Mathematics in English Toán học bằng Tiếng anh

24 45 Vol XXXIV No September 2016 Corporate Office: Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-6601200 e-mail : info@mtg.in website : www.mtg.in 69 66 Regd Office: 406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029 Managing Editor : Mahabir Singh Editor : Anil Ahlawat 43 64 39 CONTENTS 67 Concept Boosters Class XI 24 Brain @ Work 56 34 Ace Your Way (Series 5) 39 Challenging Problems 34 Subscribe online at Class XII 56 Ace Your Way (Series 5) Mathematics Today Chemistry Today Physics For You Biology Today 64 MPP-3 Competition Edge 66 Maths Musing Problem Set - 165 67 Math Archives 69 Mock Test Paper - WB JEE 78 Olympiad Corner 83 You Ask We Answer 85 Maths Musing Solutions www.mtg.in Individual Subscription Rates 43 MPP-3 45 Concept Boosters 83 yr yrs yrs 330 330 330 330 600 600 600 600 775 775 775 775 Combined Subscription Rates yr PCM PCB PCMB 900 900 1000 yrs yrs 1500 1500 1800 1900 1900 2300 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 Owned, Printed and Published by Mahabir Singh from 406, Taj Apartment, New Delhi - 29 and printed by Personal Graphics and Advertisers (P) Ltd., Okhla Industrial Area, Phase-II, New Delhi Readers are advised 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 Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd All rights reserved Reproduction in any form is prohibited MATHEMATICS TODAY | SEPTEMBER ‘16       
 This column is aimed at Class XI students so that they can prepare for competitive exams such as JEE Main/Advanced, etc and be also in command of what is being covered in their school as part of NCERT syllabus The problems here are a happy blend of the straight and the twisted, the simple and the difficult and the easy and the challenging THE COMPLEX NUMBER SYSTEM (a) Solutions of equation x2 + = are not real so they are imaginary i was regarded as a fictitious or imaginary number which could be manipulated algebrically like an ordinary real number, except that its square was –1 The letter i was used to denote −1 , possibly because i is the first letter of the Latin word ‘imaginarius’ (b) To permit solutions of such polynomial equations, the set of complex numbers is introduced We can consider a complex number of the form a + ib, where a and b are real numbers It is denoted by z i.e z = a + ib ‘a’ is called as real part of z which is denoted by Re(z) and ‘b’ is called as imaginary part of z which is denoted by Im (z) 
 
 



 
   
 
     
        Remarks : (a) The set R of real numbers is a proper subset of the Complex Numbers Hence the complete number system is N ⊂ W ⊂ I ⊂ Q ⊂ R ⊂ C (b) Zero is purely real as well as purely imaginary (c) i = −1 is the imaginary unit and called 'iota' Also i2 = −1; i3 = − i ; i4 = etc (d) a ⋅ b = ab only if atleast one of a or b is non-negative MATHEMATICS TODAY | SEPTEMBER ‘16 (e) If z = a + ib, then a – ib is called complex conjugate of z and written as z = a − ib (f) Real numbers satisfy order relations whereas imaginary numbers not satisfy order relation i.e i > 0, + i < are meaningless ALGEBRAIC OPERATIONS Fundamental operations on complex numbers In performing operations with complex numbers we can proceed as in the algebra of real numbers, replacing i2 by –1 when it occurs (i) Addition : (a + bi) + (c + di) = a + bi + c + di = (a + c) + (b + d)i (ii) Subtraction : (a + bi) – (c + di) = a + bi – c – di = (a – c) + (b – d)i (iii) Multiplication : (a + bi) (c + di) = ac + adi + bci + bdi2 = (ac – bd) + (ad + bc)i (iv) Division : a + bi a + bi c − di ac − adi + bci − bdi = ⋅ = c + di c + di c − di c − d 2i = ac + bd + (bc − ad )i = ac + bd + bc − ad i c +d c +d c2 + d Inequalities in imaginary numbers are not defined There is no validity if we say that imaginary number is positive or negative e.g., z > 0, + 2i < + 4i are meaningless In real numbers if a2 + b2 = then a = = b however in complex numbers, z12 + z22 = does not imply z1 = = z2 2 2 Equality In Complex Numbers : Two complex numbers z1 = a1 + ib1 & z2 = a2 + ib2 are equal if and only if their real and imaginary parts are equal i.e z1 = z2 ⇔ Re(z1) = Re(z2) and Im(z1) = Im(z2) REPRESENTATION OF A COMPLEX NUMBER Cartesian Form (Geometric Representation) To each complex number there corresponds one and only one point in plane, and conversely to each point in the plane there corresponds one and only one complex number Because of this we often refer to the complex number z as the point z Ever y complex number z = x + iy can be represented by a point on the Cartesian plane known as complex plane (Argand diagram) by the ordered pair (x, y) Length OP is called modulus of the complex number which is denoted by |z| & θ is called the argument or amplitude ⎛y⎞ | z | = x + y and tan θ = ⎜ ⎟ (angle made by ⎝x⎠ OP with positive x-axis) Note : (i) Argument of a complex number is a many valued function If θ is the argument of a complex number then 2nπ + θ; n ∈ I will also be the argument of that complex number Any two consecutive arguments of a complex number differ by 2nπ (ii) The unique value of θ such that −π < θ ≤ π is called the principal value of the argument Unless otherwise stated, amp z implies principal value of the argument (iii) By specifying the modulus & argument a complex number is defined completely For the complex number + 0i the argument is not defined and this is the only complex number which is only given by its modulus Trigonometric/Polar Representation z = r(cosθ + i sinθ), where |z| = r; arg z = θ; z = r (cos θ − i sin θ) Note : cosθ + i sinθ is also written as CiS θ 10 MATHEMATICS TODAY | SEPTEMBER ‘16 Euler’s Formula z = reiθ, |z| = r, arg z = θ z = re −iθ Note : If θ is real then cos θ = e iθ + e − iθ e iθ − e − iθ ; sin θ = 2i Vectorial Representation Every complex number can be considered as the position vector of a point If the point P represents the complex number z, then OP = z and |OP | = | z | ARGUMENT OF A COMPLEX NUMBER (a) Argument of a non-zero complex number P(z) is denoted and defined by arg (z) = angle which OP makes with the positive direction of real axis (b) If OP = |z| = r and arg (z) = θ, then obviously z = r(cosθ + isinθ), called the polar form of z 'Argument of z' would mean principal argument of z(i.e argument lying in (–π, π]) unless the context requires otherwise (c) Argument of a complex number z = a + ib = r(cosθ + isinθ) is the value of θ satisfying r cosθ = a and r sinθ = b b Let θ = tan −1 a a > 0, b > P.V arg z = θ a = 0, b > P.V arg z = π/2 a < 0, b > P.V arg z = π – θ a < 0, b = P.V arg z = π a < 0, b < P.V arg z = – (π – θ) a = 0, b < P.V arg z = – π/2 a > 0, b < P.V arg z = – θ a > 0, b = P.V arg z = P.V stands for principal value GEOMETRICAL REPRESENTATION OF FUNDAMENTAL OPERATIONS (a) Geometrical representation of addition of complex numbers Let P, Q be represented by z1 = r1 eiθ1 , z2 = r2eiθ2 respectively To find point R representing complex number z1z2 , we take a point L on real axis such that OL = and draw triangle OQR similar to triangle OLP Therefore OR OP = ⇒ OR = OP ⋅ OQ OQ OL i.e OR = r1r2 and ∠QOR = θ1 ∠LOR = ∠LOP + ∠POQ + ∠QOR = θ1 + θ2 – θ1 + θ1 = θ1 + θ2 Hence, R is represented by z1z2 = r1r2 ei(θ1 + θ2 ) (d) Geometrical representation of the division of complex numbers Let points P, Q be represented by z1 = r1 eiθ1, z2 = r2 eiθ2 respectively To find point R representing complex number z1 , we take a point z2 L on real axis such that OL = and draw a triangle OPR similar to OQL If two points P and Q represent complex numbers z1 and z2 respectively in the Argand plane, then the sum z1 + z2 is represented by the extremity R of the diagonal OR of parallelogram OPRQ having OP and OQ as two adjacent sides (b) Geometrical representation of subtraction of complex numbers Therefore r OP OR = ⇒ OR = OQ OL r2 and ∠LOR = ∠LOP – ∠ROP = θ1 – θ2 z1 r1 i(θ1 − θ2 ) = e Hence, R is represented by z2 r2 (c) Geometrical representation of multiplication of complex numbers 12 MATHEMATICS TODAY | SEPTEMBER ‘16 Modulus and argument of multiplication of two complex numbers For any two complex numbers z1, z2, we have |z1z2| = |z1||z2| and arg (z1z2) = arg(z1) + arg(z2) Notes : (i) P.V arg(z1z2) ≠ P.V arg(z1) + P.V arg(z2) (ii) |z1 z2 zn| = |z1||z2| |zn| (iii) arg (z1z2 zn) = arg z1 + arg z2 + + arg zn Modulus and argument of division of two complex numbers If z1 and z2 (≠ 0) are two complex numbers, we z |z | have = and z2 | z2 | ⎛z ⎞ arg ⎜ ⎟ = arg(z1 ) − arg(z2 ) ⎝ z2 ⎠ Note : ⎛z ⎞ P.V arg ⎜ ⎟ ≠ P.V arg(z1 ) − P.V arg(z2 ) ⎝z ⎠ CONJUGATE OF A COMPLEX NUMBER Conjugate of a complex number z = a + ib is denoted and defined by z = a − ib In a complex number if we replace i by –i, we get conjugate of the complex number z is the mirror image of z about real axis on Argand’s Plane Geometrical representation of conjugate of complex number : |z | = |z | arg(z ) = − arg(z ) Properties z+z z−z , y= (i) If z = x + iy, then x = 2i (ii) z = z ⇔ z is purely real (iii) z + z = ⇔ z is purely imaginary (iv) | z |2 = zz (v) z = z (vi) (z1 ± z2 ) = z1 ± z2 (vii) (z1 z2 ) = z1z2 ⎛ z ⎞ (z ) (viii) ⎜ ⎟ = , (z2 ≠ 0) ⎝ z ⎠ (z ) Imaginary roots of polynomial equations with real coefficients occur in conjugate pairs |z1 ± z2|2 = |z1|2 + |z2|2 ± (z1z2 + z1z2 ) = |z1|2 + |z2|2 ± Re(z1z2 ) = |z1|2 + |z2|2 ± 2|z1||z2| cos (θ1 – θ2) Note : If w = f (z ), then w = f (z ) DISTANCE BETWEEN COMPLEX POINTS If z1 = x1 + iy1, z2 = x2 + iy2, then distance between points z1, z2 in argand plane is | z1 − z2 | = (x1 − x2 )2 + ( y1 − y2 )2 INEQUALITIES IN COMPLEX NUMBERS If z1 and z2 are two complex numbers such that |z1| ≤ 1, |z2| ≤ 1, then (i) |z1 – z2|2 ≤ (|z1| – |z2|)2 + (Arg(z1) – Arg(z2))2 (ii) |z1 + z2|2 ≥ (| z1| + |z2|)2 – (Arg(z1) – Arg(z2))2 (iii) |z1 ± z2| ≤ |z1| + |z2| In general |z1 ± z2 ± z3 ± ± zn | ≤ |z1| + |z2| + |z3| + + |zn| (iv) |z1 ± z2| ≥ || z1| – |z2|| ⇒ ||z1| – |z2|| ≤ |z1 + z2| ≤ |z1| + |z2| ROTATION (a) Important results (i) arg z = θ represents points (non-zero) on ray eminating from or i g i n m a k i ng an angle θ with positive direction of real axis (ii) arg(z – z1) = θ represents points (≠ z1) on ray eminating from z1 making an angle θ with positive direction of real axis (b) Rotation theorem (i) If P(z1) and Q(z2) are two complex numbers such that |z1| = |z2|, then z2 = |z1| eiθ where θ = ∠POQ (ii) If P(z1), Q(z2) and R (z ) are t hre e complex numbers and ∠PQR = θ, then ⎛ z − z ⎞ z − z iθ ⎜⎝ z − z ⎟⎠ = z − z e 2   (iii) If P(z1), Q(z2), R(z3) and S(z ) are four complex numbers and ∠STQ = θ, then z3 − z z −z = e iθ z1 − z2 z1 − z2 MATHEMATICS TODAY | SEPTEMBER ‘16 13  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 Continuity & Differentiability Total Marks : 80 Time Taken : 60 Min Only One Option Correct Type tan x log x , then f(x) is discontinuous at − cos x ⎧ nπ ⎫ (a) ⎨ ; n ∈Q ⎬ (b) (2n + 1)π /2 : n ∈Z ⎩ ⎭ ⎫ ⎧ nπ : n ∈ N ⎬ ∪ (−∞, 0) (c) ⎨ ⎭ ⎩ (d) none of these Let f (x ) = { } Let f : R→R be a function defined by f (x) = max {x, x3} The set of all points where f (x) is not differentiable is (a) {–1, 1} (b) {–1, 0} (c) {0, 1} (d) {–1, 0, 1} dy If y = log10x + logx10 + logxx + log1010, then ⎛⎜ ⎞⎟ ⎠ x =10 ⎝ dx is equal to (a) (b) (c) dy is dx dx (b) – n2y (c) – y (d) 2n2y y = (x + + x )n then (1 + x ) (a) n2y (d) –1 ⎧ ⎪ ⎡⎣ f ( x )⎤⎦ , ⎪ If g(x) = ⎨ ⎪ 3, ⎪⎩ d y +x ⎛ π⎞ ⎛π ⎞ x ∈ ⎜ 0, ⎟ ∪ ⎜ , π ⎟ ⎝ 2⎠ ⎝2 ⎠ π x= where [x] denotes the greatest integer function and f (x ) = (a) (b) (c) (d) 64 2(sin x − sinn x )+ |sin x − sinn x | ,n n ∈ R, then 2(sin x − sinn x )− |sin x − sinn x | g(x) is continuous and differentiable at x = π/2, when < n < g(x) is continuous and differentiable at x = π/2, when n > g(x) is continuous but not differentiable at x = π/2, when < n < g(x) is continuous but not differentiable at x = π/2, when n > MATHEMATICS TODAY | SEPTEMBER ‘16 x Let f ( x ) = p Then d3 dx (a) p (c) p + p3 sin x cos x where p is a constant −1 p2 p3 ( f (x )) at x = is (b) p + p2 (d) Independent of p One or More Than One Option Correct Type ⎧ x −3 ,x ≥1 ⎪ The function f ( x ) = ⎨ x 3x 13 ,x (1 + t ) − (1 − t ) 11 If y = (1 + t ) + (1 − t ) and x = (1 − t ), then equal to (a) t {1 + −1 (1 − t ) { (c) t + (1 − t ) Matrix Match Type } (b) } (d) { dy dx } (1 − t ) − t 16 Match the columns: Column I y = sin–1(3x–4x3), P , ([.] denotes the greatest integer sin [ x] function) then (a) domain of f(x) is (2nπ + π, 2nπ + 2π) ∪ {2nπ + π/2}, where n ∈ I (b) f(x) is continuous, when x ∈ (2nπ + π, 2nπ + 2π), where n ∈I (c) f(x) is differentiable at x = π/2 (d) none of these 13 If F(x) = f(x)g(x) and f ′(x)g′(x) = c, then Comprehension Type If D * f ( x ) = lim f (x + h) − f (x ) h→0 h here f 2(x) = {f(x)}2 14 If u = f(x), v = g(x), then the value of D* (u.v) is (a) (D*u) v + (D* v)u (b) u2(D*v) + v2(D*u) (c) D*u + D*v (d) uv(D* (u + v) * ⎧u ⎫ 15 If u = f(x), v = g(x) then the value of D ⎨ ⎬ is ⎩v ⎭ * * u(D v ) − v(D u) (a) u (D v ) − v (D u) (b) v v2 (c) * * v (D *u) − u2 (D *v ) v (d) then (a) (b) (c) (d) ,x < ⎛ 1⎞ , x ∈⎜ − , ⎟ ⎝ 2⎠ − x2 ( dy is dx Q R 2 Integer Answer Type ) 17 If f(x) is a continuous function ∀ x ∈ R and the range ⎡ f (x ) ⎤ of f(x) is 2, 26 and g ( x ) = ⎢ ⎥ is continuous ⎣ c ⎦ ∀ x ∈ R, then the least positive integral value of c is (where [.] denotes the greatest integer function) ⎛ log e / x ⎞ ⎞ ⎛ e −1 ⎜ ⎟ + tan −1 + log e x , y = tan 18 If ⎜⎝ − log x ⎠⎟ ⎜ log ex ⎟ e e ⎝ ⎠ d2 y then is dx ⎛x −2⎞ 19 If y = 2sin–1 ⎜ − + x − x , then at x = the ⎝ ⎟⎠ ( ) ( ( ) ) ( value of ) dy must be dx ⎧ x − x − x + 1x , ⎪ 20 If f ( x ) = ⎨ x2 ⎪e x sin x + πx + λ ln , ⎩ u(D *u) − u(D *v ) v P 3 −2 + x2 y=tan–1 ⎜ 12 Let f ( x ) = F ′′ f ′′ g ′′ 2c ⎡ f g⎤ = + + (a) F ′ = c ⎢ + ⎥ (b) F f g fg ⎣ f ′ g′⎦ F ′′′ f ′′′ g ′′′ F ′′′ f ′′′ g ′′ = + (c) (d) = + F ′′ f ′′ g ′′ F f g ⎛ 3x − x ⎞ ⎟, ⎝ − 3x ⎠ R ⎛ 1 ⎞ , x ∈⎜ − , ⎟ ⎝ 3⎠ 1+ x 2⎞ ⎛ y = cos–1 ⎜ − x ⎟ ⎝ + x2 ⎠ dy then is dx t6 dy is then dx Q − (1 − t ) Column II x >0 is continuous at x ≤0 λ x = 0, then the value of e must be ”” Keys are published in this issue Search now! ☺ Check your score! If your score is > 90% EXCELLENT WORK ! You are well prepared to take the challenge of final exam You can score good in the final exam You need to score more next time No of questions attempted …… 90-75% GOOD WORK ! No of questions correct …… 74-60% SATISFACTORY ! Marks scored in percentage …… < 60% NOT SATISFACTORY! Revise thoroughly and strengthen your concepts MATHEMATICS TODAY | SEPTEMBER ‘16 65 M aths Musing was started in January 2003 issue of Mathematics Today with the suggestion of Shri Mahabir Singh 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 pattern by changing the style of problems Some of the Maths Musing problems have been adapted in JEE benefitting thousand of our 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 JEE MAIN Let P be a point on the parabola y = x2 with focus S If A = (1, 2), then the minimum value of (PS + PA) is (a) (b) (c) (d) 4 A five-digit number is formed with the digits 0, 1, 2, 3, 4, without repetition The probability that it is divisible by 11 is 2 (a) (b) 50 25 3 (c) (d) 50 25 10 m , reduced fraction, n r =1 then the sum of digits of (m + n) is (a) (b) 10 (c) 11 (d) 13 If ∑ (2r − 1)(2r + 1)(2r + 3) = If the equation x2 + ax + 6a = has integer roots, then the number of values of a is (a) (b) (c) (d) 10 c +i If z = a + ib, a, b ∈ R, b ≠ and |z| = 1, then z = , c −i where c = a a −1 (a) (b) b b a +1 a +1 (c) (d) b +1 b JEE ADVANCED tan3θ + cot3θ = 12 + 8cosec32θ if θ = 11π 7π (a) (b) 12 12 23π 19 π (c) (d) 12 12 66 MATHEMATICS TODAY | SEPTEMBER ‘16 COMPREHENSION dy = y(0) = Then Let dx x + y , dy At y = ln 3, = dx (a) (b) –1 (c) 1/2 (d) –2 ∫ xdy = e (d) e − 2 INTEGER MATCH Let a1, a2, a3, , a10 be a permutation of the set {1, 2, 3, ., 10} such that the sequence decreases first and then increases like 8, 6, 4, 1, 2, 3, 5, 7, 9, 10 If N is the number of such permutations, then the sum of digits of N is (a) (b) (c) e − MATRIX MATCH 10 Match the following columns Column I (P) If dice are rolled once, the number of ways of getting the sum 10 is (Q) If a, b, c, d are odd positive integers, the number of solutions of the equation a + b + c + d = 16 is (R) The coefficient of x2y in the expansion of (1 + x + 2y)5 is (S) 10 r ⋅ Cr ⎛10 ⎞ If Cr = ⎜ ⎟ , then ∑ is r ⎝ ⎠ C r =1 (r −1) (a) (b) (c) (d) P 4 Q 3 R 3 Column II 84 55 60 80 S 4 See Solution Set of Maths Musing 164 on page no 85     Math Archives, as the title itself suggests, is a collection of various challenging problems related to the topics of JEE (Main and Advanced) and other PETs This section is basically aimed at providing an extra insight and knowledge to the candidates preparing for competitions In every issue of MT, challenging problems are offered with detailed solutions The readers’ comments and suggestions regarding the problems and solutions offered are always welcome π⎞ ⎛ Let p = ⎜1 + cos ⎟ ⎝ 10 ⎠ 3π ⎞ ⎛ ⎜⎝1 + cos 10 ⎟⎠ 7π⎞ ⎛ 9π⎞ ⎛ ⎜⎝1 + cos 10 ⎟⎠ ⎜⎝1 + cos 10 ⎟⎠ 5π ⎞ 3π ⎞ ⎛ π⎞⎛ ⎛ + cos and q = ⎜1 + cos ⎟ ⎜1 + cos ⎟ ⎜ ⎝ ⎠ ⎟⎠ ⎠⎝ ⎝ 7π⎞ ⎛ ⎜⎝1 + cos ⎟⎠ then (b) 2p = q (d) p + q = 1/4 (a) p = q (c) p = 2q Let 2x – 3y = be a given line and P(sinθ, 0) and Q(0, cos θ) be two points Then P and Q lie on the same side of the given line if θ lies in the (a) 1st quadrant (b) 2nd quadrant rd (c) quadrant (d) 4th quadrant If sin x cos 3x = n ∑ ak cos kx Then the value of n is k=0 (a) (c) Let f : R → R be a function defined by, f (x) = (a) (b) (c) (d) (b) (d) none of these 2x − x + then f is x + x + 10 injective but not surjective surjective but not injective injective as well as surjective neither injective nor surjective If α and β are two distinct roots of the equation a tan x + b sec x = c, then tan (α + β) is equal to (a) (c) ac a +c 2ac a2 − c (b) 2ac a + c2 (d) none of these π⎞ π⎞ ⎛ 2⎛ If g ( θ ) = sin θ + sin ⎜ θ + ⎟ + cos θ.cos ⎜ θ + ⎟ ⎝ ⎝ 3⎠ 3⎠ ⎛5⎞ and f ⎜ ⎟ = then find (fog)(x) ⎝4⎠ If {x} and [x] denote the fractional and integral parts of a real number x respectively, then solve 2x + {x + 1} = 4[x + 1] – Find the domain of the following functions (a) cos−1 x where [·] denotes the greatest integer [x] function −1 1 + 2sin x + x x−2 − cos x cos x Evaluate Lt x→ x2 L’ Hospital Rule (b) p ⎧ sin x ⎪ ⎪(1 + sin x ) ⎪ ⎪ q 10 If f ( x ) = ⎨ ⎪ tan x ⎪ ⎪ e tan 5x ⎪⎩ is continuous at x = Find the ;− without using π