MATHEMATICS Time allowed : 3 hours Maximum marks : 100 General Instructions : (i) All questions are compulsory. (ii) Question number 1 to 15 are of 2 marks each. Question number 16 to 25 are of 4 marks each. Question number 26 to 30 are of 6 marks each. (iii) If you wish to answer any question already answered for any reason, cancel the previous answer. (iv) Use of calculator is not permitted, however you may ask for logarithmic tables Question Paper Code 65/1/1 1. A coin is tossed 12 times. What is the probability of getting exactly 8 tails ? 2. Three coins are tossed simultaneously. List the sample space for the event. 3. Two cards are drawn without replacement from a well shuffled pack of 52 cards. What is the probability that one is a red queen and the other is a king of black colour ? 4. Find the unit vector perpendicular to both a +3 + j-2k and b= 2+3j-k 5. Find if a =4-j+k and -2j+2k are perpendicular to each other. 6. Find the regression co-efficient bxy and byx given that n=7 ∑x=24 ∑y=12 ∑x2=374 ∑y2=97 and ∑xy=157 7. Solve: (1+x) (1+y2) dx+1 (x+y) (1+x2) dy=0 8. Evaluate: 9. Evaluate: 10.Evaluate: 11.If y= tan-1x show that: 12.Discuss the applicability of Rolle’s theorem for the function f(x) = x2/3 on [-1,1] 13. Evaluate: 14. If find the matrix C such that A+B+C is a zero matrix 15. Construct a 2 X 3 matrix whose element in the in the ith row and the jth column are given by aij = 3i-j/2. 16. Show that the points with position vectors 6- 7, 6-19 -4k,3j - 6k and 2-5+10 k are coplanar. 17. Find the root of the perpendicular from (0,2,7) on the line 18. Three balls are drawn without replacement from a bag containing 5 white and 4 red balls. Find the probability distribution of the number of red balls drawn. 19. Evaluate: 20. Evaluate: ∫ x cos-1x dx 21. Sketch the region common to the circle x2+y2 = 16 and the parabola x2 = 6y. Also, find the area of the region, using integration. 22. Evaluate ∫3 0f(x) dx, when f(x)=| x | + | x-1 | + | x-2 |

23. Prove, using the properties of determinants 24. A particle moves along the curve 6y = x3+2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as x-coordinate 25. Find the derivative of cos(3x+2) w.r.t x from first principle. 26. Given the following pairs of values of the variable x and y x : 3 5 7 12 20 22 24 y : 30 25 24 16 11 9 5 (i) Find the Karl Pearson’s coefficient of correlation. (ii) Interpret the result. (iii) confirm your interpretation with the help of a scatter diagram. 27. Find the Cartesian as well as vector equations of the planes through the intersection of the planes: which are at unit distance from the origin. 28. The slops of the tangent at any point of a curve is times the slope of the straight line joining the point of contract to the origin. Formulate the differential equation representing the problem and hence find the equation of the curve. 29. 30. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of the material will be least when the depth of the tank is half of its width. Question Paper Code 65/1 1. A coin is tossed 12 times. Find the probability of getting exactly 10 tails. 2. Two dice are thrown simultaneously. List the sample space for this event. 3. Two cards are drawn without replacement from a well-shuffled pack of 52 cards. Find the probability that one is a spade and the other is a queen of red colour. 4. Find a unit vector perpendicular to both the vectors a=4-j-3k and b=2+2j-k. 5. Find if the vectors a=1-j+3k and b=4-5j+2k are perpendicular to each other. 6. Find the regression coefficients b and b given that n=5, _x=15, _y=15, _x2=55, _y2=83 and _xy=53 7. Solve cos x .cosy dy + sin x dx=0 8. Evaluate: 9. Evaluate: 10. Evaluate: 11. 12. Discuss the applicability of Rolle’s theorem for the function f(x) = X1/3 on [-1,1] 13. Evaluate: 14. 15. Construct a 2 x 3 matrix whose elements in the ith row and the jth column are given by: 16. Find the foot of the perpendicular form (1,6,3) on the line: 17. Show that the points with position vectors: 18. Four bad eggs are mixed with 10 goods ones. If 3 eggs are drawn one by one without replacement, find the probability distribution of the number of bad eggs drawn. 19. Evaluate: 20. Evaluate:

21. Sketch the region lying in the first quadrant and bounded by y=9x2, x=0, y=1 and y=4, Find the area of the region, using integration 22. Evaluate: 23. Prove, using properties of determinants: 24. At what points of the ellipse 16x2+9y2 = 400, does the ordinate decrease at the same rate at which the abscissa increases ? 25. Find the derivative of sin(3x+2) w.r.t.x from first principle. 26. Given the following pairs of values of the variables x and y : x : 3 5 7 12 20 22 24 y : 30 25 24 16 11 9 5 (i) Find the Karl Pearson’s coefficient of correlation. (ii) Interpret the result. (iii) Confirm your interpretation with the help of a scatter diagram. 27. Find the Cartesian and the vector equations of the planes thought the intersection of the planes which are at unit distance from the origin. 28. The normal lines to a given curve at each point (x,y) on the curve pass through the point (2,0). The curve passes through the points (2,3). Formulate the differential equation representing the problem and hence find the equation of the curve. 29. 30. Show that a closed right circular cylinder of given total surface area S and maximum volume V is such that its height h is equal to the diameter d of the base. Marking Scheme - Mathematics Expected Answer/ Value Points General Instructions : 1. Marking scheme is to be followed strictly in order to maintain uniformity. 2. Alternative method accepted. Proportional marks are to be awarded. 3. Marks may not be deducted in questions on integration if constant of integration is not indicated. 4. If a question is attempted twice and the candidate has not crossed any answer only first attempt is to be evaluated. Write EXTRA with second attempt. 5. No marks are to be awarded if data is wrongly copied from the question paper. Question Paper Code 65/1/1 1. Probability of success (getting a tail) = ½ prob. (failure)= ½ ½ mark P(exactly 8 tails)=12C8 (½)8 (½)4 1 mark

= 495/4096 of 495/212 ½ mark 2. S={HHH,HHT,HTH,THH,THT,HTT,TTT} ½ mark for every two correct [All eight out comes must be mentioned] 3. Prob (one red queen & one black king) =

C CC 2 52 1 2 1 2

1 mark = 2.2.2 / 52.51 = 2 / 663 1 mark Alternative Method Required Prob. =P(Red queen & Black King) + P (Black King & Red Queen) = 2/52 x 2/51 +2/52 x 2/51 1 mark = 2/663 1 mark 4. Unit vector perpendicular to both a and b = bXa rr rr +

a x b 1 mark ½ mark for numerator or ½ mark for denominator 5. 1 mark (for Writing or Using) 4λ+ 2 + 2= 0 to get λ= -1 1 mark 6. bxy= N )y( y N yx ) xy ( 2

2∑

−∑ ∑∑ −∑

½ mark = 811 / 535 or 1.51 ½ mark bxy = N )y( y N yx ) xy ( 2 2∑

−∑ ∑∑ −∑

[ ] ba

½ mark =811/2042 or 0.39 ½ mark 7. Writing 0 dx x1 x1 dy y1 )41( 22= + + + + + 1 mark

Integrating to get tan-1 y+ ½ log (1+y2) + tan-1 x+ ½ log (1+x2) = C 1 mark (Deduct ½ mark if C is not mentioned) 8. Writing 1 mark =2x4x 2 2 x 2+− −

- log l (x-2)+ C I 2 x 4 x2 + + − 1 mark 9. Putting x2+4x+3 = t ½ mark to get (2x+4)dx =dt ∴∫ ∫ = + + + dt t dx 3 4x x ) 4 x 2 ( 2 ½ mark = 2/3 t3/2 + C =2/3 (x2+4x+3)3/2 + C ½ mark 10. Writing I = ½ ∫ dx / 25x+x2 ½ mark Writing (or using) ∫ + = +

−C

a x tan a 1 dx xa 11

½ mark ⇒ I = 1/10 tan-1 x/5 + C 1 mark 11. y=tan-1 x Differentiating to get ½ mark = 1/1+x2 ⇒ (1+x2) 22

=1 Differentiating again to get 0 dx dy . dx yd )x1( 2 2

mark 12. (i) Showing or stating f(x) =x2/3 is continuous in [-1,1] ½ mark (ii) Since f (x) = 2=

+1

3 1

x3 2

∴ if (x) if not differentiable at x=0 in (-1,1) 1 mark Hence Rolle’s theorem is not applicable ½ mark 13. x 2 sin x 4 x 7 x 3 sin 0x lim + + →

1 mark Dividing numerator and denominator by x to get x2 x 2 sin 2 4 7 x3 x 3 sin 3 0x lim + + → 3 5 24 73= + + 1 mark

14. A+B+C = O ⇒ C=O-A-B ½ mark = −

−− − − − 101 112 202 231 000 000

½ mark = −− −− 101 143

1 mark 15. A=

23 22 21 13 12 11

aaa aaa

½ mark 2/322/5 02/11

½ mark for every two correct (=1 ½ ) 16. Writing or Using “A,B,C,D are coplanar if [AB,AC,AD] = 0 1 mark Getting AB=10I - 12j-4k 1 ½ marks AC = 6I+10j+6k AD = -1 4I+2j+10k or any three vectors using four points Getting = − −− −−

10 2 4 6 10 6 4 12 10

1120-1008-112=0 1 ½ marks 17. Any point (say B) on the given line 1 mark is (- -2, 3λ +1,-2λ +3) if point A is (0,2,7) ⇒AB = (-λ -2, 3λ - 1, -2λ -4) 1 mark AB. (-1,3,-2)=0 to get λ = - ½ Getting foot of ^ B = (-3/2, -1/2, 4) 1 mark 18. Getting P(O) = 3 9 3 5

C C ½ mark 42 5

½ mark 3 9 1 4 3 5

C C.C )1(P=½ 42 20

mark

½ mark 42 15 ½ mark 3 9 3 4

C C )3(P=½ 42 2

mark

½ mark ∴Required Probability distribution is x:0123 p 5 20 15 2 42 42 42 42 19. Writing dx ) 3x4x 3 x 4 1 ( dx

3x4x x 22 2

∫ ∫ +−

−+= +−

1 mark dx xxx x x

∫

−− + +− −+ 1)2( 1 5 34 42 2 1 2 ½+½ = 1 mark x+2 log |x2-4x+3I+5/2 log 1x 3x − −

+C ½+½ + 1 = 2 mark OR Using Partial Fraction Method, Ans is x-½ log |x-1| + 9/2 log |x-3| +C 20. Substituting x=cosx = dx=-1 sinx dx to get 1 mark I= ½ 0(0-sin20) d0 1 mark = ¼ 0 cos 2 0-sin2 0 +C 2 = ¼ |cos x(2x-1) -x 1-x + C 1 mark OR X2 /2 cos x+1 sin x-x 1-x+C 1 mark 44

21. Solving x2+y2 = 16 & x2=6y 1 mark to get point of intersection as y=2

Req. area 2[_6 ∫

∫ −+

2 0 4 2 2 dy

y 16 dx y 1

mark

=2 +−+ − 2 4 12 0 2 2/3

4 y sin 8 y 16 2 y y 3 6 2 1 mark

=2/3 (8p+2_3)sq. units (if area is not doubled, deduct 1 Mark) ½ mark 22. 1 mark 2 mark 1 mark 23. Operating C-C -(C+C) to get -2 a c + a a+b b a + b b+c c b+c c+a Using C2--------->C2 - C1 to get = -2 a b c C3--------->C3-C1 b c a c a b 1 mark for each operation = 4 marks 24. For writing or using dy/dt=8 dx/dt 1 mark Getting 6

=3x2 1 mark =>x2 = 16 x= ± 4 ½+½ = 1 mark points are (4,11) −− 3 31 , 4 ½+½

= 1 mark 25. Writing x 2 x 3 cos( ) x 3 2 x 3 cos( x y ∆ +−∆++ = ∆ ∆

1 mark x 2 x 3 sin 2 x 3 2 x 3 sin 2 ∆ ∆ ∆++− = 1 mark 2 x 3 2 x 3 sin 2 x 3 2 x 3 sin 2 3.2 0x lim dx dy ∆

∆ ∆++− →∆ = 1 mark

= -3sin (3x+2) 1 mark 26. (i) ∑x=93, ∑y=120, ∑(xy)=1113, ∑x2=1687, ∑y2=2588 ½ x5 = 2 ½ marks r(x,y) = − − −

∑∑∑∑ ∑∑∑ N )y( y N )x( x N yx ) xy ( 2 2 2 2

½ mark Correct Substitution ½ mark = -0.98 ½ mark (Alternative Method) (ii) Taking Assunmed Means as 12,16 respectively, ∑dx=9, ∑dy=8 ∑ (dx dy) = -471 ½ x5 = 2 ½ marks ∑dx2 = 463 ∑dy2=536

r(x,y)= − − −

∑∑∑∑ ∑∑∑ N ) dy ( dy N ) dx ( dx N dy dx ) dxdy ( 2 2 2 2

½ mark Correct substituting ½ mark = -0.98 ½ mark (i) Interpretation : High degree of -ve correlation 1 mark [½ mark for writng ‘ y decreases when x increases’ ] (iii) for scatter diagram ½ mark Scatter diagram confirms interpretation ½ mark 27. Eqn. of plane through the intersection of two planes is 2x+6y+12+ λ (3x-y+4z) = 0 1 mark or (2+3λ)x+(6- λ) y+4λz+12 = 0 1½ mark length of ^ from origin p = 2 2 2 16

)6()32( 12 λ+λ−+λ+

1½ mark putting p=1 & solving to get λ= +_2 1½ mark Eqn. of planes are 1 mark

2x+y+2z+3 = 0 x-2y+2z-3 = 0 vector equations of these two planes are r.(2i+j+2k) +3 = 0 1 mark and r.(i-2j+2k) -3 = 0 28. Writing = λ y/x 3 marks

∫ ∫ λ= x dx y dy

1 mark log |y| = log |x| + log|C| 1 mark log |y| = log Cxλ I y=Cxλ 1 mark 28. Getting |A| = 5 1 mark Getting A-1 =1/5 − − − 322 232 223

3 marks (2 marks for correct Co-factors (1 marks for every 4 correct) 1 mark for writing A-1 =1/|A| (adj A) Writing L.H.S =A (A-4i-5A-1) ½ mark Substituting to show A-4i - 5A-1 = 0 Hence A2 - 4A-5i = 0 ½ mark 29. Let length = x units width or breadth = x units depth = y units volume V=x2y3 (unit)3 (given) 1 mark Surface Area = 4xy+x2 1 mark Substituting y=V/x2 to get A=4V/x+x2 1 mark solving dA=0 or -4V+2x = 0 1 mark dx x2 x3 = 2V or x3 = 2x y y= x/2 1 mark

Showing 0 dx Ad 2 2

⟩1

mark Hence area is least when y=x/2 Question Paper Code 65/1 1. Prob. of success (p) = ½ , q= ½, n=12 ½ mark P (getting 10 Tails) = 2 10 10 12

2 1 2 1 C 1 mark

= 33 or 66 ½ mark 2048 2 66 65 64 63 62 61 56 55 54 53 52 51 46 45 44 43 42 41 36 35 34 33 32 31 26 25 24 23 22 21 16 15 14 13 12 11

For mentioning that they are 36 outcomes ½ mark For writing 36 outcomes correctly 1½ mark [deduct 1 mark if some entries out of 36 are incorrect] P (one spade and a red queen) = 2 52 1 2 1 13

C C.C

1 mark =1/51 1 mark

4. Unit vector | to a and b = a x b 1 mark la x bl = - 5i+10j+10k 1 mark 15 = - i + 2 j + 2 k or i -2j -2k ½ mark for numerator 333333 ½ mark for Denominator 5. a ^ b = a . b = 0 (writing or Using) 1 mark ∴4+5ë+ 6 = 0=> ë = -2 1 mark 6. bxy = (

)

N y y N yx xy 2

∑∑ ∑∑∑ 2

− −

½ mark (Writing or Using) =4/19 ½ mark bxy = (

)

N x x N yx xy 2

∑∑ ∑∑∑ 2

− −

½ mark =4/5 ½ mark 7. For separating variables to get 1 mark cot y dy= -tan x dx Getting log sin y = log cos x+log c => OR sin y = c cos x 1 mark 8. Writing I = dx =1/2 dx ½ mark

32-2x 16-x writing or using dx = 1 log a+x + C ½ mark a- x 2a a-x I= 1 log 4+x +C 1 mark 16 4-x 9. Writing I = òÖ(x+4)2 –(2 Ö3) dx 1 mark x +4 Öx2+8x+4-6 log (x+4) + Öx2+8x+4 +C 1 mark 2 10. for setting x2+x+1 = to get (2x+1) dx = dt 1 mark =>I = 2/3 x (x2+x+1) 3/2+C 1 mark 11. Writing = -cosec x cot x-cosec2 x 1 mark = - coses x(cot x+cosec x) = -(cosec x). y 1 mark =-cosec x + cosec x cot x y sin x = cosec x.y + cosec x cot x.y =y (cosec x + cot x) = y2 1 mark 12. (i) if (x) =x1/3 is defined for all [-1,1] continuous in [-1,1] ½ mark (ii) f (x) = 3 / 2 x 3 1

=> f (x) is not differential be at x=0 in (-1, 1) 1 mark Hence Rolle’s Theorem is not applicable ½ mark 13. for x5 x 5 sin 54 3 x2 x 2 sin 2 0x lim sin5x - 4x 3x sin2x 0x lim − + → = + →

1 mark =-5 14. A + B + C = 0 Þ C = O – A – B 1½ marks (1/2 mark for any two correct elements) C = 003-

-2 -1 1

15. A = a23 a22 a21 a13 a12 a11

½ mark = 2 / 13 5 2 / 7 2 / 11 4 2 / 5

1½ marks 16. Any point (say, B) on the line is (l, 2l+1, 3 l+2) 1 mark Let A be the given point (1,6,3) AB = (l -1,2 -5, 3 l -1) 1 mark AB = (1,2,3) = 0 Þ l = 1 1 mark => foot of perpendicular =(1,3,5) 1 mark 17. Getting three vectors from four points as AB = 6J-6K, AC=6 + 2J+6K, AD=3 + 3J+3K 1 ½ mark Writing or Using that “A,B,C,D are coplaner if [AB, AC, AD] = 0 1 mark Getting -3 1 3 -6 2 6 -6 6 0

=-6 (0) -6(0) = 0 1 ½ mark 18. P(o) = 10C3 / 14C3 = 30 / 91 ½ + ½ = 1 mark P(I) = 10C2 4C1 / 14C3 = 45 / 91 ½ + ½ = 1 mark P(2) = 10C1 4C2 / 14C3 = 15 / 91 ½ + ½ = 1 mark P(3) = 4C3 / 14C3 = 1 / 91 ½ + ½ = 1 mark 19. Writing I= dx 12 x 6 x 12 x 6 1 dx 12 x 6 x x 22 2

∫ ∫

−+ − −= −+

1 mark = dx ) 21 ( ) 3 x ( 1 30 12 x 6 x 6x2 3 1 222

∫

−+ + −+ + −½

+ ½ = 1 mark =x-3 log I x2 +6x-12 I + 15/Ö21 log 21 3 x 21 3 x ++ −+

+C ½ + ½ +1 = 2 mark 20. I = òtan-1x.xdx Integrating by Parts =x2/2 tan -1 x – ½ òx2 / 1+x dx 1 mark x tan -1 x -1/2 I dx+1/2 1 dx 1½ mark 2 2 1+c x/2 tan -1x -1/2x + ½ tan -1 x+c 2 OR x+1 tan-1x-x+c 1½ mark 22 21. Required area =1/3 ∫

4

1

dy y 1

=1 4 2/3y

9 2

mark

1

mark =14/9 sq.units 1 mark 22. I= ∫ π π−

− 2/ 2/

dx ) IxI cos IxI (sin

Since IxI is an even function 1 mark I= 2 ∫ π

− 2/ 0

dx ) x cos x (sin 1

mark =2 [ -cos x - sin x]π/2 0 1 mark =0 1 mark Note : Full credit should be given to those who gave taken the function to be integrated as Greatest integer function & solved correctly. 23 1 mark each for one operation and taking one factor (a+b+c) common [Minimum 3 operations required] 3x1 = 3 marks R ----- R1 -----R3 to get O -(a+b+c) R2 -----R2--------R3 a+b+c a+b+c -(a+b+c) 0 a c+a+2b C (a+b+c) 1 0 -1 0 1 -1 1 mark to get final answer c a c+a+2b C C3+C2+C1 (a+b+c) 1 0 0 010 c a 2(a+b+c) 2(a+b+c) 1 0 0 0 1 0 =2(a+b+c) ca1 24. For = - 1 mark Differentiating the given equation w.r.t.x and substituting to get, y=16 x 9 1 mark Substituting in the given equations to get x= +_3 1 mark y= ± 16/3 ` ½ mark 25. For

1 mark 1 mark 1 mark =3cos (3x+2) 1 mark 26. X= 93 y=120 (xy) = 1113, x = 1687 y=2584 ½ x 5 = 2½ marks Correct Substitution to get ½ marks r(x, y) = -0.98 ½ marks (Alternative Method) (i) Taking Assumed Means as 12,16 respectively dx = 9 , dy = 8 dxdy = -471 ½ x 5 = 2½ marks dx = 463 , dy = 536 ½ marks Correct substitution to get r(x,y)= -0.98 ½ marks (ii) Interpretation : high degree of negative correlation 1 mark (If written “y decreases as x increases, may be given ½ mark) ½ marks (iii) For scatter diagram ½ marks Scatter diagram confirms interpretation 27. Equation of required plane is (x-4+6) (3x+3y-4z) = 0 1 mark or (1+3) x+1 (-1+3) y-4 z + 6=0 1 mark =161 =1 (1+3) + (-1+3) + (-4) 1 mark Solving to get = +_1 1 mark Getting two equations as 2x+y-2z+3 = 0 1 mark x+2y-2z-3 = 0 Getting equations in vector from as r.(2I+j-2k) + 3 =0 1 mark & r (I+2j-2k) - 3 =0 28. Writing dy = (x-2) 2 marks dx (y-0) y dy =1(x-2) dx 1 mark Integrating to get y + (x-2) =C 1 mark 22 Putting x=2, y=3 to get C=9/2 1 mark Equation of curve is (x-2) +y = 9 1 mark 29. Getting |A| = 2 1 mark Adj A -1 1 1 (1 mark for any for four correct co factors) 2 marks 1 -1 1 1 1 -1 A -I = ½ -1 1 1 1 mark 1 -1 1 1 1 -1 Getting A I=1/2 -1 1 1 1 mark 1 -1 1

1 1 -1 Showing 1 A2 - 3 0 0 2 0 3 0 =A-1 1 mark 003 30. Writing S= 2 + 2 1 mark Volume V = h V = (S-2 ] = 1 [Sr-2 ] ½ mark 2 dV = 1 [S-6 ] 1 mark dr 2 Putting dV = 0 to get 6 =5 ½ mark dr Putting the value of S to get 6=2+2 to get h = 2r= diameter 1 mark Showing d2 V < 0 d2 1 mark Hence h= diameter will give maximum volume. Physics Economics Contents

23. Prove, using the properties of determinants 24. A particle moves along the curve 6y = x3+2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as x-coordinate 25. Find the derivative of cos(3x+2) w.r.t x from first principle. 26. Given the following pairs of values of the variable x and y x : 3 5 7 12 20 22 24 y : 30 25 24 16 11 9 5 (i) Find the Karl Pearson’s coefficient of correlation. (ii) Interpret the result. (iii) confirm your interpretation with the help of a scatter diagram. 27. Find the Cartesian as well as vector equations of the planes through the intersection of the planes: which are at unit distance from the origin. 28. The slops of the tangent at any point of a curve is times the slope of the straight line joining the point of contract to the origin. Formulate the differential equation representing the problem and hence find the equation of the curve. 29. 30. An open tank with a square base and vertical sides is to be constructed from a metal sheet so as to hold a given quantity of water. Show that the cost of the material will be least when the depth of the tank is half of its width. Question Paper Code 65/1 1. A coin is tossed 12 times. Find the probability of getting exactly 10 tails. 2. Two dice are thrown simultaneously. List the sample space for this event. 3. Two cards are drawn without replacement from a well-shuffled pack of 52 cards. Find the probability that one is a spade and the other is a queen of red colour. 4. Find a unit vector perpendicular to both the vectors a=4-j-3k and b=2+2j-k. 5. Find if the vectors a=1-j+3k and b=4-5j+2k are perpendicular to each other. 6. Find the regression coefficients b and b given that n=5, _x=15, _y=15, _x2=55, _y2=83 and _xy=53 7. Solve cos x .cosy dy + sin x dx=0 8. Evaluate: 9. Evaluate: 10. Evaluate: 11. 12. Discuss the applicability of Rolle’s theorem for the function f(x) = X1/3 on [-1,1] 13. Evaluate: 14. 15. Construct a 2 x 3 matrix whose elements in the ith row and the jth column are given by: 16. Find the foot of the perpendicular form (1,6,3) on the line: 17. Show that the points with position vectors: 18. Four bad eggs are mixed with 10 goods ones. If 3 eggs are drawn one by one without replacement, find the probability distribution of the number of bad eggs drawn. 19. Evaluate: 20. Evaluate:

21. Sketch the region lying in the first quadrant and bounded by y=9x2, x=0, y=1 and y=4, Find the area of the region, using integration 22. Evaluate: 23. Prove, using properties of determinants: 24. At what points of the ellipse 16x2+9y2 = 400, does the ordinate decrease at the same rate at which the abscissa increases ? 25. Find the derivative of sin(3x+2) w.r.t.x from first principle. 26. Given the following pairs of values of the variables x and y : x : 3 5 7 12 20 22 24 y : 30 25 24 16 11 9 5 (i) Find the Karl Pearson’s coefficient of correlation. (ii) Interpret the result. (iii) Confirm your interpretation with the help of a scatter diagram. 27. Find the Cartesian and the vector equations of the planes thought the intersection of the planes which are at unit distance from the origin. 28. The normal lines to a given curve at each point (x,y) on the curve pass through the point (2,0). The curve passes through the points (2,3). Formulate the differential equation representing the problem and hence find the equation of the curve. 29. 30. Show that a closed right circular cylinder of given total surface area S and maximum volume V is such that its height h is equal to the diameter d of the base. Marking Scheme - Mathematics Expected Answer/ Value Points General Instructions : 1. Marking scheme is to be followed strictly in order to maintain uniformity. 2. Alternative method accepted. Proportional marks are to be awarded. 3. Marks may not be deducted in questions on integration if constant of integration is not indicated. 4. If a question is attempted twice and the candidate has not crossed any answer only first attempt is to be evaluated. Write EXTRA with second attempt. 5. No marks are to be awarded if data is wrongly copied from the question paper. Question Paper Code 65/1/1 1. Probability of success (getting a tail) = ½ prob. (failure)= ½ ½ mark P(exactly 8 tails)=12C8 (½)8 (½)4 1 mark

= 495/4096 of 495/212 ½ mark 2. S={HHH,HHT,HTH,THH,THT,HTT,TTT} ½ mark for every two correct [All eight out comes must be mentioned] 3. Prob (one red queen & one black king) =

C CC 2 52 1 2 1 2

1 mark = 2.2.2 / 52.51 = 2 / 663 1 mark Alternative Method Required Prob. =P(Red queen & Black King) + P (Black King & Red Queen) = 2/52 x 2/51 +2/52 x 2/51 1 mark = 2/663 1 mark 4. Unit vector perpendicular to both a and b = bXa rr rr +

a x b 1 mark ½ mark for numerator or ½ mark for denominator 5. 1 mark (for Writing or Using) 4λ+ 2 + 2= 0 to get λ= -1 1 mark 6. bxy= N )y( y N yx ) xy ( 2

2∑

−∑ ∑∑ −∑

½ mark = 811 / 535 or 1.51 ½ mark bxy = N )y( y N yx ) xy ( 2 2∑

−∑ ∑∑ −∑

[ ] ba

½ mark =811/2042 or 0.39 ½ mark 7. Writing 0 dx x1 x1 dy y1 )41( 22= + + + + + 1 mark

Integrating to get tan-1 y+ ½ log (1+y2) + tan-1 x+ ½ log (1+x2) = C 1 mark (Deduct ½ mark if C is not mentioned) 8. Writing 1 mark =2x4x 2 2 x 2+− −

- log l (x-2)+ C I 2 x 4 x2 + + − 1 mark 9. Putting x2+4x+3 = t ½ mark to get (2x+4)dx =dt ∴∫ ∫ = + + + dt t dx 3 4x x ) 4 x 2 ( 2 ½ mark = 2/3 t3/2 + C =2/3 (x2+4x+3)3/2 + C ½ mark 10. Writing I = ½ ∫ dx / 25x+x2 ½ mark Writing (or using) ∫ + = +

−C

a x tan a 1 dx xa 11

½ mark ⇒ I = 1/10 tan-1 x/5 + C 1 mark 11. y=tan-1 x Differentiating to get ½ mark = 1/1+x2 ⇒ (1+x2) 22

=1 Differentiating again to get 0 dx dy . dx yd )x1( 2 2

mark 12. (i) Showing or stating f(x) =x2/3 is continuous in [-1,1] ½ mark (ii) Since f (x) = 2=

+1

3 1

x3 2

∴ if (x) if not differentiable at x=0 in (-1,1) 1 mark Hence Rolle’s theorem is not applicable ½ mark 13. x 2 sin x 4 x 7 x 3 sin 0x lim + + →

1 mark Dividing numerator and denominator by x to get x2 x 2 sin 2 4 7 x3 x 3 sin 3 0x lim + + → 3 5 24 73= + + 1 mark

14. A+B+C = O ⇒ C=O-A-B ½ mark = −

−− − − − 101 112 202 231 000 000

½ mark = −− −− 101 143

1 mark 15. A=

23 22 21 13 12 11

aaa aaa

½ mark 2/322/5 02/11

½ mark for every two correct (=1 ½ ) 16. Writing or Using “A,B,C,D are coplanar if [AB,AC,AD] = 0 1 mark Getting AB=10I - 12j-4k 1 ½ marks AC = 6I+10j+6k AD = -1 4I+2j+10k or any three vectors using four points Getting = − −− −−

10 2 4 6 10 6 4 12 10

1120-1008-112=0 1 ½ marks 17. Any point (say B) on the given line 1 mark is (- -2, 3λ +1,-2λ +3) if point A is (0,2,7) ⇒AB = (-λ -2, 3λ - 1, -2λ -4) 1 mark AB. (-1,3,-2)=0 to get λ = - ½ Getting foot of ^ B = (-3/2, -1/2, 4) 1 mark 18. Getting P(O) = 3 9 3 5

C C ½ mark 42 5

½ mark 3 9 1 4 3 5

C C.C )1(P=½ 42 20

mark

½ mark 42 15 ½ mark 3 9 3 4

C C )3(P=½ 42 2

mark

½ mark ∴Required Probability distribution is x:0123 p 5 20 15 2 42 42 42 42 19. Writing dx ) 3x4x 3 x 4 1 ( dx

3x4x x 22 2

∫ ∫ +−

−+= +−

1 mark dx xxx x x

∫

−− + +− −+ 1)2( 1 5 34 42 2 1 2 ½+½ = 1 mark x+2 log |x2-4x+3I+5/2 log 1x 3x − −

+C ½+½ + 1 = 2 mark OR Using Partial Fraction Method, Ans is x-½ log |x-1| + 9/2 log |x-3| +C 20. Substituting x=cosx = dx=-1 sinx dx to get 1 mark I= ½ 0(0-sin20) d0 1 mark = ¼ 0 cos 2 0-sin2 0 +C 2 = ¼ |cos x(2x-1) -x 1-x + C 1 mark OR X2 /2 cos x+1 sin x-x 1-x+C 1 mark 44

21. Solving x2+y2 = 16 & x2=6y 1 mark to get point of intersection as y=2

Req. area 2[_6 ∫

∫ −+

2 0 4 2 2 dy

y 16 dx y 1

mark

=2 +−+ − 2 4 12 0 2 2/3

4 y sin 8 y 16 2 y y 3 6 2 1 mark

=2/3 (8p+2_3)sq. units (if area is not doubled, deduct 1 Mark) ½ mark 22. 1 mark 2 mark 1 mark 23. Operating C-C -(C+C) to get -2 a c + a a+b b a + b b+c c b+c c+a Using C2--------->C2 - C1 to get = -2 a b c C3--------->C3-C1 b c a c a b 1 mark for each operation = 4 marks 24. For writing or using dy/dt=8 dx/dt 1 mark Getting 6

=3x2 1 mark =>x2 = 16 x= ± 4 ½+½ = 1 mark points are (4,11) −− 3 31 , 4 ½+½

= 1 mark 25. Writing x 2 x 3 cos( ) x 3 2 x 3 cos( x y ∆ +−∆++ = ∆ ∆

1 mark x 2 x 3 sin 2 x 3 2 x 3 sin 2 ∆ ∆ ∆++− = 1 mark 2 x 3 2 x 3 sin 2 x 3 2 x 3 sin 2 3.2 0x lim dx dy ∆

∆ ∆++− →∆ = 1 mark

= -3sin (3x+2) 1 mark 26. (i) ∑x=93, ∑y=120, ∑(xy)=1113, ∑x2=1687, ∑y2=2588 ½ x5 = 2 ½ marks r(x,y) = − − −

∑∑∑∑ ∑∑∑ N )y( y N )x( x N yx ) xy ( 2 2 2 2

½ mark Correct Substitution ½ mark = -0.98 ½ mark (Alternative Method) (ii) Taking Assunmed Means as 12,16 respectively, ∑dx=9, ∑dy=8 ∑ (dx dy) = -471 ½ x5 = 2 ½ marks ∑dx2 = 463 ∑dy2=536

r(x,y)= − − −

∑∑∑∑ ∑∑∑ N ) dy ( dy N ) dx ( dx N dy dx ) dxdy ( 2 2 2 2

½ mark Correct substituting ½ mark = -0.98 ½ mark (i) Interpretation : High degree of -ve correlation 1 mark [½ mark for writng ‘ y decreases when x increases’ ] (iii) for scatter diagram ½ mark Scatter diagram confirms interpretation ½ mark 27. Eqn. of plane through the intersection of two planes is 2x+6y+12+ λ (3x-y+4z) = 0 1 mark or (2+3λ)x+(6- λ) y+4λz+12 = 0 1½ mark length of ^ from origin p = 2 2 2 16

)6()32( 12 λ+λ−+λ+

1½ mark putting p=1 & solving to get λ= +_2 1½ mark Eqn. of planes are 1 mark

2x+y+2z+3 = 0 x-2y+2z-3 = 0 vector equations of these two planes are r.(2i+j+2k) +3 = 0 1 mark and r.(i-2j+2k) -3 = 0 28. Writing = λ y/x 3 marks

∫ ∫ λ= x dx y dy

1 mark log |y| = log |x| + log|C| 1 mark log |y| = log Cxλ I y=Cxλ 1 mark 28. Getting |A| = 5 1 mark Getting A-1 =1/5 − − − 322 232 223

3 marks (2 marks for correct Co-factors (1 marks for every 4 correct) 1 mark for writing A-1 =1/|A| (adj A) Writing L.H.S =A (A-4i-5A-1) ½ mark Substituting to show A-4i - 5A-1 = 0 Hence A2 - 4A-5i = 0 ½ mark 29. Let length = x units width or breadth = x units depth = y units volume V=x2y3 (unit)3 (given) 1 mark Surface Area = 4xy+x2 1 mark Substituting y=V/x2 to get A=4V/x+x2 1 mark solving dA=0 or -4V+2x = 0 1 mark dx x2 x3 = 2V or x3 = 2x y y= x/2 1 mark

Showing 0 dx Ad 2 2

⟩1

mark Hence area is least when y=x/2 Question Paper Code 65/1 1. Prob. of success (p) = ½ , q= ½, n=12 ½ mark P (getting 10 Tails) = 2 10 10 12

2 1 2 1 C 1 mark

= 33 or 66 ½ mark 2048 2 66 65 64 63 62 61 56 55 54 53 52 51 46 45 44 43 42 41 36 35 34 33 32 31 26 25 24 23 22 21 16 15 14 13 12 11

For mentioning that they are 36 outcomes ½ mark For writing 36 outcomes correctly 1½ mark [deduct 1 mark if some entries out of 36 are incorrect] P (one spade and a red queen) = 2 52 1 2 1 13

C C.C

1 mark =1/51 1 mark

4. Unit vector | to a and b = a x b 1 mark la x bl = - 5i+10j+10k 1 mark 15 = - i + 2 j + 2 k or i -2j -2k ½ mark for numerator 333333 ½ mark for Denominator 5. a ^ b = a . b = 0 (writing or Using) 1 mark ∴4+5ë+ 6 = 0=> ë = -2 1 mark 6. bxy = (

)

N y y N yx xy 2

∑∑ ∑∑∑ 2

− −

½ mark (Writing or Using) =4/19 ½ mark bxy = (

)

N x x N yx xy 2

∑∑ ∑∑∑ 2

− −

½ mark =4/5 ½ mark 7. For separating variables to get 1 mark cot y dy= -tan x dx Getting log sin y = log cos x+log c => OR sin y = c cos x 1 mark 8. Writing I = dx =1/2 dx ½ mark

32-2x 16-x writing or using dx = 1 log a+x + C ½ mark a- x 2a a-x I= 1 log 4+x +C 1 mark 16 4-x 9. Writing I = òÖ(x+4)2 –(2 Ö3) dx 1 mark x +4 Öx2+8x+4-6 log (x+4) + Öx2+8x+4 +C 1 mark 2 10. for setting x2+x+1 = to get (2x+1) dx = dt 1 mark =>I = 2/3 x (x2+x+1) 3/2+C 1 mark 11. Writing = -cosec x cot x-cosec2 x 1 mark = - coses x(cot x+cosec x) = -(cosec x). y 1 mark =-cosec x + cosec x cot x y sin x = cosec x.y + cosec x cot x.y =y (cosec x + cot x) = y2 1 mark 12. (i) if (x) =x1/3 is defined for all [-1,1] continuous in [-1,1] ½ mark (ii) f (x) = 3 / 2 x 3 1

=> f (x) is not differential be at x=0 in (-1, 1) 1 mark Hence Rolle’s Theorem is not applicable ½ mark 13. for x5 x 5 sin 54 3 x2 x 2 sin 2 0x lim sin5x - 4x 3x sin2x 0x lim − + → = + →

1 mark =-5 14. A + B + C = 0 Þ C = O – A – B 1½ marks (1/2 mark for any two correct elements) C = 003-

-2 -1 1

15. A = a23 a22 a21 a13 a12 a11

½ mark = 2 / 13 5 2 / 7 2 / 11 4 2 / 5

1½ marks 16. Any point (say, B) on the line is (l, 2l+1, 3 l+2) 1 mark Let A be the given point (1,6,3) AB = (l -1,2 -5, 3 l -1) 1 mark AB = (1,2,3) = 0 Þ l = 1 1 mark => foot of perpendicular =(1,3,5) 1 mark 17. Getting three vectors from four points as AB = 6J-6K, AC=6 + 2J+6K, AD=3 + 3J+3K 1 ½ mark Writing or Using that “A,B,C,D are coplaner if [AB, AC, AD] = 0 1 mark Getting -3 1 3 -6 2 6 -6 6 0

=-6 (0) -6(0) = 0 1 ½ mark 18. P(o) = 10C3 / 14C3 = 30 / 91 ½ + ½ = 1 mark P(I) = 10C2 4C1 / 14C3 = 45 / 91 ½ + ½ = 1 mark P(2) = 10C1 4C2 / 14C3 = 15 / 91 ½ + ½ = 1 mark P(3) = 4C3 / 14C3 = 1 / 91 ½ + ½ = 1 mark 19. Writing I= dx 12 x 6 x 12 x 6 1 dx 12 x 6 x x 22 2

∫ ∫

−+ − −= −+

1 mark = dx ) 21 ( ) 3 x ( 1 30 12 x 6 x 6x2 3 1 222

∫

−+ + −+ + −½

+ ½ = 1 mark =x-3 log I x2 +6x-12 I + 15/Ö21 log 21 3 x 21 3 x ++ −+

+C ½ + ½ +1 = 2 mark 20. I = òtan-1x.xdx Integrating by Parts =x2/2 tan -1 x – ½ òx2 / 1+x dx 1 mark x tan -1 x -1/2 I dx+1/2 1 dx 1½ mark 2 2 1+c x/2 tan -1x -1/2x + ½ tan -1 x+c 2 OR x+1 tan-1x-x+c 1½ mark 22 21. Required area =1/3 ∫

4

1

dy y 1

=1 4 2/3y

9 2

mark

1

mark =14/9 sq.units 1 mark 22. I= ∫ π π−

− 2/ 2/

dx ) IxI cos IxI (sin

Since IxI is an even function 1 mark I= 2 ∫ π

− 2/ 0

dx ) x cos x (sin 1

mark =2 [ -cos x - sin x]π/2 0 1 mark =0 1 mark Note : Full credit should be given to those who gave taken the function to be integrated as Greatest integer function & solved correctly. 23 1 mark each for one operation and taking one factor (a+b+c) common [Minimum 3 operations required] 3x1 = 3 marks R ----- R1 -----R3 to get O -(a+b+c) R2 -----R2--------R3 a+b+c a+b+c -(a+b+c) 0 a c+a+2b C (a+b+c) 1 0 -1 0 1 -1 1 mark to get final answer c a c+a+2b C C3+C2+C1 (a+b+c) 1 0 0 010 c a 2(a+b+c) 2(a+b+c) 1 0 0 0 1 0 =2(a+b+c) ca1 24. For = - 1 mark Differentiating the given equation w.r.t.x and substituting to get, y=16 x 9 1 mark Substituting in the given equations to get x= +_3 1 mark y= ± 16/3 ` ½ mark 25. For

1 mark 1 mark 1 mark =3cos (3x+2) 1 mark 26. X= 93 y=120 (xy) = 1113, x = 1687 y=2584 ½ x 5 = 2½ marks Correct Substitution to get ½ marks r(x, y) = -0.98 ½ marks (Alternative Method) (i) Taking Assumed Means as 12,16 respectively dx = 9 , dy = 8 dxdy = -471 ½ x 5 = 2½ marks dx = 463 , dy = 536 ½ marks Correct substitution to get r(x,y)= -0.98 ½ marks (ii) Interpretation : high degree of negative correlation 1 mark (If written “y decreases as x increases, may be given ½ mark) ½ marks (iii) For scatter diagram ½ marks Scatter diagram confirms interpretation 27. Equation of required plane is (x-4+6) (3x+3y-4z) = 0 1 mark or (1+3) x+1 (-1+3) y-4 z + 6=0 1 mark =161 =1 (1+3) + (-1+3) + (-4) 1 mark Solving to get = +_1 1 mark Getting two equations as 2x+y-2z+3 = 0 1 mark x+2y-2z-3 = 0 Getting equations in vector from as r.(2I+j-2k) + 3 =0 1 mark & r (I+2j-2k) - 3 =0 28. Writing dy = (x-2) 2 marks dx (y-0) y dy =1(x-2) dx 1 mark Integrating to get y + (x-2) =C 1 mark 22 Putting x=2, y=3 to get C=9/2 1 mark Equation of curve is (x-2) +y = 9 1 mark 29. Getting |A| = 2 1 mark Adj A -1 1 1 (1 mark for any for four correct co factors) 2 marks 1 -1 1 1 1 -1 A -I = ½ -1 1 1 1 mark 1 -1 1 1 1 -1 Getting A I=1/2 -1 1 1 1 mark 1 -1 1

1 1 -1 Showing 1 A2 - 3 0 0 2 0 3 0 =A-1 1 mark 003 30. Writing S= 2 + 2 1 mark Volume V = h V = (S-2 ] = 1 [Sr-2 ] ½ mark 2 dV = 1 [S-6 ] 1 mark dr 2 Putting dV = 0 to get 6 =5 ½ mark dr Putting the value of S to get 6=2+2 to get h = 2r= diameter 1 mark Showing d2 V < 0 d2 1 mark Hence h= diameter will give maximum volume. Physics Economics Contents