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-199390-14287500KENDRIYA VIDYALAYA SANGATHAN HYDERABAD REGION QUESTION BANK MATERIAL CLASS-XII 2015-16 -2952753111500Class: XIISub: MATHS-29527513716000KENDRIYA VIDYALAYA KHAMMAMK.V.NO.1 PANAMBUR, MANGALOREChapter - Inverse Trigonometric FunctionsAreas to be revised: Principal value branch table.Properties of Inverse Trigonometric functions.Properties:tan-1(x+y1-xy )ifxy< 11. tan-1x +tan-1y=π+tan-1x+y1-xy if x > 0, y > 0, xy> 1-π+tan-1x+y1-xyifx < 0, y < 0, xy> 1tan-1(x-y1+xy )ifxy>-12.tan-1x -tan-1y=π+tan-1x-y1+xyifx > 0, y < 0, xy<-1-π+tan-1x-y1+xyifx < 0, y> 0, xy<-1Problems1. Prove that tan-134+ tan-135- tan-1819=π4Sol: L.H.S = tan-134+ tan-135- tan-1819=tan-134 + 351 - 34 . 35- tan-1819[tan-1x+tan-1y=tan-1x+y1-xy if xy<1]= tan-12711-tan-189= tan-1[2711 - 891 + 2711 * 89-]( xy> -1)= tan-1425425 = tan-1(1) =π4 = R.H.S.2. If sin(sin-1(12)+cos-1x)=1,then find the value of x.Sol.: We have sin(sin-1(12)+cos-1x)=1,= sin(sin-1(12)+cos-1x)=sinπ2= sin-1(12)+cos-1x=π2 = x=123. Write the value of .tan-1[2sin(2cos-132 )]Sol:tan-1[2sin(2xπ6 )] since cos-132=π6= tan-1[2sinπ3]=tan-1(2 x32)=tan-1(3)=π34. Prove that 2tan-112+ tan-117=sin-131252=2tan-112+ tan-117=tan-12x121-122+ tan-117 [2tan-1x= tan-12x1-x2]=tan-143+ tan-117=tan-143+171-43 x17 [tan-1x+tan-1y=tan-1x+y1-xy,xy<1=tan-13117Let tan-13117=θ =>tanθ=3117sin θ=31252 =>θ=sin-1(31252)tan-13117 =sin-1(31252)= R.H.S5. Find the value of tan12[sin-12x1+x2+cos-11-y21+y2] |x|<1 ,y>0, xy<1Sol: tan12[sin-12x1+x2+cos-11-y21+y2]=tan12 [2tan-1x+2tan-1y] [2tan-1x= sin-12x1+x2 cos-11-x21+x2=tan12 x 2 [ tan-1x+tan-1y]=tan [ tan-1(x+y1-xy)]= x+y1-xy6. Prove that tan-115+ tan-117+ tan-113+tan-118=π4Sol. L.H.S tan-115+ tan-117+ tan-113+tan-118= tan-115+171-15x17+ tan-113+181-13x18= tan-11234+ tan-11123= tan-1617+ tan-11123= tan-1617+11231-617x1123 (617x1123<1)=tan-1325325= tan-11 =π47. Prove that cot-17+cot-1 8+ cot-1 18= cot-1 3Sol. LHS =cot-17+cot-1 8+ cot-1 18 =tan-1(17)+tan-1(18)+ tan-1(118) since cot-1x=tan-11x=tan-117+181-17x18+tan-1118 since 17x18<1 =tan-1311+tan-1118=tan-1311+1181-311x118=tan-165195= tan-113= cot-13=RHS8. Solve tan-12x+tan-1 3x = π4Sol.=tan-12x+tan-1 3x = π4=tan-1(2x+3x1-2x x 3x) = π4 =>5x1-6x2=tanπ4=>5x1-6x2=1 = 6x2 +5x-1=0 => (6x-1)(x+1) = 0x=16or x=-1sincex= -1 doesn’t satisfy the equation, x=1/6 is the only solution of the given equation.9. Solve for x, tan-1x-1x-2+ tan-1x+1x+2=π4Sol. Given tan-1x-1x-2+ tan-1x+1x+2=π4=tan-1x-1x-2+x+1x+21- x-1x-2+x+1x+2=π4= x-1x+2+(x+1)(x-2)x-2x+2-(x-1)(x+1)=tanπ4= x2+x-2+x2-x-2x2-4-x2+1=1 =>2x2-4-3=1= 2x2=4-3 =>x2=12 => x= ±1210. If 0<x<1, then solve the following for x tan-1x+1+ tan-1x-1= tan-1(831)Sol. Given tan-1x+1+ tan-1x-1= tan-1(831)= tan-1x+1+x-11-x+1x-1=tan-1(831)= tan-12x1-x2+1=tan-1(831)= 2x2-x2=831 =>16-8x2=62x=4x2+31x-8=0 =>x+84x-1=0x= -8 &x=14As 0 <x <1 ,x ≠ -8 therefore x=1/4Matrices & Determinants.1. Let A=325 413067express A as a sum of two matrices such that one is symmetric and other is skew symmetric.Sol. A can be expressed as A=12 [A+A1] + ? [A-A1]-------------(1) Where A+A1 and A-A1 are symmetric and skew symmetric matrices respectively.A+A1=325 413067+340 216537=6656295914A1=325 413067+340 216537=0-2520-3-530Putting the values of A+A1 and A - A1 in equation (1) we getA= ? 6656295914+120-2520-3-5302.Using properties of determinants, prove that -a2abacba-b2bccacb-c2=4a2b2c2Sol.LHS let?=-a2abacba-b2bccacb-c2takinga, b, c common from R1, R2 and R3 respectively?=abc-abca-bcab-cNow taking a, b, c common from C1, C2 and C3 respectively ?=a2b2c2-1111-1111-1applying R1 R1 + R2?=a2b2c20021-1111-1expanding along first row, we get ?=a2b2c20+0+21+1 = 4a2b2c2=RHS3. Using properties of determinants, show that 1xx2x21xxx21=(1-x3)2Sol. LHS = let ?=1xx2x21xxx21 applying R1R1+R2 + R3 we get ?=1+x+x21+x+x21+x+x2x21xxx21Taking 1+x+x2 common from R1 ?=1+x+x2111x21xxx21applyingC2C2– C1,C3C3– C1?=1+x+x2100x21-x2x-x2xx2-x1-xexpanding along R1?=1+x+x2[11-x21-x-(x-x2)(x2-x)]=1+x+x2[1-x2(1+x+x2)]= [(1+x+x2)(1-x)]2=(1-x3)2=RHS4. Using properties of determinants, show thata2+1abacabb2+1bccacbc2+1=1+a2+b2+c2Sol. LHS ?=a2+1abacabb2+1bccacbc2+1Multiplying C1, C2 and C3 by a, b and c respectively , we get ?=1abca(a2+1)ab2ac2a2bb(b2+1)bc2a2ccb2c(c2+1)taking a, b, c common from R1, R2 and R3 respectively ?=abcabca2+1b2c2a2b2+1c2a2b2c2+1applying C1C1+ C2 + C3?=1+a2+b2+c2b2c21+a2+b2+c2b2+1c21+a2+b2+c2b2c2+1Taking common (1+a2+b2+c2)from C1?=(1+a2+b2+c2)1b2c21b2+1c21b2c2+1Applying R2R2 – R1 , R3R3 – R1?=(1+a2+b2+c2)1b2c2010001expanding along C1 , we get?=1+a2+b2+c211-0+0+0=1+a2+b2+c2=RHS5. Prove thata+b+2cabcb+c+2abcac+a+2b=2(a+b+c)3Sol. let ?=a+b+2cabcb+c+2abcac+a+2b applying C1C1+C2+C3?=2(a+b+c)2(a+b+c)ab2(a+b+c)b+c+2ab2(a+b+c)ac+a+2bTaking 2a+b+c common from C1 , we get?=2a+b+c1ab1b+c+2ab1ac+a+2bapplying R2R2-R1 , R3R3-R1?=2a+b+c1ab0a+b+c000a+b+cexpanding along C1 we get?=2a+b+c[1a+b+ca+b+c-0]= 2(a+b+c)3= RHS6. Using properties of determinants prove thata+xyzxa+yzxya+z=a2(a+x+y+z)Sol. LHS let ?=a+xyzxa+yzxya+z applying C1C1+C2+C3?=a+x+y+zyza+x+y+za+yza+x+y+zya+zTaking (a+x+y+z) common from C1 , we get?=(a+x+y+z)1yz1a+yz1ya+zapplying R2R2-R1 , R3R3-R1 and expanding along C1 we get?=a+x+y+z1a2-0=a2a+x+y+z=RHS7. Prove that1+a1111+b1111+c=abc1a+1b+1c=ab+bc+ca+abcSol.LHS= let ?=1+a1111+b1111+cTaking a,b,c common from R1,R2 and R3 respectively?=abc1+1a1a1a1b1+1b1b1c1c1+1capplying R1R1+R2+R3?=abc1+1a+1b+1C1+1a+1b+1C1+1a+1b+1C1b1+1b1b1c1c1+1ctaking1+1a+1b+1C common from R1?=abc1+1a+1b+1C1111b1+1b1b1c1c1+1capplying C2C2– C1 , C3C3– C1?=abc1+1a+1b+1C1001b101c01expanding along R1, we get?=abc1+1a+1b+1C11-0= abc1+1a+1b+1C = ab+bc+ca+abc=RHS8. If a, b, c are real numbers and b+cc+aa+bc+aa+bb+ca+bb+cc+a=0show that either a+b+c=0 ora=b=cSol. Let ?=b+cc+aa+bc+aa+bb+ca+bb+cc+a applying C1C1+C2+C3?=2(a+b+c)c+aa+b2(a+b+c)a+bb+c2(a+b+c)b+cc+a = 2(a+b+c) 1c+aa+b1a+bb+c1b+cc+aapplying R2R2-R1 , R3R3-R1 and expanding along C1 and on simplification we get?=2a+b+ca2+b2+c2-ab-bc-ca= -(a+b+c)[a-b2+b-c2+(c-a)2]given?=0 either a+b+c=0 or a-b=0 or b-c=0 or c-a=0Either a+b+c=0 or a=b=c11. 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 equation2x +2y + z = 150003x + y + 2z= 19000x + y + z = 10,000Rewriting the above equations in matrix form221312111xyz=150001900010000 AX=BWhere A=221312111 X=xyz B=150001900010000|A| = -2-2+2=-2≠0, so A-1 exists and have unique solutionsadjA=-1-13-11-120-4 A-1=adjAA=1-2-1-13-11-120-4X=A-1B=1-2-1-13-11-120-4150001900010000=>xyz=200030005000=>x=2000y=3000z=5000Hence the award for honesty =Rs 2000, award for regularity =Rs 3000 and award for handwork=Rs 5000Value: 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 underMenWomenChildrenFamily A231Family B213Family C426Daily 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 X=family Afamily Bfamily C231213426 and daily expenses of men, women and children can be represented by 3 x 1 matrix as Y= Menwomenchildren200150200Daily expense of each family is given by the product XY XY = family Afamily Bfamily C231213426200150200=family Afamily Bfamily C105011502300 Hence daily expense of i) Family A = Rs 1050ii) Family B = Rs 1150iii) Family C = Rs 2300VALUE: More children in the family will increase the expenses of family, which will affect the economy of society.13. For the matrix A=2-11-12-11-12show that A2-5A+4I=0 hence find A-1Sol. Given A=2-11-12-11-12A2=AA= 2-11-12-11-122-11-12-11-12=6-55-56-55-56A2-5A+4I=6-55-56-55-56-10-55-510-55-510+400040004=000000000=0A2-5A+4I=0 Premultiplying by A-1 both sides, we getA-1A2-5A-1A+4A-1I=A-10 => A-5I+4A-1=0 => 4A-1=5I-A => A-1=14(5I-A) A-1 = 14500050005-2-11-12-11-12=1431-1131-113Try TheseIf A = 2-1-12and I is the identity matrix of order 2 then show that A2-4A+3I=0 hence find A-1Ans. 23131323To 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 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 970Values: helping the orphans, use of recycle paper.Find non-zero values of x satisfying the matrix equation x2x23x+285x44x=2(x2+8)24106xLibrarian 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.If A=102021203prove that A3-6A2+7A+2I=0Using matrix, solve 3x-2y+3z=8 , 2x+y-z=1 , 4x-3y+2z=4Ans: x=1 , y=2 , z= 3 Find A-1, if A=1251-1-123-1 hence solve the following system of linear equation x+2y+5z=10 , x-y-z=-2 , 2x+3y-z=-11 Solve using matrix,2x-3y+3z=10 , 1x+1y+1z=10 , 3x-1y+2z=13Using properties of Determinants, show thatb+caabc+abcca+b=4abcSol: L.H.S Let ?=b+caabc+abcca+bApplying R1-> R1+R2+R3?=2(b+c)2(c+a)2(a+b)bc+abcca+bTaking 2 common from R1?=2(b+c)(c+a)(a+b)bc+abcca+bApplying R2-> R2-R1 , R3-> R3-R1?=b+cc+aa+b-c0-a-b-a0Applying R1->R1+R2+R3?=0cb-c0-a-b-a0Expanding along R1, we get20-c0-ab+bca-0=4abc =RHSIf x, y ,z are all different and ?=xx21+x3yy21+y3zz21+z3=0, then show that 1+xyz=0.Sol: Let ?=xx21+x3yy21+y3zz21+z3 = xx21yy21zz21+xx2x3yy2y3zz2z3=xx21yy21zz21+ (xyz)1xx21yy21zz2 (Taking common x,y,z from R1,R2 and R3 respectively)=-121xx21yy21zz2+(xyz)1xx21yy21zz2= (1+xyz) 1xx21yy21zz2Applying R2->R2-R1, R3->R3-R1=(1+xyz) 1xx20y-xy2-x20z-xz2-x2Expanding along C1 and simplifying, we get =(1+xyz)(x-y)(y-z)(z-x)Since ?=0 and x,y,z are all different x-y ≠0, y-z≠0, we get 1+xyz=0.TRY THESEUsing properties of determinants, prove the followingx+42x2x2xx+42x2x2xx+4=(5x+4)4-x2b+cq+ry+zc+ar+pz+xa+bp+qx+y=2apxbqycrzaa+ba+b+c2a3a+2b4a+3b+2c3a6a+3b10a+6b+3c=a3ab-cc+ba+cbc-aab-cc+b=(a+b+c)(a2+b2+c2)-bcb2+bcc2+bca2+ac-acc2+aca2+abb2+ab-ab=ab+bc+ca3Using properties of determinants solve for x.3x-83333x-83333x-8=0[Ans: x=113, 113,23x-22x-33x-4x-42x-93x-16x-82x-273x-64=0[Ans: x=4]x-1111x-1111x-1=0[Ans: x=-1, 2]x+axxxx+axxxx+a=0, a≠0[Ans: x= -a3]Using properties of det. Prove thatxx+yx+2yx+2yxx+yx+yx+2yx=9y2(x+y)xx21+px3yy21+py3zz21+pz3=(1+pxyz)(x-y)(y-z)(z-x)Solutions of Linear Equations using MatricesSolve x-y+2z=7;3x+4y-5z=-5;2x-y+3z=12 using Matrix method.Sol: The given system of equations can be expressed in Matrix from A x =B, whereA=1-1234-52-13, X=xyz, B=7-1512A=1-1234-52-13=7+19-22=4≠0A≠0=>A-1exists and given system has unique solution X=A-1Badj A=71-3-19-111-11-17, A-1=adj AA=1471-3-19-111-11-17 X=A-1B=1471-3-19-111-11-177-1512=148412=213=>xyz=213=>x=2, y=1,z=3If A=2-3532-411-2, find A-1, using A-1 solve the system of equations 2x-3y+5z=11, 3x+2y-4z=-5, x+y-2z=-3.Sol: A=2-3532-411-2=0-6+5=-1≠0A is a non-singular Matrix , so A-1 exists.adj A= 0-122-9231-513, A-1=adj AA=1-10-122-9231-513=01-2-29-23-15-13The given system of equations can be expressed as A x =B 2-3532-411-2xyz=11-5-3Where A= 2-3532-411-2, X=xyz, B=11-5-3AX=B => A-1 B => X=-0-1-22-9-23-1-5-1311-5-3=123xyz =123 => x=1, y=2, z=3Determine the product -444-7135-3-11-111-2-2213 and using it solve the equations.x+y+2z=1, -x-2y+z=4, x-2y+3z=0Sol: Let A= 1-111-2-2213, C=-444-7135-3-1CA= -444-7135-3-11-111-2-2213= 800080008=8I318cA=I3 => A-1=18c [ ∵A-1A=I]A-1=18-444-7135-3-1The given system of equations can be written inmatrix form as PX=BWhere P= 112-1-211-23, X= xyz, B= 140PX=B = >P-1BBut P=112-1-211-23= ATP-1=(AT)-1=(A-1)T∴ P-1=18-4-7541-343-1X=P-1B=18-4-7541-343-1140=18-32816xyz =-412=>x=-4, y=1, z=2Solve 2x+3y+10z=4, 4x-6y+5z=1 , 6x+9y-20z=2 by using Matrix method.Sol: Rewriting the given equations in Matrix form, we get23104-6569-201x1y1z=412AX=BWhere A= 23104-6569-20, X=1x1y1z, B=412|150 + 330 + 720 = 1200 ≠0A is non-singular so A-1 exists and X= A-1B.adj A= 7515075110-10030720-24, A-1= adj A|A|=112007515075110-10030720-24X=A-1B =>112007515075110-10030720-24412=11200600400240=>1x1y1z=121315 => x=2, y=3, z=5The 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 x+y+z=12, 2x+3y+3z=33, x-2y+z=0.The above system of equations can be written in matrix for AX=B as1112331-21xyz=12330 where A= 1112331-21 , X=xyz , B=12330|A|=9+1-7=3≠0, So A-1 exists.AX=B = > A-1Badj A=9-3010-1-731, A-1=adj A|A|=139-3010-1-731X=A-1B=>X=139-3010-1-73112330= 1391215xyz =345=> x=3, y=4, z=5Number of awards for honesty = 3Number of awards for helping others= 4Number of awards for supervising = 5Value: 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.Given A=22-4-42-42-15 B=1-10234012 find BA and use this to solve the system of equations y+2z=7 , x-y=3 , 2x+3y+4z=1710. 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, 511. If A-1=3-1-1-156-55-22 and B=12-2-1300-21 then find (AB)-1use (AB)-1=B-1 A-112. Express the matrix A=42-13571-21 as the sum of a symmetric and a skew symmetric matrix.13. Find a matrix X such that 2A+B+X=0, when A=-1234, B=3-21514. 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 ? DIFFERENTIABILITYLOGARITHMIC DIFFERENTIATION :Rules of logarithmic functionlogbmn = logbm+logbnlogbm/n = logbm-logbnlogbmn =nlogbmChange of base rule logab = logeblogealoge = 1, log1 = 0, elogfx=f(x)PRACTICE QUESTIONS: Differentiate xsinx+(sinx)cosxSolution: If (cos x)y=(siny)x find dydxIf y=eacos-1x, -1≤x≤1,then show that1-x2d2ydx2-xdydx-a2y=0Differentiate xcosx+(x2+1x2-1) with respect to x:If xy=ex-y, show that dydx=logx(logxe)2If (cosx)y=(cosy)x finddydxIf xmyn=(x+y)m+n find dydxDifferentiate (logx)x+xlogxPRACTICE QUESTIONS: Find dydx for the following :If y= xy .yx= ex-yy=(sinx)cos-1xy= xtanx+(sinx)cosxy= 8xx8y=(logx)xxmyn=(x+y)m+ny=xcosx+(cosx)xlogxlogx+1+2x2x If x16y9=(x2+y)17 prove that dydx= yx Differentiate with respect to x: xsinx+(sinx)cosx If xy+yx=loga, find dydxAPPLICATIONS 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 :- 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.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.SAMPLE QUESTIONS AND SOLUTION:Find the intervals in which the function f given by fx=3x4-4x3-12x2+5 is a) strictly increasingb) strictly decreasing Find the intervals in which the function f given byfx=sinx+cosx,0≤x≤2π, is strictly increasing or strictly decreasing.SOLUTION:Find the intervals in which the function f given by fx=2x3-3x2-36x+7 is a) strictly increasingb) strictly decreasingFind the intervals in which the function f given by fx=-2x3-9x2-12x+1 is a) strictly increasingb) strictly decreasing.5. Find the intervals in which the function f given by fx=32x4-4x3-45x2+51 is a) strictly increasingb) strictly decreasingShow that y=log1+x-2x2+x , x> -1 is an increasing function throughout its domain.PRACTICE QUESTIONS:Find the intervals in which the following functions are increasing and decreasing.fx= 2x3-9x2+12x+15fx= 2x3-12x2+18x-15fx= 2x3-12x2+18x-7fx=5x3-15x2-120x+3fx=-2x3+3x2+12x+6fx=-2x3-9x2-12x+1fx=(x+1)3(x-1)3fx= x(x-2)2 fx=x3-3x2-3x-100 Prove that the function f given by fx=log cosx is strictly increasing on ( 0,π2) Prove that the function f given by fx=4sinθ2+cosθ-θ is strictly increasing on ( 0,π2) Show that the function x2-x+1 is neither increasing nor decreasing on (0,1)TANGENTS AND NORMALS :Find the equation of the tangent to the curve x=sin 3t,y=cos2t at t=π4Find the equation of the tangent to the curve y=3x-2, which is parallel to the line 4x-3y +5 = 0Find the equation of the tangent to the curve y=4x-2, which is parallel to the line 4x-2y +5 = 0Find the points on the curve y=x3 at which the slope of the tangent is equal to the y-coordinate of the point.Find the equations of the tangent and the normal to the curve x=1-cosθ , y= θ-sinθ at θ=π4Find the equation of the normal at the point ( am2 ,am3) for the curve a y2= x3Find the equation of the tangent and normal to the curve x2a2-y2b2=1 at the point ( 2a ,b)PRACTICE QUESTIONS:Find the points at which the tangent to the curve y=x3-3x2-9x+7Is parallel to the x axis .Find the slope of the normal to the curve x=1-a sinθ, y=bcos2θ at θ=π4Find the points on the curve y=x3-11x+5 at which the tangent has the equationy=x-11Prove that the tangents to the curve y=x2-5x+6 at the point 2,0and (3,0)Are at right angles.Find the equation of the tangent to the curve y=3x-2, which is parallel to the line 4x-3y +5 = 0Find the equation of the tangent and normal to the curve x=asin3θ and y=acos3θ at θ=π4Find the equations of the normals to the curve y= x3+2x+6 which is parallel to the line x+14y+4=0Find the equation of the tangent to the curve y=x-7x-2(x-3) at the point ,where it cuts the x axis. Find the equation of the tangent to the curve y=cos x+y, -2π≤x≤2πThat are parallel to the line x+2y = 0Find 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 tangentINTEGRATION & APPLICATION OF INTEGRALSPOINTS TO REMEMBER:A: Integration of standard functionsSin x dx= -Cos x+CCos x dx= Sin x+CSec2 x dx= Tan x+CCosec2 x dx=- Cot x+CSec x Tan x dx= Sec x+CCosec x Cot x dx= -Cosec x+Cex dx= ex +Cxm dx= xm+1m+1+C ; where m≠–11xdx= log x+C11+x2dx= tan-1x+C11-x2dx= Sin-1x+C1x2-1dx= Sec-1x+CB: Integration by substitutionf/xfxdx= logfx+Cf/xfxdx= 2fx+Cfgxg/xdx= ftdt+CWhere t=g(x)C: Integration using trigonometric identitiesSin2xdx=1-Cos2x2dx=12dx-12Cos 2x dx=x2-Sin 2x4+cCos2xdx=1+Cos2x2dx=12dx+12Cos 2x dx=x2+Sin 2x4+cSin3xdx=3 Sin x-Sin 3x4dx=34Sin x dx-14Sin 3x dx=-3Cos x4+Cos 3x12+cCos3xdx=Cos 3x+3 Cos x4dx=14Cos 3x dx+34Cos x dx=Sin 3 x12+3 Sin x4+cSin ax. Cos bx. dx=122Sin ax Cos bx dx=12Sin a+bx+Sin a-bx dxSin ax. Sin bx. dx=122Sin ax Sin bx dx=12Cos a-bx-Cos a+bx dxCos ax. Cos bx. dx=122Cos ax Cos bx dx=12Cos a+bx+Cos a-bx dxD: Integration of special functionsdxx2+a2=1atan-1xa+Cdxx2-a2=12alogx-ax+a+Cdxa2-x2=12aloga+xa-x+Cdxx2+a2=logx+x2+a2+Cdxx2-a2=logx+x2-a2+Cdxa2-x2=sin-1xa+Cx2+a2dx=x2x2+a2+a22logx+x2+a2+Cx2-a2dx=x2x2-a2-a22logx+x2-a2+Ca2-x2dx=x2a2-x2+a22Sin-1xa+CEvery quadratic polynomial ax2+bx+c can be expressed in one of the three formsx2+a2, x2-a2 or a2-x2 by completing square methodExample: 3x2-7x+5=3x2-73x+53 [see that x coefficient is 1][Add and subtract square of (half the x coefficient)i.e. 76]3x2-73x+4936-4936+53=3x-762+1136=3x-762+1162It is in the form of t2+a2 where t=x-76 and a=116E: Integration of quadratic equationsdxQuadratic EquationExpress Q. E as ±x2±a2 and use 1,2 and 3 formulaelinear equation Quadratic Equationdx Find A, B Such that L.E= A.ddxQ.E+B, separate integrals and proceeddxQuadratic Equation Express Q. E as ±x2±a2 and use 4,5 and 6 formulaelinear equation Quadratic Equationdx Find A, B Such that L.E= A.ddxQ.E+B, separate integrals and proceedQuadratic Equation dx Express Q. E as ±x2±a2 and use 7,8 and 9 formulaeA rational expression P(x)Q(x) is called proper if degree of P(x) is smaller than degree of Q(x)If P(x)Q(x) is proper and the polynomial Q(x) can be expressed as product of linear/quadratic factors, then it can be decomposed into small fractions called partial fractions.P(x)Q(x)=P(x)x-ax-bx-cx-d=Ax-a+Bx-b+Cx-c+Dx-d(All non-repeated linear factors only)P(x)Q(x)=P(x)x-a2x-bx-c=Ax-a+Bx-a2+Cx-b+Dx-c (Repeated but linear factors only )P(x)Q(x)=P(x)x2+a2x-bx-c=Ax+Bx2+a2+Cx-b+Dx-c(involve non-repeated quadratic factors)lin.EqQuad. Eqdx Find A, B Such that L.E= A.ddxQ.E+B, separate integrals and proceedF: Integration of a rational expressionIf it is proper, decompose it into partial fractions and then integrateIf it is not proper, divide p(x) by Q(x) P(x)Q(x)=Quotient+ R(x)Q(x)And R(x)Q(x) can be split into partial fractions as it is properG: Integration by partsfxgxdx=fxgxdx-f/(x)gxdxdx( learn ILATE rule)exfx+f/(x)dx=exfx+CH: Definite Integralsabf(x)dx+bcf(x)dx=acf(x)dx-aaf(x)dx=0 if f-x=-f(x)-aaf(x)dx=20af(x)dx if f-x=f(x)0af(x)dx=0af(a-x)dxabf(x)dx=0af(a+b-x)dx02af(x)dx=0 if f2a-x=-f(x)02af(x)dx=20af(x)dx if f2a-x=f(x)I: Definite integral by limit Sum methodTo find abf(x)dx; follow the steps mentioned below.Writenh=b-a (constant)Find f(a+rh)Substitute the in the formula abf(x)dx= limn→∞hr=1nf(a+rh)Use values : nh=b-a, 1=n, r=nn+12, r2=nn+12n+16 and r3=n2(n+1)24J: Application of Integralsabf(x)dxdenote the area under the curve y=f(x) bounded by three lines x=a, x=b and y=0.To find the area bounded by a curve and a curve/straight line, First find the points of intersection where the curve intersects the curve/line.Draw the rough sketch of the curve and curve/straight lineWrite the required area using definite integrals and then solve.Questions for practiceEvaluate Sin (x-a)Sin (x+a)dxEvaluate 3Sin x-2Cos x5-Cos2x-4Sin xdxEvaluate tan-1Cos x -Sin xCos x+Sin xdxEvaluate tan-11-Cos xSin xdxEvaluate Sin4x dxEvaluate 1Cos4x+Sin4xdxEvaluate 1x-1x-2dxEvaluate x+2x2+2x+3dxEvaluate 6x+7x-5x-4dxEvaluate 19x2+6x+5dxEvaluate 2x+1x+2x-3dxEvaluate 5x-21+2x+3x2dxEvaluate x2+2x+3dxEvaluate x+33-4x-x2dxEvaluate x2+x+1x+12x+2dxEvaluate x2x2+1x2+4dxEvaluate xsin-1xdxEvaluate x cos-1x1-x2dxEvaluate exx2+1x+12dxEvaluate e2x1-Sin 2x1-Cos 2xdxEvaluate 04x+x-2+x-4dxEvaluate 0π4x Sin x1+Cos2xdxEvaluate 0πlog1+Cos xdxEvaluate 0π2x2Sin x dxEvaluate 0π4Sin x+Cos x9+16 Sin 2xdxEvaluate π6π3Sin x+Cos xSin 2xdxEvaluate 01log1+x1+x2Evaluate π6π3dx 1+Tan xEvaluate 0π4log1+tanxdxEvaluate 0π2x Sin x Cos xCos4x+Sin4x dxEvaluate 0π22sinxCos xtan-1Sin xdxEvaluate 0πxtanxsecx+Tan x dxEvaluate 133x2+1dx by the method of limit of sums.Evaluate -127x-5dx by the method of limit of sumsEvaluate 14x2-xdx by the method of limit of sumsUsing integration, find the area of the region bounded by the curves y=x+1+1, x=-3, x=3, y=0Using the integration, find the area of the region bounded by the curve x2=4y and the line x=4y-2Sketch the graph of y=x+3and evaluate the area under the curve y=x+3above x-axis and between x=-6to x=0.Using the integration, find the area of the region bounded by the curve y=x2 and y=xFind the area of the circle 4x2+4y2=9which is interior to the parabola x2=4y.Using integration find the area of the triangle ABC, coordinates of whose vertices are A(4, 1), B(6, 6) and C(8, 4).Using the integration find the area of the triangular region whose sides have equations y=2x+1, y=3x+1 and x=4Using the integration find the area of the triangular region whose sides have equations x+2y=2, y-x=1 and 2x+y=7Using the integration, find the area of the region enclosed between the two circles x2+y2=4 and x-22+y2=4HINTS/SOLUTIONS1Sin (x-a)Sin (x+a)dxWrite x+a=t, expand Sin(t-2a)Ans: x+acos2a-Sin2alogSin x+a+C23Sin x-2Cos x5-Cos2x-4Sin xdxPut sin x = t, continue as in problem 13Ans: 3log2-Sinx+42-Sinx+C3tan-1Cos x -Sin xCos x+Sin xdxDivide Nr and Dr by cos x, replace 1 by tanπ4Ans: π4x-x22+C4tan-11-Cos xSin xdxUse identities Ans: x24+C5Sin4x dxWrite 1-Cos 2x=2Sin2xAns: 38x-Sin 2x4+Sin 4x32+C61Cos4x+Sin4xdxMultiply and divide by Sec4x, and puttanx=tObtain the form 1+t21+t4dt=1+1t2t2+1t2dt, let t-1t=yObtain the form dyy2+2=12tan-1tan2x-12Tan x+C71x-1x-2dxWrite x2-3x+2=x-322-122use formula D5Ans: logx-32+x-1x-2+C8x+2x2+2x+3dxWrite x+2=122x+3+12 Sol: x2+2x+3+logx+1+x2+2x+3+C9Evaluate 6x+7x-5x-4dxSimilar to previous problem given for practice10Evaluate 19x2+6x+5dxSimilar to previous problem given for practice11Evaluate 2x+1x+2x-3dxSimilar to previous problem given for practice12Evaluate 5x-21+2x+3x2dxSimilar to previous problem given for practice13Evaluate x2+2x+3dxSimilar to previous problem given for practice14Evaluate x+33-4x-x2dxSimilar to previous problem given for practice15Evaluate x2+x+1x+12x+2dxWrite x2+x+1x+12x+2 =-2x+1+1x+12+3x+2 and integrateAns: -2logx+1+3logx+2-1x+116Evaluate x2x2+1x2+4dxWrite x2x2+1x2+4=43x2+4-13x2+1 then integrate17Evaluate xsin-1xdxsin-1xxdx-11-x2x22dx (by parts)Ans: 2x2-1sin-1x4+x1-x24+C18Evaluate x cos-1x1-x2dxTake cos-1x=t and use by parts formulaAns: -x-Cos-1x1-x2+C19Evaluate exx2+1x+12dxx2+1x+12 =x2-1+2x+12=x-1x+1+2x+12, ddxx-1x+1=2x+12Ans: exx-1x+1+C20Evaluate e2x1-Sin 2x1-Cos 2xdx (put 2x=t)1-Sin t1-Cos t =12Sin2t2-2Sin t2Cost22Sin2t2=-cott2+12Cosec2t2Ans: 12e2x(-Cotx)+C21Evaluate04x+x-2+x-4dxFor 0<x<4, x=x; and For 0<x<2, x-2=2-x; For 2<x<4, x-2=x-2; For 0<x<4, x-4=4-x∴02x+2-x+4-xdx+24x+x-2+4-xdx026-xdx+24x+2dx=6x-x2202+x22+2x24=2022Evaluate 0π4x Sin x1+Cos2xdxUse rule H4, add both integrals,2I=4π0π Sin x.dx1+Cos2xPut cosx =t, I=2π-11dt1+t2=2πtan-1t-11 =π223Evaluate 0πlog1+Cos xdxUse rule H4, add both integrals,2I=0πlogSin x dxUse rule H7,2I=20π2Sin xdx, Use rule H4 again add integrals … AnsI=-π.log224Evaluate 0π2x2Sin x dxx2Sin x dx=-x2Cos x+2xCosxdx=-x2 Cosx+2xSin x+2Cos xAns: -x2 Cosx+2xSin x+2Cos x0π2 =π-225Evaluate 0π4Sin x+Cos x9+16 Sin 2xdx9+16Sin 2x=25-161-Sin2x=25-16t2 where Sin x – Cos x = t, after substitution,it becomes-10dt25-16t2=12.5.4log5-4t5+4t-10=140log926Evaluate π6π3Sin x+Cos xSin 2xdxSin 2x=1-1-SIn2x=1-t2 where Sin x – Cos x = t, after substitution, it becomes1-32-1+32dt1-t2Ans: 2sin-13-1227Evaluate 01log1+x1+x2Put x = Tan t, with that I=0π4log1+tantdtUse rule H4, I=0π4log1+1-tant1+tantdt=0π4log21+tantdtI=0π4log2dt-I?2I=π4log2?I=π8log228Evaluate π6π3dx 1+Tan xUse rule H5, π6π3dx 1+Tan x=π6π3dx 1+Cotx=π6π3Tan xdx 1+tanxAdd 2I=π6π3dx?I=π1229Evaluate 0π4log1+tanxdxRefer problem 2730Evaluate 0π2x Sin x Cos xCos4x+Sin4x dxUse Rule H4, simplify to getI=π40π2 Sin x Cos xdxCos4x+Sin4xSin4x+Cos4x=2Sin4x-2Sin2x+1 let Sin2x=tI=π801 dt2t2-2t+1=π1601 dtt-122+14=π23231Evaluate 0π22sinxCos xtan-1Sin xdxGiven for practice use H432Evaluate 0πxtanxsecx+Tan x dxUse rule H4, simplify to get 2I=π0πtanxsecx+Tan x dxApply H7, 2I=2π0π2tanxsecx+Tan x dx?I=π0π2Sinxdx1+SinxI=π0π2Sinxdx1+Sinx=π0π21-11+Sinxdx=π0π21-1-SinxCos2xdxI=πx-Tanx+Secx0π2Ans: π2π-2Evaluate 133x2+1dx by the method of limit of sums.Here nh=b-a=3-1=2, and fa+rh=f1+rh=31+rh2+1=3r2h2+6rh+4133x2+1dx = limn→∞hf(a+rh)=limn→∞h3r2h2+6rh+4limn→∞3h3r2+6h2r+4h1 = limn→∞3h3n(n+1)(2n+1)6+6h2n(n+1)2+4hnlimn→∞3.23(1+1n)(2+1n)6+6.22(1+1n)2+4.2=28Evaluate -127x-5dx by the method of limit of sumsHere nh=b-a=2+1=3, and fa+rh=f-1+rh=7-1+rh-5=7rh-12-127x-5dx = limn→∞hf(a+rh)=limn→∞h7rh-12limn→∞7h2r-12h1 = limn→∞7h2n(n+1)2-12hnlimn→∞7.32(1+1n)2-12.3=-92Evaluate 14x2-xdx by the method of limit of sumsHere nh=b-a=4-1=3, and fa+rh=f1+rh=1+rh2-1+rh=r2h2+rh14x2-xdx = limn→∞hf(a+rh)=limn→∞hr2h2+rhlimn→∞h3r2+h2r = limn→∞h3n(n+1)(2n+1)6+h2n(n+1)23968750112395limn→∞33(1+1n)(2+1n)6+32(1+1n)2=9+92=272Using integration, find the area of the region bounded by the curves y=x+1+1, x=-3, x=3, y=0Draw the lines y=x+2, y=-x, x=-3, x=3, y=0Area of Shaded region:=-3-1-x.dx+-13x+2dx=-x22-3-1+x22+2x-13=163969385186690Using the integration, find the area of the region bounded by the curve x2=4y and the line x=4y-2The curves x2=4y and x=4y-2 intersect at (-1,14), (2,1)Area of Shaded region:=-12x+24-x24dx=14x22+2x-x33-12=1442+4-83-12-2+13=98374205566675Sketch the graph of y=x+3and evaluate the area under the curve y=x+3above x-axis and between x=-6to x=0.Area of Shaded region:=-6-3-x-3.dx+-30x+3dx=-x22-3x-6-3+x22+3x-30=9Using the integration, find the area of the region bounded by the curve y=x2 and y=x409575026035Points of intersection (0,0), (1,1)Area of Shaded region:=01x-x2dx=x22-x3301=16Find the area of the circle 4x2+4y2=9which is interior to the parabola x2=4y.403225010160Circle and parabola intersect at -2,12, 2,12Area of Shaded region:=-2294-x2-x24dx=x294-x2+98sin-12x3-x312-22414909017145Using 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 :y=52x-9, Eq to BC : y=-x+12, Eq to AC : y=34x-2Required area = 46yAB+68yBC-46yAC =4652x-9dx+68-x+2dx-4834x-2dx = 5x24-9x46+-x22+2x68-3x28-2x48=7Using the integration find the area of the triangular region whose sides have equations y=2x+1, y=3x+1 and x=4Given for practice do Same as aboveUsing the integration find the area of the triangular region whose sides have equations x+2y=2, y-x=1 and 2x+y=7Given for practice do Same as aboveUsing the integration, find the area of the region enclosed between the two circles x2+y2=4 and x-22+y2=4Given for practice Find the particular solution of the differential equation 3xy+y2dx+xy+x2dy = 0 for x=1, y=1.Given for practice ALL THE BESTDIFFERENTIAL EQUATIONSPOINTS TO REMEMBER:An equation involving derivative (derivatives) of the dependent variable with respect to independent variable (variables) is called a differential equationOrder 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 equationThe 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.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.The solution which contains arbitrary constants and satisfy the given differential equation is called the general solution (primitive) of the differential equationThe solution of a differential equation independent from arbitrary constants is called a particular solution of the differential equationTo obtain differential equation when the general solution(Family of curves) is givenIdentify the number of arbitrary constants involved in general solution, say ‘n’Derivate the general solution for ‘n’ times let y, y1,y2,y3….ynEliminate the arbitrary constants by using y, y1,y2,y3….ynThe equation then obtained involve differentials in place of arbitrary constants, and it is required differential equation.To obtain general/particular solution when the differential equation is given(VARIABLE SEPARABLE)If the given differential equation can be expressed as dydx=f(y)g(x)OR dydx=f(x)g(y) then it can be solved by separating the variables and integrating both sides.dydx=f(y)g(x)?dyf(y)=dxg(x)?dyf(y)=dxg(x) ORdydx=f(x)g(y)?gydy=fxdx?gydy=fxdx(HOMOGENEOUS)If the given equation is expressed as dydx=f(x,y)g(x,y) such that both the functions f and g are homogeneous then; substitute y=Vxand dydx=V+xdVdx . Then the differential equation can be solved by using method discussed in type (i).(LINEAR D E)If the given differential equation is in the form dydx+Py=Q where P and Q are functions of x, OR dxdy+Px=Q where P and Q are functions of y; then it is called linear differential equation(LDE). It can be solved by multiplying both sides of the LDE by integrating factor I.F.ePdx, and integrating both sides.ePdxdydx+Pydx=ePdxQdxand write ePdxdydx+Pydx=y.ePdxQuestions for practiceSolve the differential equation: 1+x2+y2+x2y2+xydydx=0Solve the differential equation 1+e2xdy+1+y2exdx=0, given that x=0, y=1Find the particular solution of the differential equation dydx=1+x+y+xy, given that y=0 when x=1If y(x) is a solution of the differential equation 2+Sin x1+ydydx=-Cos x and y0=1, then find the value of y(π2)Find the particular solution of the differential equation x1+y2dx-y1+x2dy=0, given that y=1 when x=0Find the general solution of the differential equation x-ydydx=x+2ySolve the differential equation xdydx+y=x Cos x+Sin x, given that yπ2=1.Solve the differential equation 1+x2dydx+y=etan-1xSolve the differential equation x2-1dydx+2xy=1x2-1;x≠1Solve the differential equation xdy-ydx=x2+y2dxSolve the differential equation y+3x2dxdy=xFind the particular solution of the differential equation 3xy+y2dx+xy+x2dy = 0 for x=1, y=1.Obtain the differential equation of all circles of radius ‘r’Show that the differential equation 2yexydx+y-2xexydy=0 is a homogeneous. Find the particular solution of this differential equation, given that x=0 when y=1.HINTS/SOLUTIONS1. given differential equation: 1+x2+y2+x2y2+xydydx=01+x21+y2dx=-xydyimpliesy dy1+y2+ 1+x2xdx=0 integrating both sidesy dy1+y2+ 1+x2x1+x2dx=0implies1+y2+1x1+x2dx+x1+x2dx=01+x2+1+y2-log1+1+x2x=cGiven differential equation 1+e2xdy+1+y2exdx=0, given that x=0, y=1dy1+y2+exdx1+ex2=0implies General Solution is tan-1y+tan-1ex=Cgiven when x=0, then y=1implies C = π2 Particular Solution: tan-1y+tan-1ex=π2Given differential equation dydx=1+x1+yimplies dy1+y= 1+xdxGeneral Solution: log1+y=x+x22+Cgiven that y=0 when x=1implies C= 32Particular Solution: log1+y=x+x22+32Given differential equation 2+Sin x1+ydydx=-Cos x implies dy1+y=-Cos x2+Sin xdxlog1+y=-log2+Sin x+logC General Solution :2+Sinx1+y=Candy0=1 implies C = 4 , Particular solution: 2+Sinx1+y=4the value of y(π2) is 13given differential equation x1+y2dx-y1+x2dy=0, x1+x2dx-y1+y2dy=0integrating we get General Solution12log1+x21+y2=Cgiven that y=1 when x=0 that implies 12log12=C=-log2 P.S. y2-2x2=1Given differential equation x-ydydx=x+2yOR dydx=x+2yx-y put y=VxV+xdVdx=1+2V1-Vimplies1-VV2+V+1dV=dxx integrate both sides Ans:logx2+xy+y2=23tan-12y+x3x+CGiven differential equation xdydx+y=x Cos x+Sin xdydx+yx=Cos x+Sin xxintegrating factor is edxx=elogx=x, Multiply both sides by xxdydx+yx=xCos x+Sin xintegrating both sides G.S. xy=x Sin x+Cosx+Cgiven that yπ2=1 C=0, PS.= xy=x Sin x+CosxGiven differential equation 1+x2dydx+y=etan-1xdydx+y1+x2=etan-1x1+x2integrating factor is edx1+x2=etan-1x, Multiply both sides by etan-1xetan-1xdydx+y1+x2=etan-1x21+x2integrating both sides GS etan-1x.y=e2tan-1x2+CGiven differential equation x2-1dydx+2xy=1x2-1dydx+2xx2-1y=1x2-12integrating factor is e2xdxx2-1=elogx2-1, Multiply both sides by x2-1dydx+2xx2-1y=1x2-1integrating both sides G.S. x2-1y=logx-1x+1+CGiven differential equation xdy-ydx=x2+y2dx (Homogeneous)Put y = Vx… Ans G.S. = y+x2+y2=Cx2Given differential equation y+3x2dxdy=xcan be written as dydx-yx=3x (LinearDE)Ans: General Solution yx=3x+CFind the particular solution of the differential equation 3xy+y2dx+xy+x2dy = 0 for x=1, y=1.Given for practice Obtain the differential equation of all circles of radius ‘r’Equation to a circle with radius r is given byx-a2+y-b2=r2 ………..(1) where a, b are arbitrary constantsDerivating (1) both sides with respect ‘x’ we get 2x-a+2y-b.dydx=0 ………….(2)Derivating (2) both sides with respect ‘x’ we get 2+2y-b.d2ydx2+2dydx2=0 ………….(3)So, y-b=-1+dydx2d2ydx2; x-a=1+dydx2d2ydx2.dydxsubstituting these in eq 1, we get1+dydx2d2ydx2.dydx2+-1+dydx2d2ydx22=r2 OR 1+dydx22=r2d2ydx2Show that the differential equation 2yexydx+y-2xexydy=0 is a homogeneous. Find the particular solution of this differential equation, given that x=0 when y=1.(Put x = Vy )=========================================================================================================================================Hint: a2-a1= i-3 j-2k and b1×b2= -3i+ 3k : a2-a1. b1×b2 = -3-6=-9 and b1×b2 =92 d = -992 = 12 Hint: Convert in to Vector form and proceed.The vector form of the lines are r = - i- j-k+ λ (7 i-6 j+k) r = 3 i+5 j+7k+ λ ( i-2 j+k)=========================================================================================================================================================d= a2-a1×bb===============================================================================1. Find the distance between two parallel lines Hint : a2-a1 = 2i+ j-k , ( a2-a1) × b= 9i-14j+4k ( a2-a1) × b = 293 , b =7 ∴ d = 2937 STRATEGIC ACTION PLAN FOR SLOW LEARNERSLINEAR PROGRAMMINGSOME 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 byax + 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. SOLVED PROBLEMS.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 25To maximize Z = 40x + 30y 20Subject to constraints,C(0, 20)B(,)X+y ≤ 20 400x + 250y ≤ 6000 X0 A(15, 0),=>8x +5y ≤ 120, X ≥ 0, y ≥ 0Possible points for maximum Z are A(15, 0), B(,), C(0, 20)POINTZ = 40x + 30yVALUE A(15, 0)600600B(,)666.66[MAXIMUM VALUE]C(0, 20)0 + 600600Z is maximum for B(,) , ie., 6 tables and 13 chairs must be purchased and sold for a maximum profit of Rs 666.6One 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) XAnd 25x + 50y ≤ 1000 =>x + 2y ≤ 40 25 40POINT Z = x+ y VALUE A(25, 0)25 + 025B(20, 10) 20 + 10 30 maximumC(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.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 YMaximize Z = 8000x + 12000ySubject to the constraints: 9x + 12y ≤ 180 C(0,10) B(12,6)=> 3x + 4y ≤ 60X + 3y ≤ 30 0 A(20,0) X X ≥ 0, y ≥ 0 CORNER POINT Z = 8000x + 12000y 0(0, 0)0A(20, 0) 160000B(12, 6)168000C(0, 10)120000We 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. 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 = 12So, the point of intersection of the lines is B(8,12). Graphical representation of the lines is given below YC(0,20)3X + 2Y = 48B(8, 12)X + Y = 20 A(16,0) X 0:- Feasible region is OABCAThe 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 POINTSZ = 22x + 18y0(0,0)Z = 22(0) + 18(0) = 0A(16,0)Z = 22 x 16 + 0 = 352B(8,12)Z = 22 x 8 + 18 x 12 = 392C(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.PRACTICE PROBLEMS1)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 1,400calories. Two foods X and Y are available at a cost of Rs. 4 and Rs. 3 per unit respectively. One unit ofthe food X contains 200 units of vitamins, 1 unit of mineral and 40 calories, whereas one unit of foodY contains 100 units of vitamins, 2 units of minerals and 40 calories. Find what combination of X andY should be used to have least cost? Also find the least cost.PROBABILITYBAYE’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, thenP(Ei|A) = for any i = 1, 2, 3, ..., nProblems with solutionsQ1.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= 7And 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) = and P (E2) = P (A/E1) =P (getting both red balls from 2nd bag, when transfer ball is red) = 5C2/10C2==P (A/E2) =P (getting both red balls from 2nd bag, when transfer ball is black) = 4C2/10C2==Therefore, by Baye’s theoremP (E1/A) = =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) = Also, let A be the event that ‘the coin drawn is of gold’Thus, P (A/E1) = P (a gold coin from bag1) =P (A/E2) = P (a gold coin from bag2) = 0P (A/E3) = P (a gold coin from bag3) = 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) = =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 = P (E2) = Probability that six does not occurs = P (A/E1) = Probability that the man reports that six occurs when six has actually occurred on the die. Probability that the man speaks the truth = 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 = 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) = = 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)=P(A/E1)=1, P(A/E2)=75%=, P(A/E3)=Required probability= p(E1/A) == 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=2000Car drivers = 4000Truck drivers=6000Total drivers= 12000Let E1,E2 and E3 be the events that scooter driver , car driver and truck driver are selecting respectivelyLet A be the event of meeting with an accidentP(E1)= , P(E2)= ,P(E3)= P(A/E1)=0.01 and P(A/E2)=0.03 and P(A/E3)=0.15By Bayes theorem P(E1/A) ==PROBABILITY DISTRIBUTION MEAN & VARIANCE OF RANDOM VARIABLEThe probability distribution of a random variable X is the system of numbers X : x1 x2….... xn P(X): p1 p2….... pnWhere, pi > 0, =1 Problem with solutionsFrom 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 4Probability of getting a defective bulb=Probability of getting a non defective bulb=P(X=0)=P(no defective bulb)=p( all 4 good ones)= ()4 =P(x=1)=4c1 ( ) ()3 = P(x=2)=4c2 () 2 ()2 = P(x=3)=4c3 () 3 () =P(x=4)= () 4 = Probability distribution of X isX01234P(X)PRACTICE QUESTIONSSuppose 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?In answering a question on amultiple choice test, a student either knows the answer or guesses. Let ? be the probability that he knows the answer and ? be the probability that he guesses. Assuming that a student who guesses the answer will be correct with probability ?. What is the probability that a student knows the answer given that he answered it correctly?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.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. ................
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