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LINEAR INEQUALITY, , TOPIC-1- SOLUTIONS OF SYSTEM OF LINEAR, INEQUALITIES IN TWO VARIABLES, , 1. The region containing all the solutions of an, inequality is called the solution region., 2. In order to identify the half plane represented by an, inequality, it is just sufficient to take any point (a, b), (not online) and check whether it satisfies the, inequality or not. If it satisfies, then the inequality, represents the half plane and shade the region, which contains the point, otherwise, the inequality, represents that half plane which does not contain the, point within it. For convenience, the point (0, 0) is, preferred., 3. If an inequality is of the type ax + by ≥ c or, ax + by≤ c, then the points on the line ax + by = c are, also included in the solution region. So draw a dark, line in the solution region., 4. If an inequality is of the form ax + by > c or, ax + by < c, then the points on the line ax + by = c, are not to be included in the solution region. So draw, a broken or dotted line in the solution region., , 6. The graph of x  2 and y  2 will be situated in the, (a) First and second quadrant, (b) Second and third quadrant, (c) First and third quadrant, (d) Third and fourth quadrant, Solution:(a), Y, , x =2, , 7. The position of points O (0,0) and P (2, – 2) in the, region of graph of in equation 2 x  3 y  5 , will be, (a) O inside and P outside, (b) O and P both inside, (c) O and P both outside, (d) O outside and P inside, Solution:(a), Y, , O, , (5/2, 0), , X, , A, , 4x–2y = –3, A(0,3/2), , B(-3/4,0) X, O, , (a) 4 x  2 y  3, (b) 4 x  2 y  3, (c) 4 x  2 y  3, (d) 4 x  2 y  3, Solution:(b) Origin is not present in given shaded area,, so 4 x  2 y  3 satisfy this condition., 5., The necessary condition for third quadrant region, in xy-plane, is, (a) x  0, y  0, (b) x  0, y  0, , n   2,, , x4 y2 z, , y, 1, 1, , (0,–5/3) x, , Y, , b   3, 4,5, , n   2,1, 2  , , C, O, , 8. The true statement for the graph of inequations, 3 x  2 y  6 and 6 x  4 y  20 , is, x  4 graphs, y  2 zare, 4, (a) Both, , y disjoint, 1 do 1not contain, 2, (b) Both, origin, (c) Both contain point (1, 1), (d) Noneb of 3,these, 4,5 , Solution:(a)The equations, corresponding to, Inequalities 3 x  2 y  6 and 6 x  4 y  20 , are, 3 x  2 y  6 and 6 x  4 y  20 . So the lines represented, by these equations are parallel. Hence the graphs are, Y, disjoint., (0,5), , A, , n  2, , x4 y2, , y, 1, 1, , (3,0), (10/3,0), O (2,0), 3x+2y=6, , CHETHAN M G, , A, , 2x–3y=5, , 4. Shaded region is represented by, , (d) x  0, y  0, , b   3, 4,5 , , X, , O, , LEVEL-I, , (c) x  0, y  0, Solution:(b), , x4 y2 z, , y, 1, 1, 2, , y=2, , SOLVED PROBLEMS, , Shaded, region, , n   2,1, , A, , X, , 6x+4y=2, [email protected], Phon number-81054187620 / 7019881906, , b   3, 4,, , 1

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CONCEPTS OF MATHEMATICS FOR KCET JEE MAIN & ADVANCED, , 9. The vertex of common graph of inequalities, 2 x  y  2 and x  y  3 , is, 5, 3, , (b)  , , , (a)(0, 0), 5 4 , , 3 3, , (c)  ,, , (d), , Solution:(b), , 4, , 3, ,  3, 0, , Y, , 12. The feasible region for the following constraints, L1  0, L2  0, L3  0, x  0, y  0 in the diagram, shown is, (a) Area DHF, (b) Area AHC, (c) Line segment EG, (d) Line segment GI, x IC, (e) Line segment, Y, , 2x+y=2, A, , n   2,1, 2  , , C, O, , F, , (0,2), (1,0), O, , L1 = 0, , x–y=3, , x  4E y  2 G, z4, , y, 1 D 1, 2, , X, , (3,0), (5/3,–4/3), , H, , I, , , , 10. The graph of in equations x  y and y  x  3 is, located in, (a) II quadrant, (b) I, II quadrants, (c) I, II, III quadrants, (d) II, III, IV quadrant, Solution:(c) The shaded area is the required area given, in graph as below., , A, , X, , Y, 5x+3y=15, , n  2x+3y=6,  2,1, 2  , , A, , C, O, (0,2), , x4 y2 z4, , y, 1, 1, 2, , 11. The area of the feasible region for the following, constraints 3y  x  3, x  0, y  0 will be, (a) Bounded, (b) Unbounded, (c) Convex, (d) Concave, Solution:(b), Y, , (3,0), , x4, 1, , X, , O, , Here (0, 2); (0, 0) and (3, 0) all are vertices of feasible, , region. Hence, option (d) is correct., b   3, 4,5 , 14. The solution of set of constraints x  2 y  11,, 3 x  4 y  30 , 2 x  5 y  30 , x  0, y  0 includes, the point, (a)(2, 3), (b)(3, 2), (c)(3, 4), (d)(4, 3), x, Solution:(c), Y, n   2,1,(0,15/2), 2  , , A (0,6), , C, O, , (3, 0), , X, , Hence, it is unbounded., , O, , x4 y2 z, , y, 1, 1, 2, , 2x+5y=30, , x+2y=11, , (10,0), , b   3, 4,5 , , (15,0), (11,0), , X, , Obviously, solution set, , 2, , [email protected], Phon number-8105418762 / 7019881906, , n   2,1,, , 3x+4y=30, , x4 y2 z4, , y, 1, 1, 2, b   3, 4,5 , , b, , A, , (0,11/2), , (0, 1), O, , b   3,, , x, (0,5), , x =y, , Hence it is in I, II and III quadrant., , X, , (b)(0, 0), (d)None of these, , y =x+3, , O, , L2= 0, , C, A B, Solution:(c) Line segment EG, 13. Which of the following is not a vertex of the, positive region bounded by the inequalities, 2 x  3 y  6 , 5 x  3 y  15 and x , y  0, (a)(0, 2), (c)(3, 0), Solution:(d), , n, , x4 y2, , 1, 1, , L3 = 0, , b   3, 4,5 , , (0,–3), , Y, , A, , MCP-MATHEMATICS

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LINEAR INEQUALITY, , origin (0, 0), 2 x  y  2 . Again for x  y  1, 0  0  1 ,, , 15. For the following feasible region, the linear, constraints are, ,  x  y  1 Similarly, x, for, , Y, ,  x  2y  8 ., n   2,1, 2  , C, O, 18. The, shown in the fig is given by the, A shaded region, yx, in equation, , y, , x4 y2 z4, , y, 1, 1, 2, , X, x+3y=11, , O, 3x+2y=12, , b   3, 4,5 , , L.R, , ,  0,14 , , (a) x  0, y  0, 3 x  2y  12, x  3y  11, (b) x  0, y  0, 3 x  2y  12, x  3y  11, , [KCET-2015-1M] [One option correct type], , Solution:(a), 19. The shaded region shown in the fig is given by, the in-equation, x y, , A, , n   2,1, 2    0,5 , , C, O, ,  0,3, , 4x–3y+2=0, x4 y2 z4, , y, 1, 1, 2, , 2x+3y–5=0, X, , , , b   3, 4,5 , , (a), (b), (c), (d), , For the following shaded area, the linear, constraints except x  0 and y  0 , are, Y, ,  4, 0 , ,  6, 0 , , x, , 5 x  4 y  20, x  6, y  3, x  0, y  0, 5 x  4 y  20, x  6, y  3, x  0, y  0, 5 x  4 y x 20, x  6, y  3, x  0, y  0, 5 x  4 y  20, x  6, y  3, x  0, y  0, , n   2,1, 2 [One, , [KCET-2017-1M], C, O option correct type], , x–y=1, , A, , 2x+y=2, , O, , x, , (a) 14 x  5 y  70, y  14, x  y  5, (b) 14 x  5 y  70, y  14, x  y  5, (c) 14 x  5 y  70, y  14, x  y  5, (d) 14 x  5 y  70, y  14, x  y  5, , Y, , O, , 19,14, , 5,0, , (c) x  0, y  0, 3 x  2y  12, x  3y  11, (d) None of these, 16. The region represented by 2 x  3 y  5  0 and, 4 x  3 y  2  0 , is, (a) Not in first quadrant, (b) Bounded in first quadrant, (c) Unbounded in first quadrant, (d) None of these, Solution:(b), , 17., , x  2 y  8,0  0  8 ;, , y, , Solution:(c), ,  0,5, , x4 y2 z4, , y, 1, 1, 2, , X, x+2y=8, , (a) 2 x  y  2, x  y  1, x  2y  8, , b   3, 4,5 , , (b) 2 x  y  2, x  y  1, x  2y  8, , , ,  0,3, x6, , 5 x  4 y  20, ,  4, 0 , , (c) 2 x  y  2, x  y  1, x  2y  8, , y3, ,  6, 0 , , x, , (d) 2 x  y  2, x  y  1, x  2y  8, Solution:(b) To test the origin for 2 x  y  2, x  y  1, and x  2 y  8 in reference to shaded area, 0  0  2 is, true for 2 x  y  2 . So for the region does not include, CHETHAN M G, , 21. The shaded region in the figure is the solution set, of the in-equations., , [email protected], Phon number-8105418762 / 7019881906, , 3, , yx

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CONCEPTS OF MATHEMATICS FOR KCET JEE MAIN & ADVANCED, , a) x + y  8, x + y  4, b) x + y  8, x + y  4, c) x + y  8, x + y  4, 0  x  5, 0  y  5, d) x + y  8, x + y  4 x  0, y  0, Solution:(c) x + y  8, x + y  4, 0  x  5, 0  y  5, 2. Shaded region is represented by, (0,20) Y, C(0,16), , a) 4 x  5 y  20, 3x  10 y  30,, x  6, x, y  0, b) 4 x  5 y  20, 3x  10 y  30,, x  6, x, y  0, c) 4 x  5 y  20, 3x  10 y  30,, x  6, x, y  0, d) 4 x  5 y  20, 3x  10 y  30,, x  6, x, y  0, , y, , (40,0)X, O, (a) 2 x  5 y  80, x  y  20, x  0, y  0, (b) 2 x  5 y  80, x  y  20, x  0, y  0, (c) 2 x  5 y  80, x  y  20, x  0, y  0, (d) 2 x  5 y  80, x  y  20, x  0, y  0, Solution:(c) In all the given equations, the origin is, present in shaded area. Answer (c) satisfy this, condition, , Solution:(d), 19. For the constraints of a L.P. problem given by, x1  2 x 2  2000 , x1  x 2  1500 , x 2  600 and, x 1 , x 2  0 , which one of the following points does, not lie in the positive bounded region, (a) (1000, 0), (b)(0, 500), (c) (2, 0), (d)(2000, 0), Solution:(d) Clearly point (2000, 0) is outside., x1+x2=1500, (0,1500), y, (0,1000), x2=600, , (1000,500), (2000,0), (1500,0), , X1, , x1+2x2=2000, , 1. The linear inequalities for which the shaded region, in the given figure is the solution set are, , 4, , A(20,0), , TOPIC-2- SOLUTIONS OF L.P.P BY GRAPHICALY, , X2, , O, , 2x+5y=80, , Shaded, region, x, , [KCET-19], , (900,600), , x+y=20, , 52. If an LPP admits optimal solution at two, consecutive vertices of a feasible region, then, a) the required optimal solution is at the midpoint, of the line joining two points., b) the optimal solution occurs at every point on the, x, line joining these two points, c) the LPP under consideration is not solvable, C, O, d) the LPP under consideration must be, reconstructed, Solution:(b) If an LPP admits optimal solution at two,  consecutive vertices of a feasible region, then, the optimal solution occurs at every point on the, line joining these two points, 3. The maximum value of Z  x  y subject to, x  2 y  70 and 2 x  y  95, x  0, y  0 is, a) 65, b) 45, c) 55, d) 50, Solution:(c), 4. The minimum value of linear objective function, Z x 5 x  2 y subject to 10 x  2 y  20,, 5 x  5 y  30, x  0 , y  0 is, C, O, a) 10, b) 15, c) 20, d) 25, Solution:(b), 5. yThe maximum value of Z  40x  50 y subject, to 3 x  y  9, x  2 y  8, and x  0, y  0 is, , a) 220, b) 240, c) 260, d) 230, Solution:(d), , [email protected], Phon number-8105418762 / 7019881906, , MCP-MATHEMATICS

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6. The maximum value of Z  4 x  3 y subjected to, the constraints 3x  2 y  160, 5 x  2 y  200,, x  2 y  80, x, y  0 is, (a) 320, (b) 300, (c) 230, (d) None of these, Solution:(d)Obviously, it is unbounded. Therefore, its maximum value does not exist., , LINEAR INEQUALITY, , 9. The maximum value of   3 x  4 y , subject to the, conditions x  y  40, x  2 y  60, x, y  0 is, (a) 130, (b)120, (c)40, (d)140, Solution:(d), , Y, , x+y=4 (0, 40), x, 0, (0,30) (20,20), C, O, (60,0), O (40,0) x+2y=60 X, , 3x+2y=160, , x+2y=80, , 5x+2y=200, , 1. If x  y  2, x  0, y  0 the point at, which maximum value of 3 x  2 y attained will, be, a)  0, 0 , , b)  1 , 1 , , c)  0, 2 , , d)  2, 0 , , 54. The feasible region of an LPP is shown in the, y, figure. Z  3 x  9 y , then the minimum value of z, , occurs at, , 2 2, , [KCET-14-1m] [Only one option correct], , Solution:(d) Let Z  3 x  2 y, At  0, 0   Z  3  0   2  0   0, At  1 , 1   Z  3  1   2  1   1.5  1  2.5, 2, 2, 2 2, At  0, 2   Z  3  0   2  2   4, , a)  5, 5, , b)  0,10 , , c)  0, 20 , , d) 15,15 , , Solution:(a), , At  2, 0   Z  3  2   2  0   6, , Corner, points, A(0 , 20), C(0 , 10), G(5 , 5 ), E(15 , 15), ,  Z  3 x  2 y is maximum at  2, 0 , 7. The minimum value of the objective function, Z  2 x  10 y for linear constraints, x  y  0, x  5 y  5 and x, y  0 is, (a)10, (b)15, (c)12, (d)8, Solution:(b)Required region is unbounded whose, vertex is  5 , 5  Hence the minimum value of, , , 4 4, , 5, 5, objective function is  2   10   15 ., 4, 4, Y, , CHETHAN M G, , (0,0,0), , Minimum, , y, , x, Solution:(d), , (0,4), x  2y  8, , (0,1), , y, , , O, , A, , [KCET-2018-1M] [One option correct type], [NCERT-DIRECT QUESTION], , C, O, , X, , 180, 90, 60, 180, , 55. For the LPP, maximize Z  x  4 y subjected to, the constraints 3 x  2 y  2, x  2 y  8, x, y  0, (a) 4, (b) 8, (c) 16, (d) Has no feasible solution, , x – y=0, x – 5y = –5, , x, (–, , Z  3x  9 y, , (8,0), , (2, 0), x  2y  2, , [email protected], Phon number-8105418762 / 7019881906, , 5, , P(1,1,0), 

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(0,4), , CONCEPTS OF MATHEMATICS FOR KCET JEE MAIN & ADVANCED, , 18. The feasible region of an LPP is shown in the, figure. Z  11x  7 y, then the maximum value of z, occurs at, Y, , n   2,1, 2  , , (0,1), (2, 0), x, O, , x  3y  9, , x y 5, a)  3, 2 , , b)  0,5, , c)  3,3, , d)  5,0 , , (8,0), , C, , x4 y2 z4, , y, 1, 1, 2, b   3, 4,5 , , [KCET-2020-1M] [One option correct type], , n   2,1, 2  , , Solution:(a), , Y, , C, , (0,5), (3, 2), , (0,3), , x  3y  9, , x y 5, , x, (–/2, 0), , Z=11x+7y, 47, 35, 21, , maximum, , (8,0), , x4 y2 z4, , y, 1, 1, 2, , x, , O, Corner, points, A(3 , 2), B(0 , 5), C(0 , 3), , , , b   3, 4,5 , , , , A, (0,0,0), , y, , 19. corner points of the feasible region determined by, the system of linear constraints are  0,3 , 1,1, , P(1,1,0), , , &  3, 0  . Let Z  px  qy, where p, q  0., Condition on p and q so that the minimum of Z, occurs at  3, 0  & 1,1 is, a) p  q, q, c) p , 2, , b) p  2q, d) p  3q, , [KCET-2020-1M] [One option correct type], , Solution:(c) Given minimum of Z occurs at,  3, 0  & 1,1 is, So min Z  3 p & min Z  p  q, ,  3p  p  q  2p  q  p , , 6, , q, 2, , [email protected], Phon number-8105418762 / 7019881906, , MCP-MATHEMATICS