Tiêu đề Math KCET Solution.pdf ✅

Mathematics - CET 1 KCET MATHEMATICS Ch No. Chapter Name Pg No. 1 Sets, Relations and Function 2-6 2 Complex Numbers and Quadratic Equations 6-12 3 Permutations and Combinations 12-14 4 Binomial Theorem & Principle of Mathematical Induction 14-17 5 Trigonometry Ratios and Identities 18-20 6 Trigonometrical Functions and Properties 20-22 7 Trigonometrical Equations and Inequations, Properties of Triangles, Height and Distance 22-26 8 Straight Lines 26-30 9 Conic Sections 31-34 10 Probability 34-41 11 Statistics 42-43 12 Mathematical Reasoning 44-46 13 Matrices and Determinants 46-57 14 Vector Algebra 57-65 15 Three-Dimensional Geometry 65-73 16 Inverse Trigonometric Functions 73-79 17 Continuity and differentiability 79-84 18 Limits & Derivatives 84-99 19 Application of Derivatives 100-113 20 Integrals 113-126 21 Application of Integrals 126-132 22 Differential Equations 132-138 23 Linear Programming 138-142
2 Mathematics - CET 1. (a) n A B n B ( ) ( )  = A B  As A is a subset of B so all the elements of A are present in B   =  = n A B n B n A B n A ( ) ( ) & ( ) ( ) 2. (d) Given Since Now, 3. (b) We have, Total number of subsets of Number of subsets of A which contains atleast two elements subsets of A having none of 1 element are 4. (b) Let total number of families be x. Then according to question, Thus, there are 30,000 families in the town 5. (b) Number of odd numbers in the given set A 5 6. (d) Consider option (a) Option (b), Option (c), Option (d), There does not exist any for which . Hence, this is an empty set 7. (b) Given Now, but f is not one-one Also, co-domain of f is R and range of f is f is not onto 8. (b) We have, but Thus * is not commutative Now, and * is not associative [infact, * is not a binary operation !!!] 9. (a) Let Clearly, But . Hence 10. (c) 11. (c) 12. (c) Total number of functions from A to Number of bijective function Number of functions which are not bijective 13. (a) We have 2 3( 1) (2) 2(4) 3 4 8 9 − + + = − + + = 14. (c) Here and is defined by and Now, (2) ( (2)) ( 2) 2) 2 2 fog = f g = f = = , which is not defined 15. (b) We have A ={x|xN,x  5} and and Now number of onto functions from a set containing 5 elements to a set containing 2 elements is given by n(U)=100,n(A) = 50,n(B)= 60,n(AB)= 20 n(A B) = n(A)+ n(B)−n(A B)  n(A B) = 50 + 60 − 20 = 90 n(A'B') = n(A'B') = n(U)−(A B) =100−90 =10 n(A) = 6 6 A= 2  2 7 6 = − [ {},{1},{2],{3},{4},{5},{6} = 64 −7 = 57 15000 30,000 100 2 15 15000 100 65 = + −  =  x = x x x x { : −9 = 0,  }= 3,−3} 2 x x x R { : = −1= 0,  }={−1,1} 2 x x x x R { : = + 2 = 0,  }={−1,1} 2 x x x x R { : 1 0, } 2 x x + = xR xR 1 0 2 x + = 4 f(x) = x f(1) = f(−1) 1  −1  [10,)  1 * + = b a a b 3 1 1* 2 = 1 2 2 2*1 = = 1* 2  2*1  12 1 4 1/ 3 *3 3 1 (1* 2)*3 = = = 3 2 1 2 1 1 2 1 1*(2* 3) 1* = + = =  2 2 2 y = 9 − x  y = 9 − x  2 2 2 x − 9 − y  x = 9 − y 9 0 9 2 2 − y   y   −3  y  3 y  0 0  y  3 fog(x) f(g(x)) f(x ) 8(x ) 8x 1/ 3 1/ 3 3 = = = = (fog)(x) = f(g(x))= f(|x|−x) =||x|−x||+|x|−x =|x|−x+|x|−x = −4x 6 A = 6 = 6!  6 6! 6 = − f(−1)+ f(2)+ f(4) f : R → R g :[0,)→R 2 f(x) = x g(x) = x gof(4) = g(f(4)) = g(16) = 16 = 4 fog(−4) = f(g(−4)) gof(−2) = f(f(−2)) = g(4) = 4 = 2 { | , 5 6 0} 2 B= x xZ x − x + =  A ={1, 2, 3, 4, 5} B ={x|xZ,(x −3)(x − 2) = 0}B={2,3} 2 2 32 2 30 5 − = − = Sets, Relations and Function 1 Level 1
Mathematics - CET 3 16. (d) Here * is defined by Now we know that Also, Now, 17. (a) Given defined by Now, will be defined if 7 12 0 2 x − x +  Domain of 18. (c) We have, Let number of elements in set B be p Then, number of possible relations from A to 19. (a) Let So, range of 20. (b) R is symmetric and transitive but not reflexive as (2, 2) and (3, 3) 21. (c) We have, So, total number of binary operations on 22. (d) For to be defined, Domain of f R= −{1,1} 23. (d) 24. (a) We have, f is strictly decreasing f is one-one Let for f is not onto 25. (a) For to be defined, [ ] [ ] 6 0 2 x − x −  or [x]  3 x(−, −2)[4, ) 26. (a) So, there can’t be any bijections Number of bijections from x to y will be zero because only sets with the same cardinality have bijections between them and it is given that number of elements in x and y are not same. 27. (a) 2 27 3 b a = − 27 3 2 a b − = R = {(1, 2),(3, 9),(5, 6),(7, 3)} 28. (a) We have, Either or 29. (d) For domain, ------- (i) Also, ------- (ii) ------- (iii) From (i), (ii) and (iii), we get 30. (d) We have, Put x = – 1 Put x = 3, we get ------ (i) Add equation (i) and (ii), we get 5 2 * ab a b = a ae a e = a  = 5 2 * 2 5  e = a a aa a a e 4 25 2 5 5 2 * 1 1 1 =  −  = − − − 4(3) 25 5 2(2 ) 2* 3 1 =  = − x x       − a a 4 1 25  48 125  x = f : R → R ( ) 7 12 2 f x = x − x + f(x)  (x −4)(x −3)  0 x(−,3][4,)  f(x) = −(, 3][4, ) n(A) = 2 p B 2 = 2  2p 10 2p 1024 = 2  2 = 2  p = 5  n(B) = 5 4 5 2 y = x − x +  2 y −1 = (x − 2)  x − 2 = y −1  x = 2+ y −1  y −1  0  y  1 f(x) =[1, )  R n(A) = 3 3 9 3 3 2 A = = f(x) 1−|x| 0  |x| 1  ( 2) (3) (4) 3( 2) 3 2(4) 2 f − + f + f = − + + = −6+9+8 =11 0 ( 1) 2 '( ) 1 2 ( ) 2  − −  = − = x f x x x f x   A y y y x x x  − =  = −1 2 2 y = 2  f(x) ([x]−3)([x]+ 2)  0 [x]  −2  n(x)  n(y) n(A) = p,n(B) = q n(A B) = p q  7 = pq  p =1, q = 7 q =1, p = 7 1 7 50 2 2 2 2 p + q = + = x + 2  0  x  −2 1− x  0  x  1 log10(1− x)  0  1− x  1 x  0 x[−2, 0), (0, 1) f(x) = ax + b f(−1) = a(−1) + b = −5  a −b = 5 f(3) = 3a + b = 3 4a = 8  a = 2 b = 2−5 = −3
4 Mathematics - CET 31. (b) 2 Given: ( ) 3 5 and ( ) 2 1 x f x x g x x = − = + ( )( ) ( ( )) gof x g f x = 2 = − g x (3 5) 2 2 2 3 5 (3 5) 1 x x − = − + 2 2 2 2 2 3 5 (3 ) 5 2(3 )(5) 1 x x x − = + − + 2 4 2 3 5 9 25 30 1 x x x − = + − + 2 4 2 3 5 9 30 26 x x x − = − + 32. (d) We have and Since, will be invertible in i.e.,  33. (a) 16 2 x =  x = 4  2x = 6  x = 3 There is no value of x which satisfies both the above equations. Thus, A =  . 34. (c) Number of subsets of n n n n n A = C0 + C1 + ......... + C = 2 . 35. (a) A  B  A . Hence A (A  B) = A. 36. (a,b) ( ) [( ) ( ) ] c c c c c c c R P Q = R P  Q = R(P Q) = (R P)(RQ) = (RQ)(R P) . 37. (a) If A ={a,b,c},B ={b,c,d} and C = {a,d,c} A−B ={a}, BC ={c,d} Then (A−B)(BC) ={a}{c,d} ={(a,c),(a,d)} 38. (c) n(A B) = pq . 39. (b) Since 1 0, 2 x + = gives 1 2 x = −  x = i  x is not real but x is real (given)  No value of x is possible. 40. (a) The number of non- empty subsets = 2 − 1 n 2 1 16 1 15 4 = − = − = . 41. (b) Since, 4 − 3n−1=(3 +1) − 3n−1 n n 3 3 3 ..... 3 3 1 2 2 1 1 0 = C0 + C + C + + C − n− n n n n n n n n n n n C 3 C .3 ... C 3 3 3 2 = 2 + + + ,( , etc.) 0 = 1 = n−1 n n n n n C C C C 9[ (3) .... 3 ] 2 2 3 − = + + + n n n n n C C C  4 − 3n −1 n is a multiple of 9 for n  2 . For n = 1, 4 − 3n −1 n = 4 −3−1= 0 , For n = 2, 4 − 3n −1 n = 16−6 −1= 9  4 − 3n −1 n is a multiple of 9 for all nN  X contains elements, which are multiples of 9, and clearly Y contains all multiples of 9.  X  Y i.e., X Y = Y . Trick : Put the values of n=1, 2, 3.... we will get X and Y so X  Y i.e. X Y = Y . 42. (c) (24, 92) = 4, by verification, we get (4, −1). 43. (d) Since A  A .  Relation '' is reflexive. Since A  B, B  C  A  C  Relation '' is transitive. But A  B,  B  A ,  Relation is not symmetric. 44. (c) We have, , 2 2 4 ( ) 2  − − = = x x x y f x 2, 2 2 ( 2)( 2) = +  − − + = x x x x x . It follows from the above relation that y take all real values except when x takes values in the set R − {2}  Range (f) = R −{4} 45. (a) 2 f(x) = 3x − 2, g(x) = x ( ) ( ( )) 3 ( ( )) 2 3 2 2 fog x = f g x = g x − = x − 46. (c) f : R → R, f(x) = x Now, f(1) = 1 = 1 and f(−1) = −1 =1 = f(1) = f(−1) , but 1 1 −  f is not one-one. Also, −1 R, but f(x) = x is non-negative.  f is not onto,  f is neither injective nor surjective. f(x) = sin 2x + cos 2x ( ) 1 2 g x = x − ( ( )) [sin2 cos2 ] (sin2 cos2 ) 1 2 g f x = g x + x = x + x − −1+ 2sin2xcosx −1= sin4x sinx       − 2 , 2          − 2 , 2 4   x        − 8 , 8   x Level 2