Class: XII Session: 2020-21 Subject: Mathematics Value .

Class: XII Session: 2020-21Subject: MathematicsValue points, Practice Paper 3S. No.1SolutionsMarks1YesOrR is transitive but not symmetric2Yes13A and 1or 24mxn1531640517𝑒𝑥313 c , where c is arbitrary constant832/3 sq units19order is 2 and degree is 11OrC 5105 6𝑠𝑞 𝑢𝑛𝑖𝑡3111 21112K 41

131𝑝214(1731, 0,2331)150.12116False117(i) c1(ii) b1(iii) d1(iv) a1(v) b1(i) b1(ii) a1(iii) c1(iv) c1(v) b11819 1 1𝑡𝑎𝑛 1 (1) 𝑐𝑜𝑠 1 ( 2 ) 𝑠𝑖𝑛 1 ( 2 ) 𝜋411 ( 𝜋 𝑐𝑜𝑠 1 (2)) - 𝑠𝑖𝑛 1 (2)𝜋 4 (𝜋 𝜋3) 𝜋6 3𝜋411

20(𝑥2 𝑥1) (4) 03 11 0)( 5 1 0 22 0𝑥2𝑥 8) (4) 01 (𝑥 2 101 𝑥 ( 𝑥 2 ) 40 2𝑥 8 01 x 4 3OR1 𝑦3 47) () (5 𝑥100 12( (7108 𝑦7) (102𝑥 10)50)511 x 2 and y -8 and x – y 1021𝑓(𝑥) {𝑘𝑥 1,𝑐𝑜𝑠𝑥,𝑖𝑓 𝑥 𝜋} is continuous at 𝑥 𝜋𝑖𝑓 𝑥 𝜋If LHL RHL f(𝜋) k 𝜋 1 cos 𝜋 k 1 21𝜋𝑥 1 𝑎𝑠𝑖𝑛𝜃, 𝑦 𝑏𝑐𝑜𝑠 2 𝜃22𝑑𝑥 -a cos 𝜃 ,𝑑𝜃𝑑𝑦𝑑𝜃𝑑𝑥 𝑎So , slope of the normal - 𝑑𝑦 2𝑏 sin 𝜃 at 𝜃 231 2b cos 𝜃 ( -sin 𝜃 ) 𝑎2𝑏𝑠𝑖𝑛4 𝑥 𝑐𝑜𝑠 4 𝑥 𝑒𝑑𝑥𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥𝑥𝜋21

𝑠𝑖𝑛4 𝑥 𝑐𝑜𝑠 4 𝑥 (𝑠𝑖𝑛2 𝑥 𝑐𝑜𝑠 2 𝑥) (𝑠𝑖𝑛2 𝑥 𝑐𝑜𝑠 2 𝑥) (sin 𝑥 cos 𝑥) (sin 𝑥 cos 𝑥)( 𝑠𝑖𝑛4 𝑥 𝑐𝑜𝑠 4 𝑥) / (sin 𝑥 cos 𝑥) (sin 𝑥 cos 𝑥)So , 𝑒 𝑥𝑠𝑖𝑛4 𝑥 𝑐𝑜𝑠4 𝑥𝑠𝑖𝑛𝑥 𝑐𝑜𝑠𝑥𝑑𝑥 𝑒 𝑥 ( sin x cos x ) dx 𝑒 𝑥 sin x c11OR𝜋 2𝜋 𝑠𝑖𝑛7 𝑥 𝑑𝑥 0 (being an odd function)1 12242Area 0 4 – 𝑥 2 dx 𝑥 2 4 – 𝑥 2 41𝑥sin 1 22𝑃𝑢𝑡𝑡𝑖𝑛𝑔 𝑡ℎ𝑒 𝑢𝑝𝑝𝑒𝑟 𝑎𝑛𝑑 𝑙𝑜𝑤𝑒𝑟 𝑙𝑖𝑚𝑖𝑡𝑠 , 𝑤𝑒 𝑔𝑒𝑡, 𝐴𝑟𝑒𝑎 𝜋 𝑠𝑞 𝑢𝑛𝑖𝑡𝑠25𝑑𝑦.𝑑𝑥1 𝑥 3 𝑐𝑜𝑠𝑒𝑐 𝑦, given that 𝑓(0) 0 sin 𝑦 𝑑𝑦 𝑥 3 𝑑𝑥 𝑐-cos y 𝑥44 c1As, 𝑓(0) 0 so c - 1𝑥4426½½ cos y 1The vector equation of a plane passing throughA(2, 5, -3), B(-2,-3, 5) and C(5, 3, -3) is given by( 𝑟⃗ - 𝑎⃗ ) [ ( 𝑏⃗⃗ - 𝑎⃗ ) x ( 𝑐⃗ - 𝑎⃗ ) ] 0( 𝑟⃗ – ( 2𝑖̂ 5𝑗̂ - 3𝑘̂ )[ ( -4𝑖̂ -8𝑗̂ 8𝑘̂ )x ( 3𝑖̂ -2𝑗̂ )] 0𝑟⃗ 𝑖̂ 2𝑗̂ 4𝑘̂ 𝜆(2𝑖̂ 3𝑗̂ 6𝑘̂)27And𝑟⃗ 3𝑖̂ 3𝑗̂ 5𝑘̂ 𝜇(2𝑖̂ 3𝑗̂ 6𝑘̂).Since, the drs of the above lines are in same ratios11

So, lines are parallel⃗⃗( ⃗⃗⃗⃗⃗⃗𝑎2 ⃗⃗⃗⃗⃗⃗𝑎1 ) x 𝑏 ⃗⃗ 𝑏S.D 28 2937Here ( ⃗⃗⃗⃗⃗𝑎2 ⃗⃗⃗⃗⃗𝑎1 ) 2𝑖̂ 𝑗̂ 𝑘̂ , 𝑏⃗⃗ 7units½ 1½ pi 1 6k 1 𝑘 1/6P( X 2 ) 6k 12911A {𝑥 𝑍 0 𝑥 12}, given byR {(𝑎, 𝑏) 𝑎 𝑏 𝑖𝑠 𝑎 𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑒 𝑜𝑓 4}R is reflexiveLet a A , a R a a A as a – a 0 0 x 4 clearly a multiple of 4 .½R is symmetricLet a , b A such that a R bi.e. a – b is a multiple of 4 .Since a – b b – a 1So , b – a is a multiple of 4So , b R aR is transitiveLet a , b , c A such that a R b and b R c a – b 4m and b – c 4n where m , n 𝑁 a – b 4m and b - c 4n a–c (a–b) (b–c) 4(m n) a – c 4 ( m n ) , clearly a multiple of 4 .So , a R cHence R is an equivalence relation .1

½30𝑦 𝑥 𝑎 𝑥 sin 𝑥 u v𝑑𝑦𝑑𝑥Here ,𝑑𝑢 𝑑𝑢𝑑𝑣½ 𝑑𝑥𝑑𝑥½ a xa – 1𝑑𝑥V 𝑥 sin 𝑥log v sin x log x1 𝑑𝑣𝑣 𝑑𝑥 sin 𝑥𝑥𝑑𝑣 log x cos x 𝑥 sin 𝑥 (𝑑𝑥sin 𝑥𝑥𝑑𝑦So, 𝑑𝑥 a xa – 1 𝑥 sin 𝑥 (311½ log x cos x )sin 𝑥𝑥 log x cos x )½y 3𝑒 2𝑥 2𝑒 3𝑥.𝑑𝑦𝑑𝑥𝑑2 𝑦𝑑𝑥 2𝑑2 𝑦1 6𝑒 2𝑥 6𝑒 3𝑥𝑑𝑦-5𝑑𝑥 2 𝑑𝑥 12𝑒 2𝑥 18𝑒 3𝑥1 6y 12𝑒 2𝑥 18𝑒 3𝑥 -5(6𝑒 2𝑥 6𝑒 3𝑥 ) 6(3𝑒 2𝑥 2𝑒 3𝑥 ) 01ORWe have x a(cos sin ) , y a(sin - cos )𝑑𝑦𝑑 𝑑𝑥𝑑 a(cos sin -cos ) a sin a(-sin cos sin .1) a cos 𝑑𝑦 𝑎 sin 𝑑𝑥 𝑎 cos tan 11

𝑑𝑦𝑑𝑑𝑦𝑑𝑑 1 𝑑𝑥 (𝑑𝑥 ) 𝑑𝑥 (𝑡𝑎𝑛 ) 𝑠𝑒𝑐 2 𝑑𝑥 𝑠𝑒𝑐 2 x𝑎 cos 𝑑𝑥𝑠𝑒𝑐 3 321𝑎 . f(x) 4𝑥 3 - 6𝑥 2 -72x 30𝑓 ′ (𝑥) 12𝑥 2 -12x-72112𝑥 2 -12x-72 012(𝑥 2 -x-6) 0½X -2,3Sign of𝑓 ′ (𝑥)Interval33Nature of f(x)(- ,-2) f is strictly increasing(-2,3)-f is strictly decreasing(3, ) f is strictly increasingLet I 𝑥21½dx(𝑥 2 1)(𝑥 2 4)Put 𝑥 2 t𝑥2(𝑥 2 1)(𝑥 2 4)𝑡 (𝑡 1)(𝑡 4) 𝐴𝑡 1 𝐵1𝑡 4A -1/3 , B 4/3 13(𝑥 2 1)I 13(𝑥 2 1)413(𝑥 2 4)𝑑𝑥 43(𝑥 2 4)dx

I 132𝑥321tan 1 𝑥 tan 1 c𝑥 2 9𝑦 2 3634𝑥2𝑦2 364 1Y6A 4 0 𝑦 𝑑𝑥64A 3 0 62 𝑥 24 𝑥A 3 (2 36 𝑥 2 O362𝑥6𝑠𝑖𝑛 1 6) 6X1104A 3 [18𝑠𝑖𝑛 1 (1) 18𝑠𝑖𝑛 1 (0)]4𝜋A 3 X 18 X 2 12𝜋135𝑒 𝑥 𝑡𝑎𝑛𝑦 𝑑𝑥 (1 𝑒 𝑥 ) 𝑠𝑒𝑐 2 𝑦 𝑑𝑦 0𝑒𝑥1 𝑒 dx 𝑥𝑒𝑥1 𝑒 𝑥𝑠𝑒𝑐 2 𝑦𝑡𝑎𝑛𝑦dx 1dy 0𝑠𝑒𝑐 2 𝑦𝑡𝑎𝑛𝑦dy 0-log 1-𝑒 𝑥 log tany ctany c(1-𝑒 𝑥 )½ORx logx𝑑𝑦𝑑𝑥1½2 𝑦 logx𝑥

𝑑𝑦𝑑𝑥 1𝑥𝑙𝑜𝑔𝑥𝑦 1I.F. 𝑒 𝑥𝑙𝑜𝑔𝑥𝑑𝑥y logx 2𝑥22𝑥2 𝑒 log (𝑙𝑜𝑔𝑥)𝑑𝑥 logx𝑙𝑜𝑔𝑥𝑑𝑥 2 (𝑙𝑜𝑔𝑥)𝑥 2 dx c 2[ 𝑙𝑜𝑔𝑥ylogx 𝑥 2𝑥 𝑥 2 dx] c(1 logx) c36x 2y-3z -42x 3y 2z 23x-3y-4z 11𝑥1 2 3 4𝑦(2 3)(() 22)𝑧3 3 411AX BX 𝐴 1 B𝐴 1 1/21𝑎𝑑𝑗 𝐴 𝐴 ½1 2 3 A 2 32 3 3 4 1(-12 6)-2(-8-6)-3(-6-9) 67 0 6Adj A ( 14 15 6𝐴 67 ( 14 15 111759175913 8) 113 8) 11½

X 𝐴 1B1 67 6( 14 1517 13 4)(5 82)9 11120131 67 ( 134) 26711½X 3, y -2, z 1OR22 4A ( 4 2 4)2 1 51 1 0B (2 3 4)0 1 21 1 022BA (2 3 4) ( 4 20 1 22 1 4 4)51½6 0 0 (0 6 0) 6I0 0 6y 2z 7x-y 32x 3y 4z 17Let’s rearrange the equationsx-y 32x 3y 4z 17Y 2z 71

1 0 𝑥3𝑦)() (3 417)1 2 𝑧71 (2026 34 281 6 ( 12 34 28)6 17 35121 6 ( 6)242 ( 1)4½ 𝑥 2,374 𝑥2𝑥 4 2𝑦𝑧 41 𝑧 6 𝑦𝑦 1,3𝑧 1 6 3 k (say)1½X -2k 4 , y 6k, z -3k 1D.R.’s are (-2k 4-2 , 6k-3, -3k 1 8)½ (-2k 2 , 6k-3, -3k 9)(-2k 2)(-2) ( 6k-3)(6) (-3k 9)(-3) 04k-4 36k-18 9k-27 049k-49 0K 11½X -2x1 4 , y 6x1 , z -3x1 1X 2 , y 6 , z -2½1

Distance (2 2)2 (6 3)2 ( 2 8)2 9 36 45 3 5ORThe equation of plane passing through the intersection of two given planesis,1(x 3y-6) t(3x-y-4z) 0(1 3t)x (3-t)y -4tz-6 0Distance of this plane from the origin (0,0,0,) is unity.1So,1 6 (1 3𝑡)2 (3 𝑡)2 ( 4𝑡)2 61 10 26𝑡 2 10 26𝑡 2 361t 1putting t 1 in (A) we get,4x 2y-4z-6 01putting t -1 in (A) we get,1-2x 4y 4z-6 0

38Max 𝑧 𝑥 𝑦𝑥 4𝑦 8s,t2𝑥 3𝑦 123𝑥 𝑦 9x ,y 0𝐵(28 15, )11 11C(0,2)OA (3,0)Region OABC is the feasible regionAt O(0,0)3Z 0 0 01At A(3,0) , Z 3 0 328 15At B(,11 11) , Z 2811 151143 ,11At C(0,2) , Z 0 2 2Therefore, Optimal solution is (function is43128 15, )11 11and maximum value of the11OR1) Pointsz x 2y

32451P( 13 , 13)3Q( 2 ,1513)47 159 maxR (2 , 4 )181½5222S ( 7 , 7)73Max z 9 at Q( 2 ,154) and min z is22718min12𝑎𝑡 𝑆 ( 7 , 7)2) Z px qy3 157 3If max Z occurs at Q (2 , 4 ) and R (2 ,4 ), then3𝑝215𝑞 4 7𝑝23𝑞 42Simplifying, we get, 2p 3qThis is the required condition.Also, the number of optimal solutions in this case will beinfinite solutions lying on the line segment QR.½

Class: XII Session: 2020-21 Subject: Mathematics Value points, Practice Paper 3 S. No. Solutions Marks 1 Yes Or R is transitive but not symmetric 1 2 Yes 1 3 A and or 2 1