaaaaa dd xx 3 (xx 3 +1) bbbbxx ] NNNNNN pppppp tt = xx 3 SSSS, = 3xx xx TTheeeeeeeeeeeeee II = tt(tt + 1) 3xx

51 = 1 3 tt(tt + 1) = tt 1 tt + 1 = tt tt + 1 = 1 3 log tt 1 log(tt + 1) + CC wwheeeeee CC iiii aa iiiiiiiiiiiiiiii cccccccccccccccc. 3 = 1 3 log xx3 1 3 log(xx3 + 1) + CC = log xx 1 3 log(xx3 + 1) + CC wwheeeeee CC iiii aa iiiitttttttttttt cccccccccccccct. 15. Evaluate, ssssss66 xx+cccccc 66 xx ssssss xx cccccc xx. Solution: Let II = ssssss6 xx+cccccc 6 xx ssssss xx cccccc xx = (ssssss xx+cccccc xx) 3 3ssssss xx cccccc xx(ssssss xx+cccccc xx) ssssss xx cccccc xx. [ssssssssss ssssss xx+cccccc xx = 1] 1 = 3 ssssss xx cccccc xx ssssss xx cccccc xx ssssss xx cccccc xx = ssssss xx + cccccc xx ssssss xx cccccc xx 3 1. = ssssss xx + ssssss xx cccccc cccccc xx ssssss xx cccccc 3 = (ssssss xx + cccccccccc xx) 3 = ssssss xx + cccccccccc xx 3 = tttttttt cccccccc 3xx + CC wwheeeeee CC iiii aa cccccccccccccccc oooo iiiiiiiiiiiiiiiiiiiiii. 16.Evaluate, cccccccccc cccccccccc cccccccc cccccccc Solution: let,ii = ccccccxx ccccccαα cccccccc cccccccc = cccccc xx 1 (cccccc αα 1) cccccccc cccccccc [ssssssssss cccccc = cccccc 1.34T]

52 = (cccccc xx cccccc αα) cccccccc cccccccc = (cccccccc + cccccccc) = cccccccc + cccccccc =ssssssss + xxxxxxxxxx + cc II = (ssssssss + xxxxxxxxxx) + cc wwheeeeee cc iiii aa cccccccccccccccc oooo iiiiiiiiiiiiiiiiiiiiii. 17. Evaluate, Solution: Let II = II = xx (xxxxxxxxxx+cccccccc) xxxxxxxxxx (xxxxxxxxxx+cccccccc). xx (xxxxxxxxxx+cccccccc). xxxxxxxxxx Again let, xxxxiinnnn + cccccccc = tt => ( xxxxxxxxxx + ssssssss ssssssss) = =>xxxxxxxxxxxxxx = xxxxxxxxxx NNNNNN, (xxxxxxxxxx+cccccccc) = tt = 1 = 1 (1) tt xxxxxxxxxx+cccccccc Now integrating equation (1) by parts we get, xxxxxxxxxx I= xxxxxxxxxx. (xxxxxxxxxx+ccoooooo) 1 =xxxxxxxxxx. - (1. ssssssss + xxxxxxxxxx tttttttt). xxxxxxxxxx+cccccccc = xxxxxxxxxx xxxxxxxxxx + ssssssss(1 + ) xxxxxxxxxx+cccccccc cccccccc xxxxxxxxxx+cccccccc = xxxxxxxxxx xxxxxxxxxx+cccccccc + ssssss xxxxxx = xxxxxxxxxx xxxxxxxxxx+cccccccc xxxxxxxxxx+cccccccc [BBBB uuuuuuuuuu (1)] + tttttttt + cc WWheeeeee CC iiii cccccccccccccccc oooo iiiiiiiiiiiiiiiiiiiiii. 18. Evaluate, xx +44 xx Solution: let, II = xx +4 xx On dividing numerator and denominator by xx, We get II = (1+ 4 xx ) (xx + 16 xx ) = (1+ 4 xx ) (xx 4 xx ) +8.(1)

53 Put xx 4 4 =t =>( 1 + xx xx )dx=dt.ii = = tt +8 tt +( ) = 1 tan 1 ( tt ) + cc [. = 1 aa +xx aa tan 1 ( xx )] aa tan 1 ( xx 4 xx = 1 )+c II = 1 tan 1 ( xx 1 ) + cc where c is constant of integration xx APPLICATIONS OF INTEGRALS 1. Find the area between the curves y = x and y = x Sol. The required area is represented by the shaded area OBAO as The points of intersection of the curves, y = x and y = x, is A (1, 1). We draw AC perpendicular to x-axis. Area (OBAO) = Area ( OCA) Area (OCABO) (1). Find the area enclosed between the parabola y = 4ax and the line y = mx Sol. The area enclosed between the parabola, y = 4ax, and the line, y = mx, is represented by the shaded area OABO as

54 . The points of intersection of both the curves are (0, 0) and We draw AC perpendicular to x-axis. Area OABO = Area OCABO Area ( OCA)

55 3. Find the area of the region enclosed by the parabola x = y, the line y = x + and x- axis Sol. The area of the region enclosed by the parabola, x = y, the line, y = x +, and x-axis is represented by the shaded region OABCO as The point of intersection of the parabola, x = y, and the line, y = x +, is A ( 1, 1). OABCO = Area (BCA) + Area COAC Area 4. Using the method of integration find the area bounded by the curve x + y =1 Sol. The area bounded by the curve,, is represented by the shaded region ADCB as The curve intersects the axes at points A (0, 1), B (1, 0), C (0, 1), and D ( 1, 0).

56 It can be observed that the given curve is symmetrical about x-axis and y-axis. Area ADCB = 4 Area OBAO 5. Using the method of integration find the area of the triangle ABC, coordinates of whose vertices are A (, 0), B (4, 5) and C (6, 3) Sol. The vertices of ABC are A (, 0), B (4, 5), and C (6, 3). Equation of line segment AB is Equation of line segment BC is

57 Equation of line segment CA is Area ( ABC) = Area (ABLA) + Area (BLMCB) Area (ACMA) 6. Using the method of integration find the area of the region bounded by lines: x + y = 4, 3x y = 6 and x 3y + 5 = 0 Sol. The given equations of lines are x + y = 4 (1) 3x y = 6 () And, x 3y + 5 = 0 (3) The area of the region bounded by the lines is the area of ABC. AL and CM are the perpendiculars on x-axis. Area ( ABC) = Area (ALMCA) Area (ALB) Area (CMB)

58 7. Find the area of the region Sol. The area bounded by the curves,, is represented as The points of intersection of both the curves are. The required area is given by OABCO. It can be observed that area OABCO is symmetrical about x-axis. Area OABCO = Area OBC Area OBCO = Area OMC + Area MBC Now apply appropriate formulas to get the solution. 8. Find the area of the circle x + y = 16 exterior to the parabola y = 6x. Sol. The given equations are x + y = 16 (1) y =6x ()

59 Area bounded by the circle and parabola Area of circle = π (r) = π (4) = 16π units 9. The area bounded by the y-axis, y = cos x and y = sin x when Sol. The given equations are y = cos x (1) And, y = sin x ()

60 Required area = Area (ABLA) + area (OBLO) Integrating by parts, we obtain 10. Find the area bounded by curves Sol. The area bounded by the curves, shaded region as, is represented by the

61 It can be observed that the required area is symmetrical about y-axis. 11. Find the area of region bounded by the ellipse xx + xx = xx Solution: the given equation of the ellipse is, + xx = It can be observed that the ellipse is symmetrical about x-axis and y-axis. Area bounded by ellipse =4 area in 1 st quadrant Now area in 1 st quadrant = YY = 3 1 xx aa 0 = 3 aa xx = 3 4 [xx 16 xx + 16 sin 1 xx ] = [.0+8sin 1 1 0] = 3 [8. ππ ] = 3π. 4 Thus, Area of region bounded by the ellipse=4 3=1π unit. aa 0 16 DIFFERENTIAL EQUATIONS 1. dy x + dx 1+ x y = ( x 1 + 1) I. F = e x dx 1+ x = e log(1+ x ) = 1+ x 1 y.(1 + x ) = (1 + x ) dx (1 + x ) When x=0, y=0 1 dx = tan 1+ x 0 = tan c => c=0 1 x + c

62 y(1 + x ) = tan 1 x dy = dx x y x + y 3 3 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 y x 3. dy dx = ( x y xy ) Let y = vx dy dx = v + dv x dx The D.E. reduces to v dv x (1 v x = dx x v ) v 1+ v + dv = dx x v dx dv = log(1 + v ) = log x + c log( x + y ) x = c.(1) 1+ v x log Now put x=1,y=1, then C=log From (1), x + y log x = log ( x + y ) = x. 4. y = a e x + b e -x

63 dy = a e x - b e -x dx d y = 4 a e x +4 b e -x = 4y dx d y dx - 4y =0 dy 5. + y = x log x dx x Identifying P & Q Finding I.F. = x.. dx + C Writing the formula y I.F. = ( Q I F. ) Simplifying & getting the answer 1 1 y = x log x x + Cx 4 16 dy 6. + y cot x = 4xcosecx dx I.F e cot xdx = e log sin x = sin x Sol. y. sin x = 4x.cosecx. sin xdx x y sin x = 4. + c π Given x=, y = 0 Hence 7. I.F = y.sin x = x e tan 1 x π then c= π Solution is given by 1 1 tan x tan y. e =. e 1+ x e t t. t e dt x 1 tan x dx

64 c y= tan-1x 1+ e 8. Take Putting y = vx We have. 1 tan x y + x + y = xx = (1+ vv ) xx Log I v + (1 + v ) = logx + log C Returning the value of v we have, 9. Reduce to 1 st order linear equation. Find I.F. as ee ssssssssssss = ssssssss + tttttttt, y + + x y = c x Then find the value if the integrating constant C=1 yy(ssssssss + tttttttt) = ssssssss + tttttttt xx The given differential equation can be rewritten as dy 1 1 cot y cosec y + cosecy = dx x x On putting dv v = cos ecyand = cosecy cot dx dy y dx dv dx 1 v = x 1 x 1 1 Here P = and Q = x x I.F = e 1 pdx dx 1 x = e = x Required solution is v = dx + c = c x. x x x + x + (1 cx )sin y = dy y y = + tan( ) dx x x

65 dy dv Put y = vx =v + x dx dx dv v + x = v+tan v dx cot vdv = dx x sin yy =cx. xx x dy e dx 1. Writing = x 1 + y 1+ e Integrating and getting tan e x + tan y = C 1 1 Getting C =π/ 1 1 Getting required solution tan e x + tan y = π / VECTORS Q.1 Solution: squaring both side of given relation magnitude of sum of vectors a+b=vector a Q. Then apply condition vector A.B=0 if A perpendicular to B SOLUTION.

66 Q3. Q.4

67 Q.5. SOLUTION Q.6. SOLUTION Q.7. SOLUTION

68 Q.8. SOLUTION Q.9. SOLUTION

69 Q.10. SOLUTION Q.11. SOLUTION

70 Q.1 SOLUTION Q.13. SOLUTION

71 Q.14. SOLUTION Q.15. SOLUTION Q.16. SOLUTION

72 Q.17. SOLUTION

73 Q.18. SOLUTION 3 D GEOMETRY THREE DIMENSIONAL GEOMETRY 1.Find the coordinates of the foot of perpendicular and the perpendicular distance of the point P(3,,1) from the plane x y + z + 1 = 0. Find also, the image of the point in the plane. SOLUTION: Direction ratios of the normal to the given plane x y + z +1 = 0 are (,-1,1). 1 Let M(x,y,z) be the foot of perpendicular from the point P to the given plane. Equation of PM is xx 3 = yy = zz = λ (say).. 1 Coordinates of the point M are ( λ + 3, λ +, λ + 1).(1) But, M lies on the plane x y + z +1 = 0 λ + ) ( λ + ) + ( λ + 1) + 1 = 0 ( 3

74 4λ +6 +λ - +λ +1 =0 6λ +6 =0 λ = From equation (1) P(3,,1) Coordinates of the point M are (-+3, 1+, -1+1) =(1,3,0) Perpendicular distance PM = (1 3) + (3 ) + (0 1) M = = 6 units 1 Let,P / ( αα, ββ, γγ ) be the required image of P (3,,1) Mid point of PP / = Point M.. 1 P / (αα, ββ, γγ) 3+α, +β, 1+γ =(1,3,0) 3+α =1, +β = 3, 1+γ =0 αα =-3 =-1,ββ = 6 - =4,γγ =0-1 =-1 Image of the point P(3,,1) is P / (-1,4,-1) 1 1.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(,1,), Q(3,1,0),and R(4,-,1) SOLUTION: Equation of the line is xx 3 =yy 4 3 =zz 1 5 1m Equation of the plane is xx yy 1 zz =0 1m ( xx )(0-6) +( yy 1)(-4+1) +( zz )(- 3-0) =0 (-6)( ) +(- 3)( yy 1) +(- 3)( zz ) = 0-6x +1 3y + 3 3z + 6 = 0-6x - 3 y - 3z +1 = 0

75 x + y + z 7 = 0..(i) 1m General point on given line is ( λ+ 3, 3 λ+ 4,5 λ+ 1) lies on line (i) 1m ( λ + 3) + ( 3λ + 4) + (5λ + 1) - 7 = 0 4λ+ 6 3λ+ 4+ 5λ+ 1 7= 0 λ = 3 1m The point of intersection is. ( 3 ) +3, -3. ( 3 )+4,5.( 3 )+1 i.e ( 5 3, 6, 7 3 ) 1m 3.If lines xx 11 = yy =zz 11 xx 33 and =yy kk =zz intersect, then find the value of k and hence find 11 the equation of the plane containing these lines. SOLUTION: Any point on line xx 11 is ( λ +1, 3 λ -1, 4 λ +1) and any point on line xx is (μμ +3, μμ + k, μμ ) = yy =zz 11 = λ. (1) 44 =yy kk =zz =μμ.. () 11 If lines (1) and () intersect,then for some values of λ and μμ λ +1 = μμ +3, 3 λ -1 = μμ + k and 4 λ +1 = μμ Solving these equations we get λ +1 = (4 λ +1) +3 and μμ = 4 (- 33 ) + 1 = = - 5 λ = -3 λ = λ -1 = μμ + k 3( - 3 ) -1 = (-5) +k k -10 = - 11 k = = 9 The lines (1) and () are parallel to the vectors b1 and b given by

76 b1 = iˆ+ 3ˆj+ 4kˆ and b = iˆ+ 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 = b1 x b = 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 ( rˆ aˆ). Nˆ = 0 r. ( 5 iˆ + 3 ˆj+ kˆ ) = - 6 r. ( 5 iˆ + 3 ˆj+ kˆ ) ˆ ˆ = ( iˆ ˆj+ k).( 5iˆ+ 3 ˆj+ k) ˆ ˆ or, ( xiˆ+ yj ˆ+ zk).( 5iˆ+ 3 ˆj + k) = -6 or,5x +y +z +6 = 0 4.Find the equation of the plane through the line of intersection of the planes x +y + z =1 and x + 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(x +3y + 4z -5 ) =0 1m or, (1 + 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 + λ) 1 (1 +3 λ) + 1(1+4 λ) =0 1 1 m 3λ = -1 λ = 1 ½ m 3 Equation of the plane is ( 1 - ) x + (1-1) y + ( 1-4 ) z = 0 ½ m I.e x z + =0 1m Distance of the above plane from the origin = = unit 1m LINEAR PROGRAMMING PROBLEMS 1. Solve the LPP graphically Maximize ZZ = 60xx + 15yy

77 Subject to, xx + yy 50 3xx + yy 90 xx, yy 0 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 OABC are O (0, 0), A (30, 0), B (0, 30) and C (0, 50). 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+15y O(0,0) Z= =0 A(30,0) Z= =1800 B(0,30) Z= =1650 C(0,50) Z= =750. Solve the LPP graphically Maximize ZZ = xx + yy Subject to, xx + 3yy 15 3xx 4yy 1 xx, yy 0 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 OABC are O (0, 0), A (4, 0), B(96/13,57/13) and C(0,5).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+y

78 O(0,0) A(4,0) B(96/13,57/13) C(0,5) Hence maximum z=49/13 at B (96/13, 57/13) Z= 0+1 0=0 Z= 4+1 0=8 Z= 96/ /13=49/3 Z= 0+1 5=5 3. Solve the LPP graphically Minimize ZZ = 3xx + 5yy Subject to, xx + 3yy 3 xx + yy xx, yy 0 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 upper half of ABC are A(3,0), B(3/,1/) and C(0,). 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=3x+5y A(3,0) Z= =9 B(3/,1/) Z=3 3/+5 1/=7 C(0,) Z=3 0+5 =10 Hence minimum z=7 at B (3/,1/) 4. 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 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 5 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 3 1 1

79 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 + 3yy 1 Similarly, 3xx + yy 1 Let Z denote the total profit. Then ZZ =.5xx + yy Clearly xx 0, yy 0 Thus, Maximize ZZ =.5xx + yy Subject to, xx + 3yy 1 3xx + yy 1 And, xx 0, yy 0 To solve the LPP graphically, we first convert the inequations into equations and draw the corresponding lines. Thus the shaded region OA PB1 represents the feasible region of the given LPP. The values of the objective function are given in the following table: Point(x,y) Value of the objective function Z=.5x+y O(0,0) Z= =0 A(4,0) Z= =10 P(3,3) Z= =10.50 B1(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 0 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.

80 Solution:-Suppose the dealer buys x fans and y sewing machines. Since the dealer has space for at most 0 items. Therefore, xx + yy 0 A fan costs 360 and a sewing machine costs 40, but the dealer has only 5760 to invests. Therefore, 360xx + 40yy 5760 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 + 18yy Subject to, xx + yy 0 360xx + 40yy 5760 And xx 0, yy 0 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 OAPB1 are O(0, 0),A(16,0), P(8,1) and B1(0,0). 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(16,0) Z= =35 P(8,1) Z= =39 B1(0,0) Z= =360 Hence Z is maximum at x=8, y=1 and maximum value is 39 Hence, maximum profit is obtained when dealer will purchase 8fans and 1 sewing machines. 6. Solve the following LPP graphically: Minimize ZZ = 30xx + 0yy Subject to, xx + yy 8 xx + 4yy 1 5xx + 8yy 0 xx, yy 0 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(0/3,4/3) and. These points have been obtained by solving the corresponding interesting lines, simultaneously.

81 7. 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 1men 60 Machine B 100sq.m 8men 40 He has an area of 7600 sq.m available and 7 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 = 60xx + 40yy Subject to, 1000xx + 100yy xx + 8yy 7 xx, yy 0 i.e 5xx + 6yy 38 3xx + yy 18 xx, yy 0 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). These points have been obtained by solving the corresponding interesting lines, simultaneously. The values of the objective function are given in the following table:

82 Point(x,y) O(0,0) A(6,0) B(4,3) C(0,38/5) Value of the objective function Z=60x+40y Z= =0 Z= =360 Z= =360 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. 8. 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 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 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 = 5xx + 7yy Subject to, xx + yy 8 xx + yy 10 xx, yy 0 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(,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

83 A(10,0) B(,4) C(0,8) Z= =50 Z=5 +7 4=38 Z= =3 Hence Z is minimum at C(0,8) 9. 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 4 hours of craftsman s time. In a day, the factory has the availability of not more than 4 hours of machine time and 4 hours of craftsman s time. If the profit on a racket and on a bat is 0 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, 1.5xx + 3yy 4 And tennis racket takes 3 hours and cricket bats 1 hours of craftsman s time. Then, 3xx + 1yy 4 Clearly x,y 0 Let Z be the maximum profit. Then the formulated LPP is given below: Maximize ZZ = 0xx + 10yy Subject to, 1.5xx + 3yy 4 3xx + yy 4 xx, yy An oil company requires 1,000, 0,000, and 15,000 barrels of high grade, medium grade and low grade oil, respectively. Refinery A produces 100,300 and 00 barrels per day of high grade, medium grade and low grade oil respectively, while refinery B produces 00,400 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 1,000 0,000 15,000