L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 PAPER B IIT JEE () (Trigonometry and Algebra) TOWARDS IIT JEE IS NOT A JOURNEY, IT S A BATTLE, ONLY THE TOUGHEST WILL SURVIVE TIME: 6 MINS MAX. MARKS: 76 MARKING SCHEME In Section I (Total Marks: 4), for each question you will be awarded marks if you darken ONLY the bubble corresponding to the correct answer and zero marks if no bubble is darkened. In all other cases, minus one () mark will be awarded. In Section II (Total Marks: 6), for each question you will be awarded 4 marks if you darken ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero marks otherwise. There are no negative marks in this section. In Section III (Total Marks: 4), for each question you will be awarded 4 marks if you darken ONLY the bubble corresponding to the correct answer and zero marks otherwise. There are no negative marks in this section. In Section IV (Total Marks: ), for each question you will be awarded marks for each row in which you have darkened ALL the bubble(s) corresponding to the correct answer(s) ONLY and zero marks otherwise. Thus, each question in this section carries a maximum of 6 marks. There are no negative marks in this section. NAME OF THE CANDIDATE CONTACT NUMBER L.K. Gupta (Mathematics Classes) FOR SOLUTIONS KINDLY VISIT www.pioneermathematics.com (In latest Updates) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c), (d) out of which ONLY ONE is correct.. The equation sin (cos x) = cos (sin x) has (a) only one real solution (c) no real solution Ans. (c) sin (cos x) = cos (sin x) cos (sin x) sin (cos x) = cos sin x cos cos x (b) infinitely many solution (d) none of the above cosx sin x cos x sin x cos. cos 4 4 cosx sin x If cos 4 cos x sin x n 4 sin x 4n, n I 4 4n sin x 4 sin x [,], which does not certify the original solution, 4 cos x sin x cos 4 cos x sin x Then, cos 4 cosx sin x r 4 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 cos x 4r 4 cos x 4r, r I 4 cos x [, ] 4 cos x sin x Also, cos 4 LHS and RHS Hence, no real solution.. If x x is factors of x x 4, then ( ) equals (a) (b) (c) 5 (d) 7 Sol: (b) x x x, 4 Then x, are the roots of x x 4 then () () 4 6 4 and ( 4) ( ) we get 5 and 4 ( ) 56 9. The number of solutions of 5 cos r x = 5 in the interval r, is (a) (b) (c) 5 (d) Ans. (b) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 5 r cos r x 5 cos x + cos x + cos x + cos 4x + cos 5x = 5 which is possible only, when cos x = cos x = cos x = cos 4x = cos 5x = and is satisfied by x = only. Hence number of solution =. 4. Set a, b, be such that cos (a b) = and cos (a + b) =. e of a, b satisfying the above system of equation is (a) (b) (c) (d) 4 Ans. (d) cos a b cos a b n, n I a b,, a b a, a, a The number of pairs cos a b cos a, cos a, cos a y cos a e e Hence, number of solutions is 4. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 4
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 5. The number of values of x for which sin x + cos 4x = is (a) (b) (c) (d) infinite Ans. (a) sin x + cos 4x = It is possible only, when sin x = and cos 4x = x n and x m x n and x m m, n I 4 Then, solution n, n I m, m I 4 = 6. If, ω, ω are the three cube roots of unity, then for α βω γω δω α, β, γ, δ R, the expression is β αω γωδω (a) (b) ω (c) ω (d) ω Ans. (b) α βω γω δω β αω γω δω ω α βω γω δω βω αω γω δω ω ω. 7. The number of positive integral solutions of 5 6 x 5 x 7 4 x x 4 x is (a) four (b) three (c) two (d) only one Ans. (b) 5 6 x 5 x 7 4 x x 4 x Since, PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 5
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 4 7 x, 7/, 5 x,, 4. 8. If tan cos θ cot sin θ, then the value of cos θ is 4 (a) (b) (c) (d) none of these Ans. (c) tan cos θ cot sin θ tan cos θ tan sin θ cos θ sin θ cos θsin θ cos θ sin θ cos θ 4 Section II (Total Marks: 6) (Multiple Correct Answer (s) Type) This section contains 4 multiple choice questions. Each question has four choices (a), (b), (c), and (d) out of which ONE or MORE may be correct. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 6
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 9. If tan α and tan β are the roots of the equation (a) sin α β p sinα β cos α β qcos α β q (b) tan α β p/ q (c) cos α β q (d) sin α β p Ans. (a, b) tan α tan β p, tan α tan β q q tan α tan β p tan α β tan α tan β q p [Alternate. (b)] q Alternate. (a) : LHS cos α β tan α β p tan α β q = q tan α β tan α β p tan α β q p p q q p q q q q p q p q q q p q p q Alternate. (c) : tan α β cos α β q p q Alternate. (d) : sin α β p q p p q x px q p, then PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 7
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677. The argument and the principal argument of the complex number i, where i are 4i i (a) tan (b) tan (c) Ans. (a, b) i i i i 4i i 4i i i 4i i 6i i 6 6 θ tan tan tan imaginary part real part and principal arg ument θ tan Argument tan tan tan. The roots of the equation, (x + ) = x(x + 4x + ), are given by (a) i /, i (c) (b) (d) Ans. (a, b, c, d) Given equation is x x x 4x 4 x x x x x x x x x x x x x x x x 4 x x i /, i (d) tan PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 8
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 x 4 x x x or x 4x x x or x h x 4 i x,.. In a triangle tan A + tan B + tan C = 6 and tan A tan B =, then the values of tan A, tan B and tan C are (a),, (b),, (c),, (d) none of these Ans. (a, b) In a triangle tan A + tan B + tan C = tan A tan B tan C or 6 = tan C tan C = tan A + tan B =, tan A tan B = tan A = or and tan B = or. (i) Section III (Total Marks: 4) (Integer Answer Type) This section contains 6 questions. The answer to each of the questions is a single-digit integer, ranging from to 9.The bubble corresponding to the correct answer is to be darkened in the Answer sheet.. cos 9 sin5 4 If λ, then the value of 9λ 8λ 97 78 must be Ans. Here, cos 9 = cos (7 + ) = sin and sin 5 = sin (7 ) = cos The given expression λ sin cos PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 9
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 cos 6 λ sin sin6 cos sin 6 cos cos 6 sin λ sin cos sin6 sin 6 sin4 λ 4 λ 6 λ 4 56 6 Then, 9λ 8λ 97 9 8 97 9 = 56 + 4 + 97-78 = 4. The least degree of a polynomial with integer coefficient whose one of the roots may be cos is Ans. 4 5 6 let θ 5θ 6 θ θ 6 cos θ θ cos 6 cos θ cos θ sin θ sin θ 4 cos θ cos θ cos θ sin θ 4sin θ sin θ cos θ PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 Let cot θ x 4x xx x 4 x x 8x x x x x 4x 5 5 6x x 6x 4x 4x 4x 5 x 4x x 4 x x 6x x 6x 4 but x, 6x 8x 6x 8x Degree is 4. 5. The three angles of a quadrilateral are 6, 6 g and 5, 6 the value of λ -9 must be Ans. 6 First angle =6,second angle=6 g = if fourth angle is then λ, 9 6 54 5 8 Third angle 5 6 Fourth angle=6 -(6 +54 +5 ) =96 = -9=6 6. The number of solutions of sec54 tan in [,4 ] must be Sol. 8 The given equation can be written as PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 4sin 5 cos cos 4 sin5 sin,cos 4 5 sin cos (i) 4 4 4 Let 5 4 cos, then sin 4 4 Now, from Eq. (i), cos cos 4 n, where cos or or n n Now, in [, ], n So, in total in [,4 ] 4 gives two solutions. we have, + + + = 8 solutions 7. The sum of the roots of equation cos 4x + 6 = 7 cos x over the interval [, 4] is λ, then the numerical quantity λ 4949 must be Ans. On putting cos x = we get t + 6 = 7t 5 t, 5 t t (impossible) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 cos x x n x n, n I The roots over [, 4] are,,,..., 99 4 Sum of roots... 99 495 λ 495 8. The sides of a cyclic quadrilateral are in AP, the shortest is 6 and the difference of the longest and the shortest is also 6. The square of the area of the quadrilateral is n-5756,find the value of n. Ans.4 Let sides are a, a + d, a + d, a + d Given, a = 6 and a + d a = 6 d Sides 6, 8,, s = 6 + 8 + + = 6 s 8 Then, (Area) = (8 6) (8 8) (8 ) (8 ) = 8 6 = 576 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 Section IV (Total Marks: ) (Matrix-Match Type) This section contains questions. Each question has three statements (a, b and c) given in Column I and five statements (p, q, r, s and t) in Column II. Any given statements in Column I can have correct matching with ONE or MORE statements(s) given in Column II. For example, if for a given question, statement B matches with the statements given in q and r, then for the particular question, against statement B, darken the bubbles corresponding to q and r in the ANSWER SHEET. 9. Observe the following columns : Column I (A) If maximum and minimum values of 76tanθ tan θ for tan θ all real values of θ are λ and μ respectively, then (B) If maximum and minimum values of 5cosθcos θ for all real values of θ are λ and μ respectively, then (C) If maximum and minimum values of sin θ cos θ 4 4 for all real values of θ are λ and μ respectively, then Column II (P) λ μ (Q) λ μ 6 (R) λ μ 6 (S) λ μ (T) λ μ 4 Ans. A R, S ; B R, T ; C P, Q A Let y 7 6 tan θ tan θ tan θ 7 cos θ 6 sin θ cos θ sin θ cos θ cos θ 7 sin θ sin θ 4 cos θ 4 sin θ 4 cos θ 4 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 4
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 y 8 λ 8, μ λ μ 6, λ μ R, S B Let y 5 cot θ cos θ / 5 cos θ cos θ sin θ cos θ sin θ cos θ sin θ 7 y 7 4 y λ, μ 4 λ μ 6, λ μ 4 R, T C Let y sin θ cos θ 4 4 cos θ cos θ 4 4 cos θ cos θ 4 4 cos θ 4 cos θ 4 cos θ 4 cos θ 4 y 4 λ 4, μ λ μ, λ μ 6 P,Q PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 5
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677. Observe the following columns : Column I (A) If α,β are the solutions of sinx in, and α, γ are the solutions of cos x = in,, then If α,β are the solutions of cot x in, and (B) (C) α, γ are the solutions of cosec x in,, then If α, β are the solutions of sinx in, and α, γ are the solutions of tan x = in,, then Ans. A Q, S ; B P,T ; C R, S, T A sin x sin 6 Column II (P) α β (Q) (R) β γ α γ (S) α β (T) β λ sin, sin 6 6 7 x,..(i) 6 6 and cos x cos 6 cos, cos 6 6 5 7 x,..(i) 6 6 From Eqs. (i) and (ii). It is clear that 7 5 α, β, y 6 6 6 α β S, β γ Q B cot x cot 6 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 6
L.K. Gupta (Mathematic Classes) www.pioneermathematics.com MOBILE: 985577, 4677 cot, cot 6 6 5 x,.(i) 6 6 and cosec x cosec 6 cosec, cosec 6 6 7 x, (ii) 6 6 From Eqs. (i) and (ii), It is clear that 5 7 α, β, γ 6 6 6 β γ T, α β P C sin 6 sin x sin, sin 6 6 7 x,.(i) 6 6 and tan x tan 6 tan, tan 6 6 7 x, (ii) 6 6 From Eqs. (i) and (ii), it is clear that 7 α, β, γ 6 6 6 α β, S, β γ T, α γ R PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O., SECTOR 4 D, CHANDIGARH 7