1 65/1/3 P.T.O.Series:OSR/1RollNo.Code No. 65/1/3CandidatesmustwritetheCodeonthetitlepageoftheanswer-book. Please check that this question paper contains 15 printed pages. Code number given on the right hand side of the question paper should be written on the title page of the answer-book by the candidate. Pleasecheck that this question paper contains 29 questions. Please write down the Serial Number of the questions before attempting it. 15 minutes time has been allotted to read this question paper. The question paper will be distributed at 10.15 a.m.From 10.15 a.m. to 10.30 a.m., the student will read the question paper only and will not write any answer on theanswer script during this period.Mathematics[Time allowed : 3 hours] [Maximum marks : 100]General Instructions:(i) All questions are compulsory.(ii) The question paper consists of 29 questions divided into three sections A, B and C. Section -A comprises of 10 questions of one mark each, Section - B comprises of 12 questions of fourmarks each and Section - C comprises of 7 questions of six marks each.(iii) All questions in Section - A are to be answered in one word, one sentence or as per the exactrequirementofthequestion.(iv) There is no overall choice. However, internal choice has been provided in 4 questions of fourmarks each and 2 questions of six marks each. You have to attempt only one of the alternativesin all such questions.(v) Use of calculators is not permitted. You may ask for logarithmic tables, if required.Studymate Solutions to CBSE Board Examination 2013-2014DISCLAIMER: All model answers in this Solution to Board paper are written by Studymate Subject Matter Experts.This is not intended to be the official model solution to the questionpaper provided by CBSE.The purpose of this solution is to provide a guidance to students.2 65/1/3 P.T.O.1. If 23 4 1 7 0,5 0 1 10 5 (((+ = ((( yx find (x y)Sol.3 4 1 7 025 0 1 10 5yx (((+ = ((( 6 1 8 7 010 0 2 1 10 5yx+ +((= ((+ + y = 8 x = 2 10 = x y2. Solve the following matrix equation for x :, [x 1] 1 0.2 0O ( = ( Sol.| |1 22 21 012 0x O (= ( | | 2 0 x O =x 2 = 0 2 x =3.2 5 6 2,8 7 3=xx write the value of x.Sol.2 5 6 28 7 3xx=2x2 40 = 18 + 146 x = 4. Write the antiderivative of 13 .| |+ |\ .xxSol.13 I x dxx= +}32 23 23xx c| | | + + | |\ . 2 2 x x x c + +( ) 2 1 I x x c = + +5. If 1 11sin sin cos 1.5 | |+ = |\ .xthen find the value of x.Sol.1 11sin sin cos 15x | |+ = |\ .1 1 11sin cos sin 15x + =1 11cos sin2 5x = 1 11cos cos5x =15x =3 65/1/3 P.T.O.6. Let * be a binary operation, on the set of all non-zero real numbers, given by a * b = 5ab for alla, b e R {0}. Find the value of x, given that 2 * (x * 5) = 10.Sol.*5aba b =2* (x * 5) = 1052* 105x | | = |\ .2 * x= 10210255xx = =7. Find the projection of the vector 3 7 + + i j kon the vector 2 3 6 . + i j kSol. 3 7 , 2 3 6a i j k b i j k = + + = + Projection of or | |a ba bb= = 58. Write the vector equation of the plane, passing through the point (a, b, c) and parallel to theplane .( ) 2 + + =r i j k .Sol. Equation of plane( ). 0 r a n = ( ) ( ) ( ). . r i j k ai b j ck i j k + + = + + + + ( ). r i j k a b c + + = + +9. Evaluate :/20(sin cos ) .t}xe x x dxSol. ( )20sin cosxI e x x dx= }( ) cos and'( ) sin f x x f x x = =

20cosxI e x (= = 1.10. Write a unit vector in the direction of the sum of the vectors 2 2 5 = + a i j kand 2 7 . = + b i j kSol. 2 2 52 7 a i j k b i j k = + = + Unit vector = 4 3 12 4 3 1213 13 13 | | 16 9 144a b i j ki j ka b+ + = = + + + + SECTION-BQuestion numbers 11 to 22 carry 4 marks each.11. Prove that, for any three vectors, , a b c[ , , ] 2[ , , ] + + + = a b b c c a a b cSol. a b b c c a (+ + + ( ) ( ) ( )a b b c c a (+ + + 4 65/1/3 P.T.O. ( )a b b c b a c c c a (+ + + + ( ) ( ) + a b c b c aa b c b c a ((+ 2 a b c ( OR11. Vectors, a band c are such that0 a b c + + = and | | 3,| | 5 = =a band | | 7. =cFind the angle between a and b.Sol.0 a b c + + = | | 3, | | 5&||=7 a b c = = a b c + = ( ) ( ) ( ) ( ). . a b a b c c + + = 2| | a a a b a b b b c + + + = 2 2 2| | 2| || |cos | | | | a a b b c + + = 9 + 2(3)(5) cosu + 25 = 493 =12. Solve the following differential equation :222( 1) 21dyx xydx x + =Sol. ( )2221 21dyx xydx x + =( )2 222 211dy xydx xx+ =It is linear differential equation of typedyPy Qdx + =where 221xPx=

( )2221Qx=221.xdxPdxxI F e e= =}}( )2log| 1| 21xe x= = Solution is given by ( )( )( )2 22221 11y x x dx cx = +}( )211 log1xy x cx = ++13. Evaluate : 6 62 2sin cossin .cos+}x xdxx xSol.6 62 2sin cossin cosx xdxx x+}5 65/1/3 P.T.O.( ) ( )3 32 22 2sin cossin cosx xdxx x+}( ) ( )2 2 4 4 2 22 2sin cos sin cos sin cossin cosx x x x x xdxx x+ + }( )22 2 2 22 2sin cos 3sin cossin cosx x x xdxx x+ }2 213sin cosdxx x| | |\ .}2 22 2sin cos3sin cosx xdx x cx x+ +}2 2sec cosec 3 xdx xdx x c + +} } tanx cot x 3x + cOR13.2( 3) 3 18 + }x x x dxSol. ( )23 3 18I x x x dx = + }Put x2 + 3x 18 = t(2x + 3)dx = dtLet x 3 = A(2x + 3) + BComparecoefficients:We get 12A =and 92B=I ( )2 21 92 3 3 183 18 2 2x x x dx x x dx = + + + } }=221 9 3 93 182 2 2 4tdt x x dx| | + + |\ .} }=32 22 1 2 9 3 92 3 2 2 2tx dx (| | | | ( + || (\ . \ . }= ( )( )32 2 221 9 729 33 18 2 3 3 18 log 3 183 8 16 2x x x x x x x x c| |+ + + + + + + + |\ .14. Find the intervals in which the function f (x) = 3x4 4x3 12x2 + 5 is(a) strictlyincreasing(b) strictlydecreasingSol.4 3 2( ) 3 4 12 5 f x x x x = +3 2'( ) 12 12 24 f x x x x = ( )212 2 x x x = ( ) ( ) ( ) ' 12 2 1 f x x x x = +Put'( ) 00, 1,2 f x x = = Intervalsare( ) ( ) ( ) , 1 ,( 1,0), 0,2 , 2, 6 65/1/3 P.T.O.Intervals Sign of( ) f x Nature of( ) f x(1, 0 )( ) , 1 (0, 2)( ) 2, ve + ve ve+ veStrictly Dec. Inc Dec. IncStrictly StrictlyStrictlyf(x) is strictly increasing in( ) ( ) 1,0&2, , andf(x) is strictly decreasing in( ) ( ) , 1&0,2 ORFind the equations of the tangent and normal to the curve x = a sin3u and y = a cos3u at 4tu =Sol. x =a sin3uy =a cos3u23 sin cosdxad =23 cos sindyad = 223 cos sincot3 sin cosdy adx a = = 41dydx =(= (Equation of tangent at 4=y a cos34 = 1 (x a sin34)2

2 2 2a ay x y x + = + =equation of normal at 4=12 2 1 2a ay x| | = |\ .0 y x =15. Let A = {1, 2, 3,..:.., 9} and R be the relation in A A defined by (a, b) R (c, d) if a + d = b + c for(a, b), (c, d) in A A. Prove that R is an equivalence relation. Also obtain the equivalence class[(2, 5)].Sol. First Let us prove that the relation is(i) Reflexive (ii) Symmetric(iii) Transitive(i) Reflexive :(a, b) R (a, b) a + b = b + a which is true (additioniscommutative) Givenrelationisreflexive.(ii) Symmetric :7 65/1/3 P.T.O.If (a, b) R (c, d) a + d = b + cthen (c, d) R (a, b) c +b = d + a b + c = a + d ( addition is commutative) Givenrelationissymmetric.(iii)Transitive :(a, b) R (c, d) a + d = b + c ...(i)and (c, d) R (e, f) c + f = d + e ....(ii)Adding (i) & (ii) we get a + f =b + e (a, b) R (e, f) Givenrelationistransitive Givenrelationisequivalancerelation.The equivalence class for [(2, 5)] is [(2,5),(5,2)].16. Prove that 11 sin 1 sincot ; 0, .2 4 1 sin 1 sinx x xxx x| |+ + t | |= e | | |+ \ .\ .Sol.11 sin 1 sincot1 sin 1 sinx xx x| |+ + | |+ \ .1cos sin cos sin2 2 2 2cotcos sin cos sin2 2 2 2x x x xx x x x (+ + ( ( (+ + 1cot cot2 2x x( = ( OR16. Prove that 1 1 11 5 2 12tan sec 2tan .5 7 8 4 | |t | | | |+ + = | || |\ . \ .\ .Sol.1 1 11 5 2 12tan sec 2tan15 7 8 | || | | |+ + | || |\ . \ .\ .1 11 15 25 82 tan sec17140 ( | |+ (| | |+ (|| | (| \ . ( \ . 1 11 5 22 tan sec3 7 | | ( | |+ | |( |\ . \ .1 13 5 2tan sec4 7 | |+ | |\ .( )1 1 13 1tan tan tan 14 7 4 | | | |+ = = ||\ . \ .8 65/1/3 P.T.O.17. If y = xx, prove that 22210.| | = |\ .d y dy yy dx x dxSol. y = xx log y = x log x( )11 logdyxy dx= +( ) 1 logdyy xdx= +....(i)( )21211 logd yy x yx dx= + +2112.y d y yyx y dx= +( )221210d y yyy x dx =18. Assume that each born child is equally likely to be a boy or a girl. If a family has two children,what is the conditional probability that both are girls ? Given that(i) the youngest is a girl.(ii) atleast one is a girl.Sol. S = {GG, GB, BG, BB}(i) A = Both are GirlsB = Youngest is a GirlP (A/B) = 1( ) 142( ) 24= =P A BP B(ii) A : Both are GirlsB : atleast one GirlP (A/B) = 1( ) 143( ) 34= =P A BP B19. Using properties of determinants, prove the following :22 2 2 2211 11++ = + + ++x xy xzxy y yz x y zxz yz zSol. Applying R1 xR1, R2 yR2,

R3 zR3 and taking x,y,z common from C1, C2 , C3 we get2 2 22 2 22 2 2111x x xxyzy y yxyzz z z+A = ++R1 R1+ R2 + R32 2 2 2 2 2 2 2 22 2 22 2 21 1 111x y z x y z x y zy y yz z z+ + + + + + + + += ++9 65/1/3 P.T.O.( )2 2 2 2 2 22 2 21 1 11 11x y z y y yz z z= + + + ++C3 C3 C1 and C2 C2 C1 ( )2 2 2 221 0 01 1 00 1x y z yzA = + + +=( )2 2 21 x y z + + +20. Differentiate 211 1tan| |+ | |\ .xxwith respect to 122sin ,1| | |+\ .xx when x = 0.Sol. Let y =211 1tan put tan| |+ | = |\ .xxx1sec 1tantany | |= |\ .11 costansiny | |= |\ .1tan tan2y| |= |\ .11 1 tan2 2y y x = =21 12 1dydx x| |= |+ \ .again122sin ,puttan1xz xx| |= = |+ \ .( )1 1sin sin2 2 2tan z x = = =

221dzdx x=+Now,/ 1/ 4dy dy dxdz dz dx= =21. Find the particular solution of the differential equation (2log 1),sin cos+=+dy x xdx y y y given that 2t= ywhen x = 1.Sol. (sin cos ) (2log 1) y y y dy x x dx + = +} }2cos cos 2 log2 + = + +} }xy y y dy x xdx c2 2 2cos sin cos 2log .2 2 2x x xy y y y x c| | + + = + + |\ .2sin log y y x x c = +When; 12t= = y x2ct= 2sin log2y y x xt= +10 65/1/3 P.T.O.22. Show that lines ( ) (3 ) = + + r i j k i jand (4 ) (2 3 ) = + +r i k i kintersect. Alsofindtheirpoint of intersection.Sol.1 1 13 1 0 += = = x y z.... (1)4 0 12 0 3 += = = x y z... (2)Point on 1st line[3 + 1, + 1, 1] ... (A)point on 2nd line[2 + 4, 0, 3 1] ... (B)For point of intersection, A = BWe get = 1 and = 0 point of intersection (4, 0, 1)SECTION-CQuestion numbers 23 to 29 carry 6 marks each.23. A dealer in rural area wishes to purchase a number of sewing machines. He has only ` 5,760to invest and has space for at most 20 items for storage. An electronic sewing machine costhim ` 360 and a manually operated sewing machine ` 240. He can sell an electronic sewingmachine at a profit of ` 22 and a manually operated sewing machine at a profit of ` 18. Assumingthathecansellalltheitemsthathecanbuy,howshouldheinvesthismoneyinordertomaximize his profit ? Make it as a LPP and solve it graphically.Sol. Let the dealer buy x-electronic swing machine and y- manually operated sewing machines x, y > 0Also x + y s 20[storage constraints]360x + 240 y s 5760i.e. 3x + 2y s 48objective function is z = 22 x + 18yPoint z x y = 22+ 18A(16,0) 352B(8,12) 392 (maxima)C(0,20) 36048121620240481216202428xyA (16, 0)B (8, 12)C (0, 20)Thedealer shouldby 8 electronicand12 manual sewing machines.He willmake a maximumprofit of ` 392.24. Acardfrom apackof52 playingcardsislost.From theremainingcardsofthepackthreecardsaredrawnatrandom(withoutreplacement)andarefoundtobeallspades.Findtheprobability of the lost card being a spade.Sol. Let A : getting three spade cardsE1 : lost card is a spade P (E1) = 131521CCE2 : lost card is a diamond 1312 521P(E ) =CC11 65/1/3 P.T.O.E3 : lost card is a club 1313 521P(E ) =CCE4 : lost card is a heart 1314 521P(E ) =CC41( ) ( ) . ( | )== i iiP A P E P A E12 13 13 133 3 1 152 51 52 511 3 1 33 = + C C C CC C C CRequired probability = P (E1 | A)= 1 1P(E ).P(A/E )P(A)12 133 152 511 312 13 13 133 3 1 152 51 52 511 3 1 3C CC CC C C C3C C C C= + = 1049OR24. From a lot of 15 bulbs which include 5 defectives, a sample of 4 bulbs is drawn one by one withreplacement.Findtheprobabilitydistributionofnumberofdefectivebulbs.Hencefindthemean of the distribution.Sol. Let success be defined as getting a defective bulb. If x indicates success then x = {0, 1, 2, 3, 4}Probability of success, 5 1p15 3= = and 1 2q 13 3= =We use P (x) = nCx px qn x, where n is the number of trials.x 0 1 2 3 416 2 24 8 1P(x)81 81 81 81 813Mean, = 51( )= i iix P x = 4325. Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x and thecircle x2 + y2 = 32.Sol. Solving y = x and x2 + y2 = 32we get x = 4Required area = Area of region OCABO= 4 4 220 032 + } }x dx x dx(4, 0)C(4 2, 0)Coyxy = x= 4 2 422 10 43232 sin2 2 2 4 2 (+ + ( x x xx2 21 1(4) (0) 4 116sin 1 16 16sin2 2 2 2 (( | |= + + (|(\ . 8 16 8 162 4 t t ( | |= + + |(\ . = 4t sq. units.26. Findthedistancebetweenthepoint(7,2,4)andtheplanedeterminedbythepointsA (2, 5, 3), B (2, 3, 5) and C (5, 3, 3).12 65/1/3 P.T.O.Sol. Equation of plane 2 5 34 8 8 03 2 0 + =x y zor 2x + 3y + 4z = 7Distance from (7, 2, 4) 292929= =OR26. Findthedistanceofthepoint(1,5,10)fromthepointofintersectionoftheline 2 2 (3 4 2 ) = + + + +r i j k i j kand the plane .( ) 5. + =r i j kSol. ThelineisQ (1,5,10)P( 2) ( 1) ( 2)3 4 2 + = = = x y zany point on it is P (2 + 3, 1 + 4, 2 + 2) P lieson x y + z = 5 2 + 3 + 1 4 + 2 + 2 = 5i.e. = 0 the point of intersection is : P (2, 1, 2) required distance from point Q = 13 units27. Two schools P and Q want to award their selected students on the values of Discipline, PolitenessandPunctuality.TheschoolPwantstoaward`xeach,`yeachand`zeachforthethreerespective values to its 3, 2 and 1 students with a total award money of ` 1,000. School Q wantsto spend ` 1,500 to award its 4, 1 and 3 students on the respective values (by giving the sameaward money forthethreevaluesasbefore).If thetotalamountofawards for oneprizeoneachvalueis`600,usingmatrices,findtheawardmoneyforeachvalue.Apartfromtheabove three values, suggest one more value for awards.Sol. From thegiven situationthe setof related equationsis3x + 2y + z = 10003x + y + 3z = 1500x + y + z = 600The corresponding system ofmatrix equations isAX = B where 3 2 1 10004 1 3 , and 15001 1 1 600xA X y Bz ((( (((= = = ((( ((( 3 2 1| | 4 1 3 51 1 1= = A

12 1 51 11 2 5| | 53 1 5A adjAA ( (= = ( ( 1100200300 ( (= =( ( X A Bx = ` 100, y = ` 200, z = ` 30028. Evaluate : /24 40sin cossin cost+}x x xdxx x13 65/1/3 P.T.O.Sol./24 40sin cossin cost=+}x x xIx x... (i)/24 40cos sin2cos sintt | | |\ .=+}x x xI dxx x... (ii)Adding (i) and (ii) we get/24 40sin cos22 sin costt=+}x xI dxx xPut sin2x = t 2 sinx cosx dx = dt sinx cosx dx = 2dt12 20122 2 (1 )t= + }dtIt t

1208 2 2 1t= +}dtIt t 1021162t= +}dtIt t1211200112tan1 116 16 161 12 22 4| | |t t t= = = || || | + |\ .\ .}tdtt29. Of all the closed right circular cylindrical cans of volume 128t cm3, find the dimensions of thecanwhichhasminimumsurfacearea.Sol. Given, V = 128t = tr2hMinimize : S = 2trh + 2tr2Put 2128= hrhr

2256( ) 2t= + t S r rr2256( ) 4t' = + t S r rr32 256( ) 4 t'' = + t S rrFor CPs : S' (r) = 0 22564 0t + t = rrr = 4 cmNow S" (4) > 0 at r = 4 is point of minima1288cm4 4t= =t h For minimum surface area the can should have a radius of 4 cm and a height of 8 cm.