CLASS XII (2019-20) MATHEMATICS (041) SAMPLE PAPER-1
Mathematics XIISample Paper 1 Solvedwww.rava.org.inCLASS XII (2019-20)MATHEMATICS (041)SAMPLE PAPER-1Time : 3 HoursMaximum Marks : 80General Instructions :(i) All questions are compulsory.(ii) The questions paper consists of 36 questions divided into 4 sections A, B, C and D.(iii) Section A comprises of 20 questions of 1 mark each. Section B comprises of 6 questions of2 marks each. Section C comprises of 6 questions of 4 marks each. Section D comprises of4 questions of 6 marks each.(iv) There is no overall choice. However, an internal choice has been provided in three questionsof 1 mark each, two questions of 6 marks each. You have to attempt only one of thealternatives in all such questions.(v) Use of calculators is not permitted.7 83 63. A , B H, 2A 3B ?H5 65 4Section-ADIRECTION : (Q 1-Q 10) are multiple choicetype questions. Select the correct option.1. The operation ) is defined as a ) b 2a b, then 2 ) 3h ) 4 is.[1](a) 18(b) 17(c) 19(d) 21Ans : (a) 1827 24(a) 22 10H27 36(b) 25 10H27 36(c) 25 15H27 36(d) 35 10H27 36Ans : (b) 25 10HWe have,We have,3 67 8 3 H2A 3B 2 H5 45 6a ) b 2a b 2 ) 3h ) 4 2 # 2 3h ) 4 7 ) 4 2 # 7 4 1822. cos 1 1 x2 [1]1 x(a) 2 cos 1 x(c) 2 tan 1 x(b) 2 sin 1 x(d) cos 1 2xAns : (c) 2 tan 1 x.(i)f (x) cos 1 1 x21 xLet, x tan θ and substituting in equation(i)2Let,f (x) cos 1 1 tan2 θ1 tan θ21 tan θ cos 2θ1 tan2 θ2ButTherefore,[1]f (x) cos 1 cos 2θ 18cos cos t h tB 2θ 2 tan 1 x6 1221 24 H10 815 18H27 36 25 10H4. Ifx2773x6and 2 are two roots of the equation7[1]2 0 then the third root isx(a) 9(c) 12Ans : (a) 9(b) 14(d) None of thesex 3 7We have 2 x 2 07 6 xx x2 12h 3 2x 14h 7 12 7x h 0x3 12x 6x 42 84 49x 0x3 67x 126 0To Get 20 Solved Paper Free PDF by whatsapp add 91 89056 29969 in your class GroupPage 1
Mathematics XIISample Paper 1 Solved x 7h x2 7x 18h 0dy y sec x tan xdxI.f. e # secx dx e log sec x tan x h x 7h x2 9x 2x 18h 0 x 7h x 2h x 9h 0 sec x tan xx 7, 2, 95.xd (2 ) d (3x )[1]x 1(b) 2x 13x(d) b 2 l log 2 33x(a) b 2 l3x(c) b 2 l log 3 23f x h 8. xit 3tj 5kt, xit xjt 2kt are perpendicularto each other then x [1](b) 2, 5(a) 2 , 5(d) 2, 5(c) 2 , 5Ans : (d) 2, 5andFor av bv,d 2x hd 3x hav bv x2 3x 10.(i) x2 3x 10hd 2x 2x log 2dx hd 3x h 3x log 3 d ax ax log a:dx hDdxPutting values in eq.(i),f x h av bv 0x2 3x 10 0x2 5x 2x 10 0x x 5h 2 x 5h 02x log 23x log 3(x 2) (x 5) 0 b 2 l log 3 23xlog a; log b log baE6. If y x2 3x 4 , then the slope (gradient)of the normal to the curve at (1, 1) is[1]1(a) 5(b) 5(c) 8(d) 181Ans : (b) 5We have,y x2 3x 4Differentiating both sides w.r.t. x , we getdy 2x 3dxdy 2 3 5dx 1, 1Hence, Slope of normal dy 1 15dx ix 2, 59. The coordinates of the midpoint of the linesegment joining the points (2, 3, 4) and[1](8, 3, 8) are(a) (10, 0, 12)(c) (6, 5, 0)7. Integrating factor of the differential equationdy[1] y sec x tan x isdx(a) sec x tan x(b) sec x tan x(c) sec x(d) tan x sec xAns : (a) sec x tan x(b) (5, 6, 0)(d) (5, 0, 6)Ans : (d) (5, 0, 6)We haveA (2, 3, 4) B (8, 3, 8)Mid point of line segment joining (x1, y1, z1)and (x2, y2, z2) isy y 2 z1 z 2P b x1 x 2 , 1,222 lHere,x1 2 y1 3 z1 4x2 8 y2 3 z2 8P b 2 8, 3 3, 4 8 l222P (5, 0, 6)(1, 1)We have,av xit 3tj 5ktbv xit xjt 2ktav bv 0Let,xAns : (c) b 2 l log 3 23We havewww.cbse.online10. If A and B are two events such that P (A) ! 0[1]and P b B l 1A(a) B 1 A(c) B φ(b) A 1 B(d) A B φAns : (b) A 1 BDownload 20 Solved Sample Papers pdfs from www.cbse.online or www.rava.org.inPage 2
Mathematics XIISample Paper 1 SolvedPb B l AP B AhP Ahwww.rava.org.in 1 3sin b 2 sin 5 l sin 2xNow,[Conditional probability]P B Ah 1P Ah 2 sin x cos x 2 # 3 # 4 242555B;P b A l 1E&P B Ah P ]Ag&A1BHence,:a sin 1 3 x D5sin b 2 sin 1 3 l 24255(1/2)14. cos 1 (2x 1) .(a) 2 cos 1 x(c) 2 cos 1 x11. If the binary operation * defined on Q , isdefined as a * b 2a b ab , 6a, b d Q ,then the value of 3 * 4 is .[1]We have, sd a * b 2a b ab3 * 4 2 3h 4 3h 4h orA cos 1 xThen given equationvalue of λ is . .[1]Ans :Given that, av 2it tj kt, bv it 2tj 3ktand cv 3it λtj 5kt are coplanar.7a 6av bv cv@ 0AorThe principal value of sin 1 cLet,sin θ sin 6020 6λ 4 λ 6 013. sin b 2 sin 1 3 l is equal to .5Ans :We have, sin b 2 sin 1 3 l5Let,sin 1 3 x5sin x 35andcos x 1 sin2 x1 9 42525[1]3 θ2 m3 sin θ2[expanding along R1 ](1/2)sin 1 c3 is .2 m(b) π6(d) π3(a) 2π3(c) π4Ans : (d) π32 10 3λh 1 5 9h 1 λ 6h 0λ 2 cos 1 cos 2Ah 2A .(ii)cos 1 (2x 1) 2 cos 1 x(1/2)10 5λ.(i)cos 1 (2x 1) cos 1 2 cos2 A 1hFrom Eq.(i),(1) 6 4 12 212. If the vectors: av 2it tj kt, bv it 2tj 3ktand cv 3it λtj 5kt are coplanar, then the x cos2 ALet,Ans : (b) cos 1 x(d) None of theseAns : (c) 2 cos 1 xQ. 11-15 (Fill in the blanks)2 1 11 2 3 03 λ 5[1]orθ 6060 # π π180315. If y x2 3x 4 , then the slope (gradient)of the normal to the curve at (1, 1) is .[1](a) 5(b) 15(c) 8(d) 181Ans : (b) 5We have,y x2 3x 4Differentiating both sides w.r.t. x , we get(1/2)To Get 20 Solved Paper Free PDF by whatsapp add 91 89056 29969 in your class GroupPage 3
Mathematics XIISample Paper 1 Solveddy 2x 3dxdy 2 3 5dx 1, 1Hence, Slope of normal dy 1 15dx ior(1, 1)The angle which the tangent to curve y x2at (0, 0) makes with the positive direction of[1]x -axis is .(a) 90c(b) 0c(c) 45c(d) 30cwww.cbse.onlinecos θ sin θA sin θ cos θHcos θ sin θAT sin θ cos θHWe have,2 cos θ01 0AT A HH02 cos θ0 12 cos θ 1cos θ 12 π/2# x sin xdx17. EvaluateAns :Given:y xDifferentiating both sides w.r.t., we getdy 2xdxdy 0dx (0, 0)Thus, tangent is parallel to x axis.So, angle between tangent at (0,0) with x-axis is 0c.4 63 616. Find matrix X , if X [1] H 3 75 8HAns :Let3 64 6H and B A 5 8H 3 7Then, the given matrix equation is X A B X A B&0I II#0π/21 # cos x h dx[Integrating by parts]π/2 b π cos π 0 cos 0 l c sin π sin 0 m222 118. Examine the continuity of the functionZ sin x]x!0,f x h [ x] 1,x 0 at x 0\Ans :[1]f 0 h 1We have,lim f x h lim f 0 h hX B Ah3 6 4 6 H5 83 7Hx " 0 h"0sin 0 h hh"0 0 h hsin h lim lim sin hh"0h"0 hh lim 1lim f x h lim f 0 h hx " 0 1 12X 8 15Hh"0sin h hh"0 h sin h limh"0 h lim sin hh"0 h limcos θ sin θ, then find the value ofA sin θ cos θHθ a where, θ ε a0, π kk satisfying the equation2A A I2 .Ans :x sin xdx 6 x cos x @0π/2 6sin x @0orTπ/2 6 x cos x @0π/2 3 4h 6 6h H5 3 8 7h 1 12 8 15HHence,#I Let[1]0Ans : (b) 0c2θ π3 1 lim f x h ! lim f x hh " 0 x " 0 Download 20 Solved Sample Papers pdfs from www.cbse.online or www.rava.org.inPage 4
Mathematics XIISample Paper 1 SolvedSo, lim f x h does not exist.Hence, f x h is discontinuous at x 0 .7it 5tj 3kt.Ans :av 3it 7tj 5ktbv 5it 7tj 3ktandcv 7it 5tj 3kt.We know that, the volume of a parallelopipedwhose three adjacent edges are, av, bv, cv, isequal to 6av bv cv@ . 3 7 5vvvHere, 6abc @ 5 7 37 5 3Let6abcv v v@ 3 21 15h 7 15 21h 5 25 49h6abcv v v@ 108 252 120 264Hence, volume of the parallelopipedv v v@ 264 6abcConsider the eventB "HH ,A k B "HH ,or#Evaluate26x2@dx , where 6 @ is the greatest0integer function.Ans :Z]0, when 0 x 1]1, when 1 x 2Here, 6x2@ [2, when 2 x 3]]3, when 3 x 2\26x2@dx #010dx #211dx #32[2](1/2)2# 3dx2dx 3(1/2) 0 6x @1 62x @2 63x @322 2 1h 2 3 32 h 3 2 3h(1/2)2 1 2 3 2 2 6 3 32 (1/2)3Give vertices are k, 0h, (4, 0) and (0, 2).Also, Area of a triangle 4 sq units.k 0 11 4 0 1 ! 420 2 1(1)6k 0 2h 1 8 0h@ ! 8Section B0 1 6log 6 log 1@ 1 log 655(1)Ans :P Ah 3/4P B h 1 , P A k B h 144 Required probability1P A k B hP B/Ah 43 13P Ah42x dx .5x2 1(1)22. Find the value of k , if the area of a triangleis 4 sq units and vertices are k, 0h, 4, 0h and[2] 0, 2h.A "HT, TH, HH ,1t 5x2 1 & t 1When, x 1, t 5x2 1 & t 61dt2x dx 6 2x ## 10x#2t5x1 106 1 # dt 1 6log t @165 1 t5 5 B Getting two heads#When, x 0 , A Getting at least one head02x dx5x2 110xdx dt020. Two coins are tossed. What is the probabilityof coming up two heads if it is known that atleast one head comes up.[1]Ans :1Then, d 5x2 1h dt# 264 cu units#5x2 1 tLet19. Find the volume of a parallelopiped whosesides are given by[1]tttttt 3i 7j 5k, 5i 7j 3k and21. EvaluateAns :I Given,x"0&www.rava.org.in[expanding along R1 ][2] 2k 8 ! 8 2k 8 8 or 2k 8 8To Get 20 Solved Paper Free PDF by whatsapp add 91 89056 29969 in your class GroupPage 5
Mathematics XIISample Paper 1 Solved 2k 0 or 2k 16www.cbse.onlinecos x andk 0 or k 8Hence, the value of k is 0 or 8.(1)23. Using properties of determinants, prove thata b c2a2a2bb c a2b a b c h3[2]2c2cc a bWe know that,On applying R1 " R1 R2 R 3 , we geta b c a b c a b cT 2bb c a2b2c2cc a b tan b 1 sin 1 3 l 24On taking a b c h common from C1 andC2 we get0 01T a b c h 1 12b0 1 c a b3orProve that, cot d π 2 cot 1 3 n 7 .4Ans :Let,LHS cot d π 2 cot 1 3 n4We have,2 cot 1 3 cot 1 b 9 1 l62 cot 1 3 cot 1 43 cot d π 2 cot 1 3 n cot b π cot 1 4 l344cot π4 cot cot 1 43 i 1cot cot 1 43 h cot π44 1 7 RHS 343 1Hence proved. (1)25. Rajeev appears for an interview for two postsA and B for which selection is independent.The probability of his selection for post Ais 1/5 and for post B is 1/6. He preparewell for two posts by getting all the possibleinformations. What is the probability that heis selected for atleast one of the post?[2]Given that,(1)[2]P (selection for post A) P Ah 15andP (selection for post B ) P B h 16 P (not selected for post A) P A h 1 P Ah 1 1 455LHS tan b 1 sin 1 3 l24sin 1 3 x & sin x 344(1) T a b c h3 60 1 0 1h 0@We have,7434Ans :On expanding along C1 , we get24. Show that, tan b 1 sin 1 3 l 4 7 .234Ans :1 21 a 2 cot 1 x cot 1 c x 2 x mFOn applying C1 " C1 C2 and C2 " C2 C 3, we get0012bT a b c h b c a a b c h0c a b c a b3(1) 4 7 R.H.S.3Hence proved. (1)aOn taking a b c h common from R1 , weget111(1)T a b c h 2b b c a2b2c2cc a b a b c h1 9 7164tan x 1 cos x2sin xAns :To prove,a b c2a2a2bb c a2b a b c h22c2cc a ba b c2a2aLet,T 2bb c a2b2c2cc a b 1 sin2 xand P (not selected for post B ) P ]B gDownload 20 Solved Sample Papers pdfs from www.cbse.online or www.rava.org.inPage 6
Mathematics XIISample Paper 1 Solvedwww.rava.org.in 1 P ]B g 1 165(1) 6 P (Rajeev is selected for atleast one post) 1 P (Rajeev is selected for none of them)On differentiating both sides of w.r.t. x , wegetdu e sinx d sin xhdx dxdu e sinx cos x.(iii)&dx 1 P A h P B h(1) 1 4 #5 1 2 1563326. If av it tj kt, bv 4it 2tj 3kt andcv it 2tj kt, then find a vector of(1)log v x logtan xOn differentiating both sides w.r.t. x , we get1 dv x 1 sec2 x logtan x 1v dxtan x[by product rule of derivative] 1 P A k B hmagnitude 6 units, which is parallel to thevector 2av bv 3cv.[2]Ans :Let the vector, rv λ 2av bv 3cvh&rv λ (2it 2tj 2kt 4it 2tj 3kt 3it 6tj 3kth rv λ it 2tj 2kthrv λ it 2tj 2ktAs,rv 6 λ(1)[given]1 4 4 6 & λ 3 6λ !2Therefore,rv ! 2 it 2tj 2kth ! 2it 4tj 4kthwhich is the required vector.&&dv tan x x log tan x 2x cosec 2x h hdx.(iv)6put v tan x hx@On putting the values of du and dv fromdxdxEqs. (iii) and (iv) in Eq. (ii), we getdy esinx cos x tan x hx 6log tan x 2x cosec 2x @dx(1)28. EvaluateAns :(1)Section C27. Differentiate the equation y e sinx tan x hxw.r.t. x .[4]Let,&#π/20I I sin2 xdx .sin x cos x#Let,π/20π/2#0;aAns :Given that,y e sinx tan x hxThen,.(i)y u vOn differentiating both side of Eq. (i) w.r.t.x , we getdy.(ii) du dvdx dxdxNow, consider u e(1)#π/22I #π/22I #.(i)sin2 c π x m2dxsin a π x k cos a π x k22#af x h dx #a0f a x h dx Ecos2 x.(ii) (1)dxsin x cos x0On adding Eqs (i) and (ii), we get&u e sinx and v tan x hxI [4]sin2 xdxsin x cos x0sinx(1)& 1 dv 2x cosec 2x log tan xv dx1:a 2 2 cosec 2x Dsin x cos xsin 2xav it tj kt,bv 4it 2tj 3ktcv it 2tj kt&2& 1 dv x sec x logtan xtan xv dxx logtan x& 1 dv v dxsin x cos xGiven that,andandv tan x hxTaking log on both sides, we get&00π/2sin2 x cos2 x dxsin x cos xdxsin x cos x6a sin2 x cos2 x 1@To Get 20 Solved Paper Free PDF by whatsapp add 91 89056 29969 in your class GroupPage 7
Mathematics XII&Sample Paper 1 Solved2I #π/2Ans :dx2 tan x/21 tan x/2 21 tan x/2 1 tan2 x/22 tan x/21 tan2 x/2 andcosx a sin x H1 tan2 x/21 tan2 x/22I 20#π/20sec2 x/2dx2 tan x 1 tan2 x222x:a 1 tan sec2 x D (1)22&2I 2 #[adding and subtracting 1 from denominator](1)1dt&2I 2 #220 t 1h 2 h12t12 &2I logG2 2t 1 2 021 dx 1 log x a G2ax a a2&I 1 log 1 1 2 log 1 2 H2 21 1 2 1 2 I 1 log 1 log2 2 1 0 log2 2f x h dx0where, f x h x x 1 x 2Now,I # #3003x dx xdx 01#x3#0#013x 1 dx x 1h dx # x 2h dx dt20 1 t 2twhen x π , then t π 124 Hand when x 0, then t tan 0 01dt2I 2 #20 6t 2t 1@ 2 a#I Let,2Put, tan x t & 1 sec2 x dx dt222 www.cbse.online2 1G2 12 1G2 17a log 1 0A#30x 2 dx# x 1hdx31# x 2hdx3(1)2x, if x 0 a x * x if x 04H32 222 12 3 ;x E ;x x E ;x x E ;2x x E2 02 022 013 ;x 2x E (1)222 9 ;b1 1 l 0 0hE ;b 9 3 l b 1 1lE2222 ;b 4 4 l 0 0hE ;b 9 6 l b 4 4 lE (1)222 9 1 3 1 4 3 4 1922 2 2 2 2 2 2(1)29. For the matrices A and B , prove thatR1VS W AB hl BlAl, if A S 4W and B 6 1 2 1@.SS 3 WW[4]T XAns :R 1VS WGiven that,A S 4WSS 3WWT XandB 6 1 2 1@To prove, AB hl BlAlR 1VS W12 12 1 log G#Now,AB S 4W 6 1 2 1@1 # 3 (1)2 22 12 1SS 3WW[multiplying numerator and denominator byT X3 # 1R 1 2 1V 2 1h]WS2&AB 4 8 4WS 2 1hSS 3 6 3WW 1 log2 12 2RT 1 4 3VXWS1(1) log 2 1hl.(i)AB 2 86WS h2SS 1 4 3WWorXT3[Interchange rows and columns](1)Evaluate # f x h dx , whereR V0S 1WNow,Bl S 2W and Al 61 4 3@f x h x x 1 x 2 .SS 1WWT XDownload 20 Solved Sample Papers pdfs from www.cbse.online or www.rava.org.inPage 8
Mathematics XIISample Paper 1 Solvedwww.rava.org.in(1)cos B .(ii)LHS cos e cos 1(1)[4]1 x .2 x26a cos cos 1 θh θ@Ans :31. EvaluateTo prove,Now,Ans :1 x22 x2Let,# 5 dx4 cos xI # #.(i)Put,LHS cos tan )sin c sin 1LHS cos tan 1 c 1(1)&1H1 x2 m3I (1)[4]# 5 dx4 cos xLHS cos 8tan 1 sin cot 1 x h.Bcot 1 x A & cot A x1sin A 1 x21A sin 1 c1 x2 mFrom Eq. (i), we get1 x2o2 x21 x2 RHS2 x2 2cos 8tan 1 sin cot 1 x h.B (1)1 x2o2 x2B cos 1 e&From eq. (ii), we get30. Prove that,cos 8tan 1 sin cot 1 x h.B 11 x21 x22 x2tan B R VS 1W BlAl S 2W 61 4 3@SS 1WWRT XVS 1 4 3W&BlAl S 2 8 6WSS 1 4 3WWTXFrom Eq. (i) and (ii), we get AB hl BlAl#dx2xa1 tan 2 k5 42xa1 tan 2 kRVS1 tan2 x W2WSa cos x x2S1 tan W2XTsec2 x dx22x5 c1 tan m 4 c1 tan2 x m22xx a 1 tan2 sec2 F222xsec dx2(1)9 tan2 x2Put, tan x t & 1 sec2 x dx dt2221m1 x2 G 18a sin sin θ h θB .(ii) (1)I 2#dt 2# 3hdt2 t29 t2 2 tan 1 t C33;a# a dx x22(1) 1 tan 1 x E (1)aatan x p2 1 f2 C tan33Again putting,1tan 1 dn B1 x2:Put t tan x D (1)2Evaluate#orsin 2xdx . 1 sin x h 2 sin x hTo Get 20 Solved Paper Free PDF by whatsapp add 91 89056 29969 in your class GroupPage 9
Mathematics XIISample Paper 1 SolvedAns :# sin x cos x sin x dx C # sin x cos x dx Cy sin x I Let,www.cbse.online2xdx# 1 sinsinx h 2 sin x h 2#y sin xy sin x 6a sin 2x 2 sin x cos x @ (1)233 sin x C3(1)t 2A B h A B h tOn comparing the coefficients of t andconstant term, we get(1)(1)[Put t sin x ]Section D33. Find the area of the region bounded by theparaboles y2 4ax and x2 4ay .[6]Ans :Given parabolas are,y2 4ax and x2 4ay .For points of intersection of these parabolas,considerA 1 and B 2I 2 ; # dt # 2dt E1 t2 t # t dt Cy sin x t C y sin x3Put,2A B 0 and A B 1On solving the above equations, we get(1)Put, sin x t & cos x dx dtsin x cos x dx1 sin x h 2 sin x h sin x t & cos x dx dtt I 2#dt 1 t h 2 t htLet, A B 1 t h 2 t h 1 t 2 tt A 2 t h B 1 t h26a x2 4ay@2 2c x m 4ax4a 2 log 1 t 4 log 2 t C 2 log 1 sin x h 4 log 2 sin x h C6put t sin x @ (1)32. Show that the differential equation:dysin x cos x h y cos x sin2 x is lineardxand also solve the differential equation. [4]x 4 64a3 x&&x x3 64a3h 0 & x 0, 4awhen x 0 , then y 0when x 4a , then y 4a(1)So, the points of intersection are (0,0) and 4a, 4a h.Ans :Given differential equation is,dysin x cos x h y cos x sin2 xdxdy cot x h y cos x sin xdx[dividing both sides by sin x ]which is a linear differential equation of theform,dy Py Q ,dxwhere P and Q are the functions of xHere,Now,(1)P cot x and Q cos x sin xIF e# Pdx e# cotxdx e# IF Q hdx C Area of the shaded region Area of region ODBCO Area of region ODBAOlog e sinx sin xTherefore, the required solution is,y IF (2)(1) #4a0y parabola y2 4ax h dx # y parabola x2 4ay h dx (1)4a0 #04a4ax dx Download 20 Solved Sample Papers pdfs from www.cbse.online or www.rava.org.in#04ax2 dx4aPage 10
Mathematics XIISample Paper 1 Solvedwww.rava.org.in21 2 6x3/2@0 ;2x x E32 12 2 a x H 1 ; x E34a 3 02 03 4a h3/2 2 a 1 ; 64a E34a 32 4 # 8a2 16 a2333/2 4a3 4a 2 1 0h ;b 4 4 l b 2 1 lE322 2 4 33 2 2 2 1 7 sq units.3 26 32 a2 16 a233 16 a2 sq units.3or(2)Find the area of the region included betweenthe parabola y2 x and the line x y 2and the X -axis.Ans :Given parabola is y2 x and line x y 2 .(2)34. A merchant plant to sell two types ofpersonal computers, a desktop model anda portable model that will cost 25000 and 40000, respectively. He estimates that thetotal monthly demand of computers will notexceed 250 units. Find the number of units ofeach type of computers which the merchantshould stock to get maximum profit, if hedoes not want to invest more than 70 lakhand his profit on the desktop model is 4500and on the portable model is 5000. Make anLPP and solve it graphically.[6]Ans :Let merchant stock x units of desktop modeland y unit of portable model.Total profit to maximise,Z 4500x 5000ySubject to constraints are,x y # 250(i)25000x 40000y # 7000000(2)y2 2 y6a x y 2@5x 8y # 1400&(1)andx 0, y 0Firstly, draw the graph of line x y 250&y2 y 2 0x0250&y2 2y y 2 0y2500&& y y 2h 1 y 2h 0On putting (0,0) in the inequality. y 1h y 2h 0y 1, 2x y # 250 , we getWhen y 1, then x 1 and when y 2 ,then x 4 .So, the points of intersection are (1,1) and 4, 2h.(1) Area of the shaded region Area of region OACO Areaof region ABCA # #1001y parabolah dx x dx # y linehdx21# 2 x hdx2.(ii)(1)0 0 # 250 & 0 # 250 , which is true.So, the half plane is towards the origin. (1)Secondly, draw the graph of line 5x 8y 1400x0280y1750On putting (0,0) in the inequality 5x 8y# 1400 , we get5 0 h 8 0 h # 1400&0 # 1400 . which is true. (1)So, the half plane is towards the origin.Also, x 0 , y 0 , it means feasible region1To Get 20 Solved Paper Free PDF by whatsapp add 91 89056 29969 in your class GroupPage 11
Mathematics XIISample Paper 1 Solvedlies in 1st quadrant.The point of intersection of lines (1) and (2)is E 200, 50h.Then,www.cbse.onlinec2 Surface area of the open box x2 4xy&Now,&(1)22y c x4xV Volume of the box22V x2 y x2 c c x m4x.(i) (1)V 1 c2 x x3h4On differentiating both sides twice, w.r.t. x, we getdV 1 c2 3x2h4 dx&(2)Now, we draw all the lines on a graph paper,the feasible region is OAECO . The cornerpoints of the feasible region are O 0, 0h,A 250, 0h, E 200, 50h and C 0, 175h.CornerPointsA 250, 0hE 200, 50hC 0, 175hO 0, 0hZ 4500x 5000yZ 4500 # 250 5000 # 0 1125000Z 4500 # 200 5000 # 50 11500002and d V2 1 6x h 3 x.(ii) (1)24dxFor maximum or minimum value, putdV 0 .dx(1)& c2 3x2 0 & x ! c3As x (length of side) can never be negative,so x c .3On putting x c in Eq.(ii), we get32d V 3 c2dx23Z 4500 # 0 5000 # 175 875000Z 4500 # 0 5000 # 0 0Clearly, the maximum profit is 1150000 atE 200, 50h, i.e., when 200 desktops and 50portable models are in stock.(1)35. An open box with a square base is to be madeout of a given quantity of metal sheet of areac2 . Show that maximum volume of the box is[6]c3 /6 3 .Ans :Let, x be the side of the square base and ybe the height of the open box.6a c 0@ (1) 3c 02So, V is maximum at x c33 Maximum value of V 1 ;c2 c b c l E433[from eq. (i)]3 c6 6(1)orFind all the points of local maxima and localminima of f x h x 2 sin x on 60, 2π@. Also,find local maximum and minimum values.Ans :We have,f x i x 2 sin xOn differentiating both sides w.r.t. x , we get.(i) (1)f l x h 1 2 cos xFor local maxima and local minima, putf l x h 0 .& 1 2 cos x 0 & cos x 12Download 20 Solved Sample Papers pdfs from www.cbse.online or www.rava.org.inPage 12
Mathematics XII&Sample Paper 1 Solvedx 2n π ! π & x π ,33system of equations by matrix method.5π d 0, 2π(1)6@3On differentiating both sides of Eq. (i) w.r.t.x , we getf ll x h 2 sin xAtx π , f lld π n 2 sin π333 3 0(1) x π is a point of local maxima3At,x 5π , f llb 5π l33 2 sin 5π 3www.rava.org.in3 0 x 5π is a point of local minima.3Hence, the points of local maxima is π and3(1)local minima is 5π .3On putting x π in f x h, we get3f b π l π 2 sin π333 π 2 # 332(1) π 33which is the required local maximum value.On putting x 5π in f x h, we get3f b 5π l 5π 2 sin 5π333 5π 2 # c 3 m32 5π 33which is the required local minimum value. (1)36. A manufacturer produces three models oftoys in the form of bikes say X, Y and Z. Model X takes as 10 man-hour to makeper unit, Model Y takes 5 man-hour per unitand model Z takes 4 man-hour per unit.There are a total 212 man-hour available perweek. Handling and marketing costs are 20, 30 and 40 per unit for models X, Yand Z respectively. The total funds availablefor these purposes are 920 per week. Profitsper unit for models X, Y and Z are 40, 10and 70 respectively, but at the end of theweek, company get a profit of 810. Solve the[6]Ans :let x, y and z denote the number of bikes ofmodels, X, Y and Z , respectively.(i) Given conditions in linear equations are,10x 5y 4z 212 ,20x 30y 40z 920and 40x 10y 70z 810Above system of equation can be written inmatrix form as.RVR V R VS10 5 4 WSx W S212WS20 30 40WSy W S920WSS40 10 70WWSSz WW SS810WWTXT X T Xi.e.(i)AX B & X A 1 BRx VR10 5 4 VS WSWwhere,A S20 30 40W, X Sy WSSz WWSS40 10 70WWT XTR VX212S W(1)andB S920WSS810WWT10 X5 4Now,A 20 30 4040 10 70 10 2100 400h 5 1400 1600h 4 200 1200h [expanding along R1 ] 10 1700h 5 200h 4 ] 1000g 17000 1000 4000(1) 14000Now, the cofactors of determinant A areA11 30 40 2100 40010 70 1700A12 20 4040 70 1400 1600h 200A13 20 30 200 120040 10 1000A21 5 4 350 40h10 70 31010 4 700 16040 70To Get 20 Solved Paper Free PDF by whatsapp add 91 89056 29969 in your class GroupPage 13A22
Mathematics XIISample Paper 1 Solvedwww.cbse.online 540A23 10 5 100 200h40 10 100A 31 5 4 200 12030 40 80A 32 10 4 400 80h20 40 320A 33 10 5 300 10020 30(2) 200VTRSA11 A12 A13W adj. Ah SA21 A22 A23WSSA A A WW313233RT 1700 200X 1000V TWS100W S 310 540SS 80 320200WWRT 1700 31080VXWS S 200 540 320WSS 1000 100 200WWXadj AhT 1Now, A AR 1700 31080VWS 1 S 200 540 320W (1)14000 SS 1000 100 200WWTXFrom Eq. (i), we getRx VR 1700 31080VWRS212VWS WSSy W 1 S 200 540 320WS920WSSz WW 14000 SS 1000 100 200WWSS810WWT XTXT XR 360400 285200 64800VSW 1 S 42400 496800 259200W14000 SS 212000 92000 162000WWRTV R VXS140000W S10W 1 S280000W S20W14000 SS 42000 WW SS 3 WWTX T X(1) x 10 , y 20 and z 3WWW.CBSE.ONLINEDownload unsolved version of this paper fromwww.cbse.onlineThis sample paper has been released by website www.cbse.online for the benefits of the students. This paper has been prepared by subject expertwith the consultation of many other expert and paper is fully based on the exam pattern for 2019-2020. Please note that website www.cbse.online isnot affiliated to Central board of Secondary Education, Delhi in any manner. The aim of website is to provide free study material to the students.Download 20 Solved Sample Papers pdfs from www.cbse.online or www.rava.org.inPage 14
Mathematics XII Sample Paper 1 Solved www.rava.org.in To Get 20 Solved Paper Free PDF by whatsapp add 91 89056 29969 in your class Group Page 3 Now, P A blB PA PB A h h [Conditional probability] PA PB A h h 1 P A;EblB 1 & PB A h PA]g & AB1 Q. 11-15 (Fill in the blanks) 1
Number of Clusters XII-9. C) Overlap XII-10. D) An Example XII-10. 5. Implementation XII-13. A) Storage Management XII-14. 6. Results XII-14. A) Clustering Costs XII-15. B) Effect of Document Ordering XII-19. Cl. Search Results on Clustered ADI Collection . XII-20. D) Search Results of Clustered Cranfield Collection. XII-31. 7. Conslusions XII .
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CLASS - L K G St Pauls Vallabh Nagar Year - 2020-21 . 8 Laboratory Manual Math IV APC 9 Learning Science IV Cordova 10 Sahaj Hindi Gram IV Learnwell . 7 Lab Manual Physics XII Elite 8 Lab Manual Chemistry XII Elite 9 Mathematics I XII NCERT 10 Mathematics II XII NCERT 11 Info. Practics XII NCERT
TEKNIK PEMESINAN GERINDA 1 Program Studi: Teknik Pemesinan Kode: TM.TPM-TPG 1 (Kelas XII-Semester 5) . Teknik Pemesinan Frais (TM.TPM-TPF) Teknik Pemesinan Bubut (TM.TPM-TPB) TM.TPM- TPB1 (XI-3) (XII-5) TM.TPM- TPB 2 (XI-4) TM.TPM- TPB 3 TM.TPM- TPB 4 (XII-6) TM.TPM-TPF 1 (XI-3) TM.TPM-TPF 2 (XI-4) TM.TPM-TPF 3 (XII-5) TM.TPM-TPF 4 (XII-6) TM.TPM-TPG 1 (XII-5) TM-MK/EM 1 (X-1) TM.TPM-TPC)1 .
Mathematics for class - XII (R.S. Agarwal) Bharati Bhawan Physics Physics 1 & 2 for Class XII NCERT Concept in Physics Part 1 & 2 Bharati Bhawan Lab Manual in Physics for XII Part II Arya REFERENCE BOOK New Era Physics Part 1 & 2 GRB Bathala Pub Chemistry Chemistry for Class XII NCERT
Practical Work in Geography Part I, Class XI, Published byNCERT 4. Fundamentals of Human Geography, Class XII, Published byNCERT 5. India - People and Economy, Class XII, Published byNCERT 6. Practical Work in Geography Part II, Class XII, Published byNCERT Note: The above
(ii) Class XI/IX examinations shall be conducted internally by the schools themselves. (iii) The Board will conduct the external examinations at the end of Class XII/Class X. (iv) Class XII/Class X examinations will be based on the syllabi as prescribed by the Board for class XII/X respectively from time to time.
HOTEL MANAGEMENT SYSTEM A Project Report Submitted in Partial Fulfillment of the Requirements AISSCE - All India Senior School Certificate Examination 2019-2020: SCIENCE – XII A In COMPUTER SCIENCE (083) By: 1. SAHISTHA – XII A – ROLL NO: 20 2. PALLAVI – XII A – ROLL NO: 14 3. SABNAM – XII A – ROLL NO: 19
