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- 1. Sample Question Paper MATHEMATICS Class XII Time: 3 Hours Max. Marks: 100 General Instructions 1. All questions are compulsory. 2. 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 four marks each and section C comprises of 07 questions of six marks each. 3. All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. 4. There is no overall choice. However, internal choice has been provided in 04 questions of four marks each and 02 questions of six marks each. You have to attempt only one of the alternatives in all such questions. 5. Use of calculators is not permitted. You may ask for logarithmic tables, if required. SECTION – A 1. Construct a 2 × 2 matrix whose elements are given by a= 2i − j . ij 2. If A is a square matrix of order 3 such that adj A = 144 , write the value of A . K 3 3. For what value of k,the matrix 2 has no inverse. 4 4. If f : R → R is defined byf(x) = 2x + 3, write the value of f ( f ( x)) . 5. Write the value of 6. Evaluate tan −1 (tan 5π ). 6 1 + cos x ∫ x + sin x dx 7. Write the order and degree of the differential equation ( a + x) d2y +y=0. dx 2
- 2. 8 .Find the value of λ ,such that 2i -λ j + k and i + j + k are perpendicular 9. Write the value of a − b , if two vectors a and b are such that = 2, b 3 and = 4 . a = a .b 10.Let A =(1,2,3) and B = (2,-3,5).Find the direction ratios of AB. SECTION – B 11. Show that the function f: R → R defined by f(x) = 2x −1 , x ∈ R, is one –one and onto 3 function .Also find the inverse of the function f. OR Prove that the relation R on the set Z of all integers defined by ( a , b )∈ R ⇔ a − b is divisible by 5 is an equivalence relation. 12. Prove that: 2 tan −1 5 2 1 π −1 1 . + sec −1 7 + 2 tan 8 = 5 4 13. Using properties of determinants show that: −a + b −a + c 3a −b + a 3b −b + c 3 ( a + b + c ) ( ab + bc + ca ) . = −c + a −c + b 3c 14. Find all the points of discontinuity of the function f defined by x ≤1 x + 2 f(x) = x − 2 1 < x < 2 0 x≥2 15. Find dy , if y = x cos x .+ 2 x dx OR
- 3. 1− x 2 dy + y= 0. , show that 1 − x dx 1+ x ( If y = ) 16. Find the equation of the normal to the curve y = x 3 + 2 x + 6 at point (0, 6). 3x + 2 ∫ ( x − 1)(2 x + 3)dx 17. Evaluate π /2 ∫ 18 Evaluate 0 19 sin x sin x + cos x dx Solve the following differential equation: dy 2x + = y dx x 2 + 1 (x 1 2 + 1) 2 = 0 ; y (0) OR Solve the following differential equation: (x 3 + y 3 ) dy − x 2 y dx = 0. 20. Find the area of the parallelogram whose diagonals are determined by the vectors ∧ ∧ ∧ ∧ ∧ ∧ 2 i − j + k and 3 i + 4 j − k .. 21. Find the shortest distance between the lines: r = + − k + λ 3 − and r = 4 − k + µ 2 + 3k . i j i j i i ( ) ( ) ( ) ( ) 22. A factory has two machines A and B. Past record shows that machine A produced 60% of the items of output and machine B produced 40% of the items. Further, 2% of the items produced by machine A and 1% produced by machine B were defective. All the items are put into one stockpile and then one item is chosen at random from this and is found to be defective. What is the probability that it was produced by machine B? SECTION – C 23. Using matrix method, solve the following system of linear equations: x + 2 y + z = 4, x − y − z = 0, x − 3 y + z = 2.
- 4. 24.. . A wire of length 36cm is cut into two pieces is turned in the form of a circle and the other in the form of an equilateral triangle .find the length of each piece so that the sum of area of the two be minimum. OR Show that the right circular cylinder of given volume open at the top has minimum total surface area, provided its height is equal to the radius of its base. 4 25. Evaluate the following integral as limit of sums: 1 ∫ (x 2 − x)dx 26. Using integration, find area of ∆ ABC whose vertices have coordinates A ( 2, 0 ) , B ( 4,5 ) and C ( 6,3) . 27. Find the equation of the plane which contains line of intersection of planes r . + 2 + 3k − 4 =, i j 0 ( ) r . 2 + − k + 5 = and which passes through the i j 0 ( ) Point (1,0,-2). OR 28. Equation of the plane containing the lines → ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ r = i + j + λ (− i + 2 j − k )and r = i + j + λ (− i + j − 2 k ).Find the distance of this plane from the origin and also from the point (1,1,1). 28. A furniture firm manufactures chairs and tables, each requiring the use of three machines A, B and C. Production of one chair requires 2 hours on machine A, 1 hour on machine B and 1 hour on machine C. Each table requires 1 hour each on machine A and B and 3 hours on machine C. The profit obtained by selling one chair is Rs. 30 while by selling one table the profit is Rs. 60. The total time available per week on machine A is 70 hours, on machine B is 40 hours and on machine C is 90 hours. How many chairs and tables should be made per week so as to maximize profit? Formulate the problem as L.P.P. and solve it graphically. 29. If two cards are drawn at random from a deck of 52 cards and X is the number of aces obtained then find the value of mean and variance.
- 5. SOLUTION PAPER -A SECTION A a 1. Required matrix is 11 a 21 3−1 a12 2 * 1 − 1 = a 22 2 * 2 − 1 2. adjA =144 => A 3. 2 * 1 − 2 1 0 = 2 * 2 − 2 3 2 =144 => A =+12 or -12 k 2 3 =o =>4k-6=0 => k= . 3 4 2 4. f(x)= 2x+3 => f(f(x))= f(2x+3) = 2(2x +3) + 3 =4x +6+3 = 4x+9 5. tan −1 (tan Π Π Π 5Π ) = tan −1 tan(Π − ) = tan −1 tan(− ) = − 6 6 6 6 6. Let x+sinx = t => (1+cosx)dx = dt I= ∫ dt = log t + c = log x + sin x + c t 7. Order -2 deg- 1 8. 2- λ + 1 + 0 ⇒ λ = 3 . → →2 →2 →2 → → 9. a − b = a + b − 2 a ⋅ b . =4 +9 - 2× 4 =5 10. Direction ratios are (2-1, -3-2, 5-3) That is SECTION B 11. Let x1 , x2 ∈ R such that f ( x1 ) = f ( x 2 ) (1,-5, and 2).
- 6. 2 x1 − 1 2 x2 − 1 = 3 3 ⇒ x1 = x2 ⇒ ⇒ f is1-1 Let ‘y’ be any arbitrary element of co domain R Then 2x −1 =y 3 ⇒ x= 3y +1 2 3y + 1 x ∈ R Such that f(x) = f 2 3y +1 2 −1 2 = 3 =y ∴ f is onto. f-1 (x)= 3x + 1 2 (OR) Reflexive : (a,a) ∈ R a − a = 0isdivisibleby5 Symmetric: (a,b) ∈ R => (a-b) is divisible by 5 => (b-a) is divisible by 5 => (b,a) ∈ R Transitive: let (a,b),(b,c) ∈ R => a-b and b-c are divisible by5 => (a-b) + (b-c) = a-c is divisible by 5 As R is reflexive ,symmetric and transitive then then R is equivalence. 12. ( 2(tan −1 1 1 5 2 1/ 5 + 1/ 8 1 + tan −1 ) + sec −1 = 2(tan −1 ) + tan −1 5 8 7 1 − 1 / 40 7
- 7. = 2 tan −1 25 28 1 3 1 2 ×1/ 3 1 1 Π = tan −1 1 = × + tan −1 = tan −1 + tan −1 = tan −1 + tan −1 = tan −1 4 28 25 7 4 7 1 − 1/ 9 7 3 13 a+b+c −a+b −a+c a+b+c 3b − b + c (c1 → c1 + c 2 + c3) a+b+c −c+b 3c 1 −a+b −a+c 1 −a+b −a+c 1 3b − b + c (a + b + c) = (a + b + c) 0 a + 2b a − b ( R 2 → R 2 − R1)and ( R3 → R3 − R1) 1 −c+b 3c 0 a − c a + 2c (a + b + c) a + 2b a − b = (a + b + c)[(a + 2b)(a + 2c) − (a − b)(a + 2c)] a − c a + 2c (a + b + c)[3(ab + bc + ca)] = 3(a + b + c)(ab + bc + ca) 14. For x ≤ 1andx ≥ 2 f(x) is a polynomial function so f(x) is continuous At x=1 L.H.L is lim ( x + 2) = lim(1 − h + 2) = 3 − h →0 x →1 R.H.L is lim+ ( x − 2) = lim(1 + h − 2) = −1 h →0 x →1+ As L.H.L ≠ R.H.L X=1 is point of discountinity At x = 2 L.h.L is lim− ( x − 2) = lim+ (2 − h − 2) = 0 x→2 x→2 R.H.L is lim+ 0 = 0 x→2 As L.H.L = R.H.L X=2 is f(x) is continuous. ∴x =1 is the only point of discontinuity. 15. y = x cos x +2 x Let u = x cos x ⇒ ⇒ log u = cos x log x 1 du 1 du 1 = x cos x ( cos x − sin x × log x) = − sin x log x + cos x ⇒ u dx x dx x
- 8. 1 dy du d x = + (2 ) = x cos x ( cos x − Sinx × log x) + 2 x log 2 x dx dx dx OR y2= 1+ x 1− x => (1+x) y 2 = (1 − x) => (1+x) . 2y =>2y(1+x) dy + y 2 = −1 dx dy 1 − x + = −1 dx 1 + x 1 1− x 2 dy =− × = −y dy 1− x 2 ⇒ (1 − x ) dx 1+ x y => 2y(1+x) = −1 − =− dx x +1 1+ x => (1-x 2 ) dy +y=0 dx 16. y = x 3 +2 x + 6 dy dy = 3x 2 + 2 ⇒ dx dx Slope of the normal is - 1 2 Equation of the normal is y-6 = 17. =2 ( 0, 6 ) −1 ( x − 0) ⇒ 2 y − 12 + x = 0 2 3x + 2 A B = + ( x − 1)(2 x + 3) x − 1 2 x + 3 3x+2=A(2x+3)+B(x-1) ⇒ A=1 B=1 I= dx dx ∫ x −1 + ∫ 2x + 3 1 ⇒ log x − 1 + log 2 x + 3 + c 2 ∴
- 9. π sin( 2 18 I = ∫ 0 cos( π π 2 − x) + sin( 2 π − x) 2 π 2 − x) sin x + cos x 0 π π 2 2I = cos x dx = ∫ 2 ∫ 0 sin x sin x + cos x dx π cos x dx + ∫ sin x + cos x 0 2 dx = ∫ 0 sin x + cos x sin x + cos x π π 2 2I = ∫ 1dx = x 02 = 0 19. π 2 => I = π 4 dy 2x 1 + y= 2 2 dx 1 + x ( x + 1) 2 2x ⇒ I .F = e ∫ 1+ x 2 dx ⇒ y.(1 + x 2 ) = ∫ ⇒∫ = e log(1+ x ) = 1 + x 2 2 1 × (1 + x 2 )dx (1 + x 2 ) 2 1 dx = tan −1 x + c 1+ x2 When x=0 ,y=0 0 = tan −1 0 + c => c=0 ∴ y (1 + x 2 ) = tan −1 x (OR) dy x2 y = 3 dx x + y 3 Let y=vx ⇒ ⇒v+x dv dy =v+x dx dx dv v dv v = ⇒x = −v 3 dx 1 + v dx 1 + v 3 1 + v3 1 1 dv v3 =− ⇒ ∫ 3 dx = − ∫ dx ⇒ ∫ 3 dv + ∫ dv = − log x + c 3 dx x 1+ v v v −2 1 − x2 y v ⇒ + v = − log x + c ⇒ − 2 + v = − log x + c ⇒ + = −x + c −2 y x 2v ⇒x dx
- 10. → → ∧ Λ ∧ ∧ ∧ 20. d1 = 2 i − j + k and d 2 = 3 i + 4 j − k ∧ ∧ ∧ i j k ∧ ∧ ∧ 1 → → Area of parallelogram = d1 × d 2 = 2 − 1 1 = −3 i + 5 j + 11 k 2 3 4 −1 21. → → → → → → → S.D = → → b1 = 3 i − j → a2 = 4 i − k → → → a1 = i + j − k → b2 = 2 i + 3 k → → → (a − a ) • (b × b ) 1 2 1 → 2 → (b × b ) 2 1 → → → → a2− a1 = 3 i − j → → → i j k → → → → → × b2 = 3 − 1 0 = −3 i − 9 j + 2 k b1 2 0 3 → → b1× b2 = 94 . → → → → → S.D = (3 i − j ) • ( − 3 i − 9 j + 2 k ) / 94 =0 22. E - Items from Machine A and 1 E - Items from Machine B 2
- 11. E- Choosing a defective item P( E1) = 3 5 P ( E 2) = P(E/ E1) = 1 50 P( E 2 / E ) = 2 5 P(E/ E 2) = 1 100 P(E/ E2 )P(E2 ) P(E/ E1 )P(E1 ) + P(E/ E2 )P(E2 ) 1 2 × 100 5 = 2 1 3 1 × + × 5 100 5 50 = 1 4 SECTION C 1 1 2 1 − 1 − 1 Let A = 1 − 3 1 24. 23. A = -10 − 4 1 − 1 AdjA = − 2 0 − 2 − 2 − 5 − 3 − 4 1 − 1 −1 A = − 2 0 2 10 − 2 5 − 3 −1 X = A −1 B
- 12. − 4 1 − 1 −1 = − 2 0 2 10 − 2 5 − 3 4 0 2 − 18 −1 X= −4 10 − 14 9 5 2 5 7 5 = 24.. 2πr+3x=36 ⇒ x = s = π r2 + 36 − 2πr 3 3 2 x 4 3 36 − 2πr 3 4 =π r + 2 2 3 ds −144π + 8π 2 r = = 2 π r+ 0 36 dr d 2s 3 2 = 2π + 8π 0 ∨ r 2 36 dr for minimum x= ds 18 =0⇒r = dr 3 3 +π 108 3 π +3 3 (OR)V = πr 2
- 13. r h = V πr 2 S = 2πrh + πr 2 S = 2V + πr 2 r ds 2V = 2πr − 2 dr r ds = 0 ⇒ V = πr 3 dr h = V =r πr 2 d 2S 4V = 2π + 3 > 0 2 dr r ∴ h= r is a minima. 25. = 1, b = 4 and h = 3 n f(x) = x2 – x. f(a) = f(1) = 12- 1 = 0 f(a+h)= f(1+h) = (1+h)2 – (1+h) = 12h2 + h f(a+2h) = f(1+2h) = (1+2h)2-(1+2h) =22h2 +2h … …. f(a+(n-1)h) = (1+(n-1)h)2-(1+(n-1)h) = (n-1)2 h2 – (n-1)h 1mark
- 14. b ∫ f ( x)dx = Lim h{f(a)+ f(a+h)+ f(a+2h)+…..+ f(a+(n-1)h)} n →∞ a = 3 12h2 + h+22h2 +2h+…. +(n-1)2 h2 – (n-1)h) lim n (0 + n →∞ = 3 2 lim n {h (12+22+32+… (n-1)2) + h (1+2+3+….. (n-1)} n →∞ 3 lim n { h = (n − 1)(n )(2n − 1) + h (n − 1)n 2 6 n →∞ = lim n →∞ 3 3 n n 2 2 1 1 (1 − )(2 − ) n n + n.n.n 6 } 3 3 n n 1 (1 − ) n n.n. 2 1mark = 27 × 2 + 9 = 6 2 27 2 5 ( x − 2) 26. Equation of AB is y = 2 , equation of BC is y = -x + 9, equation of AC is 3 ( x − 2) y= 4 4 6 5 3 ∫ 2 ( x − 2)dx + ∫ (− x + 9)dx − ∫ 9 ( x − 2)dx =7 square unit Required area = 2 4
- 15. y f(x)=2.5(x-2) f(x)=(-x+9) f(x)=(1/3)(x-2) B(4,5) 5 C(6,3) x A(2,0) -8 -6 -4 -2 2 4 6 8 -5 27. Equations of plane in Cartesian form are x+2y+3z-4=0 and 2x+y-z+5=0 Required equation is (x+2y+3z-4)+k(2x+y-z+5)=0----------(1) It passes through (1,0,-2) => (1+0-6-4) + k ( 2+0+2+5)=0 ⇒ k =1 equation is (x+2y+3z-4) + 1(2x+y-z+5)=0 3x+3y+2z+1=0 (OR) Given lines are → ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ ∧ → ∧ r = ( i + j ) + λ ( i + 2 j − k )and r = ( i + j ) + µ (− i + j − 2 k ) Normal perpendicular to the plane containing two lines is ∧ → ∧ ∧ i j k ∧ ∧ ∧ n = 1 2 − 1 = −3 i + 3 j + 3 k −1 1 − 2
- 16. ∧ → → Plane is → ∧ ∧ ∧ ∧ ∧ ∧ ∧ r . n = {( i + j ) + λ (i + 2 j − k )}.(−3 i + 3 j + 3 k ) ∧ ∧ ∧ r .(−3 i + 3 j + 3 k ) = 0 ---------- (1) Distance from origin is 0 ∧ Distance from (1, 1, 1) to the plane (1) is {( 0}/ . 1+1+1 1 = 3 28. x chairs &y tables are made per week LPP is max z=30x+60y x≥0 , y≥0 2x+y ≤ 70, x+y ≤ 40 , x+3y ≤ 90 FIGURE &SHADING A(35,0) Z=1050 B(30,10) Z=1500 C(15,25) Z=1950 D(0,30) Z=1800 Profit is max at C(15,25) ie x=15, y=25 Max profit is 1950 29. Range of the random variable X is {0, 1, and 2} P(x = 0) C (48,2) 188 = C (52,2) 221 P (x = 1) = C (4,1) × C (48,1) 32 = C (52,2) 221 ∧ ∧ ∧ − i + j+ k ) (. ∧ ∧ i + j+ k )-
- 17. P(x = 2) = C (4,2) 1 = C (52,2) 221 X 0 1 2 P(X) 188 221 32 221 1 221 Mean = E(X) = E(x 2 ) = 0 × 188 32 1 34 ∑ x.p = 0 × 221 + 1 221 × 1 + 2 × 221 = 221 188 2 +1 221 32 221 + 22 × 1 36 = 221 221 Variance = E(X 2 ) − E ( X ) = 36/221 – (34/221) 2 =0.372

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