26 32 Vol XXXVII No 73 18 March 2019 Corporate Office: Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-6601200 e-mail : info@mtg.in website : www.mtg.in Regd Office: 406, Taj Apartment, Near Safdarjung Hospital, Ring Road, New Delhi - 110029 Managing Editor : Mahabir Singh Editor : Anil Ahlawat 50 46 47 10 CONTENTS Class XII Class XI Competition Edge 84 Maths Musing Problem Set - 195 10 JEE Main Solved Paper 2019 18 JEE Work Outs 26 Practice Paper - JEE Advanced 32 Mock Test Paper JEE Main 2019 82 42 Subscribe online at www.mtg.in Individual Subscription Rates (Series 8) 40 Math Archives 42 Challenging Problems 48 You Ask We Answer 49 Maths Musing Solutions 50 Practice Paper - BITSAT 70 Olympiad Corner 46 Concept Map 84 MPP 47 Concept Map 73 CBSE Drill Practice Paper 82 MPP 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 | MARCH ‘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 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 Set 195 JEE MAIN COMPREHENSION x f(x) Domain of the function f(x), if + = minimum of f(t), where f(t) = minimum of {2t3 – 15t2 + 36t – 25, + |sin t| ; ≤ t ≤ 4} is (a) (– ∞, 1) (b) (– ∞, log3 e) (c) (0, log3 2) (d) (– ∞, log3 2) If in a triangle ABC, a = 5, b = and cos(A – B) 31 = , then the third side c is equal to 32 (a) (b) (c) (d) The values of x for which the angle between the vectors x i + x j + k and i − j + xk are obtuse and the angle between the z-axis and i − j + xk π is given by is acute and less than 1 (a) < x < (b) > or x < 2 (c) < x < 15 (d) there is no such value for x A straight line touches the rectangular hyperbola 9x2 – 9y2 = and the parabola y2 = 32x An equation of the line is (a) 9x + 3y – = (b) 9x – 3y + = (c) 9x + 3y + = (d) 9x – 3y – = π/ ∫0 sin x d(x − [x]) is equal to (where [⋅] denotes the greatest integer function) (a) 1/2 (b) − (c) (d) none of these JEE ADVANCED If θ = tan −1(2 tan2 θ) − tan −1 tan θ , then tanq 3 is equal to (a) –2 (b) (c) (d) MATHEMATICS TODAY | MARCH ‘19 A is a set containing 10 elements A subset P of A is chosen at random and the set A is reconstructed by replacing the elements of P Another subset Q of A is now chosen at random Then, the probability that P ∪ Q = A, is 10 10 10 10 2 1 3 2 (a) (b) (c) (d) 3 2 4 5 P ∩ Q contains atleast elements, is 31 15 30 41 (a) (b) 20 (c) (d) 20 20 2 220 NUMERICAL ANSWER TYPE z and z are two complex numbers such that z1 − 2z2 is unimodular, while z2 is not unimodular, − z1z2 then |z1| = MATRIX MATCH 10 In the following [x] denotes the greatest integer less than or equal to x Column-I (P) (Q) x|x| Column-II (1) continuous in (–1, 1) (2) differentiable in (–1, 1) |x| (R) x + [x] (3) strictly increasing in (–1, 1) (S) |x – 1| + |x + 1| (4) not differentiable at atleast one point in (–1, 1) (a) (b) (c) (d) P 2, 1, 1, 2, 2, Q 3, 3, 1, 3, R S 1, 2, 1, 3, 2, 3, 3, 1, 1, 3, 2, See Solution Set of Maths Musing 194 on page no 49 Held on 12th January (Evening Shift) Let S and S be the foci of an ellipse and B be any one of the extremities of its minor axis If S BS is a right angled triangle with right angle at B and area ( S BS) = sq units, then the length of a latus rectum of the ellipse is (d) 2 (a) (b) (c) 2 The expression ~(~p q) is logically equivalent to (a) p q (b) ~p q (c) ~p ~q (d) p ~q If the sum of the first 15 terms of the series 3 3 3 1 1 3 + + + 33 + + 4 is equal to 225 k, then k is equal to (a) 27 (b) (c) 108 (d) 54 [0, If sin α + cos β + = sin cos ; , ], then cos( + ) – cos( – ) is equal to (d) − (a) –1 (b) (c) In a game, a man wins Rs 100 if he gets or on a throw of a fair die and lose Rs 50 for getting any other number on the die If he decides to throw the die either till he gets a five or a six or to a maximum of three throws, then his expected gain/loss (in rupees) is 400 400 (b) (a) loss gain 400 (c) (d) loss Let S be the set of all real values of such that a plane passing through the points (– 2, 1, 1), (1, – 2, 1) and (1, 1, – 2) also passes through the point (–1, –1, 1) Then S is equal to (d) { , − } (a) {1, –1} (b) {3, –3} (c) { } The tangent to the curve y = x2 – 5x + 5, parallel to the line 2y = 4x + 1, also passes through the point 10 MATHEMATICS TODAY | MARCH ‘19 (a) , 2 4 (b) , 4 2 (c) − , 1 (d) , − 8 7 The total number of irrational terms in the binomial expansion of (71/5 – 31/10)60 is (a) 55 (b) 54 (c) 48 (d) 49 If a straight line passing through the point P(–3, 4) is such that its intercepted portion between the coordinate axes is bisected at P, then its equation is (a) 4x + 3y = (b) 4x – 3y + 24 = (c) 3x – 4y + 25 = (d) x – y + = 10 (a) lim π − sin−1 x x →1− π (b) 1− x 2π is equal to (c) π (d) π 11 If nC4, nC5 and nC6 are in A.P., then n can be (a) 12 (b) 14 (c) (d) 11 12 Let f be a differentiable function such that f(1) = and f (x) = f(x) for all x R If h(x) = f(f(x)), then h (1) is equal to (b) 4e (c) 2e (d) 4e2 (a) 2e2 13 The set of all values of for which the system of linear equations x – 2y – 2z = x x + 2y + z = y –x – y = z has a non-trivial solution (a) is an empty set (b) contains exactly two elements (c) is a singleton (d) contains more than two elements = lim ( − kh − + kh )( − kh + + kh ) −h[ − kh + + kh ] −2kh 2k = lim = =k (iii) h→0 −h[ − kh + + kh ] + 2(h) + + lim f (x ) = lim f (h) = lim = = −1 (iv) + h →0 h →0 h − −1 x →0 From (i), (ii), (iii) and (iv), for f(x) to be continuous at x = 0, we have k = –1 19 We have x2 + y2 = (i) y a circle with centre at O(0, A (1, ) 0) and radius = units and (x – 2)2 + y2 = (ii) x a circle with its centre at x (0, 0)O D C(2, 0) C(2, 0) and radius = units B Points of intersection y (1, − ) of the two circles are A(1, ) and B(1, − ) h →0 Required area = Area of shaded region 2 x − x2 x +2 + sin −1 2 1 − 3 π π π π = 2 + − − − + 2 ⋅ − − × 6 2 2 C cm D 10 cm r B A h 1 1 Now , V = πr 2h = π × h × h = πh3 3 12 dV dV dh d dh πh2 dh = × = πh ⋅ = ⋅ dt dt dh dt dh 12 dt MATHEMATICS TODAY | MARCH ‘19 [ h = (10 – 2.5) cm = 7.5 cm] Hence, the rate of change of water level at h = 7.5 cm is −16 cm / sec 45π a 1 −1 21 Given, A = ,B= 2 −1 b −1 Now, (A + B)2 = (A + B) (A + B) = AA + AB + BA + BB (A + B)2 = A2 + AB + BA + B2 AB + BA = O (i) [ (A + B)2 = A2 + B2] 1 −1 a a − b 2 AB = = 2 −1 b −1 2a − b 3 2a − b + − a 0 0 = 2a − − b 0 0 2a – b + = 0; – a = 0, 2a – = a = and b = 22 Let sin −1 = A and sin −1 = B Then, 13 −π π A, B ∈ , cos A > and cos B > 2 ∴ sin A = and sin B = 13 ⇒ cos A = − sin2 A = − 16 = = 25 25 and cos B = − sin2 B = − 25 144 12 = = 169 169 13 O From similar s OAB and OCD, we have AB CD r 1 = ⇒ = = ⇒ r = h OA OC h 10 2 78 dh ⇒ dt h=7.5cm dh −20 = dt πh2 −16 −20 = cm / sec = 45π π × (7.5) ⇒ a − b 2 a + −a − 1 0 0 From (i), 2a − b 3 + b − −b + 1 = 0 0 (x − 2) − (x − 2)2 (x − 2) =2 + sin −1 2 0 ∴ πh2 dh ⋅ dt a 1 −1 a + −a − 1 BA = = b −1 2 −1 b − −b + 1 = ∫ − (x − 2)2 dx + ∫ − x dx 4π = − sq units 20 Let r be the radius of the conical funnel, h the height of the conical funnel and V the volume of the water in the conical funnel at any instant t dV Then, = −5 (given) (i) dt ⇒ −5= 5 ∴ cos sin −1 + sin −1 = cos( A + B) 13 = cos A cos B – sin A sin B 12 48 15 33 = × − × = − = 13 13 65 65 65 23 Let A, B and C denotes the events of choosing bag I, II and III respectively and E denotes the event that one white and one red ball is drawn Now, P(A) = P(B) = P(C) = 1/3 Also, P (E / A) = C1 × 3C1 C2 1× × = = 6×5 ∫ = sin x dx (sin x + cos x ) ∫ = π /2 ∫ = (sin2 x + cos2 x ) dx = (sin x + cos x ) .(i) π /2 ∫ dx sin x + cos x tan(x / 2) − tan2 (x / 2) + 2 1 + tan (x / 2) + tan (x / 2) =∫ + 2t ) , where tan x =t = 2∫ = ⋅ log 2 2 ( ) − (t − 1) dt + (t − 1) − (t − 1) ( + 1) ( + 1) +1 = × log log −1 2 ( − 1) ( + 1) = log( + 1)2 = log( + 1) 1 ∴ I= log( + 1) = log( + 1) 2 = π /2 ∫ cos2 x dx (1 + sin x cos x ) .(ii) π /2 ∫ (sin2 x + cos2 x ) dx = (1 + sin x cos x ) π /2 ∫ sec2 x (1 + tan x + tan x ) ∞ dx = ∞ π /2 ∫ dx (1 + sin x cos x ) dt ∫ (t + t + 1) , where tan x = t ∞ 2t + dt =∫ = tan −1 2 1 + t+ π π 2π ⋅ − = 2 6 3 π Hence, I = 3 = .(1) Taking r = xi + y j + zk, we get (xi + y j + zk ) ⋅[(1 + λ)i + (1 + 3λ)j + (1 + λ)k ] = − 5λ sec (x / 2) (1 − t sin2[(π / 2) − x] dx = + sin[(π/2) − x]cos[(π/2) − x] ⇒ r ⋅[(1 + λ)i + (1 + 3λ) j + (1 + λ)k] = − 5λ 2dt .(i) r ⋅[i + j + k + λ(2i + j + 4k)] = − 5λ ∫ [1 − tan2 (x / 2) + tan(x / 2)] dx sin x dx (1 + sin x cos x ) 25 Here, n1 = i + j + k and n2 = 2i + j + 4k and d1 = and d2 = –5 Hence, using the relation r ⋅ (n1 + λn2 ) = d1 + λd2, we get dx π /2 OR sec2 x = ∫ dx (sec x + tan x ) π π/2 sin2 − x 2 cos2 x I= ∫ dx = ∫ dx (sin x + cos x ) π π sin − x + cos − x 2 (ii) Adding (i) and (ii), we get π /2 2I = π /2 π/2 2I = ∫ Adding (i) and (ii), we get 2 × 45 9 = = × = = 1 1 1 34 17 × + × + × + + 3 9 24 Let I = π /2 P (C )P (E /C ) P ( A)P (E /A) + P (B)P (E /B) + P (C )P (E /C ) π /2 ∫ I= P(E/B) = 1/3, P(E/C) = 2/9 P(C/E) = Let I = π /2 (1 + )x + (1 + )y + (1 + )z = – (x + y + z – 6) + (2x + 3y + 4z + 5) = (2) Given that, the plane passes through the point (1, 1, 1) So the point (1, 1, 1) must satisfy equation (2) i.e., (1 + + – 6) + (2 + + + 5) = ⇒ λ = 14 Putting the value of in equation (1), we get 6 15 r ⋅ 1+ i + 1+ j + 1+ k = − 14 14 or r ⋅ 10 23 13 69 i+ j+ k = 7 14 14 or r ⋅ (20i + 23 j + 26k ) = 69 which is the required vector equation of the plane MATHEMATICS TODAY | MARCH ‘19 79 OR The equation of the given plane is 4x + 12y – 3z + = .(i) Let P(–2, 3, –4) be the given point Let l be given line whose equation is x + 2 y + 3z + (ii) = = = λ(say ) λ − 5λ − , The general point on this line is 3λ − 2, For some value of , let the point λ − 5λ − Q 3λ − 2, , lie on the line (ii) such that PQ is parallel to the given plane (i) λ − 5λ + , D.r.’s of PQ are 3λ, D.r’s of the normal to the given plane are 4, 12, –3 Now, PQ is parallel to the given plane (i) PQ is perpendicular to the normal to the plane (i) (4 λ − 9) (5λ + 8) (4 × ) + 12 × −3× =0 12 + (24 – 54) – (5 + 8) = = Coordinates of Q are 4, , 5 ⇒ PQ = (4 + 2) + − + (2 + 4)2 = 8.5 units 2 26 The equation of the family of circles in the first quadrant which touch the coordinate axes is (i) (x – a)2 + (y – a)2 = a2, Y where a is a parameter Differentiating (i) with respect a (a, a) to x, we get a dy X X 2(x − a) + 2( y − a) = O dx dy Y ⇒ x − a + ( y − a) = dx dy x+y dx ⇒ a = x + py , where p = dy ⇒ a= dy 1+ p dx 1+ dx Substituting the value of a in (i), we get 2 x + py x + py x + py x − + p + y − + p = + p (xp – py)2 + (y – x)2 = (x + py)2 (x – y)2 (p2 + 1) = (x + py)2 80 MATHEMATICS TODAY | MARCH ‘19 2 dy dy (x − y ) + = x + y , which is the dx dx required differential equation OR dy dy = 1 + x We have, y − x dx dx ⇒ y − = x(2 x + 1) 1 dy dx 1 ∫ x(2x + 1) dx = ∫ y − dy ∫ x − (2x + 1) dx = ∫ y − dy log |x| – log| 2x + 1| = log |y – 2| + log C ⇒ log x − log | y − | = log C 2x + x x × = log C ⇒ =C 2x + y − (2 x + 1)( y − 2) (i) Putting x = and y = in (i), we get C = 1/3 x 1 = Putting C = in (i), we get ( x + )( y − ) 3 x ⇒ =± (2 x + 1)( y − 2) 3x 3x ⇒ y −2= ± ⇒ y =2± 2x + 2x + 3x But, y = + is not satisfied by y(1) = 2x + 3x Hence, y = − , where x ≠ − is the required 2x + solution ⇒ log 27 Let company makes x belts of type A and y belts of type B per day Now, 1000 belts of type B can be made in day 500 belts of type A can be made in day Thus, time taken to make x belts of type A and y belts of y x + days type B = 500 1000 y x ∴ + ≤ i.e., x + y ≤ 1000 500 1000 Maximize, Z = 4x + 3y Subject to x 0, y 0, x ≤ 400, y ≤ 700, x + y ≤ 800 and 2x + y ≤ 1000 The shaded region represents the feasible region, where O(0, 0), Q(400, 0), R(400, 200), S(200, 600), T(100, 700), U(0, 700) y Thus, P is maximum when x = a/ Putting x = a/ in (i), we get y = a/ (100, 700) x = 400 T y = 700 (0, 700) U (200, 600) S x+ y= 80 R (400, 200) x O y (400, 0)Q x 2x + y = 1000 Now, maximize Z = 4x + 3y At O(0, 0), Z = 4(0) + 3(0) = At Q (400, 0), Z = 4(400) + 3(0) = 1600 At R(400, 200), Z = 4(400) + 3(200) = 2200 At S(200, 600), Z = 4(200) + 3(600) = 2600 At T(100, 700), Z = 4(100) + 3(700) = 2500 At U(0, 700), Z = 4(0) + 3(700) = 2100 Z is maximum when x = 200 and y = 600 Thus, the company should make 200 belts of type A and 600 belts of type B to have a maximum profit 28 Let ABCD be a rectangle in a given circle of radius a with centre at O Let AB = 2x and AD = 2y be the sides of the rectangle In OAM AM2 + OM2 = OA2 D C x2 + y2 = a2 O y = a2 − x (i) Let P be the perimeter of the rectangle ABCD A x y M 2x 2y B P = 4x + 4y ⇒ P = x + a2 − x dP 4x ⇒ =4− dx a − x2 dP The critical points of P are given by = dx 4x dP =0 Put =0 ⇒ 4− dx a − x2 a2 − x = x ⇒ x = a2 ⇒ x = a / x (− x ) −4 a2 − x − d P −4a2 a2 − x Now, = = dx (a2 − x )3/2 a2 − x d2P −8 −4a2 = for all real x, then (a) f[g(x + 1)] > f[g(x + 5)] (b) f[g(x)] < f[g(x + 1)] (c) g[f(x)] < g[f(x + 2)] (d) Both (a) and (c) Let a, b, c be non-zero real numbers such that 0 8 ∫ (1 + cos x)(ax + bx + c)dx = ∫ (1 + cos x) × (ax + bx + c)dx Then the quadratic equation ax + bx + c = has (a) no root in (0, 2) (b) atleast one root in (1, 2) (c) double root in (0, 2) (d) two imaginary roots x ∫ x sin x + cos x dx = x sec x +c (a) cot x + x sin x + cos x x sec x +c (b) tan x + x sin x + cos x x sec x +c (c) cot x − x sin x + cos x x sec x +c (d) tan x − x sin x + cos x n 2 −1 If n is a natural number, then = 3 −2 82 MATHEMATICS TODAY | MARCH ‘19 1 (a) 0 1 (b) 0 0 if n is even 1 0 if n is odd 1 −1 0 (c) if n is a natural number 1 (d) none of these The centroid of the triangle formed by (0, 0, 0) and x −1 y −1 z −1 the point of intersection of = = with x = and y = is (a) (1, 1, 1) (b) (1/6, –1/3, 1/6) (c) (–1/6, 1/3, –1/6) (d) (1/3, 1/3, 1/3) One or More Than One Option(s) Correct Type The system of equations –2x + y + z = a; x – 2y + z = b; x + y – 2z = c has (a) no solution if a + b + c (b) unique solution if a + b + c = (c) infinite number of solutions if a + b + c = (d) none of these The angle formed by the positive y-axis and the −3 is tangent to y = x2 + 4x – 17 at , π −1 (b) − tan (9) (a) tan–1(9) π (d) none of these (c) + tan −1 (9) If f (x ) = lim (sin x )2n , then f is (a) (b) (c) (d) n→∞ continuous at x = discontinuous at x = discontinuous at x = – discontinuous at an infinite number of points 10 If A and B are two independent events such that 1 P(A) = , P(B) = , then A A (a) P = = (b) P B A ∪ B A = (d) none of these (c) P A ∪ B 10 dy ax + h = 11 The solution of represents a parabola if dx by + k (a) a = – 2, b = (b) a = – 2, b = (c) a = 0, b = (d) a = 0, b = 12 Let f(x) be a non-constant twice differentiable function defined on (–∞, ∞) such that f(x) = f(1 – x) 1 and f ′ = 0, then 4 (a) f′(x) vanishes at least twice on [0, 1] 1 (b) f ′ = 2 1/2 1 f x + sin x dx = (c) ∫ −1/2 2 (d) 1/2 ∫0 f (t )e sin πt dt = ∫ 1/2 Matrix Match Type 16 Match the following Column-I x P Q Let f (x) = sin (sin x) + cos (cos x), ≤ x ≤ 2π and g(x) = x2, ≤ x ≤ 2π 14 f (x) increases in π 3π (a) (0, π) (b) (π, 2π) (c) 0, (d) π, 2 2 15 f (g(x)) decreases in π π (b) , π (a) 0, 2 2 3π (d) , 2π = (a) (b) (c) (d) P 1 2 3 4 + y2 = and sec x − cos x dx = Q R S 4 Numerical Answer Type 17 The value of 3a2 if the curves Comprehension Type 3π (c) π, 2 ∫ π π − x sin x e sin x d F (x ) = ,x>0 dx x 2 sin x dx = F(k) – F(1), If ∫1 e x then 2k = S The function y defined by the equation xy – log y = satisfies x(yy′′ + y′2) – y′′ + kyy′ = The value of k is f (1 − t )e sin πt dt –1 x →0 sec t dt Column-II 1 R Let 13 The vectors ai + 2a j − 3ak ,(2a + 1)i + (2a + 3)j + (a + 1) k and (3a + 5)i + (a + 5)j + (a + 2)k are non-coplanar for a in (a) {0} (b) [0, ∞) (c) (–∞, 1) (d) (1, ∞) –1 ∫ lim x2 a2 y = 16x intersect orthogonally is 18 If a = (0, 1, – 1) and c = (1, 1, 1) are given vectors, then | b |2 , where b satisfies a × b + c = and a ⋅ b = is 19 The area bounded by the curves x = y2 and x = – 2y2 is 20 A fair coin is tossed n times If the probability that head occurs six times is equal to the probability that head occurs eight times, then the value of n is Keys are published in this issue Search now! J 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 | MARCH ‘19 83 MPP T Class XI 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 Total Marks : 80 Time Taken : 60 Min Only One Option Correct Type The mirror image of the directrix of the parabola y2 = 4(x + 1) in the line mirror x + 2y = is (a) x = –2 (b) 4y – 3x = 16 (c) x – 3y = (d) x + y = If in a ABC, cosA + cosB + cosC = 2, then a, b, c are in (a) A.P (b) G.P (c) H.P (d) none of these A multiple choice examination has questions Each question has three alternative answers of which exactly one is correct The probability that a student will get or more correct answers just by guessing is 13 10 11 (a) 17 (b) (c) (d) 5 35 3 If z is a complex number such that |z| 2, then the minimum value of z + (a) is equal to (b) lies in the interval (1, 2) (c) is strictly greater than (d) is strictly greater than but less than 2 Let and be the roots of equation px2 + qx + r = 0, p If p, q and r are in A.P and 1 + = 4, then the value of | – | is α β (a) 84 61 17 (b) MATHEMATICS TODAY (c) 34 | MARCH ‘19 13 (d) The eccentric angles of the extremities of latus2 y2 rectum to the ellipse x + = are given by a b2 ae (a) tan −1 ± b −1 be (b) tan ± a b (c) tan −1 ± ae a (d) tan −1 ± be One or More Than One Option(s) Correct Type If C0, C1, C2, , Cn are coefficients in the binomial expansion of (1 + x)n, then C0C2 + C1C3 + C2C4 + + Cn – 2Cn is equal to 2n ! 2n ! (a) (b) (n − 2) !(n + 2) ! {(n − 2) !}2 2n ! (c) (d) 2nCn –2 {(n + 2) !}2 EXAM ALERT 2019 Exam Date JEE Main II to 20th April VITEEE 10th to 21st April SRMJEEE 15th to 25th April UPSEE 21st April KEAM 22nd & 23rd April Karnataka CET 23rd & 24th April AMU (Engg.) 28th April COMEDK (Engg.) th 12th May th BITSAT 16 to 26th May JEE Advanced 19th May WB JEE 26th May If ex + e f(x) = e, then for f(x) (a) domain = (–∞, 1) (b) range = (–∞, 1) (c) domain = (–∞, 0] (d) range = (–∞, 1] If 1, , 2, ., then (2 – ) (2 – n–1 are the n, nth roots of unity, ) (2 – n – 1) equals (a) 2n – (b) nC1 + nC2 + + nCn (c) [2nC0 + 2n + 1C1 + 2n +1C2 + + 2n + 1Cn]1/2 – (d) 2n+1 10 The pth term Tp of H.P is q(p + q) and qth term Tq is p(p + q) when p > 1, q > 1, then (a) Tp + q = pq (b) Tpq = p + q (c) Tp + q > Tpq (d) Tpq > Tp + q If we isolate the power of each prime contained in any number N, then N can be written as N = · · · , where i are whole numbers 14 The last non zero digit in 20 ! must be equal to (a) (b) (c) (d) 15 The number of prime numbers among the numbers 105 ! + 2, 105 ! + 3, 105 ! + 4, , 105! + 104 and 105 ! + 105 is (a) 31 (b) 32 (b) 33 (d) none of these Matrix Match Type 16 A root of the equation on L.H.S satisfies R.H.S : Column-I Column-II P 7cos2x + sinx cosx –3 = tan x = 2π , then the value of 2 cos A + cos B – cosA · cosB is equal to 3 (b) (c) (d) (a) 4 11 In triangle ABC, ∠C = 2n + 12 If n is a positive integer and (3 + 5) where is an integer and < < 1, then (a) is an even integer (b) ( + )2 is divisible by 22n + 2n + (c) the integer just below (3 + 5) by (d) is divisible by 10 13 Which of the following is/are correct (a) 10150 – 9950 > 10050 (b) 10150 – 10050 > 9950 (c) (1000)1000 > (1001)999 (d) (1001)999 > (1000)1000 Comprehension Type = π Q sin2 cos2 x 2 tanx = –1 6sec2x – 11tanx – = tan x = S 2cos2x – sin2x = 2sin2x cos x = = – cos( sin2x) R + , is divisible (a) (b) (c) (d) P 1, 1, 2, 3, Q R S 2, 1, 3, 3, 3, 1, 3, 2, 3, 3, 1, 3, 2, 1, 1, 2, 2, Numerical Answer Type 17 If (10)9 + 2(11)1 (10)8 + 3(11)2 (10)7 + + 10(11)9 = k(10)9, then k is equal to 1/ x Let p be a prime number and n be a positive integer, then exponent of p in n ! is denoted by Ep (n !) and is given by n n n n E p (n !) = + + + + k p p2 p3 p where pk ≤ n < pk + and [x] denotes the integral part of x (1 + x )1/ x 18 The value of 786 e lim e x→0 must be 19 The least value of cosec2x + 25sec2x is cos A cos B cos C 20 If and the side a = 2, then = = a b c area of triangle is k The value of 2k2 is Keys are published in this issue Search now! J 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 | MARCH ‘19 85 86 MATHEMATICS TODAY | MARCH ‘19 Registered R.N.I No 40700/1983 Published on 1st of every month Postal Regd No DL-SW-01/4045/18-20 Lic No U(SW)-29/2018-20 to post without prepayment of postage N.D.P.S.O.N.D-1 on 2-3rd same month ... rights reserved Reproduction in any form is prohibited MATHEMATICS TODAY | MARCH ‘19 M aths Musing was started in January 2003 issue of Mathematics Today The aim of Maths Musing is to augment the... an annual showcase of collaboration between Microsoft and academia 30 MATHEMATICS TODAY | MARCH ‘19 MATHEMATICS TODAY | MARCH 19 31 ... sin α + cos β + + (sin4 ⋅ 4cos4 ⋅ ⋅ 1)1/4 14 MATHEMATICS TODAY | MARCH ‘19 B = Probability that outcome is other than or = (c) : Let A = Probability that outcome is or = B (0, b) 45° O S (ae,