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'$"3: '$"3: '$"3: '$"3: 10 Vol XXXVII No 69 February 2019 &RUSRUDWH2൶FH Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-6601200 e-mail : info@mtg.in website : www.mtg.in 5HJG2൶FH 406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029 Managing Editor : Mahabir Singh Editor : Anil Ahlawat 31 19 40 48 55 46 Class XII Class XI Competition Edge CONTENTS ̀ Maths Musing Problem Set - 194 10 JEE Main Solved Paper 2019 19 Target JEE 31 Mock Test Paper JEE Main 2019 (Series 7) 40 Gear Up for JEE Main 48 JEE Work Outs 55 Mock Drill for JEE Main 63 Challenging Problems 66 You Ask We Answer 67 Math Archives 85 Maths Musing Solutions 46 Concept Map 82 MPP-10 47 Concept Map 69 CBSE Drill Practice Paper 80 MPP-10 82 80 66 Subscribe online at 47 www.mtg.in Individual Subscription Rates Mathematics Today Chemistry Today Physics For You Biology Today months 15 months 27 months 300 300 300 300 500 500 500 500 850 850 850 850 Combined Subscription Rates PCM PCB PCMB months 15 months 27 months 900 900 1200 1400 1400 1900 2500 2500 3400 Send D.D/M.O in favour of MTG Learning Media (P) Ltd Payments should be made directly to : MTG Learning Media (P) Ltd, Plot 99, Sector 44 Institutional Area, Gurgaon - 122 003, Haryana We have not appointed any subscription agent Printed and Published by Mahabir Singh on behalf of MTG Learning Media Pvt Ltd Printed at HT Media Ltd., B-2, Sector-63, Noida, UP-201307 and published at 406, Taj Apartment, Ring Road, Near Safdarjung Hospital, New Delhi - 110029 Editor : Anil Ahlawat Readers are adviced to make appropriate thorough enquiries before acting upon any advertisements published in this magazine Focus/Infocus features are marketing incentives MTG does not vouch or subscribe to the claims and representations made by advertisers All disputes are subject to Delhi jurisdiction only Copyright© MTG Learning Media (P) Ltd All rights reserved Reproduction in any form is prohibited MATHEMATICS TODAY | FEBRUARY‘19 M aths Musing was started in January 2003 issue of Mathematics Today The aim of Maths Musing is to augment the chances of bright students seeking admission into IITs with additional study material During the last 10 years there have been several changes in JEE pattern To suit these changes Maths Musing also adopted the new SDWWHUQ E\ FKDQJLQJ WKH VW\OH RI SUREOHPV 6RPH RI WKH 0DWKV 0XVLQJ SUREOHPV KDYH EHHQ DGDSWHG LQ -(( EHQH¿WWLQJ WKRXVDQG RI RXU readers It is heartening that we receive solutions of Maths Musing problems from all over India Maths Musing has been receiving tremendous response from candidates preparing for JEE and teachers coaching them We hope that students will continue to use Maths Musing to boost up their ranks in JEE Main and Advanced Set 194 JEE MAIN COMPREHENSION The cosine of obtuse angle formed by medians drawn from the vertices of the acute angles of an isosceles right angled triangle is (a) − (b) − (c) − (d) − 5 5 The shortest distance from the point (2, –7) to the circle x2 + y2 – 14x – 10y – 151 = is (a) (b) (c) (d) x − x +1 for x R, g(x) = ex for x R and x2 + x + π π h(x) = tan x for − < x < 2 The difference between the greatest and the least x values of the function f (x ) = ∫ (at + + cos t ) dt , a > for x [2, 3] is (a) 19 a + + (sin − sin 2) 18 18 a + + sin (c) a − + sin (b) 3 (d) None of these If 2x 2+ xy – 3y = is the equation of pair of x2 y2 conjugate diameters of an ellipse + =1 , a b2 then its eccentricity is 2 (a) (b) (c) (d) 3 3 π cos x π/2 sin x If A = ∫ dx is equal to dx , then ∫ ( x + 2)2 ( x + 1) 1 1 −A (a) A − − (b) + π+2 π+2 1 −A −A (d) + (c) π+2 π+2 JEE ADVANCED If z1, z2 and z3 are vertices of an equilateral triangle whose orthocentre is the origin, then (a) z1 + z2 + z3 = (b) |z1| = |z2| = |z3| (c) z2 z3 + z3 z1 + z1 z2 = (d) Oz1 is perpendicular to z2 z3 MATHEMATICS TODAY | FEBRUARY‘19 Let f (x) = Range of the function foh is 1 (a) ⎛⎜ , ⎞⎟ (b) ⎡⎢ , 3⎤⎥ ⎝3 ⎠ ⎣3 ⎦ (c) ⎢⎡ , 1⎤⎥ (d) [1, 3] ⎣3 ⎦ Range of the function goh is (a) R (b) [0, f) (c) (–f, 0] (d) (0, f) NUMERICAL ANSWER TYPE The position vectors of the points A, B, C and D a r e 3iˆ − jˆ − kˆ , 2iˆ + jˆ − 4kˆ , −iˆ + jˆ + 2kˆ a n d 4iˆ + jˆ + λkˆ It is known that these points are coplanar, then the value of –17O = MATRIX MATCH 10 A root of the equation on L.H.S satisfies R.H.S Column-I Column-II (P) cos x + sin x cos x – = (1) tan x = 4/3 (Q) sin2 ((S/2) cos2 x) = – cos (S sin 2x) (2) tan x = –1 (R) sec2 x – 11 tan x –2 = (3) tan x = 1/2 (S) cos 2x – sin 2x = sin2 x (4) cos 2x = 3/5 (a) (b) (c) (d) P 2, 1, 1, 2, Q 3, 3, 3, 3, R S 1, 2, 1, 3, 2, 3, 2, 1, 3, 1, 3, 2, See Solution Set of Maths Musing 193 on page no 85 Held on 12th January (Morning Shift) 2019 The sum of the distinct real values of P, for which the vectors, μi + j + k, i + μ j + k, i + j + μ k are co-planar, is (a) (b) (c) –1 (d) The area (in sq units) of the region bounded by the parabola, y = x2 + and the lines, y = x + 1, x = and x = 3, is (a) 15/2 (b) 17/4 (c) 21/2 (d) 15/4 Let C1 and C2 be the centres of the circles x2 + y2 – 2x – 2y – = and x2 + y2 – 6x – 6y + 14 = respectively If P and Q are the points of intersection of these circles, then the area (in sq units) of the quadrilateral PC1QC2 is (a) (b) (c) (d) Consider three boxes, each containing 10 balls labelled 1, 2, , 10 Suppose one ball is randomly drawn from each of the boxes Denote by ni, the label of the ball drawn from the ith box, (i = 1, 2, 3) Then, the number of ways in which the balls can be chosen such that n1 < n2 < n3 is (a) 120 (b) 164 (c) 240 (d) 82 Considering only the principal values of inverse functions, the set π⎫ ⎧ A = ⎨ x ≥ : tan−1 (2 x ) + tan−1 (3x ) = ⎬ 4⎭ ⎩ (a) contains two elements (b) contains more than two elements (c) is an empty set (d) is a singleton The perpendicular distance from the origin to the x +2 y −2 z +5 plane containing the two lines, and = = x −1 y − z + , is = = 11 (a) 11 (b) 11 (c) (d) 11 6 10 MATHEMATICS TODAY | FEBRUARY‘19 z −α (α ∈R) is a purely imaginary number and z+α |z| = 2, then a value of D is (c) (d) (a) (b) If If the straight line, 2x – 3y + 17 = is perpendicular to the line passing through the points (7, 17) and (15, E), then E equals 35 35 (c) − (d) –5 (a) (b) 3 The boolean expression ((p q) (p ~q)) (~p ~q) is equivalent to (a) (~p) (~q) (b) p (~q) (c) p q (d) p (~q) + + + + k If S12 + S22 + + S10 10 Let Sk = k = A, then A is equal to 12 (a) 303 (b) 283 (c) 301 (d) 156 11 An ordered pair(D, E) for which the system of linear equations (1 + D)x + Ey + z = 2, Dx + (1 + E)y + z = 3, Dx + Ey + 2z = has a unique solution, is (a) (–3, 1) (b) (2, 4) (c) (1, –3) (d) (–4, 2) 12 The maximum area (in sq units) of a rectangle having its base on the x-axis and its other two vertices on the parabola, y = 12 – x2 such that the rectangle lies inside the parabola, is (a) 36 (b) 18 (c) 20 (d) 32 13 Let S = {1, 2, 3, , 100} The number of non-empty subsets A of S such that the product of elements in A is even is (b) 250 – (a) 250(250 – 1) 50 (c) + (d) 2100 – 26 The equation of the given plane is x–y+z=5 (i) The equation of the given line is x −1 y − z + = = (ii) −6 The equation of the line passing through the point P(1, –2, 3) and parallel to line (ii) is given by x −1 y + z − = λ (say) .(iii) = = −6 Any point say Q, on line (iii) is (2O + 1, 3O – 2, –6O + 3) For some value of O, let the point Q(2O + 1, 3O – 2, –6O + 3) lie on plane (i) ? (2O + 1) – (3O – 2) + (–6O + 3) = ⇒ −7 λ = −1 ⇒ λ = ⎛ −11 15 ⎞ , So, the coordinates of Q are ⎜ , ⎝ 7 ⎟⎠ 2 ⎛ ⎞ ⎛ −11 ⎞ ⎛ 15 ⎞ Distance PQ = ⎜ − 1⎟ + ⎜ + 2⎟ + ⎜ − 3⎟ ⎠ ⎝7 ⎠ ⎝ ⎠ ⎝7 2 = 62 + (12)2 + (−3)2 = 36 + 144 + = 189 = 21 units Let R(x, y, z) be the image of P in the plane (i) Then, Q is the midpoint of PR 14 + y 5+ z 7+x ∴ = 1, = and =8 2 x = –5, y = –10, z = 11 Hence, the image of P(7, 14, 5) in the given plane is R (–5, –10, 11) 27 Let x be the number of units of the product A and y be the number of units of the product B produced Let z be the total profit According to question x and y must satisfy the following conditions : 200x + 300y t 10000 or 2x + 3y t 100 x + y ≤ 40 or x + 2y d 80 x t 14, y t 16, x t 0, y t 2 ⎛ ⎞ ⎛ ⎞ ⎛ −6 ⎞ = ⎜ ⎟ +⎜ ⎟ +⎜ ⎟ = + + 36 ⎝ 7⎠ ⎝ 7⎠ ⎝ ⎠ = ⎛1 ⎞ × 49 = ⎜ × 7⎟ = unit ⎝ ⎠ OR The given plane is 2x + 4y – z = (i) P(7, 14, 5) The d.r.’s of the normal to this plane are 2, 4, –1 The equation of the line passing Q through the point P(7, 14, 5) and the perpendicular to the given plane (i) is given by R(x, y, z) x − y − 14 z − = = = λ (say) −1 Any point say Q on this line is (2O + 7, 4O + 14, –O + 5) For some value of O, let the point Q(2O + 7, 4O + 14, –O + 5) lie on the plane (i) ? 2(2O + 7) + 4(4O + 14) – (–O + 5) = 21O = –63 O = –3 ? The coordinates of Q are[2 × (–3) + 7, × (–3) + 14, –(–3) + 5] i.e., Q (1, 2, 8) ∴ PQ = (7 − 1)2 + (14 − 2)2 + (5 − 8)2 78 MATHEMATICS TODAY | FEBRUARY ‘19 Mathematical formulation of the LPP will be Maximize z = 20x + 30y subject to constraints : 2x + 3y t 100 x + 2y d 80 x t 14, y t 16 x t 0, y t On plotting, we get the corner points of the feasible region as S(26, 16), P(48, 16), Q(14, 33) and R(14, 24) Maximize z = 20x + 30y ? At S(26, 16), z = 1000 At P(48, 16), z = 1440 At Q(14, 33), z = 1270 At R(14, 24), z = 1000 Hence, the maximum profit is ` 1440 and it is attained when 48 units of product A and 16 units of product B are produced 28 Let x be the radius of the inscribed right circular cone and h be its height, then (Pythagoras theorem) OM2 + MC2 = OC2 2 (h – r) + x = r (' OA = r) x2 = 2hr – h2 (i) Let S is the curved surface area of the cone, then A S = π xl = πx h2 + x S2 = S2x2(h2 + x2) = S2(2hr – h2) (h2 + 2hr – h2) = S2(2hr – h2)2hr B = 2S2r (2h2r – h3) M r x C Since S > 0, therefore, S is maximum iff S2 is maximum, so we need to find the value of h for which S2 is maximum Let S2 = f(h) ? f(h) = 2S2r(2h2r – h3), < h < 2r (i) Differentiating (i) w.r.t h, we get f c(h) = 2S2r(4hr – 3h2) and f cc(h) = 2S2r(4r – 6h) Now, f c(h) = 2S2r(4hr – 3h2) = h(4r – 3h) = but < h < 2r h = r ⎛ ⎞ 2 ⎛4 ⎞ = − 2 < Also f ′′ ⎜ r ⎟ = π r ⎜ r − r ⎟ 8π r ⎝ ⎝3 ⎠ ⎠ f(h) is maximum when h = r S2 is maximum when h = r Therefore, the curved surface area of the inscribed cone is maximum when its altitude = r 29 Let I = ∫ e x sin(3x + 1)dx = sin(3x + 1) ⋅ l O S is maximum when h = r e2 x e2 x − ∫ cos(3x + 1) ⋅ ⋅ dx 2 = e x sin(3x + 1) − ∫ cos(3x + 1) ⋅ e x dx + C1 2 3⎡ e2 x = e x sin(3x + 1) − ⎢cos(3x + 1) ⋅ 2⎣ − ∫ − sin(3x + 1) ⋅ ⋅ e2 x ⎤ dx ⎥ + C2 ⎦ = e x sin(3x + 1) − e x cos(3x + 1) − I + C2 4 ⎛1 + ⎞ I x ⎟ = e [2 sin(3x + 1) − cos(3x + 1)] + C2 ⎜⎝ 4⎠ ⇒ 13 x e sin(3x + 1)dx = e x [2 sin(3x + 1) 4∫ −3 cos(3x + 1)] + C2 ⇒ ∫e 2x sin(3x + 1) dx = 2x e [2 sin(3x + 1) 13 −3 cos(3x + 1)] + C MATHEMATICS TODAY | FEBRUARY ‘19 79 MPP-10 Class XII T his specially designed column enables students to self analyse their extent of understanding of specified chapters Give yourself four marks for correct answer and deduct one mark for wrong answer Self check table given at the end will help you to check your readiness Probability Total Marks : 80 Only One Option Correct Type 80 A signal which can be green or red with probability and respectively, is received by station A and 5 then transmitted to station B The probability of each station receiving the signal correctly is If the signal received at station B is green then the probability that the original signal was green is 20 (a) (b) (c) (d) 20 23 It is given that mean and variance of a binomial variate X are and respectively The probability that X takes a value greater or equal to is 13 15 (b) (c) (d) (a) 16 16 16 If A and B are two independent events such that P(AcB) = 2/15 and P(A Bc) = 1/6, then P(B) is (a) 1/5 (b) 1/6 (c) 2/3 (d) 5/6 A man takes a step forward with probability 0.4 and backward with probability 0.6 The probability that at the end of 11 steps he is one step away from the starting point is (b) 420(0.24)6 (a) 420(0.24)5 (c) 462(0.24) (d) 462(0.24)6 A bag contains a white and b black balls Two players, A and B alternatively draw a ball from the bag, replacing the ball each time after the draw till one of them draws a white ball and wins the game A begins the game If the probability of A winning the game is three times that of B, the ratio a : b is (a) : (b) : (c) : (d) none of these MATHEMATICS TODAY | FEBRUARY‘19 Time Taken : 60 Min The sum and product of the mean and variance of a binomial distribution are 24 and 128 respectively The distribution is 1 (a) ⎜⎛ + ⎞⎟ ⎝7 8⎠ 12 ⎛1 3⎞ (b) ⎜ + ⎟ ⎝4 4⎠ (c) ⎛⎜ + ⎞⎟ ⎝6 6⎠ 24 1 (d) ⎛⎜ + ⎟⎞ ⎝2 2⎠ 16 32 One or More Than One Option(s) Correct Type If X follows a binomial distribution with parameters n = and p = , then P(|X – 4| d2) is 117 118 121 119 (b) (c) (d) (a) 128 128 128 128 If A and B are two events such that P(A) = 1/2 and P(B) = 2/3, then (a) P(A B) t 2/3 (b) P(A Bc) d 1/3 (c) 1/6 d P (A B) d 1/2 (d) 1/6 d P(Ac B) d 1/2 An electric component manufactured by 'RASU Electronics' is tested for its defectiveness by a sophisticated testing device Let A denote the event "the device is defective" and B the event "the testing device reveals the component to be defective" Suppose P(A) = D and P(B/A) = P(Bc/Ac) = – D, where < D < 1, then (a) P(B) = 2D (1 – D) (b) P(Ac/B) = 1/2 (c) P(Bc) = (1 – D)2 + D2 (d) P(Ac/Bc) = [D/(1 – D)]2 10 Let E and F be two independent events The probability that the exactly one of them occurs is 11/25 and the probability of none of them occurring is 2/25 If P(T) denotes the probability of occurrence of the event T, then , P(F) = (c) P(E) = , P(F) = (a) P(E) = (b) P(E) = (d) P(E) = , P(F) = , P(F) = 5 11 Let < P(A) < 1, < P(B) < and P(AB) = P(A) + P(B) – P(A) P(B), then (a) P(B/A) = P(B) – P(A) (b) P(Ac – Bc) = P(Ac) – P(Bc) (c) P(AB)c = P(Ac) P(Bc) (d) P(A/B) = P(A) 12 For two events A and B, if P(A) = P(A/B) = 1/4 and P(B/A) = 1/2, then (a) A and B are independent (b) A and B are mutually exclusive (c) P (Ac/B) = 3/4 (d) P(Bc/Ac) = 1/2 13 A ship is fitted with three engines E1, E2 and E3 The engine function independently of each other 1 with respective probabilities , and For the 4 ship to be operational at least two of its engines must function Let X denote the event that the ship is operational and let X1, X2 and X3 denote respectively the events that the engines E1, E2 and E3 are functioning then which of the following is true? (a) P[X1c/X] = 3/16 (b) P [Exactly two engines of the ship are functioning/X] = 7/8 (c) P[X/X2] = 5/16 (d) P [X/X1] = 5/16 Comprehension Type A JEE aspirant estimates that she will be successful with an 80% chance if she studies 10 hours per day, with a 60% chance if she studies hours per day and with a 40% chance if she studies hours per day She further believes that she will study 10 hours, hours and hours per day with probabilities 0.1, 0.2 and 0.7, respectively 14 The probability that she will be successful, is (a) 0.28 (b) 0.38 (c) 0.48 (d) 0.58 15 Given that she does not achieve success, the probability she studied for hours, is 21 19 18 (b) (c) 20 (d) (a) 26 26 26 26 Matrix Match Type 16 A bag contains some white and some black balls, all combinations being equally likely The total number of balls in the bag is 12 Four balls are drawn at random from the bag without replacement P Q R S Column-I Probability that all the four balls are black is equal to If the bag contains 10 black and two white balls, then the probability that all four balls are black is equal to If all the four balls are black, then the probability that the bag contains 10 black balls is equal to Probability that two balls are black and two are white is P (a) (c) Q 1 R 2 S P (b) (d) Q Column-II 14/33 1/5 70/429 168/715 R S Numerical Answer Type 17 One ticket is selected at random from 100 tickets numbered 00, 01, 02, , 98, 99 If X and Y denote the sum and the product of the digits on the tickets, The value of 57 P(X = | Y = 0) is 18 The least number of times a fair coin must be tossed so that the probability of getting atleast one head is atleast 0.8, is 19 A student appears for tests I, II and III The student is successful if he passes either in test I and II or test I and III The probabilities of the student passing in tests I, II, III are p, q, respectively If the probability that the student is successful is , then p = 20 If X is a binomial variate with parameters n P( X = r ) and p, where < p < such that is P( X = n − r ) independent of n and r, then 2p equals VV Keys are published in this issue Search now! - Check your score! If your score is No of questions attempted …… No of questions correct …… Marks scored in percentage …… > 90% EXCELLENT WORK ! You are well prepared to take the challenge of final exam 90-75% GOOD WORK ! You can score good in the final exam 74-60% SATISFACTORY ! You need to score more next time < 60% NOT SATISFACTORY! Revise thoroughly and strengthen your concepts MATHEMATICS TODAY | FEBRUARY‘19 81 MPP-10 Class XI T his specially designed column enables students to self analyse their extent of understanding of specified chapters Give yourself four marks for correct answer and deduct one mark for wrong answer Self check table given at the end will help you to check your readiness Statistics and Probability Total Marks : 80 Only One Option Correct Type In a class of 10 students, probability of exactly i students passing an examination is directly proportional to i2, then the probability that exactly five students pass an examination is (a) 1/11 (b) 5/77 (c) 25/77 (d) 10/77 The first of two samples has 100 items with mean 15 and S.D If the whole group has 250 items with mean 15.6 and S.D = 13.44 , then the S.D of the second group is (a) (b) (c) (d) 3.52 The odds against the player A to win are : and odds in favour of another player to win are : If the two events are independent, find the probability that at least one player will win (a) 52/77 (b) 51/77 (c) 53/77 (d) 59/77 The mean of five observations is 4.4 and the variance is 8.24 Three of the five observations are 1, and The remaining two are (a) 9, (b) 7, (c) 6, (d) 10, If x denotes the mean of n observations x1, x2, , xn, then mean of the observation xi + 2i, where i = 1, 2, 3, , n is (a) x + (b) x + n (c) x + 2n (d) x + n + The probability that in a family of five members, exactly two members have birthday on Sunday is (a) (12 × 53) / 75 (b) (10 × 62) / 75 (c) 2/5 (d) (10 × 63) / 75 One or More Than One Option(s) Correct Type The variance of first 50 even natural numbers is 833 437 (a) 833 (b) 437 (c) (d) 4 82 MATHEMATICS TODAY | FEBRUARY‘19 Time Taken : 60 Min A natural number is chosen at random from the first 100 natural numbers The probability that 100 > 50 is x+ x (a) 1/10 (b) 11/50 (c) 11/20 (d) none of these Mean of the numbers 1, 2, 3, , n with respective weights 12 + 1, 22 + 2, 32 + 3, , n2 + n is 3n(n + 1) 3n2 + 7n + (b) 2(2n + 1) 2(2n + 4) 3n + 3n + (c) (d) 10 The probability that a 50 yr old man will be alive at 60 is 0.83 and the probability that a 45 yr old woman will be alive at 55 is 0.87 Then, (a) the probability that both will be alive after 10 year is 0.7221 (a) 5H1=1
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