Page 1 :
Inequalities, , , A statement involving variable (s) and the sign of inequality viz, > or, <, or,  or,  is called an, Inequation., , , , |x|  a  a  x  a ie, x  [–a, a], , , , |x|  a  x  a or x  a i.e., x  (, a]  [a, ), , , , |x  a|  r  a  r  x  a + r i.e., x  [a  r, a + r], , , , |x  a|  r  x  a  r or, x > a + r i.e., x  (, a r]  [a + r, ), , Solving Rational Algebraic Inequations, , If P(x) and Q(x) are polynomials in x, then the inequations, , P(x), P(x), P(x), P(x),  0,,  0,,  0, and,  0 are, Q(x), Q(x), Q(x), Q(x), , known as rational algebraic inequations. To solve the inequations we use the sign method as explained in the, following steps, , , Obtain P(x) and Q(x), , , , Factorize P(x) and Q(x) into linear factors, , , , Make the coefficient of x positive in all factors, , , , Obtain critical points by equating all factors to zero, , , , Plot the critical points on the number line. If there are n critical points, they divide the number line into, (n + 1) regions., , , , In the right most region the expression, , P(x), bears positive sign and in other regions the expression, Q(x), , bears alternate negative and positive signs., , Time Saving Results (TSR), Following results are useful in answering the objective type questions quickly., , , , , , , , xa, 0, xb, xa, The solution set of, 0, xb, xa, The solution set of, 0, xb, xa, The solution set of, 0, xb, , The solution set of, , is (– , a]  (b, ). With a < b, is (– , a)  (b, ). With a < b, is [a, b). With a < b, is (a, b). With a < b, , The solution set of the inequality, , NITTE PU COLLEGE, ,  x  a   0 , with a > b > c, is given by (– , c)  (b, a),  x  b  x  c , , SANTHOSH BHAT, , 1, , I PU LINEAR INEQUALITY

Page 2 :
 x  a   0 , with a > b > c, is (c, b)  (a, ).,  x  b  x  c ,  x  a   0, ,c  (b,a] and for  x  a   0, c,b  [a, )., , ,  ,  x  b  x  c ,  x  b  x  c , , whereas solution set of, For, , Example : solution set for, ,  x  3  0 is ,3  4,5, ,   ,  x  4  x  5 , ,  x  3  0 is[3, 4)  (5, ).,  x  4  x  5, Guidance Corner, , 4, 1. The inequality  5 is true, when x belongs to, x, 4, 4 , , (A)  ,  , (B)  , , 5, 5 , , , 4 , (C)  ,0    ,  , 5 , , (D) R, , Ans (C), 4, 4,  5  4  5x  x , x, 5, 4 ,  x  , , 5 , 4, 4, When x  0,  5  4  5x  x , x, 5, , When x  0,, , ( x is negative), 4 ,  solution set is  ,0    ,  , 5 , x<0, , 2. The solution of the inequation (x  R) 4x + 3 < 6x + 7 is, (A) x < –2, , (B) x > –2, , (C) –2 < x < 2, , (D) {2}, , Ans (B),  3 – 7 < 6x – 4x, , 4x + 3 < 6x + 7, , –2<xx>–2, 3. The solution set of the inequation 6  –3 (2x – 4)  12 is, (A) (0, 1], , (B) (0, 1), , (C) [0, 1], , (D) (–1, 0], , Ans (C), 6  –3 (2x – 4)  12,  6  – 6x + 12  12,  – 6  – 6x  0  0  x  1, 1, (multiplying by  ), 6, Solution set is [0, 1], NITTE PU COLLEGE, , SANTHOSH BHAT, , 2, , I PU LINEAR INEQUALITY

Page 3 :
4. Solution of the inequation, (A) –5  x < 2, , x5,  0 is, x2, , (B) –5  x  – 2, , (C) –2  x  5, , (D) 2  x  5, , (C) (– 6, 1], , (D) [– 6, –1), , Ans (A), xa,  0  x [a, b), xb, x   5 , x5, 0, 0, x2, x2, , We have,, , x  [–5, 2) (here a = 5, b = 2),  –5  x < 2, 5. Solution set of the inequation, (A) [– 6, 1), , x4,  2 is, x 1, , (B) [–6, 1], , Ans (D), x4, x4,  2, 20, x 1, x 1, x  4  2x  2, , 0, x 1, x   6 , x  6, , 0, 0, x 1, x   1, 6. Solution set of, (A) (– , 1), ,  The solution set is [–6, –1), , 2x  1,  2 is, x 1, 3, , (B)  , , 4, , , Ans (B), 2x  1, 2x  1,  2  2 , 2, x 1, x 1, 2x  1 2x  1, 2 , , 20, x 1, x 1, 4x  3, , 0, x 1, 3, , 4 x  , 4,  , 0, x 1, 3, ,   ,   1,  , 4, , 2x  1, 2x  1, 2, 20, x 1, x 1, 1, ,  0  x  1  0  x  1  x  (,1), x 1, , (C) (1, ), , 3, , (D)  ,   1,  , 4, , , 3, ,  Solution set is  , , 4, , , 7. Solution set of x  4  5, 2x  5  7 is, (A) (–1, 9), , (B) (–1, 9)  (– , – 6), , (C) (1, 9), , (D) (–, – 6)  (1, ), , NITTE PU COLLEGE, , SANTHOSH BHAT, , 3, , I PU LINEAR INEQUALITY

Page 4 :
Ans (C), x  4  5  5  x  4  5,  1  x  9  x   1,9 , 2x  5  7  2x  5  7 or 2x  5  7, ,  x < – 6 or x > 1,  (– , –6)  (1, ), Solution is (–1, 9)  {(–, –6)  (1, )} = (1, 9), 3x  4 x  1, 8. The solution set of, ,  1 is, 2, 4, (B) [1, ), , (A) [0, 1], , (C) (1, ), , (D) (–, 1], , (C) [–4, 4], , (D) [4, ), , Ans (B), 3x  4 x  1, 3x  4 x  3, , 1 , , 2, 4, 2, 4,  2 (3x – 4)  x – 3,  5x – 5  0, x1,  x  [1, ), 9. The solution set of x 2  8  8 is, (A) (–,), , (B) (–, – 4], , Ans (C), x 2  8  8  8  x 2  8  8,  0  x2  16, Now, x2  0 for all x  R, x2  16 for – 4  x  4  x  [– 4, 4], solution set is R  [– 4, 4] = [– 4, 4], 10. The solution set of the inequation x  1  x  2  x  3  6 is, (A) (– , – 2], , (B) (– , 0]  [4, ), , (C) [4, ), , (D) [6, ), , Ans (B), We have, x  1  x  2  x  3  6 ……(1), For,, x < 1,, ,  –x + 1 – x + 2 – x + 3  6, x  0, , 1 < x < 2,, ,  x –1 – x + 2 – x + 3  6,  x  –2, , 2 < x < 3,, NITTE PU COLLEGE, , x–1+x–2–x+36, SANTHOSH BHAT, , 4, , I PU LINEAR INEQUALITY

Page 5 :
x6, x–1+x–2+x–36, , x > 3,, ,  3x – 6  6, x4,  solution set is (–, 0]  (– , –2]  [6, )  [4, ) which is (–, 0]  [4. )., 11. The solution set of 5x  6  8 is, 2  14 , , (A)  ,     ,  , 5  5, , , 2   14 , , (C)  ,     ,  , 5  5, , , , 2   14 , , (B)  ,     ,  , 5  5, , , 2   14 , , (D)  ,     ,  , 5  5, , , , Ans (A), 5x  6  8  5x  6  8 or 5x  6  8, ,  5x  2 or 5x  14, 2, , 14 ,  x   ,   or x   ,  , 5, , 15 , 2  14 , ,  Solution set is  ,     ,  , 5  5, , , , 12. The solution set of the inequation, , x2  7,  1 is, 8x, , (A) (–, –7]  [7, ), , (B) ( – , –7]  [ –1, 0)  (0, 1]  [7, ), , (C) [–1, 0)  (0, 1], , (D) [–1, 0)  (0, 1]  [7, ), , Ans (B), x2  7, 1, 8x, x2  7, x2  7,  1 or, 1, 8x, 8x, x 2  8x  7, x 2  8x  7, ,  0 or, 0, 8x, 8x,  x  1 x  7   0 or  x  7  x  1  0, ,  x  0,  x  0, , ,  (– , –7]  [–1, 0), , or (0, 1]  [7, ), ,  solution set is, (–, –7]  [–1, 0)  (0, 1]  [7, ), x 2  7 x  10, 13. The solution of the inequation,  0 is, x 2  6x  9, (A) (–5, 5), , (B) (–5, –2)  (2, 3)  (3, 5), , (C) (2, 5), , (D) (–5, –2)  (4, 5), , NITTE PU COLLEGE, , SANTHOSH BHAT, , 5, , I PU LINEAR INEQUALITY

Page 6 :
Ans (B), x 2  7 x  10, , 0., x 2  6x  9, x 2  7x  10, x 0,  0;, 2,  x  3, , We have, , x 0, , x 2  7x  10, ,  x  3, , 2, , 0, , Clearly, (x – 3)2 > 0 for all x.,  x2 + 7x 10 < 0, , or, , x2 – 7x + 10 < 0, ,  (x + 5) (x + 2) < 0, , or, , (x – 5) (x – 2) < 0, ,  x  (–5, – 2), , or, , x (2, 5), , But x  3,  Solution set is (–5, –2)  (2, 3)  (3, 5), 1, 14. Solution set of x   2 is, x, (A) R – {0}, , (B) R – {–1, 0, 1}, , (C) R – {1}, , (D) R – {0, 1}, , (C) [2, 3], , (D) [1, 3], , (C) (–3, –1), , (D) (–1, 0), , Ans (B), We have, x , , , 1,  2 , clearly x  0., x, , x2 1, x2  1, 2,  2  x 2  1  0 , x, x, , Now, x < 0, we have, x2 + 1 + 2x > 0,  (x + 1)2 > 0 which is true for all x except x = – 1, Now, x > 0, we have, x2 + 1 – 2x > 0,  (x – 1)2 > 0 which is true for all x except x = 1,  solution set is R – {0, –1, 1}, 15. Solution of x  1  x  3 is, (A) x  2, , (B) x  2, , Ans (B), |x1|  |x3|, x < 1 :  x + 3  1  3  not correct, 1 < x  3 : x  1   x + 3  2x  4  x  2 , x > 3 : x  1  x  3  1  3  true, x 2  2x  3, 16. The solution of the inequation, 0, x2 1, (A) (–3, 1), , (B) (–3, 0), , Ans (A), NITTE PU COLLEGE, , SANTHOSH BHAT, , 6, , I PU LINEAR INEQUALITY

Page 7 :
x 2  2x  3,  0 . Clearly, x2 + 1 > 0 for all x  R., x2 1, x 2  2x  3, ,  0  x 2  2x  3  0, 2, x 1, ,  (x + 3) (x – 1) < 0,  –3 < x < 1,  Solution set is (–3, 1), x 2  7x  12, 17. Solution set of, 0, 2x 2  4x  5, (B) (– , 3), , (A) (3, 4), , (C) (4, ), , (D) (–, 3)  (4, ), , Ans (D),  x  3 x  4 , x 2  7x  12, 0 , 0, 2, 2x  4x  5, 2x 2  4x  5, Consider, 2x2 + 4x + 5. B2 – 4AC = 16 – 40 and A > 0.,  2x2 + 4x + 5 > 0 for all x  R.,  x  3 x  4   0  x  3 x  4  0, , , , , 2x 2  4x  5,  x < 3 or x > 4,  Solution set is (–, 3)  (4, ), 18. Solution set of, ,  x  1 x  2 , 1  x, , 2, , 0, , (A) (–, –1)  (1, ), , (B) (–, –2)  (–2, –1)  (1, ), , (C) (–1, 1)  (1, ), , (D) (– , –2 )  (1, ), , Ans (B), ,  x  1 x  2 , 1  x, , 2, ,  x  1 x  2 , 0,   x  1, , 2, , Clearly, (x + 2)2 is always positive., ,  x  1 x  2 , ,   x  1, , 2, , 0, , x 1,  0  a  0  a  0 , x 1, ,  x < –1 or x > 0, But x  –2. Thus the solution set is (–, –2)  (–2, –1)  (1, ), 7, 9, 19. The solution of the inequation, ,  1  0 is,  x  2  x  3 x  3, (A) (–5, 3), , (B) (–5, 2), , (C) (1, 2)  (3, ), , (D) (–5, 1)  (2, 3), , Ans (D), 7, 9, , 1  0,  x  2  x  3 x  3, , , 7  9  x  2    x  2  x  3, ,  x  2  x  3, , NITTE PU COLLEGE, , 0, , SANTHOSH BHAT, , 7, , I PU LINEAR INEQUALITY

Page 8 :
, ,  x  5 x  1  0, x 2  4x  5, 0,  x  2  x  3,  x  2  x  3, ,  Solution set is (–5, 1)  (2, 3), 20. The solution set of x2 + 2  3x  2x2 – 5 is, (B) [1, 2], , 5 , 5 , (C)  , 1   ,   (D)  ,  , 2 , 2 , , We have, x2 + 2  3x, , and, , 2x2 – 5  3x, ,  x2 – 3x + 2 0, , and, , 2x2 – 3x – 5  0, ,  (x – 1) (x – 2)  0, , and, , (2x – 5) (x + 1)  0, , (A) , Ans (A), , 1x2, , and, ,  5 , Clearly, 1,2   , 1   ,0    ,  2 , , , x  –1 or x  5/2, , 21. Number of pairs of consecutive odd integers both of which are larger than 8 and such that their sum is, less than 34 is, (A) 1, , (B) 2, , (C) 3, , (D) 4, , Ans (D), Let x and x + 2 be two odd integers such that x > 8 and x + (x + 2) < 34,  x > 8 and 2x + 2 < 34,  x > 8 and x + 1 < 17,  x > 8 and x < 16,  8 < x < 16,  The pairs are (9, 11), (11, 13), (13, 15), (15, 17),  There are 4 pairs., 2x, 1, , then S =, 2x 2  5x  2 x  1,  2 ,  2 1, (B)   ,0 , (C)   ,  ,  3 ,  3 2, , 22. If S is the set of all real x, such that, (A) (–2, –1), ,  2 1, (D)  2, 1    ,  ,  3 2, , Ans (D), 2x, 1, , 2x  5x  2 x  1, 2x, 1,  2, , 0, 2x  5x  2 x  1, 3x  2, , 0,  2x  1 x  2  x  1, , Now,, , , , 2, , 3x  2, 0, 2x, , 1, ,  x  2  x  1, , NITTE PU COLLEGE, , SANTHOSH BHAT, , 8, , I PU LINEAR INEQUALITY

Page 9 :
 2 1,  Solution set is  2,1    ,  ,  3 2, 3 x  2 5 2  x , 23. If, , , then x belongs to the interval, 5, 3, , (A) (2, ), , (B) [2, ), , (C) (,2], , (D) none of these, , (C)(,1), , (D) (1/3. ), , (C) (, 3/4), , (D) none of these, , (C) (4, 6), , (D) , , Ans (B), 9(x – 2)  25 (2  x), 9x  18  50  25x, 34x  68, x2, 24. If 3x  2  1 , then x belongs to the interval, (A) (1,1/3), , (B) [1,1/3], , Ans (A), 25. The solution set of the inequation, (A) (3/4, 1)  (1, ), , 2x  1,  2 , is, x 1, , (B) (3/4, ), , Ans (A), 2x  1, 2, x 1, 2x  1, 2x  1,  2and,  2, x 1, 1, 2x  1, 2x  1,  2  0 and, 20, x 1, x 1, 1, 4x  3,  0and, 0, x 1, x 1, x  (3/4, 1)  (1,), 2, 26. Solution set of,  1, x  4 is, x4, (A) (2, 4)  (4, 6), Ans (A), 2, 1, x4, 2, ,  1, x4, 2, ,  1,0, x4, x2, , 0, x4, NITTE PU COLLEGE, , or, or, or, , (B) (2, 4), , 2, 1, x 4, 2, 1  0, x4, x  6, 0, x4, SANTHOSH BHAT, , 9, , I PU LINEAR INEQUALITY

Page 10 :
, , x2, 0, x4, , or, ,  x  (2, 4), , or, , x6, 0, x4, , x  (4, 6), , Solution set is (2, 4)  (4, 6), 27. The solution set of the inequation, (A) (2, ), , x2,  0,is, x2, , (B) (,2), , (C) R, , (D) (2, 2), , Ans (B), Conceptual, 28. The number of integral solutions of x2  3x  4 < 0, is, (A) 3, , (B) 4, , (C) 6, , (D) none of these, , (C) (–2, –1)  (1, 2), , (D) (–3, 5), , Ans (B), On solving x = 0, 1, 2, 3, 29. x2 – 3|x| + 2 < 0, then x belong to :, (A) (1, 2), , (B) (–2, –1), , Ans (C), x2 – 3|x| + 2 < 0,  |x|2 – 3 |x| + 2 < 0,  (|x| – 1) (|x| – 2) < 0  1 < |x| < 2,  –2 < x < –1 or 1 < x < 2, x  (–2, –1)  (1, 2), 30. The solution set of the inequation x 2   a  b  x  ab  0,a  b, (B) (, a)  (b, ), , (A) (a, b), , (C) (b, a), , (D) (, b)  (a, ), , Ans (C), (x + a) (x + b) < 0,  (x – (–b)) (x – (– a)) < 0,  –b < x < – a, 31. If [x] denotes the greatest integer less than or equal to x, then the solution of, (A) (6, 7], , (B) [7, 8), , (C) [6, 7), , [x]  6, 0, 8  [x], , (D) [6, 7], , Ans (B), [x]  6, [x]  6, 0,   6  [x]  8  [x]  7  x  [7, 8), 8  [x], [x]  8, 32. Which of the following is a slack inequality?, (A) –2x – 3y > 0, , (B) –2x – 3y  –1, , (C) 2x + 3y < 0, , (D) –2x – 3y > 0, , Ans (B), –2x – 3y  – 1  2x + 3y  1, NITTE PU COLLEGE, , SANTHOSH BHAT, , 10, , I PU LINEAR INEQUALITY

Page 11 :
33. Which of the following is correct?, (A) 2x – 2y  0  x  y, , (B) 2x + 2y < 0  x < y, , (C) –2x – 2y  0 x  y, , (D) –2x + 2y > 0  x > y, , Ans (A), 2x – 2y  0,  2x  2y  x  y, 34. The number line represents which of the interval?, , (A) (–, 100), , (B) (–, 100], , (C) (0, 100), , (D) (100, ), , Ans (B), 35. The number line represents which of the following interval?, , (B) (3, ), , (A) (–, 3), , (C) [–, 3), , (D) [3, ), , Ans (B), 36. The inequality 4x  3 , , 10x  1, represents which of the following interval?, 3, , (A) (–, 4), , (B) (–, 4), , (C) [4, ), , (D) (4, ), , Ans (C), 4x  3 , , 10x  1,  12x  9  10x  1, 3, ,  12x – 10x  + 9 – 1  2x  8,  x  4  x  [4, ), 37. Solution of the inequation 4x + 3 < 6x + 7 when x is an integer, (A) {0, 1, 2, 3, 4, ……}, , (B) {–2, –1, 0, 1, 2, 3, 4, …..}, , (C) {–1, 0, 1, 2, 3, 4,….}, , (D) {1, 2, 3, 4,…….}, , Ans (C),  4 < 2x, , 4x + 3 < 6x + 7, ,  –2 < x,  x > –2,  Solution set is {–1, 0, 1, 2, 3,….}, 38. Given that x, y and b are real numbers and x < y, b < 0 then,, x y, x y, x y, (A) , (B) , (C) , b b, b b, b b, , (D), , x y, , b b, , Ans (C), We have, x < y, b < 0, , NITTE PU COLLEGE, , SANTHOSH BHAT, , 11, , I PU LINEAR INEQUALITY

Page 12 :
1, x y, x  y,  0  , b, b b, , 39. If –3x + 17 < – 13 then,, (A) x  (10, ), , (B) x [10, ), , (C) x  (–, 10), , (D) x  [–10, 10], , Ans (A), We have, –3x + 17 < – 13,  –3x < – 13 – 17,  –3x < –30,  3x > 30,  x > 10  x  (10, ), 40. Solution of linear inequality in variable x, represented by the line, , 9 , (A) x   ,  , 2 , , 9 , (B) x   ,  , 2 , , 9, , (C) x   , , 2, , , 9, , (D) x   , , 2, , , Ans (B), Clearly the solution is, all values of x to greater than or equal to, , 9, ., 2, , 9 , x   ,  ., 2 , , 41. Solution of linear inequality in variable x, represented by the line, , 7, , (A) x   , , 2, , , 7, , (B) x   , , 2, , , 7, , (C) x   ,  , 2, , , , 7 , (D) x   ,  , 2 , , 15 , , (C)  6,  , 2, , , 15 , , (D)  6, , 2, , , Ans (A), 7, , Clearly the solution is x   ,  ., 2, , , 42. The solution set of |4x – 3| < 27 is, 15 , ,  15 , (A)  6,  , (B)  6, , 2, 2, , , Ans (D),  –27 < (4x – 3) < 27, , |4x – 3| < 27, ,  –24 < 4x < 30,  – 6 < x < 15/2, 15 , , Solution set is  6, , 2, , , NITTE PU COLLEGE, , SANTHOSH BHAT, , 12, , I PU LINEAR INEQUALITY

Page 13 :
43. Solution set of, , x4, x4, ,  0 is, , (A) (–, 4), , (B) (4, –), , (C) (–4, 4), , (D) {4}, , Ans (A), We have,, , x4, x4, ,  0., , Now, if x – 4 > 0, then |x – 4| = x – 4 and, x4, 1 0, x4, If x – 4 , 0, then |x – 4| = –(x – 4) and, x4,  1  0, x4,  Solution set is x < 4  x  (–, 4)., 44. The set of values of x for which the inequality |x – 1| + |x +1| < 4 always holds in, (B) (–, 2)  (2, ), , (A) (–2, 2), , (C) (–, –1]  [1, ), , (D) None of these, , Ans (A), Consider the real line, , Now, x < –1, then we have, –x + 1 – x – 1 < 4,  –2x < 4  x > –2, –1 < x < 1, then x – 1 – x – 1 < 4, –2 < 4 which is true, x > 1, then x – 1 + x + 1 < 4,  2x < 4  x < 2, Solution set is (–2, 2), 45. The set of all real numbers x for which x2 + | x + 2| + x > 0 is, , , (D) , ,  , , (A) (–, –2)  (2, ), , (B) ,  2 , , (C) (–, –1)  (1, ), , 2,, , 2, , , , , , , Ans (B), If x + 2 > 0, then we have,, x2 + x + 2 + x > 0  x2 + 2x + 2 > 0, which is true for all x > – 2, ( discernment is < 0), If x + 2 < 0, then we have,, x2 – x – 2 + x > 0  x2 > 2, , NITTE PU COLLEGE, , SANTHOSH BHAT, , 13, , I PU LINEAR INEQUALITY

Page 14 :
 x   2 or x  2, , , ,  , ,  Solution set is ,  2 , , 2, , , , , 46. If |x – 1| > 5, then, (A) x (–4, 6), , (B) x [–4, 6], , (C) x (–, –4)  (6, ), , (D) x [–, –4]  (6, ), , Ans (C), We have, |x – 1| > 5,  x – 1 < –5 or x –1 > 5, x<–4, , or x > 6, ,  x  (–, – 4) or x  (6, ),  x  (–, – 4)  (6, ), 47. If |x + 2|  9, then, (A) x (–7, 11), , (B) x [–11, 7], , (C) x (–, –7)  (11, ), , (D) x (–, –7)  (11, ), , Ans (B), We have, |x + 2|  9,  – 9  ( x + 2)  9, –9–2x9–2,  –11  x  7  x  [–11, 7], 48. x and b are real numbers. If b > 0 and |x| > b , then, (A) x  (–b, ), , (B) x  [–, b), , (C) x  (–b, b), , (D) x  (–, –b)  (b, ), , Ans (D), We have,, |x| > b  x < – b or x > b,  x  (–, –b)  (b, ), x 3, 49. The solution set of,  0 is, x4, (A) (–, – 4)  (–3, ), , (B) (–, 3), , (C) (–, –4)  (3, ), , (D) (3, ), , Ans (C), xa,  0 is given by (–, b)  (a, ), xb, x 3, x3, We have,, , 0, x  4 x   4 , , The solution set of, , NITTE PU COLLEGE, , SANTHOSH BHAT, , 14, , I PU LINEAR INEQUALITY

Page 15 :
50. The solution set of the inequation, (A) (3, 0), , x, 1,   x  0 or 3, x3 x, , (B) (0, 3], , (C) (0, 3), , (D) (–, 3), , Ans (C), x, 1, x, 1, x2  x  3,  ,  0, 0, x3 x, x 3 x, x  x  3, x2 – x + 3  0 for all x  R ( a > 0 and b2 – 4ac < 0), x2  x  3, ,  0  x(x  3)  0  0  x  3  x   0,3, x  x  3, Solution set is (0, 3)., 51. The solution set of the inequation, , x, 1,  is, x2 x, , (A) [–2, –1)  [0, 2), , (B) (–2, –1]  (0, 2), , (C) (–2, –1)  (0, 2), , (D) (–2, –1)  (0, 2], , Ans (D), x, 1, x, 1,  ,  0, x2 x, x2 x, 2, x x2, , 0, x  x  2, ,  x  2  x  1  0, x  x  2,  x  2  x  1 changes sign when x passes through, the points x = – 2, x = –1, x = 0 and x = 2, Now,, x  x  2, , ,  Solution set is (–2, –1)  (0, 2], 52. The solution set of the inequation, 1, , (A)  , 2    2,    1,  , 2, , 1, , (C)  2,    1,  , 2, , , x 2  4x  4, 0, 2x 2  x  1, 1, , (B)  ,    1,  , 2, , 1  1 , , (D)  ,      ,1, 2  2 , , , Ans (A), 2, ,  x  2, x 2  4x  4, 0  2, 2, 2x  x  1, 2x  x  1, (x + 2)2 > 0 for all x  R., , ,  x  2, , 2, , 2x 2  x  1, ,  0  2x 2  x  1  0, ,  (2x + 1) (x – 1) > 0, , NITTE PU COLLEGE, , SANTHOSH BHAT, , 15, , I PU LINEAR INEQUALITY

Page 16 :
1,  x   or x  1, 2, , But x  – 2. Thus, Solution set is, 1,  , 2    2,    1,  , 2, , 1  2x  3x 2, 53. The solution of the inequation,  0 is, 3x  x 2  5, 1, , 1 , (A)   , , (B)  ,1   ,  , 3, , 3 , , 1 , (C)  ,  , 3 , ,  1, (D)  1, , 3, , , Ans (B),   3x 2  2x  1, 1  2x  3x 2, 0 , 0, 3x  x 2  5,   x 2  3x  5, , , 3x 2  2x  1,  0 ….. (1), x 2  3x  5, , Now, for x2 – 3x + 5, we have,, B2 – 4AC = 9 – 20 < 0, A = 1 > 0, Thus, x2 – 3x + 5 > 0 for all x.,  (1)  3x2 + 2x – 1 > 0,  (3x – 1) (x + 1) > 0, 1,  x  or x  1, 3, 1 , Solution set is  ,1   ,  , 3 , 1, 3, 54. Solution set of the inequation, , is, x  2 x 3,  9, , (A)   , 2    3,  ,  2, ,  9 , (C)   ,2    2,3,  2 , , 9, , (B)  ,     2,3, 2, , 9, , (D)  ,     3,  , 2, , , Ans (A), 1, 3, 1, 3, , , , 0, x  2 x 3, x  2 x 3, x  3  3x  6, , 0,  x  2  x  3, , , , ,   2x  9 , ,  x  2  x  3, , 0, , 2x  9, 0,  x  2  x  3, , NITTE PU COLLEGE, , SANTHOSH BHAT, , 16, , I PU LINEAR INEQUALITY

Page 17 :
 9, ,  Solution set is   , 2    3,  ,  2, , x 2  6x  7, 55. The solution set of the inequation,  0 is, x4, , (A) (–7, –4)  (–4, 1), , (B) (–7, 1), , (C) (–7, – 4), , (D) (–7, –4)  (4, 1), , Ans (A), We have,, , x 2  6x  7, 0., x4, , Clearly, |x + 4| > 0 for all x. Thus,, x2 + 6x – 7 < 0  (x + 7) (x – 1) < 0,  –7 < x < 1, but x  – 4,  Solution set is (–7, – 4)  (–4, 1), 56. The number of integral solutions of x2 – 5x + 4 < 0 is, (A) 1, , (B) 2, , (C) 3, , (D) 4, , Ans (B), We have, x2 – 5x + 4 < 0  (x – 4) (x – 1) < 0, 1<x<4, Since, x  Z, x = 2, 3, Thus there are 2 integral solutions., 57. Number of pairs of consecutive even integers, both of which smaller than 12 such that their sum is, greater than 14, (A) 1, , (B) 2, , (C) 3, , (D) 4, , Ans (A), Let x and x + 2 be two even integers such that x < 12 and 2x + 2 > 14,  x < 12 and x > 6,  6 < x < 12, Thus the pair of numbers is (8, 10)., There is only one pair., 58. The number of integral solutions of, (A) 1, , (B) 2, , x 1 1,  is, x2  2 4, , (C) 5, , (D) none of these, , Ans (C), x 1 1, 4x  4  x 2  2, , , 0, x2  2 4, 4  x 2  2, , NITTE PU COLLEGE, , SANTHOSH BHAT, , 17, , I PU LINEAR INEQUALITY

Page 18 :
, , x 2  4x  2, 4  x 2  2, , 0, ,  x2 – 4x – 2 < 0, ,  4  x, x, , 2, ,  2  0, , , , 4  16  8, 2 6, 2, ,  2 6  x  2 6, ,  x = 0, 1, 2, 3, 4, Thus 5 solutions., 59. The solution set of the inequation, (A) (–, 0]  (4, ), , 5x  8, 2, 4x, , (B) [0, 4), , (D) [0, ), , (C) (–, 4), , Ans (A), 5x  8, 5x  8, 2 , 20, 4x, 4x, 7x, , 0, 4x, x0, , 0, 4x, x0, , 0, 4x,  (–, 0]  (4, ), 12x, 60. If,  1 for all real values of x, the inequality being satisfied only if |x| is equal to, 4x 2  9, 3, 2, 1, 1, (A), (B), (C), (D), 2, 3, 3, 2, Ans (A), 12 x, 12x, 1  2, 1, 2, 4x  9, 4x  9, ,  4x, , 2, ,  9  0, 2, ,  4x 2  12 | x | 9  0  4x |2 12 | x | 9  0   2 x  3   0  x , , 3, 2, , 61. The pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is, more than 11., (A) (5, 7), , (B) (7, 9), , (C) (5, 9), , (D) (5, 7) (7, 9), , Ans (D), Let x and x + 2 be two consecutive odd positive integers, Given x < 10 and 2x + 2 > 10,  x < 10 and 2x + 2 > 10,  x < 10 and 2x > 8  x < 10 and x > 4, , NITTE PU COLLEGE, , SANTHOSH BHAT, , 18, , I PU LINEAR INEQUALITY

Page 19 :
 4 < x < 10  x = 5, 7, Positive pairs are (5, 7), (7, 9), Here x  9 ( 9 + 2 = 11  10), 62. Find all pairs of consecutive even positive integers, both of which are larger than 5 such that their sum is, less than 23., (A) (6, 8), , (B) (8, 10), , (C) (10, 12), , (D) all A, B, C, , Ans (D), Let x and x + 2 be the numbers (Both are greater than 5), i.e., x > 5 and x + (x + 2) < 23,  x > 5 and 2x < 21  x > 5 and x < 11  5 < x < 11, x is even,  x = 6, 8, 10 and x + 2 = 8, 10, 12, Possible pairs of numbers are (6, 8), (8, 10), (10, 12)., 63. The longest side of a triangle is 3 times the shortest side and the third side is 2 cm shorter than the, longest side. If the perimeter of the triangle is at least 61 cm, find the minimum length of the shortest, side., (A) 5, , (B) 8, , (C) 12, , (D) 9, , Ans (D), Let x , 3x, 3x – 2 are the sides of a triangle, Also x + 3x + 3x – 2  61,  7x  63  x  9,  Minimum length = 9., 64. A man wants to cut three lengths from a single piece of board of length 91 cm. The second length is to, be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the, possible lengths of the shortest board if the third piece is to be at least 5 cm longer than the second?, (A) [8, 22], , (B) (8, 22), , (C) [8, 22), , (D) (8, 22], , Ans (A), Let x be the shortest side. Then x + 3 and 2x are the remaining lengths, Given x + (x + 3) + 2 x  91 and 2x  (x + 3) + 5,  4x  88 and x  8  8  x  22,  Shortest side = 8 but not more than 22 cm., 65. Sushmitha wants to purchase two articles A and B of cost price Rs. 4 and 3 respectively. She thought, that she may earn 30 paise by selling article A and 10 paise by selling article B. She has not to purchase, NITTE PU COLLEGE, , SANTHOSH BHAT, , 19, , I PU LINEAR INEQUALITY

Page 20 :
total article worth more than Rs. 24. If she purchases the number of articles of A and B, x and y, respectively, then linear inequalities are, (A) x  0, y  0, 4x + 3y  24, , (B) x  0, y  0, 30x + 10y  24, , (C) x  0, y  0, 4x + 3y  24, , (D) x  0, y  0, 30x + 40y  24, , Ans (A), x, y  0 and 4x + 3y  24., 66. Nikitha wants to invest the total amount of Rs. 15000 in saving certificates and national saving bonds., According to rules, she has to invest at least Rs. 2000 in saving certificates and Rs. 2500 in national, saving bonds. The interest rate is 8% on saving certificate and 10% on national saving bonds per annum., She invest Rs. X in saving certificate and Rs. Y in national saving bonds. Then the equation for this, problem is, (A) 0.08x + 0.10y, , (B), , x, y, , 2000 2500, , (C) 2000x + 2500y, , (D), , x y, , 8 10, , Ans (A), 8, 10, y,  0.08x  0.10y, 100, 100, 2x  3, 2  4x, 67. The system 2(2x + 3) – 10 < 6 (x – 2) and, 6, has, 4, 3, , The function is given by profit function  x., , (A) infinite solutions, , (B) two solutions, , (C) three solutions, , (D) no solution, , Ans (A), Let 2(2x + 3) – 10 < 6 (x – 2), 2x  3, 2  4x, 6, 4, 3, (1)  4x + 6 – 10 – 6x + 12 < 0  –2x + 8 < 0,  –2x < – 8  x > 4 i.e., x  (4, ), 2x  3, 2  4x, (2), 6, 4, 3,  6x + 63  8 – 63,  6x – 16x  8 – 63, 55, 55 , , i.e., x   , , 10, 10 , , 55 , ,  55 , Solution set is given by  ,    4,     4, , 10 , ,  10 ,  10x  55  x , , 68. The inequality, , NITTE PU COLLEGE, , 2,  3 is true when x belongs to, x, , SANTHOSH BHAT, , 20, , I PU LINEAR INEQUALITY

Page 21 :
2 , (A)  ,  , 3 , , 2, , (B)   , , 3, , , 2 , (C)  ,     ,0 , 3 , , (D) none of these, , Ans (C), 2, 2, 2,  3  2  3x   x or x , x, 3, 3, 2, 2, 2, Case (ii) : when x  0,  3  2  3x   x  x . Which is satisfied when x < 0, x, 3, 3, 2 ,  x   ,      ,0 , 3 , , Case (i) : when x  0,, , 69. Solution of (x – 1)2 (x + 4) < 0 is, (A) (–, 1), , (B) (–, –4), , (C) (–1, 4), , (D) (1, 4), , Ans (B), (x – 1)2 is always positive except when x = 1 (and then it is 0),  solution is when x + 4 < 0 and x  1, i.e., x < – 4 and x  1, x  (–, –4), , NITTE PU COLLEGE, , SANTHOSH BHAT, , 21, , I PU LINEAR INEQUALITY