• MATHEMATICS - CLASS XII Time : 3 Hours Max. Marks : 100 The weightage of marks over different dimensions of the question paper shall be as follows: (A) Weightage to different topics/content units S.No. Topic Marks 1. Relations and functions 10 2. Algebra 13 3. Calculus 44 4. Vectors and three-dimensional geometry 17 5. Linear programming 06 6. Probability 10 Total: 100 (B) Weightage to different forms of questions: S.No. Form of Questions Marks for Total Number Marks each Question of Questions1.MCQ/Objective type/VSA 01 10 102.Short Answer Questions 04 12 483.Long Answer Questions 06 07 42 29 100 (C) Scheme of Option: There is no overall choice. However, an internal choice in four questions of four marks each and two questions of six marks each has been provided. Blue Print Units/Type of Question MCQ/VSA S.A. L.A. Total Relations and functions -4 (1) 6 (1) 10 (2) Algebra 3 (3) 4 (1) 6 (1) 13 (5) Calculus 4 (4) 28 (7) 12 (2) 44 (13) Vectors and three dimensional geometry 3 (3) 8 (2) 6 (1) 17 (6) Linear programming – – 6 (1) 6 (1) Probability – 4 (1) 6 (1) 10 (2) Total 10 (10) 48 (12) 42 (7) 100 (29) Section—A Choose the correct answer from the given four options in each of the Questions 1 to 3. xy 21 1 1. If , then (x, y) isxy 43 2 (A) (1, 1) (B) (1, –1) (C) (–1, 1) (D) (–1, –1) 2. The area of the triangle with vertices (–2, 4), (2, k) and (5, 4) is 35 sq. units. The value of kis (A) 4 (B) – 2 (C) 6 (D) – 6 3. The line y= x+ 1 is a tangent to the curve y2 = 4xat the point (A) (1, 2) (B) (2, 1) (C) (1, – 2) (D) (–1, 2) 4. Construct a 2 × 2 matrix whose elements aij are given by ⎧3ˆij−+ ⎪ ,if ij≠ aij =⎨ 2 ⎪ 2⎩( + ), if ij=ij . x5. Find the value of derivative of tan–1 (e) w.r.t. xat the point x= 0. 6. The Cartesian equations of a line are . Find the vector equationx3 y2 z62 53 of the line. 83 123+x dx7. Evaluate (sin x) – Fill in the blanks in Questions 8 to 10. sinx+cos xdx=_____8. ∫1+sin2x 9. If a G2iˆ 4ˆjk – ˆ and bˆ3iˆ–2 ˆjkˆ are perpendicular to each other, then λ = ______ G10. The projection of ai=+ˆ 3ˆjk ˆ along bˆ=2iˆ–3ˆj+6 ˆ is _________+ k Section—B –1 1sin x 1sin xx11. Prove that cot ,0 x 1sin x 1sin x 22 OR Solve the equation for xif sin–1x+ sin–12x= 3, x> 0 12. Using properties of determinants, prove that bc ca ab abc qrrp pq 2 pqr yz zx xy xyz 13. Discuss the continuity of the function fgiven by f(x) = |x+1|+ |x+2| at x= – 1 and x = –2. 14. If x = 2cosθ – cos2θ and y= 2sinθ – sin2θ, find 2 2 dy dx at 2 . OR dy –1 2 << 10,prove that = , where –1 x dx ()x1+ 15. A cone is 10cm in diameter and 10cm deep. Water is poured into it at the rate of 4 cubic cm per minute. At what rate is the water level rising at the instant when the depth is 6cm? OR 1 Find the intervals in which the function f given by f (x) = x3 + 3, x ≠ 0 is x (i) increasing (ii) decreasing 3x 216. Evaluate 2 dx (x 3)( x1) OR 1 Evaluate log (log x)2 dx log x) xsin x dx17. Evaluate 2 0 1cos x 18. Find the differential equation of all the circles which pass through the origin and whose centres lie on x-axis. 19. Solve the differential equation x2y dx – (x3 + y3) dy = 0 G GG GG GGG G20. If abac, a 0 and bc G, show that bc a G for some scalar . 21. Find the shortest distance between the lines GG G G r G=−++ −+(λ1) iˆ(λ 1) ˆj(1 λ)kˆ and r (1 )iˆ (2 1) ˆj ( 2) kˆ 22. A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn and found to be hearts. Find the probability of the missing card to be a heart. Section—C 23. Let the two matrices A and B be given by 1 1 0 2 2 4 A 2 34 and B 42 4 01 2 2 1 5 Verify that AB = BA = 6I, where I is the unit matrix of order 3 and hence solve the system of equations xy 3, 2 x 3y 4z 17 and y 2z 7 24. On the set R– {– 1}, a binary operation is defined by a * b = a + b + ab for all a, b ∈ R – {– 1}. Prove that * is commutative on R – {–1}. Find the identity element and prove that every element of R – {– 1}is invertible. 25. Prove that the perimeter of a right angled triangle of given hypotenuse is maximum when the triangle is isosceles. 26. Using the method of integration, find the area of the region bounded by the lines 2x + y = 4, 3x – 2y = 6 and x – 3y + 5 = 0. OR 4 (2x )Evaluate 2 xdx as limit of a sum. 1 27. Find the co-ordinates of the foot of perpendicular from the point (2, 3, 7) to the plane 3x– y– z = 7. Also, find the length of the perpendicular. OR Find the equation of the plane containing the lines rij iˆˆ ri j ˆ (ˆ ˆ 2) ˆ. G ˆˆ (2ˆjk )and G ˆ ij k Also, find the distance of this plane from the point (1,1,1) 28. Two cards are drawn successively without replacement from well shuffled pack of 52 cards. Find the probability distribution of the number of kings. Also, calculate the mean and variance of the distribution. 29. A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contains atleast 8 units of Vitamin A and 10 units of Vitamin C. Food ‘I’ contains 2 units/kg of Vitamin A and 1 unit/kg of Vitamin C. Food ‘II’ contains 1 unit/kg of Vitamin A and 2 units/kg of Vitamin C. It costs Rs 50 per kg to purchase Food ‘I’ and Rs 70 per kg to purchase Food ‘II’. Formulate this problem as a linear programming problem to minimise the cost of such a mixture and solve it graphically. Marking Scheme Section—A 1. (C) 2. (D) 3. (A) Marks 14 2 4. 5 162 15. 2 6. G (3iˆ–2 ˆ 6) ˆ (2 iˆ–5ˆ 3 ) ˆ , where is a scalar. r jk jk 7. 0 8. x + c 9. λ = –2 1 10. 1 × 10 = 107 Sections —B –1 1sin x 1 – sin x11. L.H.S. = cot 1sin x 1 – sin x = –1 ⎨⎪⎝ 2 ⎪ cos cot cos ⎝ 2 ⎛ x ⎪⎛ x ⎧ ⎜ ⎪⎜ ⎩ +sin +sin x ⎞2 ⎞2 2 ⎠ 2 ⎠ x ⎟ ⎟ – + cos ⎝ 2 cos ⎝ 2 ⎛ x ⎛ x ⎜ ⎜ – sin – sin x ⎞2 ⎞2 2 ⎠ 2 ⎠ x ⎟ ⎟ ⎫ ⎪ ⎪⎬ ⎪ ⎪⎭ 11 2 = –1cot cos 2 cos 2 x x sin sin 2 2 x x cos 2 – cos 2 x x – sin – sin 2 2 x x since 0 << ⇒ cos >sin ⎣ 24 2 ⎡ x π x ⎢ 2 ⎦ x ⎤ ⎥ = –1cot cos 2 x sin 2 x cos – sin 2 x 2 x cos 2 x sin 2 x – cos 2 x sin 2 x = –1cot 2cos 2 2sin 2 x x = cot–1 cot 2 x 2 x 11 2 since 0 24 x π⎡ ⎤<<⎢ ⎥⎣ ⎦ OR 1 sin–1x + sin–12x = 3 ⇒ sin–12x = 3 – sin–1 x ⇒ 2x = sin ( 3 – sin–1x) 1 = sin 3 cos (sin–1x) – cos 3 sin (sin–1x) = 3 2 1 2 –1 sin (sin )x 1 2 x 4x = 231– –x x , 5x = ⇒ 25x2 = 3 (1 – x2) ⇒ 28x2 = 3 231– x 112 ⇒ x2 = 3 28 ⇒ x = 13 27 1 13 1Hence x= (as x> 0 given)272 13Thus x= is the solution of given equation.27bc ca ab qrrp pq yz zx xy 12. Let Using C1 C1 +C2 +C3,we get 2(abc ) ca ab 2( )pqr rp pq 2(x yz) zx xy abc ca ab 2 pqrrp pq x yz zx xy Using C C–Cand C C–C, we get2 2 1 3 3 1 2 abc p qr ++ Δ= ++ – – b q – – c r 112 x yz++ – y –z Using C C+ C+ Cand taking (– 1) common from both Cand C1 1 2 3 2 3 abc 112 p qr 2 x yz 13. Case 1 when x < –2 f (x) = |x + 1| + |x + 2| = – (x + 1) – (x+2) = –2x –3 Case 2 When – 2 ≤ x < –1 f (x) = –x – 1 + x + 2 = 1 1 Case 3 When x ≥ –1 f (x) = x + 1 + x + 2 = 2x + 3 Thus –2x – 3 when x –2 () 1 when –2 x –1fx 2x 3 when x –1 lim fxNow, L.H.S at x = –2 , − ()= lim –2 x–3 = 4 – 3 = 1x→–2 x –2– lim f ()x lim1 1R.H.S at x = –2 , + = x→–2 x –2 Also f (–2) = |–2 + 1| + |–2 + 2| = |–1| + |0| = 1 1lim fxlim fxThus, = f (–2) = 12x –2– x –2 ⇒ The function f is continuous at x = –2 Now, L.H.S at x = –1 , –1– lim x f x = – –1 lim 1 x = 1 R.H.S at x = –1 , –1 lim x f x = –1 lim 2 3 x x = 1 Also f (–1) = |–1 + 1| + |–1 + 2| = 1 Thus, –1 lim x f x = –1– lim –1 x f ⇒ The function is continuous at x = –1 Hence, the given function is continuous at both the points x = –1 and x = –2 14. x = 2cosθ – cos2θ and y = 2 sinθ – sin2θ 112 So dy dx = dy d dx θ θ = cos – cos 2 sin 2 – sin θ θ θ θ = 3 – –2sin sin 2 2 32cos sin 2 2 θ θ⎛ ⎞ ⎜ ⎟⎝ ⎠ θ θ 3tan 2 θ = 11 2 Differentiating both sides w.r.t. x, we get 2 2 2 3 3 sec 2 2 dy d dxdx 23 3 sec 2 2 1 2 sin 2 –sin 23 3 sec 4 2 1 32cos 2 sin 2 33 3 sec cosec 8 2 2 θ θ = 11 2 2dy π 3 33ππ –3 Thus 2 at θ = is sec cosec = 1dx 284 42 OR We have x 1 yy 1 x 0 ⇒ x 1 y – y 1 x Squaring both sides, we get x2(1 + y) = y2(1 + x)1 ⇒ (x + y) (x – y) = –y x (x – y) –x ⇒ x + y = –x y , i.e., y = 21 x–1dy 1 x .1 – x 01⇒ – =2 121dx 1 x x15. Let OAB be a cone and let LM be the level of water at any time t. Let ON = h and MN = r dV Given AB = 10 cm, OC = 10 cm and dt = 4 cm3 minute, where V denotes the volume of cone OLM. Note that Δ ONM ~ Δ OCB MN ON rh h⇒ or ⇒ r = 1CBOC 510 2 12Now, V = rh .... (i)3 hSubstituting r = in (i), we get 2 11V = πh31 12 2 Differentiating w.r.t.t dV3 h2 dh dt 12 dt dh 4 dv =⇒ 2dt πh dt dh 41Therefore, when h = 6 cm, = cm/minute 1dt 9π 2 OR 1 f (x) = x3 + 3x 3⇒ f ′(x) = 3x3 – 4x6 2423 (x –1) 3 ( x –1 )( x + x +1)=4 = 41 xx As x4 + x2 + 1 > 0 and x4 > 0, therefore, for f to be increasing, we have x2– 1 > 0 1⇒ x –,–1 1, 1 2 Thus f is increasing in (–,– ∞ 1)∪(1, ∞) (ii) For f to be decreasing f ′(x) < 0 ⇒ x2 – 1 < 0 ⇒(x – 1) (x + 1) < 0 ⇒ x∈(–1, 0 )∪(0, 1 ) [ x ≠0 as f is not defined at x = 0] 12 1Thus f (x) is decreasing in (–1, 0 )∪(0, 1 ) 3–2 A Cx B 16. Let 2 21x 3 x 1 x 3 x 1 x 1 Then 3x – 2 = A (x + 1)2 + B (x + 1) (x + 3) + C (x + 3) comparing the coefficient of x2, x and constant, we get A + B = 0, 2A + 4B + C = 3 and A + 3B + 3C = –2 Solving these equations, we get A = –11 4 , B = 11 4 and C = –5 2 11 2 3 –2 x –11 11 5 ⇒ x 3 x 21 4 x 3 4 x 1 – 2 x 21 3x –2 –111 11 1 51dx = dx + dx − dxHence ∫ 2 ∫ ∫∫ 2( x + 3)( x +1) 4 x + 34 x +12 ( x +1) –11 11 5 1log x 3 log x 1C114 4 2 x 12 OR 1log log x 2 dx log x = ∫log log ( x)dx +∫ 1 dx (log x)2 Integrating log (logx) by parts, we get x 1log log xdx xlog log x – dxlog xx 11x log log x – dx 1log x 2 x –1 1 x log log x –– x dx log x log x 2 x 1 x 1 x log log x 2 dx log x log x Therefore, ∫⎛⎜log log ( x)+ 12 ⎞⎟dx =x log (log ) x – x +C 11 ⎜ log x ⎟ log x⎝ ()⎠ 2 xsin x dx17. Let I = 2 0 1cos x π aa(π–sin x) ( π– x)⎡ ⎤ = dx since () =f (a – )xdx xdx ∫ 2 ⎢∫∫ ⎥ 0 1cos + (π−x) ⎢⎣ 00 ⎥⎦ sin x dx –I21 0 1cos x sin x2I dx 0 1cos 2 x Put cos x = t for xt –1, x 0 t1 and – sin x dx dt . 1– dt 1 dt 1Therefore 2I 2 = –11 2 1 11 tt 2 –11 –1 –1 =π⎡tan t⎤−1 =π⎡tan ()1 tan () ⎤+− –1⎣⎦⎣ ⎦ ππ21⎡⎤=+π = 1⎢⎥⎣⎦22 2 π2 I = 4 18. The equation of circles which pass through the origin and whose centre lies on x – axis is ( x – a)2 + y2 =a2 ... (i) 112 Differentiating w.r.t.x, we get dy2– 2yxa 0dx dy 1 xy a 1dx 2 Substituting the value of a in (i), we get dy 2 dy 2 yy2 xydx dx 2 2 dyx y 2xy 01dx 19. The given differential equation is 2 33x ydx – x ydy 0 dy x2 y⇒= 3 3 ...(1)dx x +y dy dv Put yvx so that vx 1dx dx dv vx3 vx 3 33dx x vx dv v vxdx 1 v3 dv – v4 x dx 1 v3 1 v3 dx 4 dv –1 vx 11 dxdv dv – 41 vv x –1 log v – log x c 3v3 3–x⇒ 3 +log y =c , which is the reqd. solution. 13y 20. We have GG GGabac GGG GGab – ac 0 GGGG ab – c 01 GGG GG GGG a 0or b – c 0 or a & b – c 1 G GGGGG G⇒a & (b – c )⎡since a ≠0& b ≠c ⎤ 1⎣⎦ G bcGG for some scalar G – a, GGbc λa⇒=+1 GGGG GG21. We know that the shorest distance between the lines ra band rc dis given by GG GG (–) ca bd D GG bd Now given equations can be written as ˆ ˆˆˆ ˆ ˆ– – –r i jk i jk G and ˆˆ–r i j ˆ2 k ˆ–i ˆˆ2 jk Therefore ca GG ˆ2i ˆˆ2 3j k 1 2 and bd GG 11 –1 –1 2 1 i j k G G G 3 –0. 3i j k G G G bd GG 9 9 18 3 2 1 2 Hence D = –ca bd bd GG GG GG 6–0 9 32 15 32 5 2 5 2 2 . 2 22. Let E, E2, E3, E4 and A be the events defined as follows : E1 = the missing card is a heart card, E2 = the missing card is a spade card, E3 = the missing card is a club card, E4 = the missing card is a diamond card ½ A = Drawing two heart cards from the remaining cards. 13 1 13 1 13 1 13 1Then PE1 ,PE 2 ,PE 3 ,PE 4 ½524 52 4 524 52 4 P (A/E1) = Probability of drawing two heart cards given that one heart card is 12C2missing = 51C2 P (A/E2) = Probability of drawing two heart cards given that one spade card is 13C2missing = 51C2 13 13C2 C2Similarly, we have P (A/E3) = 51 and P (A/E4) = 51 1C2 C2 By Baye’s thereon, we have the required Probability =P (E1 A) PE PAE 1=1 1PE1 PAE 1 PE 2 PAE 2 PE3 PAE 3 PE 4 PAE 4 1 12C2 4 51C2 12 13 13 13 1C1C1C1C 1222 2 5151 51 514 C4C 4C4 C222 2 12C2 66 11 12 131313 C C C C 6678 7878 502222 Section C 23. We have 1–1 0 2 2–4 AB= 234 –4 2–4 1 012 2–1 5 600 100 =060 6010 = 6I 006 001Similarly BA = 6I, Hence AB = 6I = BA As AB = 6I, A 1 AB 6A 1I . This gives 1 2 2–4 1 1 11 –42–4 1IB=6A ,i.e., A B 16622 –1 5 The given system of equations can be written as AX = C, where x 3 X y ,C 17 z 7 1The solution of the given system AX = C is given by X = A–1C 2 634 28 2 x 2 2 –4 3 y z 1 6 –4 2 2 –1 –4 5 17 7 1 –12 34 –28 –16 6–17 34 4 Hence x = 2, y = 1 and z = 4 2 24. Commutative: For any a, b ∈ R – { – 1}, we have a * b = a + b + ab and b * a = b + a + ba. But {by commutative property of addition and multiplication on R – {–1}, we have: a + b + ab = b + a + ba . a * b = b * a Hence * is commutative on R – {–1} 2 Identity Element : Let e be the identity element. Then a * e = e * a for all a ∈ R – {–1} a + e + ae = a and e + a + ea = a e (1+a) = 0 e = 0 [since a–1) Thus, 0 is the identity element for * defined on R – {–1} 2 Inverse : Let a ∈R – {–1}and let b be the inverse of a. Then a * b = e = b * a a * b = 0 = b * a (∵e = 0) a + b + ab = 0 –a⇒= ∈R (since a ≠− 1)b a +1 –a a1.Thus bMoreover, ∈ R – {–1}.a 1 a 1 Hence, every element of R – {–1}is invertible and –a the inverse of an element a is .a 1 25. Let H be the hypotenuse AC and θ be the angle between the hypotenuse and the base BC of the right angled triangle ABC. Then BC = base = H cos θ and AC = Perpendicular = H sin θP = Perimeter of right-angled triangle 1 = H + H cos θ + H sin θ = P 1 2 For maximum or minimum of perimeter, dP =0dθH (0 – sin θ + cos θ) = 0, i.e. 14 Now d 2P Hcos H sin 2 1d d 2P π⎡ 11 ⎤⇒ at θ= =–H = 2H <0 1⎥2 ⎢dθ 4 ⎣ 22 ⎦ πThus Pis maximumatθ= . 4 HππH⎛⎞ and Perpendicular = 1For θ= ,Base=H cos =⎜⎟24 42⎝⎠Hence, the perimeter of a right-angled triangle is maximum when the 1triangle is isosceles. 2 112 x –3y +50 x –5, x 0 5 y 0, y 16 3 Finding the point of interection of given lines as A(1,2), B(4,3) and C (2,0) 1 Therefore, required Area 42 4 x 53x 6dx –42x dx – dx 3211 2 21 ⎛x ⎞⎤4 22 ⎛32 ⎞⎤41⎜ −⎤⎜ ⎥32 ⎦1 ⎝4 ⎠⎦ 2 =+5x⎟⎥– (4xx)– x −3x⎟2 ⎝ ⎠⎦⎥12 116 120– 5 –84–41–1212–36 32 2 145 7=× –1– 3=sq. units 132 2 OR 44 2x2 xdx f xdxI = 11 lim f 1 f 1 hf 12h ........ f 1 n 1 h () i 1h 0 4–1where h ,i.e., nh 3 n Now, f 1 n–1h 21 n–1 h2– 1 n–1 h 22 221 n–1 h 2 n–1 h–1 (1 + (n– 1) h) 2–1 nh23 n–1 h 1 22 22Therefore, f 12.0 h 3.0. h 1, f 1 h 2.1 h 3.1. h 1 122 22f 12h 2.2 h 3.2. h 1, f1 n–1 h 2.2 h 3.2. h 11 2 nn –1 2 n –1 23nn –1 nh – hThus, I lim hn 2 h h 06 2 2 nh nh – h 2nh – h 3 nh nh – hlim hn 2 h 06 2 23 3– h 6– h 3 3 (3– n)69 1lim 3 =1h 0622 2 The equation of line AB perpendicular to the given plane is x –2 y –3 z –7 1 = = =λ(say ) 13 –1 –1 2 Therefore coordinates of the foot B of perpendicular drawn from A on the plane 3x – y – z = 7 will be 132,– 3, 7 1 2 Since B 3 2, – 3, 7 lies on 3x – y – z = 7 , we have 332–– 3–– 77 1 Thus B = (5, 2, 6) and distance AB = (length of perpendicular) is Hence the co-ordinates of the foot of perpendicular is (5, 2, 6) and the length of perpendicular = 11 1 OR The given lines are G  --------- (i)rij i 2–jk G and ri j –ij –2 k ---------(ii) 1Note that line (i) passes through the point (1, 1, 0) 2 1and has dr..s , 1, 2, –1 , and line (ii) passes through the point (1, 1, 0) 2 and has dr..s , – 1, 1, –2 Since the required plane contain the lines (i) and (ii), the plane is parallel to the vectors GG bi 2 jk and c ij 2 k GG Therefore required plane is perpendicular to the vector bc and 1 iˆˆjkˆ GG bc 12 1 –3 iˆ 3ˆj 3kˆ1 11 2 Hence equation of required plane is GG GG –. 1ra bc 0 G GGG  rij – .333 0i jk GGGG .–rijk0 and its cartesian form is – x+ y+ z= 0 2 Distance from (1, 1, 1) to the plane is 1(–1) 1.1 1.1 1 28. Let xdenote the number of kings in a draw of two cards. Note that xis a random variable which can take the values 0, 1, 2. Now 48! 48 2 () 48×47 188 C 2! 48–2 ! P( x= 0)= P(no king) = 52 == C2 52! 5251 221 1× ( −)2!52 2! P (x= 1) = P (one king and one non-king) 4C 1 48C1 448 2 32 52 1C2 52 51 221 4C 43× 1 = 2 ==and P (x= 2) = P (two kings ) 52 1C2 52×51 221 Thus, the probability distribution of xis 1 x 0 1 2 P x 18 8 221 3 2 221 1 22 1 n Now mean of x =E( ) =∑xi P( ixx ) i =1 188 32 2 ×1 34 =×01 ++× = 221 221 221 221 n Ex22() xi pxi Also i 1 2188 232 21 36 012221 221 221 221 36 3426800 –Now var (x) = E (x2) – [E(x)2] 21221 221 221 Therefore standard deviation var ()x 6800 221 0.37 1 29. Let the mixture contains x kg of food I and y kg of food II. Thus we have to minimise Z = 50x + 70y Subject to 2x + y > 8 x + 2y > 10 x, y > 0 2 12 2 The feasible region determined by the above inequalities is an unbounded region. Vertices of feasible region are A (0, 8 ) B (2, 4 C 10, 0 )( ) 1 2 Now value of Z at A( 0, 8 = 50 ×0+ 70 8 ) ×=560 B2, 4)((=380 C 10, 0)=500 As the feasible region is unbounded therefore, we have to draw the graph of 150x + 70y < 380 i.e. 5x + 7y < 38 2 As the resulting open half plane has no common point with feasible region thus the minimum value of z = 380 at B (2, 4). Hence, the optimal mixing strategy for the dietician would be to mix 2 kg of food I and 4 kg of food II to get the minimum cost of the mixture i.e Rs 380. 1

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