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Model test paper by group CClass XII Subject MathematicsBLUE PRINT S.NOTOPIC1MARK Q4 MARKS QS6 MARKS QSTOTAL QSTOTAL MARKS1(a) (b)Relations & functionsInverse Trigonometric Functions22--4102(a) (b)MatricesDeterminants3115133(a)(b)(c)(d)(e)Continuity & ifferentiabilityApplications of DerivativesIntegrationApplication of integralsDifferential equations-8210444(a) (b)VectorsThree-dimensional geometry1124175Linear Programming--1166Probability-11210?Total613726100PRACTICE PAPER 10CLASS-XII MATHEMATICSTime: 3 hours M.Marks:100General Instructions:General Instructions:All questions are compulsory.The question paper consists of 29 questions divided into three sections A,B and C.Section A comprises of 6 questions (Q.1 to Q.6)of 1 mark each.Section B comprises of 13 questions (Q.7 to Q.19) of 4 marks each.Section C comprises of 7 questions (Q.20 to Q.26) of 6 marks each.All questions in section A are to be answered in one word, one sentence or as per the exact requirement of the question.There is no overall choice. However, internal choice has been provided. You have to attempt only one of the alternatives in all such questions.. SECTION AQ1. If f:R→R be given by fx=33-x3, find fofxQ2. Find the value of cos-1cos13π6 Q3. Construct a 2×3 matrix A=[aij] whose elements are given by aij=i-ji+jQ4. If A=α011 and B=1051, find the value(s) of α for which A2=BQ5. If A is a square matrix of order 3 and |A|= 5 , find |3A|Q6 Find a vector in the direction of a=i-2j having magnitude 7unitsSECTION BQ7. Check whether the relation R on the set of real numbers,defined asR=a,b:a≤ b2,is reflexive, symmetric or transitive. Is it an Equivalence Relation?OR Consider f: R+ [ 4 , ∞) given by f(x) = x2 + 4. Show that f is invertible with the inverse f -1 of f given by f -1(y) = ( y - 4) ? where R + is the set of all non –negative real numbers.Q8. Solve the following equation2tan-1cosx=tan-12cosecx Q9. Using properties of determinants prove that b2+c2abacbac2+a2bccacba2+b2=4a2b2c2Q10. Find the value of the constant k so that the function fx= 1-cos2x2x2 ,x≠0 k ,x=0 is continuous at x=0Q11. If x=acosθ+bsinθ and y=a sinθ-bcosθ, prove that y2d2ydx2-xdydx+y=0Q 12. Solve the differential equation ( x3 + y3 ) dy – x2 y dx =0Q13. Show that the curves x=y2and xy=k cut at right angles if 8k2=1OR Find the equation of the normal s to the curve y= x3 + 2x +6 which are parallel to the line x+ 14 y +4=0Q14. Evaluate QUOTE sinxsin4xdx ∫(x2 +1)(x2 +2)/ (x2 +3)(x2 +4) dxOREvaluate sin-1x-cos-1xsin-1x+cos-1xdxQ15. Find a unit vector perpendicular to each of the vectors a+b and a-b, where a=i+j+k , b=i+2j+3k Q 16 Evaluate 0∫π/4(sinx+cosx)/(9+16sin2x) dxQ17. Solve: x2-2y2dx+2xy dy=0, y1=1ORSolve: sec3x(dydx-2y)=1Q18. Find the mean and variance of the number of heads in a simultaneous toss of three coinsQ19. Evaluate as a limit of sum 13(2x2+5) dx SECTION CQ20. Determine the product -444-7135-3-11-111-2-2213and use it to solve the system of equations: x-y+z=4, x-2y-2z=9, 2x+y+3z=1Q21. Show that the volume of the largest cone that can be inscribed in a sphere of radius R is 827 of the volume of the sphereOR Show that the height of cylinder of maximum volume that can be inscribed in a sphere of radius ‘a’ is 2a/3Q22. Sketch the region common to the circle x2+y2=16 and the parabola x2=6y. Also, find the area of the region using integrationQ23. Find the shortest distance between the lines given by the equations r=i+j+λ 2i-j+k and, r=2i+j-k+μ 3i-5j+2k Q24. Find the equation of the plane passing through the intersection of the planes 4x-y+z=10 and x+y-z=4 and parallel to the line with direction ratios proportional to 2,1,1 . Find also the perpendicular distance of 1,1,1from this planeOR From the point (1,2,4) a perpendicular is drawn on the plane 2x+y-2z+3=0. Find the equation , the length and the co-ordinates of foot of the perpendicular.Q25. A toy manufacturer produces two types of dolls; a basic version doll A and a deluxe version doll B. Each doll of type B takes twice as long to produce as one doll of type A. The company have time to make a maximum of 2000 dolls of type A per day, the supply of plastic is sufficient to produce 1500 dolls per day and each type requires equal amount of it. The deluxe version i.e. type B requires a fancy dress of which there are only 600 per day available. If the company makes a profit of Rs 3 and Rs 5 per doll, respectively, on doll A and B; how many of each should be produced per day in order to maximize profit? Solve it by graphical method. What precaution should be taken when child is playing with plastic toys.Q26. A doctor is to visit a patient. From past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively 310,15,110 and 25. The probabilities that he will be late are 14,13, and 112, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train? Which mode of transport should be preferred and why?ANSWERSQ1. xQ2. π6Q3.0-13-1213 0-15Q4. α20α+11 =1051, so α2=1 and α+1=5which is not possible. Hence no value of αQ5. 135Q6. Let x be the radius and y be the volume. Then y=43πx3. Hence, dydx=4πx2=4π72=196π cm2SECTION BQ7 Not Reflexive: counter eg :12≤122is not true.Hence12,12does not belong to(1)Not Symmmetric: counter eg:-1≤32but 3?-12(1)Not Transitive: counter eg: 2≤(-3)2and (-3)≤12 but 2? 12(1)Hence it is not an equivalence relation(1)ORF(x1)= F(x2) => x12 +4 = x22 +4 => x1 = x2. (1)So f is one –one .For y = [4,∞), Let x ? R+ such that f(x) = y Implies, x2 +4 = y, implies x = y-4 ? R+, f is onto.(1)Therefore, f is invertible, with f -1(y) = y-4(2)Q8.We have 2tan-1cosx=tan-12cosecx ?tan-12cosx1-cos2x=tan-12cosecx (112) ?2cosxsin2x=2cosecx (1)?cosx=sinx or tanx=1?x=π4 (112)Q9. Step1-Multiplying & dividing R1,R2,R3 by a,b,c Resp & taking a,b,c common from C1,C2,C3 (12) Step2- R1→R1+R2+R3 & taking 2 common (1) Step3- R2→R2-R1, R3→R3-R1 (1) Step4-Applying R1→R1+R2+R3 & expanding (112)Q10.Since f(x) is continuous at x=0,hence lim x→0f(x)=f(0)(1)?limx→01-cos2x2x2=k (12)?limx→02sin2x2x2=k(1)?limx→0sinxx2=k (12)?12=k?k=1Hence f(x) is continuous at x=0,if k=1 (1)Q11. Solving we get x2+y2= (acosθ+bsinθ)2+ (asinθ-bcosθ)2=a2+b2(1) Differentiating we get 2x+2ydydx=0 ?dydx=-xy.........(i)(1) Differentiating againd2ydx2=-(x2+y2)y3 ..........(ii)(1) Substituting from (i) & (ii) we get y2d2ydx2-xdydx+y=0(1)Q.12. dydx= x2yx3+y3, homogeneous equation. Let y=vx, dydx=v+xdvdx.(1) Therefore, v+xdvdx.= x3vx3+x3v3(1/2) xdvdx= v1+v3-v= -v41+ v3 1+v3v4dv=-1xdx(1)(1v4+1v)dv=-1xdx(1/2) =>-13v3+logv=-logx+C=>-x33y3+logyx=-logx+C(1)Q13Finding the intersection point by solving simultaneously we get the point as P(k23,k13) (1) Differentiating x=y2 we get m1= dydxat P=12k13 (1) Differentiating xy=k we get m2= dydxat P=1k13 (1) For the curves to intersect at right angles m1 m2= -1 giving 8k2=1 (1)ORDifferentiate both sides dy/dx = 3x2 +2(1)Slope of normal = -1/ (3x2 +2)Slope of given line is -1/14(1/2+1/2)As normal is parallel to the line-1/(3x2 +2) = -1/14=> x= ±2,(1) put this values in eq. of curve we get y=18 and y=-6at point (-2,-6) eq of normal y+6 = -1/14 (x+2)at point (2,18) eq of normal y-18 = -1/14 (x-2)(1)Q14 put x2=t and using partial fractiont+1(t+2)(t+3)(t+4)=1+At+3+Bt+4(1)Comparing coff we get A= 2 and B=-6(x2+1)(x2+2)(x2+3)(x2+4)dx=1+2x2+3-6x2+4dx(2) = x+23tan-1x3-3tan-1x2 +C (1)OrI=sin-1x-cos-1xsin-1x+cos-1xdx=sin-1x-(π2-sin-1x)π2dx=4πsin-1xdx-1.dx=4πsin-1xdx-x+c (1)Put x=sin2θ and dx=2sinθcosθdθ=sin2θdθsin-1xdx=θsin2θdθ(12)Integrating by parts by taking θas the first function and sin2θ as the second function and then substituting sinθ=x, we getsin-1xdx=-12sin-1x1-2x+12x1-x (112)Therefore,I=4π-12sin-1x1-2x+12x1-x-x+cI=2π-sin-1x1-2x+x1-x-x+c(1)Q15. a+b=2i+3j+4k and a-b=-j-2k(1)a+b×a-b=ijk2340-1-2=-2i+4j-2k=c (1)c=4+16+4=26(1)cc=-16i+26j-16k (1)q.16 put cosx- sinx =t squaring we get sin 2x = 1-t2 (-sinx –cosx)dx=dt(1)now I =125-16t2 = -140log54+t54-t+C (2) =-140log5+4(cosx-sinx)5-4(cosx-sinx)+C (1)Q17. x2-2y2dx+2xy dy=0 ?dydx=2y2-x22xy Putting y=vx and dydx=v+xdvdx, we get(1)v+xdvdx=2v2-12v ?xdvdx=-12v ?2v dv=-dxx (1)?v2=-logx+c ?y2=-x2logx+cx2 (1)y1=1 ?c=1 ?y2=-x2logx+x2 (1/2+1/2)ORRewriting the equation we get . dydx+-2y=cos3xThis is of the formdydx+Py=Q, P=-2 and Q=cos3x (1)I.F.=eP dx=e-2 dx=e-2x (1)Hence the solutionye-2x=e-2xcos3xdx+c (1)Integrating the RHS by parts we get ye-2x=e-2x133sin3x-2cos3x+c (1)q.no.18 X→ No. of heads in a simultaneous toss of three coins. Then, X can take the values 0,1,2,3. (1/2)PX=0=PTTT=18 PX=1=PHTT or TTH or THT=38 PX=2=PHHT or THH or HTH=38 (1)PX=3=PHHH=18 The probability distribution of X is given byX or xi:0123PX or pi: 18383818pixi: 0386838(1/2)pixi2: 03812898 pixi=32 , pixi2=3 X=Mean=pixi=32 (1/2)VarX=pixi2-pixi2=34 (1/2)Standard Deviation =VarX=34=0.87 Q19. abfxdx=limh→0hfa+fa+h+fa+2h+…fa+n-1h, where h=b-an Here,a=1, b=3, fx=2x2+5and h=3-1n=2n(1)I=13(2x2+5) dx ? I=limh→0h212+5+21+h2+5+21+2h2+5+…21+(n-1)h2+5 49041054064000(1/2) ? I=limh→0h2n+2h.nn-12+h2nn-12n-16+5n (1)? I=limn→∞2n7n+2×2n nn-1+24n2 nn-1(2n-1)6 ? I=limn→∞14+8n-1n+83 nn-1(2n-1)n2 (1)? I=823 section cq.no.20Let A= 1-111-2-2213 and C=-444-7135-3-1Then the given product isCA=-444-7135-3-1 1-111-2-2213= 800080008?CA=8100010001=8I3(1)?A-1=18-444-7135-3-1 (1)The given system of equations can be written in matrix form as 1-111-2-2213xyz=491 AX=B where A=1-111-2-2213, X=xyz and B=491(1)Thus X=A-1B (1/2) X=18-444-7135-3-1491(1)i.e,X=18 24-16-8=3-2-1 So, x=3,y=-2 and z=-1 (112)3244215349885VBAORRCx00VBAORRCxQ21. Let VAB be a cone of greatest volume inscribed in the sphere. For max vol ,the axis of the cone must be along the diameter of the sphere. Let OC=x Then,AC=R2-x248348901968500VC=R+x (1)Volume V=13πAC2VC?V=13πR2-x22R+x (1) dVdx=13πR2-2Rx-3x2(1)For maximum or minimum, we must have dVdx=0 R2-2Rx-3x2=0 R-3xR+x=0 ?x=R3 (1)d2Vdx2=13π-2R-6x d2Vdx2x=R3=-43Rπ<0 (1)Hence V is maximum when x=R3Vmax=13πR2-R29R+R3=32πR381=827Vol of Sphere(1)OR3794760436880MBAOaLCD00MBAOaLCD35255201031875xxxxx00xxxxxLet r be the radius of the base and h be the height of the cylinder ABCD which is inscribed in a sphere of radius a. Let OL=x 51441358001000 (1)OA2=OL2+AL2 AL=a2-x2 4700270147955x00xVolume of cylinder, V=πa2-x2×2x (1) V=2πa2x-x3 dVdx=2πa2-3x2 (1)For maximum or minimum, we must havedVdx=0?2πa2-3x2=0?x=a3 (1)d2Vdx2=-12πx d2Vdx2x=a3=-12π×a3<0 (1)Hence, V is maximum at x=a3 and hence LM=2a3 (1)4780280641350047802806413500(1)2915412264160YY’XX’O00YY’XX’O271018020955B0,4A23,2C-23,2D0,2 x2+y2=16 x2=6y00B0,4A23,2C-23,2D0,2 x2+y2=16 x2=6yQ22. Putting y=x26 in x2+y2=16, we get x2+x436=16?x=±23 3942080292100044215051460500043567351460500037592001911350036245805778500For x=±23,y=2 3830320130175003942080276225004149090483235001750060717550017500607175500(112)Intersection points: -23,2 and 23,2OADO=02xdy=026ydy=833 (1)224218521145500DABD=24xdy=2416-y2dy=83π-23 22428206667500 (112) Required area =2OADO+DABD=433+16π3sq.units (1)Q23. a1=i+j, b1=2i-j+ka2=2i+j-k, b2=3i-5j+2k (1)a2-a1=i-k (1)b1×b2=ijk2-113-52=3i-j-7k (1)b1×b2=59 (1)d=b1×b2.a2-a1b1×b2=1059(2)Q.No. 24 Equation of plane passing through the intersection of the two given planes is4x-y+z-10+λx+y-z-4=0 (1)x4+λ+yλ-1+z1-λ-10-4λ=0 (12)This plane is parallel to the line with direction ratios 2,1,1. Hence,24+λ+1λ-1+11-λ=0?λ=-4 (112)Hence the equation of the plane is5y-5z-6=0 (1)Length of perpendicular from 1,1,1 on the plane is given byd=5×1-5×1-652+-52=325 (2)ORLet Q be the foot of perpendicular from point P(1,2,4)Therefore eq of PQ perpendicular to the plane is x-12=y-21=z-4-2=k(2)Therefore point Q(2k+1, k+2 , -2k+4) which lies on the plane 2x+y-2z +3=02(2k+1)+1(k+2)-2(-2k+4)+3=0(2)Therefore k=1/9And foot of perpendicular Q(11/9 ,19/9 ,34/9)Length of perpendicular PQ = 1/3(1+1)QNo.25. Let x dolls of type A and y dolls of type B be produced per day to maximize the profit. The mathematical form of the given linear programming problem is as follows: Maximize Z=3x+5y Subject to x+2y≤2000x+y≤1500 y≤600 x,y≥0 (2)-340360147955x+2y=2000YY’X’XOy=600x+y=1500B20,1500A12000,0P1000,500R0,600A21500,0Q800,60000x+2y=2000YY’X’XOy=600x+y=1500B20,1500A12000,0P1000,500R0,600A21500,0Q800,600(2)Corner points : O0,0A21500,0P1000,500Q800,600R0,600Z : 04500550054003000 (112)Hence 1000 dolls of type A and 500 dolls of type B should be produced to maximize the profit and the maximum profit is Rs.5500 (1/2)Q.NO.26.Let E1,E2,E3,E4, be the events that the doctor comes by train,bus,scooter and other means of transport resp.Then,PE1=310,PE2=15,PE3=110,PE4=25(1)Let A be the event that the doctor visits the patient late.ThenPA/E1=Probability that the doctor will be late if he comes by train?PA/E1=14PA/E2=Probability that the doctor will be late if he comes by bus?PA/E2=13PA/E3=Probability that the doctor will be late if he comes by scooter?PA/E3=112PA/E4=Probability that the doctor will be late if he comes by other means of transport?PA/E4=0 (1)To find PE1/ABy Baye’s theorem,PE1/A=PA/E1P(E1)PA/E1PE1+PA/E2PE2+PA/E3PE3+PA/E4PE4 (1)?PE1/A=310*14310*14+15*13+110*112+25*0=340*12018=12 (1/2)Hence the required probability = 12 (1/2) ................
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