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 No. of Total each Question Questions Marks 1. MCQ/Objective type/VSA 01 10 10 2. Short Answer Questions 04 12 48 3. Long Answer Questions 06 07 42 Total 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 2(2) 8 (2) – 10 (4) Algebra 3 (3) 4 (1) 6 (1) 13 (5) Calculus 2 (2) 24(6) 18(3) 44 (11) Vectors and 3-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. 1. If ∗ is a binary operation given by ∗: R × R → R, a∗ b= a+ b2, then –2∗5 is (A) –52 (B) 23 (C) 64 (D) 13 ⎡ππ3 ⎤ ⎛ 1 2. If sin–1 : [–1, 1] → ⎢, is a function, then value of sin–1 ⎜− ⎞⎟is⎣22⎥⎦ ⎝ 2 ⎠−π −π 5π 7 π (A) (B) (C) (D)6 666 ⎛96⎞⎛23⎞⎛30⎞ 3. Given that ⎜⎟= ⎜ ⎟⎜ ⎟ . Applying elementary row transformation 30 1012⎝ ⎠⎝ ⎠⎝ ⎠R1 → R1–2 R2 on both sides, we get ⎛36⎞⎛23⎞⎛1 −4⎞⎛36⎞⎛03⎞⎛30⎞ (A) ⎜⎟= ⎜⎟⎜ ⎟ (B) ⎜⎟= ⎜ ⎟⎜ ⎟30 1012 30 1012⎝⎠⎝⎠⎝ ⎠ ⎝⎠⎝⎠⎝⎠ ⎛−36⎞⎛23⎞⎛30⎞ ⎛−36⎞ ⎛−43⎞⎛30 ⎞ (C) ⎜⎟= ⎜⎟⎜ ⎟ (D) ⎜⎟= ⎜ ⎟⎜⎟⎝30⎠⎝10⎠⎝−32⎠⎝30⎠⎝10⎠⎝12⎠ 4. If A is a square matrix of order 3 and |A| = 5, then what is the value of |Adj. A|? 5. If A and B are square matrices of order 3 such that |A| = –1 and |B| = 4, then what is the value of |3(AB)|? ⎡ 3 ⎤ 22⎛dy⎞⎛dy⎞⎢+6. The degree of the differential equation 1 ⎜ ⎟⎥= ⎜ 2 ⎟ is_______.⎢ ⎝dx⎠⎥⎦ ⎝dx ⎠⎣ Fill in the blanks in each of the Questions 7 and 8: d y7. The integrating factor for solving the linear differential equation x – y= x2 dxis_______. ˆˆ2 i –j8. The value of is_______. 9. What is the distance between the planes 3x + 4y –7 = 0 and 6x + 8y + 6 = 0? GGGGGG10. If a is a unit vector and (x– a). (x + a)= 99, then what is the value of | x |? Section–B 11. Let n be a fixed positive integer and R be the relation in Z defined as a R b if and only if a – b is divisible by n, ∀ a, b ∈ Z. Show that R is an equivalence relation. 12. Prove that cot–17 + cot–18 + cot–118 = cot–13. OR −1 −1 −12Solve the equation tan(2 x) tan(2 x) tan , − >> 3.++ −= 3 x 3 x +2 x +6 x −1 x +6 x −1 x +2 =0 x −1 x +2 x +6 13. Solve for x, OR ⎛12⎞⎛1 −12 ⎞ If A = ⎜⎟and B = ⎜⎟, verify that (AB)′ = B′ A′ .34 32–3 ⎝ ⎠⎝ ⎠14. Determine the value of k so that the function: ⎧k.cos 2 x ⎪⎪ π−4 x , π ⎨ if x ≠f (x) = 4⎪5, π⎪⎩ if x = 4 π is continuous at x =.42 −1 dy dy 2a cos x15. If y = e , show that (1 – x2) – x – ay =0.2dx dx π16. Find the equation of the tangent to the curve x = sin3t , y = cos2t at t = 4 . πFind the intervals in which the function f (x) = sin4 x + cos4 x, 0 < x < , is2 strictly increasing or strictly decreasing. π 17. Evaluate 6 sin4 x cos 3 xdx ∫ 0 3x +118. Evaluate ∫ 2 dx 2x − 2x + 3 OR Evaluate ∫ x.(log )2xdx 19. Find a particular solution of the differential equation x x 2y ey dx + (y – 2 xe y ) dy = 0, given that x = 0 when y =1. G 2ˆ ˆ,ˆ j ˆ Gˆ j 4 ˆ, then find the projection20. If a = i − 2ˆj + kb G =i + 2ˆ −3k and c = 2i − ˆ + k G of bc+ Galong a G . 21. Determine the vector equation of a line passing through (1, 2, –4) and perpendicular to the two lines r G=(8 ˆ j +10 ) ˆ iˆ−16 k andi −16 ˆ k +λ (3 ˆj + 7 ) ˆ (15 iˆ + 29 ˆj + 5 kˆ) +μ (3 iˆ + 8ˆj − 5) kˆ. 22. There are three coins. One is a biased coin that comes up with tail 60% of the times, the second is also a biased coin that comes up heads 75% of the times and the third is an unbiased coin. One of the three coins is chosen at random and tossed, it showed heads. What is the probability that it was the unbiased coin? SECTION–C ⎛ 41 3 ⎞ ⎜⎟ 23. Find A–1, where A = ⎜ 21 1⎟ . Hence solve the following system of⎜⎟31 − 2⎝⎠ equations 4 x + 2 y + 3 z = 2, x + y + z = 1, 3 x + y – 2 z = 5, OR Using elementary transformations, find A–1, where ⎛ 12 −2⎞ A = ⎜⎜ −13 0 ⎟ ⎟⎜⎟0 −21⎝⎠ 24. Show that the semi-vertical angle of the cone of maximum volume and of given slant height is tan−1 25. Evaluate 3(3 x2 + 2x + 5) dx by the method of limit of sum.∫1 26. Find the area of the triangle formed by positive x-axis, and the normal and tangent to the circle x2 + y2 = 4 at (1, 3 ), using integration. 27. Find the equation of the plane through the intersection of the planes x + 3 y + 6 = 0 and 3 x – y – 4z = 0 and whose perpendicular distance from origin is unity. OR Find the distance of the point (3, 4, 5) from the plane x + y + z = 2 measured parallel to the line 2 x = y = z. 28. Four defective bulbs are accidently mixed with six good ones. If it is not possible to just look at a bulb and tell whether or not it is defective, find the probability distribution of the number of defective bulbs, if four bulbs are drawn at random from this lot. 29. 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 maximise profit? Formulate the problems as a L.P.P. and solve it graphically. Marking Scheme Section-A 1. (B) 2. (D) 3. (B) 14. 25 5. –108 6. 2 7. Marks x 8. 2 9. 2 Units 10. 10 1 × 10 = 10 Sections-B 11. (i) Since a R a, ∀ a ∈ Z, and because 0 is divisible by n, therefore 1 R is reflexive. (ii) a R b ⇒ a – b is divisible by n, then b – a, is divisible by n, so b R a. Hence R is symmetric. 1 (iii) Let a R b and b R c, for a,b,c, ∈ Z. Then a – b = n p, b – c=n q, for some p, q ∈ Z Therefore, a – c = n (p + q) and so a R c. 1 Hence R is reflexive and so equivalence relation. 1 1 1 112. LHS = tan–1 + tan–1 + tan–1 17 818 11+ 1 ⎛ 15 ⎞ 178 –1tan= tan–1 + tan–1 = ⎜⎟ + tan–1 111 18 ⎝ 55 ⎠181− .78 31+11 183 1 65 = tan–1 + tan–1 = tan–1 31 = tan–1 111 18 1− 195 11 18 1 = tan–1 = cot–13 = RHS 13OR 2 Since tan–1 (2 + x) + tan–1 (2 – x) = tan–1 3 (2 + x) + (2 − x)2 Therefore, tan–1 = tan–1 1½1(2 −+ x) (2 − x)3 42Thus = 1½ x2 − 33 ⇒ x2 = 9 ⇒ x = ± 3 1 x + 2 x + 6 x −1 x + 6 x −1 x + 2 = 013. Given, x −1 x + 2 x + 6 R → R − R2 21Using , we getR → R − R3 31 C → C − C2 21Using , we getC3 → C3 − C1 2x + 6x + 1x − 4 7− 3 0= 1½ 3− 4− 7 2x + 4 3− 4 11 − 1 0− = 1½ 3− 1− 10 Therefore, (x + 2) (–111) – 4 (37) –3 (–37) = 0 7which on solving gives x = – 13 OR ⎛12⎞⎛1 −12⎞⎛ 73 −4⎞ AB = ⎜ ⎟⎜ −⎟ = ⎜ −⎟ 13432 3 1556⎝⎠⎝⎠⎝ ⎠ ⎛715 ⎞ ⎜⎟35Therefore, LHS = (AB)′ = ⎜⎜ ⎟ 1 46⎝− −⎟⎠ ⎛13 ⎞ ⎛7 15 ⎞ ⎜ ⎟⎛13 ⎞⎜ ⎟−12 35RHS = B′ A′ = ⎜ ⎟⎜⎟= ⎜⎟and hence LHS = RHS⎜ −⎟⎝24⎠⎜4⎝23⎠ ⎝−−6⎠⎟ 1+1 π lim 14. Since f is continous at x = , we have πf (x) = 5.4 x→ 4 πk cos 2( −y)k.cos2 xlim f ( ) lim =lim πx 4 Now x→π = x→ππ−4xy→0 π , where xy,1−=44 π−4( −y)44 πk.cos( −2) ylim 2 lim (sin2) kky= == 1 y →0 ππ−+4yy →0 2.2 y 2 kTherefore,=5 k 1⇒= 10. 2 aa cos−1 xdy a cos −1 x (−)e ⇒=e15. ½y =2dx 1−x 2 dyTherefore, 1−x =−ay .......(1) ½dx Differentiating again w.r.t. x, we get 21−x 2 2 dy dx − 21− x x dy dx =− ady dx 1½ 2 2 2 2(1 ) 1⇒− − =− −dy dy dy x x a xdx dx dx = – a (– ay) [from 1] Hence 2 2 2 2(1 ) 0dy dy x x a y dx dx− − − = . ½ ½ ½ 16. 3cos3 , 2sin 2 dx dyt t dt dt =+ =− 1 Therefore, π 4 2sin 2 ,and 3cos3 t dy t dy dx t dx = ⎛ ⎞=− ⎜ ⎟⎝ ⎠ π2sin 2 π3cos3 4 − = = 2 2 2 1 33.( )2 − = − 1 Also x = sin3t = sin 3 4 π = 1 2 and y = cos2t = cos 2 π = 0. Therefore, Point is 1 ,02 ⎛ ⎞ ⎜ ⎟⎝ ⎠ 1 Hence, equation of tangent is y – 0 = 22 3 x ⎛ ⎜ − ⎝ 1 ⎞ 3 2 0x − y − = OR 1 f ′(x) = 4 sin3x cosx – 4 cos3x sinx = – 4 sinx cosx (cos2x – sin2x) = – sin 4x . Therefore, 1 f ′ (x) = 0 ⇒ 4x = nπ ⇒ x= n 4 π Now, for 0 < x < 4 π , 1 f ′ (x) < 0 Therefore, f is strictly decreasing in ( 0, 4 π) 1½ Similarly, we can show that f is strictly increasing in ( 4 π , 2 π) ½ 17. I = 4 36 0 sin x cos x dx π ∫ 4 26 0 sin 1 – sin ) cos π =∫ x ( x x dx = 1 4 22 0 (1 )t t− d t ∫ , where sin x = t 11 1 1 5 7 2 4 62 0 0 ( ) 5 7 t tt t dt ⎡ ⎤ = − =⎢ − ⎥⎣ ⎦∫ 1 = 5 711 1 1 1 1 1 23 5 2 7 2 32 5 28 4480 ⎛⎞ ⎛⎞ ⎛ ⎞− = − =⎜⎟ ⎜⎟ ⎜ ⎟⎝⎠ ⎝⎠ ⎝ ⎠ 1 35(4x 2) −+ 3x +118. I = 42 1dx = dx∫2 ∫ 222x 32 −+ x 3x −+ x 2 34x −2 51 =∫2 dx +∫ 3 dx 42 x 2x 34 2−+ xx−+2 3 25 dx− ++ 2 =log |2 x 2x 3| 44 ∫⎛−1 ⎞2 ⎛5 ⎞ 1 1x +⎟⎜ ⎟⎜⎝ 2 ⎠⎝22⎠ 3 252 −12x −11 =log |2x − ++ 2x 3 | tan +c 14 455 2 3 52x −1 =log |2x22x 3| tan −1 +c− ++ 4 25 OR ∫ 2 ∫ 2I =x(log ). = xxdx (log ) x dx 2 x21 x2 =(log x) −2log x dx ∫ 12 x 2 x = 2 (log )2 −∫log xxdx 1 x . 22 2 22x 2 ⎡ x 1 x ⎤ 1(log ) −⎢log x −∫. dx ⎥ 1= x . 22 x 2⎣⎦ 2 2 2 (log )2 2 2 x x x= − 2 log 4 x x c+ + 1 19. Given differential equation can be written as 2 2. x y x y dx x e y dy ye − = Putting x dx d v v x vy v yy d y d y = ⇒ = ⇒ = + 1 2 1 2 2Therefore, 2 v v dv v ye yvy dy ye −+ = 2 1 2 v v v e e −= 1 2 2 1 2 v v dv ve y vdy e − = − 1 Hence 2 v dyedv y =− 2 log | |ve y c⇒ =− + 1 or 2 log | | x ye y c=− + when x = 0, ⇒ C = 2 y = 1 Therefore, the particular solution is 2 x ye log | | 2y=− + 1 2 20. ˆ ˆ ˆˆ ˆ ˆ ˆ ˆ ˆ( 2 3 ) (2 4 ) 3bc i j k i j k i jk += + − + −+ = ++ G G 1 ˆˆ ˆ2 2a i jk = − + G Projection of ( +) G Gbcalong ( ). || bca a a + = G GGG G is 62 1 5 344 1 −+ = ++ units 1 1+1 21. A vector perpendicular to the two lines is given as ˆ ˆˆ ˆ ˆ ˆ(3 16 7 ) (3 8 5 ) i j k i j k − + × + − = ˆ 3 3 i ˆ 16 8 j − ˆ 7 5 k − 1 1 2 ˆ ˆˆ ˆ ˆ ˆ24 36 72 or 12(2 3 6 ) i j k i j k= + + + + 1 Therefore, Equation of required line is ˆ ˆˆ ˆ ˆ ˆ( 2 4 ) (2 3 6 ) r i j k i j k=+ − + + +G λ 1 1 2 22. Let E1: selection of first (biased) coin E2: selection of second (biased) coin E3: selection of third (unbiased) coin P(E1) = P(E2) = P(E3) = 1 3 1 2 Let A denote the event of getting a head Therefore, 1 AP E ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = 40 100 , 2 A 75 P E 100 ⎛ ⎞ =⎜ ⎟ ⎝ ⎠ , 3 AP E ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ = 1 2 1 1 2 3EP A ⎛ ⎞ ⎜ ⎟⎝ ⎠ = 1 1 AP(E ) P E ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ 3 3 2 2 AP(E ) P E AP(E )P E ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ ⎛ ⎞ + ⎜ ⎟ ⎝ ⎠ 3 3 AP(E ) P E ⎛ ⎞ + ⎜ ⎟ ⎝ ⎠ 1 2 = 1 1 .3 2 140 1 75 1 1 . . .3100 3100 3 2+ + 10 33 = 1 1 2 23. SECTION–C |A| = 4 (–3) –1 (–7) + 3 (–1) = –12 + 7 – 3 = –8 1 A11 = –3 A21 = 5 A31 = –2 A12 = 7 A22 = –17 A32 = 2 A13 = –1 A23 = –1 A33 = 2 1 1 2 Therefore, A–1 = – 1 8 3 7 1 −⎛ ⎜ ⎜⎜−⎝ 5 17 1 − − 2 2 2 −⎞ ⎟ ⎟⎟⎠ 1 2 Given equations can be written as 42 11 31 ⎛ ⎜ ⎜⎜⎝ 3 1 2 ⎞ ⎟ ⎟ −⎟⎠ x y z ⎛⎞ ⎜⎟ ⎜⎟⎜⎟⎝⎠ = 2 1 5 ⎛⎞ ⎜⎟ ⎜⎟⎜⎟⎝⎠ A.X B X =(A ′–1 1⇒′ =⇒ )B −1=(A )B′ x ⎛⎛⎞ −37 −1⎞⎛⎞ 2 ⎜⎟ −1 ⎜ ⎟⎜⎟ ⇒y = 5 −17 −11⎜⎟⎜ ⎟⎜⎟ 8⎜⎟⎜ ⎟⎜⎟ z −22 25⎝⎠⎝ ⎠⎝⎠ 1⎛⎞ ⎜⎟ 2⎛−6 +7 –5 =− 4 ⎞ ⎜⎟ 1 ⎜ ⎟⎜⎟ 3 =– 10 −17 −5 =−12 = 1⎜ ⎟⎜⎟ 8 21⎜⎝−4 +2 +10 = 8 ⎟⎠ ⎜⎟ 2−1⎜⎟⎜⎟⎝⎠ 13 1 Therefore, x = , y = ,z =−1 222 OR ⎛12 −2⎞⎛ 100⎞ 1⎜ ⎟⎜ ⎟Writing A =−13 0 = 010 A⎜ ⎟⎜ ⎟ 2⎜ ⎟⎜ ⎟0 −2 1 001⎝ ⎠⎝ ⎠ ⎛12 −2⎞⎛ 100⎞ ⎜ ⎟⎜ ⎟R2 →R2 +R1 ⇒ 05 −2 = 110 A⎜ ⎟⎜ ⎟ 1⎜ ⎟⎜ ⎟0 −2 1 001⎝ ⎠⎝ ⎠ ⎛12 −2⎞⎛ 100⎞ ⎜ ⎟⎜ ⎟R2 →R2 +2R 3 ⇒01 0 =112 A⎜ ⎟⎜ ⎟ 1⎜ ⎟⎜ ⎟0 −2 1 001⎝ ⎠⎝ ⎠ ⎛12 −2⎞⎛100⎞ ⎜ ⎟⎜⎟R3 → R3 + 2R 2 ⇒ 01 0 = 112⎜ ⎟⎜⎟ 1⎜ ⎟⎜⎟001 225⎝ ⎠⎝⎠⎛12 0 ⎞⎛ 5 410 ⎞ ⎜⎟⎜ ⎟R → R + 2R ⇒ 01 0 = 112 A11 3 ⎜⎟⎜ ⎟ 1⎜⎟⎜ ⎟001 225⎝⎠⎝ ⎠⎛10 0 ⎞⎛ 326⎞ ⎜ ⎟⎜⎟R → R − 2R ⇒ 01 0 = 112 A11 2 ⎜ ⎟⎜⎟ 1⎜ ⎟⎜⎟001 225⎝ ⎠⎝⎠⎛ 326⎞ −1 ⎜⎟ 1⇒ A = 112⎜⎟⎜⎟ 2225⎝⎠ 12 124. Volume v = v = πrh 3 211 2 1l2= h2+ r2 2 v = 1 π (l2 – h2) h = 1 π (l2h – h3)33 1dv =π (l2 −3h2)=0 11dh 3 2 l = 3h , r =2h tan α = r = 2 h α = tan–1 2 2 2 2 0dv h dh =− π < Therefore, vis maximum 25. 3 32 1 1 I (3 2 5) ( ) x x dx f xdx= + + =∫ ∫ 1 [lim (1) ho hf → = + (1 )f h+ + (1 2 ) ....... f h+ + + ](1 ( 1) ) f n h+ − ...... (i) 1 Now where 31 2h n n − = = (1) 325 10 f =+ + = 2 2(1 ) 3 3 6 22 510 8 3f h h h h h h+ = + + + + + = + + 2 2 2(1 2 ) 3 12 12 2 4 5 10 8.2. 3.2 . f h h h h h h+ = + + + + + = + + 1 1 2 (1 ( 1) ) f n h+ − 2 210 8( 1) 3( 1) . n h n h= + − + − I = 2( 1) ( 1)(2 1)lim 10 8 3 2 6n nn nn nh n h h →∞ − − −⎡ ⎤+ +⎢ ⎥⎣ ⎦ 1 1 2 = 2lim n n→∞ 2 16 ( 1) 12 10 2 nn n n n −⎡ + +⎢⎣ ( 1)(2 1) 6 nn n− − ⎤ ⎥⎦ 2 ⎡ 2 ⎤ 1lim 10 n+8( n−1) (n−1) (2 n−1) = n→∞ n ⎢⎣ n ⎥⎦ 2 ⎡ 1 11 ⎤lim2 10 +8(1 − )+ 2(1 − ) (2 − )= ⎢⎥ 1 n→∞ ⎣ n n n ⎦ 1 = 2 [10 + 8 + 4] = 44 26. Equation of tangent to x2 + y2 = 4 at (1, 3) is 4 − x x+ 3y= 4. Therefore, y= 3 Equation of normal y= 3x 1 1 44 − x+ dxTherefore, required area = ∫03 xdx ∫13 1 ⎛ x2 ⎞11 ⎛ x2 ⎞4 = 3 + 4x−⎜ ⎟ ⎜⎟ 123 2⎝⎠0 ⎝⎠1 31 ⎡ 7 ⎤ 3 33 =+ 8 −=+ =2 3 sq. units 2⎢⎥23 ⎣ 2⎦22 27. Equation of required plane is 1(x+ 3y+ 6) + λ (3x– y– 4z)=0 1 2 1⇒ (1 + 3 λ) x+ (3 – λ) y– 4 λ z+ 6 = 0 2 Perpendicular distance to the plane from origin is Therefore, 2 2 2 6 1 (1 3 ) (3 ) ( 4 ) = + +− +−λ λ λ 1 1 2 or or 36 = 1+ 9 λ2 + 6 λ+ 9 + λ2 – 6 λ+ 16 λ2 26 λ2 = 26 ⇒ λ= ±1 Equations of required planes are 4 x + 2y – 4 z + 6 = 0 and –2x + 4y + 4z + 6 = 0 1 1 2 or 2x + y – 2z + 3 = 0 and x – 2 y – 2z – 3 = 0 OR 1 Equaiton of line is 2x = y = z i.e. 1 1 1 2 x y z = = or 12 2 x y z == 1 Equation of line P Q is x −3 y −4 z −5 = ==λ 1122 ⇒λQ ( λ+3, 2+4,2+5) λ lies on plane. Therefore, 1 ++ −=0 1λ 32λ+4 +2λ+52 2 or 5λ=−10 gives λ=−2which gives the coordinates of Q(1, 0,1) 1Therefore, PQ = 4 +16 +16 =6 units 1 2 28. Let x denotes the number of defective bulbs 6C4 6.5 .4.3 1P(X =0) == = 10 1C4 10.9. 8.7 14 64CC 6.5.4.4. 31 8P(X =1) == 4 = 10 1C4 10.9.8.7 21 66CC 6.5.4.3 22 3P(X =2) == .6 = 10 1C4 10.9. 8.7 7 66CC 6.4.3.2 13 4P(X =3) == .4 = 10 1C4 10.9. 8.7 35 4C4 4.3.2.1 1P(X =4) == = 10 1C4 10.9.8.7 210 Therefore, distribution is X : 0 1 2 3 4 P ( X ): 1 14 8 21 3 7 4 35 1 210 29. Let number of chairs to be made per week be x and tables be y Thus we have to maximise P = 30 x + 60 y Subject to 2 x + y ≤ 70 x + y ≤ 40 2 x + 3y ≤ 90 x ≥ 0y ≥ 0 Vertices of feasible region are 2 1A (0,30), B (15, 25), C (30,10), D (35, 0) 2 P (at A) = 30 (60) = 1800 P (at B) = 30 (15 + 50) = 1950 1P (at C) = 30 (30 + 20) = 1500 1 2 P (at D) = 30 (35) = 1050P is Maximum for 15 chairs and 25 tables.