Sets PDF

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www.jeebooks.in Downloaded from @Freebooksforjeeneet  EBD_8344 www.jeebooks.in Downloaded from @Freebooksforjeeneet www.jeebooks.in Downloaded from @Freebooksforjeeneet  EBD_8344 www.jeebooks.in Downloaded from @Freebooksforjeeneet www.jeebooks.in Downloaded from @Freebooksforjeeneet  EBD_8344 www.jeebooks.in Downloaded from @Freebooksforjeeneet www.jeebooks.in Downloaded from @Freebooksforjeeneet  EBD_8344 www.jeebooks.in Downloaded from @Freebooksforjeeneet  www.jeebooks.in 1 Sets 6. A relation on the set A = {x : |x| < 3, x ÎZ }, Sets, Types of Sets, Disjoint Sets, Complementary Sets, Subsets, where Z is the set of integers is defined by TOPIC Ć Power Set, Cardinal Number of R = {(x, y) : y = |x|, x ¹ – 1}. Then the number of elements in Sets, Operations on Sets the power set of R is: [Online April 12, 2014] (a) 32 (b) 16 (c) 8 (d) 64 1. Set A has m elements and set B has n elements. If the total 7. Let X ={1,2,3,4,5}. The number of different ordered pairs number of subsets of A is 112 more than the total number (Y, Z) that can formed such that Y Í X , Z Í X and Y Ç Z of subsets of B, then the value of m×n is ______. is empty is : [Sep. 06, 2020 (I)] (a) 52 (b) 35 (c) 25 (d) 53 2. Let S = {1, 2, 3,... , 100}. The number of non-empty subsets 8. If A, B and C are three sets such that A Ç B = A Ç C and A of S such that the product of elements in A is even is : A È B = A È C , then [Jan. 12, 2019 (I)] (a) A = C (b) B = C (a) 2100 - 1 50 50 (b) 2 2 - 1 ( ) (c) A Ç B = f (d) A = B (c) 250 – 1 (d) 250 + 1 3. Let S = {x Î R : x ³ 0 and Venn Diagrams, Algebraic Operations on Sets, De Morgan’s TOPIC n 2| x - 3 | + x ( x - 6) + 6 = 0. Then S : Law, Number of Elements in Different Sets (a) contains exactly one element. (b) contains exactly two elements. 9. A survey shows that 73% of the persons working in an (c) contains exactly four elements. office like coffee, whereas 65% like tea. If x denotes the (d) is an empty set. percentage of them, who like both coffee and tea, then x æ1ö cannot be : [Sep. 05, 2020 (I)] 4. If f(x) + 2f ç ÷ = 3x , x ¹ 0 and èxø (a) 63 (b) 36 (c) 54 (d) 38 10. A survey shows that 63% of the people in a city read S = {x Î R : f(x) = f(–x)}; then S: newspaper A whereas 76% read newspaper B. If x% of the (a) contains exactly two elements. people read both the newspapers, then a possible value of (b) contains more than two elements. x can be : [Sep. 04, 2020 (I)] (c) is an empty set. (d) contains exactly one element. (a) 29 (b) 37 (c) 65 (d) 55 Let P = {q : sinq – cosq = 50 n 5. 2 cosq} and Q = {q : sinq + cosq = 11. Let U X i = U Yi = T , where each Xi contains 10 elements 2 sinq} be two sets. Then: i =1 i =1 [Online April 10, 2016] and each Yi contains 5 elements. If each element of the set T is an element of exactly 20 of sets Xi's and exactly 6 of (a) P Ì Q and Q - P ¹ f sets Yi's, then n is equal to [Sep. 04, 2020 (II)] (b) Q Ë P (a) 15 (b) 50 (c) 45 (d) 30 (c) P = Q (d) P Ë Q Downloaded from @Freebooksforjeeneet  EBD_8344 www.jeebooks.in M-2 Mathematics 12. Let X = {n Î N: l £ n £ 50}. If 16. Two newspapers A and B are published in a city. It is A = {n Î X: n is a multiple of 2} and known that 25% of the city population reads A and 20% B = {n Î X: n is a multiple of 7}, then the number of reads B while 8% reads both A and B. Further, 30% of elements in the smallest subset of X containing both A those who read A but not B look into advertisements and and B is __________. [Jan. 7, 2020 (II)] 40% of those who read B but not A also look into 2 advertisements, while 50% of those who read both A and 13. Let Z be the set of integers. If A = {xÎZ : 2(x + 2) ( x – 5x + 6) = 1} B look into advertisements. Then the percentage of the and B = {x Î Z : –3 < 2x – 1< 9}, then the number of subsets population who look into advertisements is: of the set A × B, is : [Jan. 12, 2019 (II)] [April. 09, 2019 (II)] (a) 215 (b) 218 (c) 212 (d) 210 (a) 13.9 (b) 12.8 (c) 13 (d) 13.5 14. In a class of 140 students numbered 1 to 140, all even 17. In a certain town, 25% of the families own a phone and numbered students opted Mathematics course, those 15% own a car; 65% families own neither a phone nor a whose number is divisible by 3 opted Physics course and car and 2,000 families own both a car and a phone. Consider those whose number is divisible by 5 opted Chemistry the following three statements : [Online April 10, 2015] course. Then the number of students who did not opt for (A) 5% families own both a car and a phone any of the three courses is: [Jan. 10, 2019 (II)] (B) 35% families own either a car or a phone (a) 102 (b) 42 (c) 1 (d) 38 (C) 40,000 families live in the town 15. Let A, B and C be sets such that f ¹ A Ç B Í C. Then Then, which of the following statements is not true ? (a) Only (A) and (C) are correct. [April 12, 2019 (II)] (b) Only (B) and (C) are correct. (c) All (A), (B) and (C) are correct. (a) B Ç C ¹ f (d) Only (A) and (B) are correct. (b) If ( A - B) Í C , then A Í C (c) (C È A) Ç (C È B) = C (d) If (A - C) Í B , then A Í B Downloaded from @Freebooksforjeeneet  www.jeebooks.in Sets M-3 1. (28) 2m = 112 + 2n Þ 2m - 2n = 112 (c) sinq – cosq = 5. 2 cosq m -n Þ 2 (2 n - 1) = 2 (2 - 1) 4 3 Þ sinq = cosq + 2 cosq \ m = 7, n = 4 Þ mn = 28 æ 2 -1 ö 2. (b) Q Product of two even number is always even and = ( 2 + 1) cosq = ç ÷ cosq è 2 -1 ø product of two odd numbers is always odd. Þ ( 2 - 1) sinq = cosq \ Number of required subsets = Total number of subsets – Total number of subsets Þ sinq + cosq = 2 sinq having only odd numbers \ P= Q = 2100 – 250 = 250(250 – 1) 6. (b) A = {x : |x | < 3, x Î Z } 3. (b) Case-I: x Î[0,9] A = {–2, – 1, 0, 1, 2} R = {(x, y) : y = |x |, x ¹ -1} 2(3 - x ) + x - 6 x + 6 = 0 R = {(–2, 2), (0, 0), (1, 1), (2, 2)} R has four elements Þ x - 8 x + 12 = 0 Þ x = 4, 2 Number of elements in the power set of R Þ x = 16, 4 = 24 = 16 Since x Î[0,9] 7. (b) Let X = {1, 2, 3, 4, 5} \ x=4 n(x) = 5 Case-II: x Î[9, ¥] Each element of x has 3 options. Either in set Y or set Z or none. (Q Y Ç Z = f) 2( x - 3) + x - 6 x + 6 = 0 So, number of ordered pairs = 35 8. (b) Q B = ( B Ç A) È B Þ x - 4 x = 0 Þ x = 16,0 Since x Î[9, ¥] = ( A Ç C) È B \ x = 16 = ( A È B) Ç ( C È B) Hence, x = 4 & 16 æ 1ö = ( A È C) Ç ( B È C) 4. (a) f (x) + 2f ç ÷ = 3x....(1) è xø = ( A Ç B) È C æ 1ö 3 f ç ÷ + 2f (x) =....(2) è xø x = ( A Ç C) È C Adding (1) and (2) =C æ1ö 1 Þ f (x) + f ç ÷ = x +...(3) 9. (b) Given, n(C ) = 73, n(T ) = 65, n(C Ç T ) = x è ø x x Substracting (1) from (2) \ 65 ³ n(C Ç T ) ³ 65 + 73 - 100 æ1ö 3 Þ 65 ³ x ³ 38 Þ x ¹ 36. Þ f (x) - f ç ÷ = - 3x...(4) èxø x 10. (d) Let n(U ) = 100, then n(A) = 63, n(B) = 76 On adding (3) and (4) n( A Ç B ) = x 2 Þ f (x) = -x x Now, n( A È B ) = n( A) + n( B) - n( A Ç B) £ 100 2 -2 2 = 63 + 76 - x £ 100 f (x) = f (- x) Þ -x = +xÞx= x x x Þ x ³ 139 - 100 Þ x ³ 39 x2 = 2 or x = 2, - 2 Downloaded from @Freebooksforjeeneet  EBD_8344 www.jeebooks.in M-4 Mathematics Q n( A Ç B) £ n( A) Q = {6n: n Î N, 1 £ n £ 23} – P Þ x £ 63 Þ n(Q) = 19 \ 39 £ x £ 63 R = {15n: n Î N, 1 £ n £ 9} – P Þ n(R) = 5 50 n 11. (d) U X i = U Yi = T S = {10n: n Î N, 1 £ n £ 14} – P i =1 i =1 Þ n(S) = 10 Q n( X i ) = 10, n (Yi ) = 5 n(T) = 70 – n(P) – n(Q) – n(S) = 70 – 33 = 37 50 n n(V) = 46 – n(P) – n(Q) – n(R) = 46 – 28 = 18 So, U X i = 500, U Yi = 5n n(W) = 28 – n(P) – n(R) – n(S) = 28 – 19 = 9 i =1 i =1 Þ Number of required students 500 5n Þ = Þ n = 30 = 140 – (4 + 19 + 5 + 10 + 37 + 18 + 9) 20 6 = 140 – 102 = 38 12. (29) From the given conditions, 15. (d) (1), (2) and (4) are always correct. n(A) = 25, n(B) = 7 and n(A Ç B) = 3 In (3) option, n(A È B) = n(A) + n(B) – n(A Ç B) If A = C then A – C = f = 25 + 7 – 3 = 29 Clearly, f Í B but A Í B is not always true. 13. (a) Let x Î A, then 16. (a) Q 2( x+2)( x2 -5 x+6) = 1 Þ (x + 2)(x – 2)(x – 3) = 0 A B x = –2, 2, 3 25 20 8 A = {–2, 2, 3} Then, n(A) = 3 Let x Î B, then –3 < 2x – 1 < 9 % of people who reads A only = 25 – 8 = 17% –1 < x < 5 and x Î Z % of people who read B only = 20 – 8 = 12% \ B = {0, 1, 2, 3, 4} % of people from A only who read advertisement n(B) = 5 = 17 × 0.3 = 5.1% n(A ´ B) = 3 ´ 5 = 15 % of people from B only who read advertisement Hence, Number of subsets of A ´ B = 2 15 = 12 × 0.4 = 4.8% % of people from A & B both who read advertisement Maths = 8 × 0.5 = 4% 14. (d) T W \ total % of people who read advertisement S Chemistry P = 5.1 + 4.8 + 4 = 13.9% Q 17. (c) n(P) = 25% R n(C) = 15% V n ( P ¢ È C ¢ ) =65% Physics Þ n(P È C)¢ = 65% n ( P È C ) = 35% P = {30, 60, 90, 120} n ( P Ç C) = n ( P ) + n (C) - n ( P È C) Þ n(P) = 4 25 + 15 – 35 = 5% x × 5% = 2000 x = 40,000 Downloaded from @Freebooksforjeeneet  www.jeebooks.in 2 Relations and Functions 5. The domain of the definition of the function Relations, Domain, Codomain and 1 Range of a Relation, Functions, f ( x) = + log10 ( x3 - x ) is: [April. 09, 2019 (II)] TOPIC Ć Domain, Codomain and Range of a 4 - x2 Function (a) (–1, 0) È (1, 2) È (3, ¥) (b) (–2, –1) È (–1, 0) È (2, ¥) 1. Let R1 and R2 be two relations defined as follows : (c) (–1, 0) È (1, 2) È (2, ¥) (d) (1, 2) È (2, ¥) R1 = {(a, b) Î R 2 : a 2 + b 2 Î Q} and 6. The range of the function R2 = {(a, b) Î R 2 : a 2 + b2 Ï Q} , where Q is the set of all x f ( x) = , x Î R , is [Online May 7, 2012] rational numbers. Then : [Sep. 03, 2020 (II)] 1+ x (a) Neither R1 nor R2 is transitive. (a) R (b) (– 1, 1) (c) R – {0} (d) [– 1, 1] (b) R2 is transitive but R1 is not transitive. (c) R1 is transitive but R2 is not transitive. 1 7. The domain of the function f ( x) = is (d) R1 and R2 are both transitive. x -x | x | +5 ö (a) (0, ¥ ) (b) (– ¥ , 0) 2. The domain of the function f ( x) = sin -1 æç ÷ is (c) (– ¥ , ¥ ) – {0} (d) (– ¥ , ¥ ) è x2 +1 ø 8. Domain of definition of the function (-¥, - a] È [a, ¥]. Then a is equal to : 3 f ( x) = + log10 ( x 3 - x) , is [Sep. 02, 2020 (I)] 4 - x2 17 17 - 1 1 + 17 17 (a) ( -1,0) È (1,2) È ( 2, ¥) (b) (a, 2) (a) (b) (c) (d) +1 2 2 2 2 (c) ( -1,0) È ( a,2) (d) (1,2) È (2, ¥) 3. If R = {(x, y) : x, y Î Z, x2 + 3 y 2 £ 8} is a relation on the Even and Odd Functions, Explicit set of integers Z, then the domain of R–1 is : and Implicit Functions, Greatest [Sep. 02, 2020 (I)] Integer Function, Periodic (a) {–2, –1, 1, 2} (b) {0, 1} TOPIC n Functions, Value of a Function, (c) {–2, –1, 0, 1, 2} (d) {–1, 0, 1} Equal Functions, Algebraic Operations on Functions x 4. Let f : R ® R be defined by f ( x ) = , x Î R. Then 1 + x2 9. Let [t] denote the greatest integer £ t. Then the equation the range of f is : [Jan. 11, 2019 (I)] in x, [x]2 + 2[x + 2] – 7 = 0 has : [Sep. 04, 2020 (I)] (a) exactly two solutions é 1 1ù (a) ê - , ú (b) R – [–1, 1] (b) exactly four integral solutions ë 2 2û (c) no integral solution é 1 1ù (d) infinitely many solutions (c) R - ê - , ú (d) (–1, 1) – {0} ë 2 2û Downloaded from @Freebooksforjeeneet  EBD_8344 www.jeebooks.in M-6 Mathematics 10. Let f (x) be a quadratic polynomial such that f (–1) + f (2) = 15. Let f be an odd function defined on the set of real numbers 0. If one of the roots of f (x) = 0 is 3, then its other root lies such that for x ³ 0, f(x) = 3 sin x + 4 cos x. in : [Sep. 02, 2020 (II)] 11p (a) (–1, 0) (b) (1, 3) (c) (–3, –1) (d) (0, 1) Then f(x) at x = - is equal to: [Online April 11, 2014] 6 x[ x] 11. Let f (1, 3) ® R be a function defined by f (x) = , (a) 3 +2 3 3 (b) - + 2 3 1 + x2 2 2 where [x] denotes the greatest integer £ x. Then the range of f is: [Jan. 8, 2020 (II)] 3 3 (c) -2 3 (d) - - 2 3 2 2 æ 2 3ö æ 3 4 ö æ 2 1 ö æ 3 4ö (a) çè , ÷ø È çè , ÷ø (b) çè , ÷ø È çè , ø÷ 16. A real valued function f (x) satisfies the functional equation 5 5 4 5 5 2 5 5 f (x – y) = f (x) f (y) – f (a – x) f (a + y) æ 2 4ö æ 3 4ö where a is a given constant and f (0) = 0, f (2a – x) is equal (c) çè , ÷ø (d) çè , ÷ø to 5 5 5 5 (a) – f (x) (b) f (x) 1- x ö æ 2x ö (c) f (a) + f (a – x) (d) f (– x) 12. If f(x) = loge æç ÷ , |x| < 1, then f ç ÷ is equal to : è 1+ x ø è 1+ x2 ø 17. The graph of the function y = f(x) is symmetrical about the line x = 2, then [April 8, 2019 (I)] (a) 2f(x) (b) 2f(x2) (c) (f(x)) 2 (d) –2f(x) (a) f ( x ) = - f (- x) (b) f (2 + x) = f (2 - x) 13. Let f(x) = ax (a > 0) be written as f(x) = f1(x) + f2(x), where (c) f ( x ) = f ( - x) (d) f ( x + 2) = f ( x - 2) f1(x) is an even function and f2(x) is an odd function. Then f1(x + y) + f1(x – y) equals : [April. 08, 2019 (II)] 18. If f : R ® R satisfies f ( x + y ) = f ( x) + f ( y ) , for all x, (a) 2f1(x) f1(y) (b) 2f1(x + y) f1(x – y) n (c) 2f1(x)f2(y) (d) 2f1(x + y) f2(x – y) y Î R and f(1) = 7, then S f (r ) is r=1 é 1 3n ù 14. Let f ( n ) = ê + n , where [n] denotes the greatest 7 n (n + 1) 7n ë 3 100 úû (a) 2 (b) 2 56 7 (n + 1) integer less than or equal to n. Then å f ( n) is equal to: (c) 2 (d) 7 n + ( n + 1) n=1 [Online April 19, 2014] (a) 56 (b) 689 (c) 1287 (d) 1399 Downloaded from @Freebooksforjeeneet  www.jeebooks.in Relations and Functions M-7 Þ 1 ³ 4y2 (a) For R1 let a = 1 + 2, b = 1 - 2, c = 8 1/ 4 1. 1 aR1b Þ a 2 + b2 = (1 + 2)2 + (1 - 2) 2 = 6 Î Q Þ |y| £ 2 bR1c Þ b2 + c 2 = (1 - 2)2 + (81/ 4 )2 = 3 Î Q 1 1 Þ - £ y£ 2 2 aR1c Þ a 2 + c2 = (1 + 2)2 + (81/ 4 )2 = 3 + 4 2 Ï Q é 1 1ù \ R1 is not transitive. Þ The range of f is ê - , ú. ë 2 2û For R2 let a = 1 + 2, b = 2, c = 1 - 2 5. (c) To determine domain, denominator ¹ 0 and x3 – x > 0 aR2b Þ a + b = (1 + 2) + ( 2) = 5 + 2 2 Ï Q 2 2 2 2 i.e., 4 – x2 ¹ 0 x ¹ ±2...(1) and x (x – 1) (x + 1) > 0 bR2 c Þ b 2 + c 2 = ( 2)2 + (1 - 2)2 = 5 - 2 2 Ï Q – + – + aR2c Þ a 2 + c2 = (1 + 2) 2 + (1 - 2) 2 = 6 Î Q x Î (– 1, 0) È (1, ¥)...(2) \ R2 is not transitive. Hence domain is intersection of (1) & (2). | x | +5 ö 2. (c) Q f ( x) = sin -1 æç i.e., x Î (–1, 0) È (1, 2) È (2, ¥) è x 2 + 1 ÷ø | x | +5 x \ -1 £ £1 6. (b) f ( x) = 1 + x , x Î R x2 + 1 Þ| x | +5 £ x2 + 1 [Q x 2 + 1 ¹ 0] x If x > 0, | x | = x Þ f ( x ) = 1+ x Þ x2 - | x | -4 ³ 0 which is not defined for x = – 1 æ 1 - 17 ö æ 1 + 17 ö x Þ ç| x | - ÷ ç| x | - 2 ÷ ³ 0 If x < 0, | x | = – x Þ f ( x ) = è 2 øè ø 1- x which is not defined for x = 1 æ 1 + 17 ö é1 + 17 ö Thus f(x) defined for all values of R except 1 and – 1 Þ x Î ç -¥, - ÷ Èê , ¥÷ è 2 ø ë 2 ø Hence, range = (– 1, 1). 1 + 17 f ( x) = 1 \a = 7. (b) , f (x) is define if | x | – x > 0 2 x -x 3. (d) Since, R = {(x, y) : x, y Î Z, x2 + 3 y 2 £ 8} Þ | x | > x, Þ x < 0 Hence domain of f (x) is (– ¥, 0) \ R = {(1, 1), (2, 1), (1, - 1), (0, 1), (1, 0)} 3 Þ DR-1 = {-1, 0, 1} 8. (a) f ( x) = + log10 ( x 3 - x) 4 - x2 x , x ÎR 4 - x 2 ¹ 0; x 3 - x > 0; 4. (a) f(x) = 1 + x2 x ¹ ± 4 and - 1 < x < 0 or 1 < x < ¥ x Let, y = 1 + x2 – + + – –1 0 1 1 ± 1- 4 y2 Þ yx2 – x + y = 0 Þ x = 2 \ D = ( -1, 0) È (1, ¥) - { 4} Þ 1 – 4y2 ³ 0 D = ( -1, 0) È (1, 2) È (2, ¥). Downloaded from @Freebooksforjeeneet  EBD_8344 www.jeebooks.in M-8 Mathematics 9. (d) The given equation æ 1 + x2 - 2 x ö æ 1- x ö 2 [ x]2 + 2[ x] + 4 - 7 = 0 = log çç 2 ÷ ÷ = log ç 1+ x ÷ è 1+ x + 2x ø è ø Þ [ x]2 + 2[ x] - 3 = 0 æ 1- x ö Þ [ x]2 + 3[ x] - [ x] - 3 = 0 = 2 log ç ÷ = 2f (x) è 1+ x ø Þ ([ x] + 3)([ x] - 1) = 0 Þ [ x] = 1 or –3 13. (a) Given function can be written as Þ x Î[ -3, - 2) È [1, 2) æ a x + a- x ö æ a x - a-x ö \ The equation has infinitely many solutions. f (x) = ax = çç ÷+ç ÷ 2 ÷ ç 2 ÷ è ø è ø 10. (a) Let f ( x) = ax 2 + bx + c Given : f ( -1) + f (2) = 0 a x + a-x where f1(x) = is even function 2 a - b + c + 4a + 2b + c = 0 Þ 5a + b + 2c = 0...(i) a x - a-x f2(x) = is odd function and f (3) = 0 Þ 9a + 3b + c = 0...(ii) 2 From equations (i) and (ii), Þ f1(x + y) + f1(x – y) a = b = c Þ a b c = = æ a x+ y + a - x - y ö æ a x - y + a - x + y ö 1 - 6 18 - 5 15 - 9 -5 13 6 = çç ÷+ç ÷ 2 ÷ ç 2 ÷ è ø è ø c -6 Product of roots, ab = = and a = 3 1é x y a 5 = a (a + a - y ) + a - x (a y + a - y ) ù 2ë û -2 Þb= Î ( -1, 0) 5 (a x + a - x )(a y + a - y ) = = 2f1(x). f1(y) 2 ì x ïï 2 ; x Î (1, 2) é 1 3n ù x +1 (d) Let f (n) = ê + n ë 3 100 úû 11. (b) f ( x) í 14. ï 2 x ; x Î[2,3) îï x 2 + 1 where [n] is greatest integer function, é 3n ù = ê0.33 + n ì 1 - x2 ë 100 úû ï ; x Î (1, 2) ï 1+ x2 For n = 1, 2,..., 22, we get f (n) = 0 f ¢( x ) í and for n = 23, 24,..., 55, we get f (n) = 1 × n 2 ï1 - 2 x ï ; x Î [2,3) For n = 56, f (n) = 2 × n î 1+ x 2 56 \ f(x) is a decreasing function So, å f (n) = 1 (23) + 1 (24) +... + 1 (55) + 2(56) n =1 æ 2 1 ö æ 6 4ù \ y Îç , ÷ Èç , ú = (23 + 24 +... + 55) + 112 è 5 2 ø è 10 5 û 33 = [46 + 32] + 112 æ 2 1 ö æ 3 4ù 2 Þ y Îç , ÷ È ç , ú è 5 2 ø è 5 5û 33 = (78) + 112 = 1399. æ1- x ö 2 12. (a) f (x) = log ç ÷ ,|x| 0 = 4cos6 - 4sin 6 - 3cos 4 + 3sin 4 8 8 8 8 Þ 2[sin q + cos q - 1] > 0 éæ p p öù = 4 êç cos 2 - sin 2 ÷ ú æ pö 1 ëè 8 8 øû Þ sin q + cos q > 1 Þ sin ç q + ÷ > è 4ø 2 éæ 4 p 4p 2p 2 p öù êç sin 8 + cos 8 + sin 8 cos 8 ÷ ú p æ p 3p ö æ pö ë è øû Þ q+ Î ç , ÷ Þ q Î ç 0, ÷ 4 è4 4ø è 2ø éæ p p öæ p p öù -3 êç cos2 - sin 2 ÷ç cos 2 + sin 2 ÷ ú 12. (a) Let the height of the tower be h and distance of the ëè 8 8 øè 8 8 øû foot of the tower from the point A is d. pé æ p pö ù By the diagram, = cos ê 4 ç1 - sin 2 cos 2 ÷ - 3ú 4ë è 8 8ø û Q 1 é 1ù 1 = ê1- ú = 2ë 2 û 2 2 2 sin a 1 1 - cos 2 b 1 9. (1) = and = 30° 2 cos a 7 2 10 B h 2 sin b 1 Þ = 30 m 2 10 45° 1 1 \ tan a = and sin b = A d P 7 10 1 h tan b = tan 45 = =1 3 d 1 2 h=d...(i) 2. 2 tan b 3 h - 30 \ tan 2b = = 3 =3= tan 30 = 1 - tan b 1 - 2 1 8 4 d 9 9 3(h - 30) = d...(ii) tan a + tan 2b tan(a + 2b) = Put the value of h from (i) to (ii), 1 - tan a tan 2b 3d = d + 30 3 1 3 4 + 21 + = 7 4 1 3 = 28 = 1 25 d= 30 3 3 -1 = 15 3 ( ) ( 3 + 1 = 15 3 + 3 ) 1-. 7 4 28 Downloaded from @Freebooksforjeeneet  www.jeebooks.in Trigonometric Functions M-17 13. (b) cos2 10 – cos10 cos50 + cos2 50 2 æ5ö 12 æ 1 + cos 20° ö æ 1 + cos100° ö 1 cos (a – b) = 1 - ç ÷ = =ç ÷+ç ÷ - (2cos10° cos50°) è 13 ø 13 è 2 ø è 2 ø 2 5 1 1 Þ tan (a – b) = = 1 + (cos 20° + cos100°) - [cos60° + cos 40°] 12 2 2 Now, tan 2a = tan ((a + b) + (a – b)) æ 1ö 1 4 5 = ç1 - ÷ + [cos20° + cos100° - cos 40°] + è 4ø 2 tan(a + b) + tan(a - b) 3 12 = 63 = = 4 5 16 3 1 1 - tan(a + b).tan(a - b) 1-. = + [2cos 60° ´ cos 40° - cos 40°] 3 12 4 2 = 3 17. (d) Q The given equation is 4 sin4 a + 4 cos4 b + 2 = 4 2 sin a × cos b, a, b Î [0, p] Then, by A.M., G.M. ineqality; 14. (a) A.M. ³ G.M. 5 sin 4 a + 4cos 4 b + 1 + 1 1 15º 10 4 ( ) ³ sin 4 a × 4 cos 4 b × 1 ×1 4 sin4a + 4cos4b + 1 + 1 ³ 4 2 sin a× |cos b | 5 Inequality still holds when cosb < 0 but L.H.S. is positive 15º than cosb > 0, then d L.H.S. = R.H.S By the diagram, 1 \ sin4 a = 1 and cos4 b = ( ) 4 5 5 5 3 +1 tan15 = Þd = = p p d tan15 3 -1 Þ a= and b = 2 4 ( 5 4+2 3 ) =5 2+ 3 \ cos (a + b) – cos (a – b) = 2 ( ) æp ö æp ö = cos çè + b÷ø - cos çè - bø÷ 1 2 2 15. (a) Q sin(60 + A).sin(60 – A) sinA = sin3A 4 p \ sin10 sin50 sin70 = sin10 sin(60 – 10) = –sinb – sinb = -2sin =- 2 4 1 1 k k sin(60 + 10) = 4 sin30 18. (a) fk(x) = (sin x + cos x ) k 1 1 1 Þ sin10 sin30 sin50 sin70 = sin230 = 4 4 f4(x) = [sin x + cos x] 4 16 4 16. (b) Q a + b and a – b both are acute angles. 1é (sin 2 x) 2 ù = 4 ê(sin x + cos x ) - 2 2 2 2 ú 3 æ3ö 4 ë 2 û cos (a + b) = , then sin (a + b) = 1 - ç ÷ = 5 è5ø 5 1 é (sin 2 x )2 ù 4 = 4 ê1 - ú tan (a + b) = ë 2 û 3 5 1 6 6 And sin (a – b) = , then f6(x) = [sin x + cos x ] 13 6 1é 3 2ù = ê(sin x + (cos x) - (sin x) ú 2 2 2 6ë 4 û Downloaded from @Freebooksforjeeneet  EBD_8344 www.jeebooks.in M-18 Mathematics 1é 3 2ù 17 æ 2 2 1 ö 17 9 17 = ê1 - (sin 2 x) ú - £ ç cos x - ÷ - £ - 6ë 4 û 16 è 4 ø 16 16 16 1 1 (sin 2 x )2 1 ìïæ 17 üï 1 2 Now f4(x) – f(6)(x) = - - + (sin 2 x )2 17 1ö 4 6 8 8 ³ -4 íç cos 2 x - ÷ - ý ³ 4 è 4ø 16 ïþ 2 îï 1 = 17 12 M= p p p p 4 19. (a) A = cos.cos 3... cos 10.sin 10 1 22 2 2 2 m= 2 1æ p p p pö = çè cos 2.cos 3... cos 9 sin 9 ÷ø 17 2 15 2 2 2 2 2 M–m= - = 4 4 4 1æ p pö 1 p = ç cos 2.sin 2 ÷ø = 9 sin 3 28 è 2 2 2 2 22. (b) Let cos a + cos b = 2 1 = a+b a -b 3 512 Þ 2 cos cos =...(i) 20. (a) We have 2 2 2 5 tan2 x – 5 cos2 x = 2 (2 cos2 x –1 ) + 9 1 Þ 5 tan2 x – 5 cos2 x = 4 cos2 x –2 + 9 and sin a + sin b = 2 Þ 5 tan2 x = 9 cos2 x + 7 Þ 5 (sec2 x – 1) = 9 cos2 x + 7 a +b a -b 1 Þ 2sin cos =...(ii) Let cos2 x = t 2 2 2 5 On dividing (ii) by (i), we get Þ - 9t - 12 = 0 t æ a + bö 1 Þ 9t2 + 12t – 5 = 0 tan ç è 2 ÷ø = 3 Þ 9t2 + 15t – 3t – 5 = 0 Þ (3t – 1) (3t + 5) = 0 a+b Given : q = Þ 2q = a + b 1 5 2 Þ t = as t ¹ –. 3 3 Consider sin 2q + cos 2q = sin (a + b) + cos (a + b) æ1ö 1 2 1 cos 2x = 2 cos2 x – 1 = 2 ç ÷ – 1 = – 1- è3ø 3 3 + 9 6 8 7 + = 2 = = æ 1ö 1 1 10 10 5 7 1+ 1+ cos 4x = 2 cos2 2x – 1 = 2 ç - ÷ - 1 = - 9 9 è 3ø 9 1 p+q p-q 21. (b) 4 + sin2 2x – 2 cos4 x 23. (b) cosecq = , sin q = 2 p-q p+q 4 + 2 (1 – cos2 x) cos2 x – 2 cos4 x 2 æ p - qö 2 pq ïì cos2 x 1 1 ïü cos q = ± 1 - sin 2 q = 1 - ç = - 4 ícos 4 x - -1 + - ý è p + q ø÷ ( p + q) îï 2 16 16 þï p q q ìïæ 1ö 2 17 üï cot cot - 1 cot - 1 - 4 íç cos 2 x - ÷ - ý æ p qö 4 2 2 è cot ç + ÷ = = îï 4ø 16 ïþ è 4 2ø p q q cot + cot cot + 1 2 0 < cos x < 1 4 2 2 - 1 1 3 £ cos2 x - £ q q cos - sin 4 4 4 2 2 = q q æ 1ö 2 9 cos + sin 0 £ ç cos2 x - ÷ £ 2 2 è 4ø 16 On rationali ing denominator, we get Downloaded from @Freebooksforjeeneet  www.jeebooks.in Trigonometric Functions M-19 Þ [sina + sin b + sin g ]2 + (cos a + cos b + cos g )2 = 0 æ q qöæ q qö cos - sin cos + sin Þ sina + sin b + sin g = 0 and cos a + cos b + cos g = 0 ç 2 2÷ ç 2 2÷ ç q q÷ç q q÷ \ A and B both are true. çè cos + sin ÷ø çè cos + sin ÷ø 27. (c) Given that p2 + q2 = 1 2 2 2 2 \ p = cos q and q = sin q satisfy the given equation cos q Then p + q = cos q + sin q = 2 q 2q q q sin + cos + 2sin cos We know that 2 2 2 2 - a 2 + b2 £ a cos q + b sin q £ a 2 + b 2 cos q 2 pq / ( p + q ) pq q = = = = \ – 2 £ cos q + sin q £ 2 1 + sin q ( p - q) p p 1+ Hence max. value of p + q is p+q 2 1 1 24. (d) A = sin 2 x + cos 4 x 28. (c) cos x + sin x = Þ 1 + sin 2 x = 2 4 = sin 2 x + cos 2 x(1 - sin 2 x ) 3 Þ sin 2 x = - , 1 4 = sin 2 x + cos 2 x - (2 sin x.cos x) 2 4 \ p < 2x < 2p 1 2 = 1 - sin (2 x ) p Þ 7 " n ³ 1 (b) an < 7 " n ³ 1 the following is true (c) an < 4 " n ³ 1 (d) an < 3 " n ³ 1 Downloaded from @Freebooksforjeeneet  EBD_8344 www.jeebooks.in M-24 Mathematics 1. (a) P(n) = n2 – n + 41 Adding 2K + 1 on both sides Þ P(3) = 9 – 3 + 41 = 47 (prime) Þ 1 + 3 + 5.... + (2 K - 1) + 2 K + 1 & P(5) = 25 – 5 + 41 = 61 (prime) \ P(3) and P(5) are both prime i.e., true. = 3 + K 2 + 2K + 1 = 3 + (K + 1)2 = S ( K + 1) 2. (b) S(K) = 1+3+5+...+(2K – 1) = 3 + K2 \ S ( K ) Þ S ( K + 1) S (1) :1 = 3 + 1, which is not true 3. (b) For n = 1, a1 = 7 < 7. Let am < 7. Q S (1) is not true. \ P.M.I cannot be applied Then am + 1 = 7 + am Let S(K) is true, i.e. Þ a2m + 1 = 7 + am < 7 + 7 < 14. S(K) : 1 + 3 + 5.... + (2 K - 1) = 3 + K 2 Þ am + 1 < 14 < 7; So, by the principle of mathematical induction an < 7, " n. Downloaded from @Freebooksforjeeneet  www.jeebooks.in Complex Numbers and Quadratic Equations M-25 5 Complex Numbers and Quadratic Equations z -i Integral Powers of lota, Algebraic 6. Let z be a complex number such that =1 Operations of Complex Numbers, z + 2i TOPIC Ć Conjugate, Modulus and Argument 5 or Amplitude of a Complex Number and z =. Then the value of |z + 3i| is : 2 [Jan. 9, 2020 (I)] 3 + i sin q 1. If , q Î [0, 2p], is a real number, then an argument 7 15 4 - i cos q (a) 10 (b) (c) (d) 2 3 2 4 of sinq + icosq is: [Jan. 7, 2020 (II)] 7. If z be a complex number satisfying |Re(z)| + |Im(z)| = 4, -1 æ 4 ö -1 æ 3 ö (a) p - tan çè ÷ø (b) p - tan çè ÷ø then |z| cannot be: [Jan. 9, 2020 (II)] 3 4 17 (a) (b) (c) (d) -1 æ 3 ö -1 æ 4ö 10 7 8 (c) - tan çè ÷ø 2 (d) tan çè ÷ø 4 3 2z - n 8. Let z Î C with Im(z) = 10 and it satisfies = 2i - 1 for 2. If the four complex numbers z, z , z - 2Re( z ) and 2z + n z - 2Re( z ) represent the vertices of a square of side some natural number n. Then : [April 12, 2019 (II)] 4 units in the Argand plane, then |z| is equal to : (a) n = 20 and Re(z) = –10 [Sep. 05, 2020 (I)] (b) n = 40 and Re(z) = 10 (a) 4 2 (b) 4 (c) 2 2 (d) 2 (c) n = 40 and Re(z) = –10 30 (d) n = 20 and Re(z) = 10 æ -1 + i 3 ö 3. The value of çç ÷÷ is : [Sep. 05, 2020 (II)] 9. The equation z - i = z - 1 , i = -1 , represents: è 1- i ø [April 12, 2019 (I)] (a) – 215 (b) 215 i (c) – 215 i (d) 65 m/2 n/3 1 æ1+ iö æ1+ iö (a) a circle of radius. If ç =ç = 1, (m, n Î N ) , then the 2 è 1 - i ÷ø è i - 1÷ø 4. (b) the line through the origin with slope 1. greatest common divisor of the least values of m and n is (c) a circle of radius 1. _________. [Sep. 03, 2020 (I)] (d) the line through the origin with slope – 1. 5. If z1, z2 are complex numbers such that Re(z1) = |z1 – 1|, (1 + i )2 2 p 10. If a > 0 and = , has magnitude , then is Re(z2) = |z2 – 1| and arg( z1 - z2 ) = , then Im( z1 + z2 ) is a -i 5 6 equal to : [April 10, 2019 (I)] equal to : [Sep. 03, 2020 (II)] 1 3 3 1 2 3 1 (a) - - i (b) - - i (a) (b) 2 3 (c) (d) 5 5 5 5 3 2 3 1 3 1 3 (c) - i (d) - + i 5 5 5 5 Downloaded from @Freebooksforjeeneet  EBD_8344 www.jeebooks.in M-26 Mathematics 11. If z and w are two complex numbers such that zw = 1 and 1 + (1 – 8a) z 18. The set of all a Î R, for which w = is a purely 1– z p arg(z) – arg(w) = 2 , then: [April 10, 2019 (II)] imaginary number, for all z Î C satisfying |z| = 1 and Re z ¹ 1, is [Online April 15, 2018] -1 + i (a) {0} (b) an empty set (a) zw = i (b) z w = 2 ì 1 1ü (c) í0, , – ý (d) equal to R 1- i î 4 4þ (c) zw = -i (d) z w = 2 2 + 3isin q 5 + 3z 19. A value ofqfor which is purely imaginary, is: 1 - 2isin q 12. Let z Î C be such that |z| < 1. If w = 5(1 - z ) , then : [April 09, 2019 (II)] -1 æ 3 ö -1 æ 1 ö (a) 5 Re (w) > 4 (b) 4 Im (w) > 5 (a) sin çç 4 ÷÷ (b) sin ç ÷ (c) 5 Re (w) > 1 (d) 5 Im (w) < 1 è ø è 3ø z -a p p 13. If (a Î R ) is a purely imaginary number and | | = 2, (c) (d) z +a 3 6 then a value of a is : [Jan. 12, 2019 (I)] 20. If is a non-real complex number, then the minimum 1 lmz 5 (a) 2 (b) 1 (d) 2 (c) value of is : [Online April 11, 2015] 2 (lmz )5 14. Let 1 and 2 be two complex numbers satisfying | 1| = 9 (a) –1 (b) –4 (c) –2 (d) –5 and | 2 | – |3|–|4i||=|4. Then the minimum value of | 1 – 2| is : [Jan. 12, 2019 (II)] 21. If z is a complex number such that z ³ 2, then the minimum (a) 0 (b) 2 (c) 1 (d) 2 1 value of z + : 15. Let be a complex number such that + =3+i 2 ( where i = -1. ) (a) is strictly greater than 5 2 Then | | is equal to : [Jan. 11, 2019 (II)] 3 5 34 5 41 5 (b) is strictly greater than but less than (a) (b) (c) (d) 2 2 3 3 4 4 Let 1 and 5 16. 2 be any two non- ero complex numbers such (c) is equal to 2 3 2 that 3 | = 1 + 2 (d) lie in the interval (1, 2) 1 | = 4 | 2 |. If 2 3 then: 2 1 22. For all complex numbers of the form 1 + ia, a Î R , if [Jan. 10 2019 (II)] 2 = x + iy, then [Online April 19, 2014] (a) y2 – 4x + 2 = 0 (b) y2 + 4x – 4 = 0 5 (a) Re( ) = 0 (b) | | = (c) y2 – 4x – 4 = 0 (d) y2 + 4x + 2 = 0 2 -i 1 17 23. Let ¹ – i be any complex number such that is a (c) | | = (d) Im( ) = 0 +i 2 2 purely imaginary number. ì æ p ö 3 + 2isin q ü + 1 17. Let A= íqÎ ç - , p ÷ : is purely imaginary ý. Then is: [Online April 12, 2014] î è 2 ø 1 - 2isin q þ (a) ero Then the sum of the elements in A is: [Jan. 9 2019 (I)] (b) any non- ero real number other than 1. 5p 3p 2p (c) any non- ero real number. (a) (b) p (c) (d) 6 4 3 (d) a purely imaginary number. Downloaded from @Freebooksforjeeneet  www.jeebooks.in Complex Numbers and Quadratic Equations M-27 24. If 1, 2 and 3, 4 are 2 pairs of complex conugate numbers, 2 2 then 30. z1 + z2 + z1 - z2 is equal to [Online May 26, 2012] æ ö æ ö arg ç 1 ÷ + arg ç 2 ÷ equals: [Online April 11, 2014] (a) 2 ( z1 + z2 ) ( (b) 2 z1 + z2 2 2 ) è 4ø è 3ø 2 2 (c) z1 z2 (d) z1 + z2 p 3p 31. Let Z and W be complex numbers such that |Z| = |W|, and (a) 0 (b) (c) (d) p 2 2 arg Z denotes the principal argument of Z. 25. Let w (Im w ¹ 0) be a complex number. Then the set of all [Online May 19, 2012] complex number satisfying the equation Statement 1:If arg Z + arg W = p, then Z = -W. w - w = k (1 - ) , for some real number k, is Statement 2: |Z| = |W|, implies arg Z – arg W = p. (a) Statement 1 is true, Statement 2 is false. [Online April 9, 2014] (b) Statement 1 is true, Statement 2 is true, Statement 2 is (a) { : = 1} (b) { : = } a correct explanation for Statement 1. (c) Statement 1 is true, Statement 2 is true, Statement 2 is (c) { : ¹ 1} (d) { : = 1, ¹ 1} not a correct explanation for Statement 1. (d) Statement 1 is false, Statement 2 is true. 26. If z is a complex n umber of unit modulus and 32. Let Z1 and Z2 be any two complex number. æ 1+ z ö Statement 1: Z1 - Z 2 ³ Z1 - Z 2 argument q, then arg ç è 1 + z ÷ø equals: Statement 2: Z1 + Z 2 £ Z1 + Z 2 [Online May 7, 2012] p (a) Statement 1 is true, Statement 2 is true, Statement 2 is (a) –q (b) –q (c) q (d) p – q a correct explanation of Statement 1. 2 (b) Statement 1 is true, Statement 2 is true, Statement 2 is 27. Let z satisfy| z | = 1 and = 1– z. not a correct explanation of Statement 1. (c) Statement 1 is true, Statement 2 is false. Statement 1 : z is a real number. (d) Statement 1 is false, Statement 2 is true. p 33. The number of complex numbers such that Statement 2 : Principal argument of is |z – 1| = |z + 1| = |z – i| equals 3 (a) 1 (b) 2 (c) ¥ (d) 0 [Online April 25, 2013] 1 (a) Statement 1 is true Statement 2 is true; Statement 2 is 34. The conugate of a complex number is then that a correct explanation for Statement 1. i –1 (b) Statement 1 is false; Statement 2 is true complex number is (c) Statement 1 is true, Statement 2 is false. –1 1 –1 1 (a) (b) (c) (d) (d) Statement 1 is true; Statement 2 is true; Statement 2 is i –1 i +1 i +1 i –1 not a correct explanation for Statement 1. 1 æx yö If z = x - i y and z 3 = p + iq, then çè p + q ø÷ ( p + q ) 2 2 35. æ 1+ z 2 ö 28. Let a = Im ç ÷ , where z is any non- ero complex ç 2iz ÷ is equal to è ø (a) –2 (b) –1 (c) 2 (d) 1 number. [Online April 23, 2013] 36. Let and w be complex numbers such that z + i w = 0 The set A = {a : | z | = 1 and z ¹ ±1 } is equal to: and arg zw = p. Then arg equals (a) (– 1, 1) (b) [– 1, 1] (c) [0, 1) (d) (– 1, 0] 5p p 3p p (a) (b) (c) (d) Z2 4 2 4 4 29. If Z1 ¹ 0 and Z2 be two complex numbers such that Z1 æ1+ i ö x 37. If ç ÷ = 1 then è1- i ø 2Z1 + 3Z 2 is a purely imaginary number, then 2Z - 3Z is equal to: (a) x = 2n + 1 , where n is any positive integer 1 2 (b) x = 4n , where n is any positive integer [Online April 9, 2013] (c) x = 2n , where n is any positive integer (a) 2 (b) 5 (c) 3 (d) 1 (d) x = 4n + 1 , where n is any positive integer.. Downloaded from @Freebooksforjeeneet  EBD_8344 www.jeebooks.in M-28 Mathematics 38. If z and w are two non- ero complex numbers such that æ z -1 ö p 46. If Re çè ÷ = 1, where z = x + iy, then the point (x, y) lies zw = 1 and Arg ( z ) - Arg (w ) = , then zw is equal to 2z + i ø 2 on a: [Jan. 7, 2020 (I)] (a) – 1 (b) 1 (c) – i (d) i æ 1 3ö (a) circle whose centre is at èç - , - ø÷. 39. If | – 4 | < | – 2 |, its solution is given by 2 2 (a) Re(z) > 0 (b) Re(z) < 0 2 (c) Re(z) > 3 (d) Re(z) > 2 (b) straight line whose slope is -. 40. z and w are two non ero complex numbers such that 3 | z | = | w| and Arg + Arg w = p then equals 3 (a) w (b) – w (c) w (d) – w (c) straight line whose slope is 2. Rotational Theorem, Square Root 5 of a Complex Number, Cube Roots (d) circle whose diameter is. 2 TOPIC n of Unity, Geometry of Complex Numbers, De-moiver’s Theorem, Powers of Complex Numbers 47. If z = 3 + i ( i = -1 ) , then (1 + iz + z5 + iz8)9 is equal 2 2 41. Let z = x + iy be a non- ero complex number such that to: [April 08, 2019 (II)] (a) 0 (b) 1 z2 = i |z|2, where i = -1 , then z lies on the: (c) (– 1 + 2i)9 (d) – 1 [Sep. 06, 2020 (II)] (a) line, y = –x (b) imaginary axis 3 x + iy (c) line, y = x (d) real axis 48. æ è 1 ö Let ç -2 - i ÷ = 3 ø 27 ( ) i = -1 , where x and y are real 42. If a and b are real numbers such that (2 + a ) = a + ba , 4 numbers then y – x equals : [Jan. 11, 2019 (I)] (a) 91 (b) – 85 (c) 85 (d) – 91 -1 + i 3 where a = , then a + b is equal to : 5 5 2 æ 3 iö æ 3 iö [Sep. 04, 2020 (II)] 49. Let z = çç + ÷÷ + çç - ÷÷. If R(z) and I(z) (a) 9 (b) 24 (c) 33 (d) 57 è 2 2ø è 2 2ø respectively denote the real and imaginary parts of z, 3 æ 2p 2p ö then: [Jan. 10, 2019 (II)] 1 + sin + i cos ç 9 9 ÷ is : (a) I(z) = 0 (b) R(z) > 0 and I(z) > 0 43. The value of ç 2p 2p ÷ (c) R(z) < 0 and I(z) > 0 (d) R(z) = – (c) ç 1 + sin - i cos ÷ è 9 9 ø n æ1+ i 3 ö [Sep. 02, 2020 (I)] 50. The least positive integer n for which çç ÷÷ = 1, is è1 – i 3 ø 1 1 (a) (1 - i 3) (b) ( 3 - i) [Online April 16, 2018] 2 2 (a) 2 (b) 6 (c) 5 (d) 3 1 1 51. The point represented by 2 + i in the Argand plane moves (c) - ( 3 - i ) (d) - (1 - i 3) 2 2 1 unit eastwards, then 2 units northwards and finally from 44. The imaginary part of (3 + 2 -54)1/ 2 - (3 - 2 -54)1/ 2 there 2 2 units in the south–westwards direction. Then its new position in the Argand plane is at the point can be : [Sep. 02, 2020 (II)] represented by : [Online April 9, 2016] (a) - 6 (b) -2 6 (c) 6 (d) 6 (a) 1 + i (b) 2 + 2i (c) –2 – 2i (d) –1 – i 100 100 52. A complex number is said to be unimodular if | | = 1. -1 + i 3. If a = (1 + a) å a and b = å a , 2k 3k 45. Let a = 2 k =0 k =0 1 -2 2 Suppose 1 and 2 are complex numbers such that 2- 1 then a and b are the roots of the quadratic equation: 2 [Jan. 8, 2020 (II)] is unimodular and 2 is not unimodular. Then the point 1 (a) x2 + 101x + 100 = 0 (b) x2 – 102 x + 101 = 0 lies on a: (c) x2 – 101x + 100 = 0 (d) x2 + 102x +101 = 0 Downloaded from @Freebooksforjeeneet  www.jeebooks.in Complex Numbers and Quadratic Equations M-29 (a) circle of radius 2. Solutions of Quadratic Equations, (b) circle of radius 2. Sum and Product of Roots, Nature (c) straight line parallel to x-axis TOPIC Đ of Roots, Relation Between Roots (d) straight line parallel to y-axis. and Co-efficients, Formation of an 2 Equation with Given Roots. 53. If z ¹ 1 and z is real, then the point represented by the z -1 61. If a and b be two roots of the equation x2 – 64x + 256 = 0. complex number z lies : (a) either on the real axis or on a circle passing through 1 1 the origin. æ a 3 ö 8