70 19 43 35 60 Vol. XXXV No. 9 September 2017 Corporate Office: 69 Plot 99, Sector 44 Institutional Area, Gurgaon -122 003 (HR), Tel : 0124-6601200 e-mail : [email protected] website : www.mtg.in Regd. Office: 406, Taj Apartment, Near Safdarjung Hospital, 41 67 Ring Road, New Delhi - 110029. 8 Managing Editor : Mahabir Singh Editor : Anil Ahlawat 72 83 46 CONTENTS 78 Class XI 8 Brain @ Work Subscribe online at www.mtg.in 19 Concept Boosters 35 Ace Your Way (Series 5) Individual Subscription Rates 41 MPP-5 Class XII 46 Concept Map Mathematics Today 1 yr. 2 yrs. 3 yrs. 43 Concept Boosters Chemistry Today 330 600 775 60 Ace Your Way (Series 5) Physics For You 330 600 775 67 MPP-5 Biology Today 330 600 775 69 Maths Musing Problem Set - 177 330 600 775 70 JEE Work Outs 72 Mock Test Paper - JEE Main 2018 Combined Subscription Rates Competition Edge (Series 3) PCM 1 yr. 2 yrs. 3 yrs. PCB 900 1500 1900 78 Challenging Problems PCMB 900 1500 1900 83 Math Archives 1000 1800 2300 85 Maths Musing Solutions 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 MTG Learning Media Pvt. Ltd. 406, Taj Apartment, New Delhi - 29 and printed by HT Media Ltd., B-2, Sector-63, Noida, UP-201307. 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. Editor : Anil Ahlawat Copyright© MTG Learning Media (P) Ltd. All rights reserved. Reproduction in any form is prohibited. MATHEMATICS TODAY | SEPTEMBER‘17 7
TWO DIMENSIONAL GEOMETRY 1. e number of points on the line 3x + 4y = 5, which 7. Equation of straight line ax + by + c = 0, where are at a distance of sec2q + 2 cosec2q, q ∈ R, from 3a + 4b + c = 0, which is at least distance from the point (1, 3), is (1, –2) is (a) 1 (b) 4 (a) 3x + y – 17 = 0 (b) 4x + 3y – 24 = 0 (c) 3 (d) none of these (c) 3x + 4y – 25 = 0 (d) x + 3y – 15 = 0 2. Let ax + by + c = 0 be a variable straight line, where 8. e curve y = 1 x2 − 2x + 27 where [x] is the a, b and c are 1st, 3rd and 7th terms of an increasing 6 A.P. respectively. en the variable straight line always passes through a xed point which lies on greatest integer less than or equal to x, 6 < x < 9 represents (a) y2 = 4x (b) x2 + y2 = 5 (a) a parabola (b) part of the parabola (c) 3x + 4y = 9 (d) x2 + y2 = 13 (c) a straight line (d) two straight line segments 3. e vertices of a triangle are A(–1, –7), B(5, 1) and 9. Equation of circle through intersection of C(1, 4). e equation of the bisector BD of ∠ABC is x2 + y2 + 2x = 0 and x – y = 0, having minimum (a) x + 7y + 2 = 0 (b) x – 7y + 2 = 0 radius is (c) x – 7y – 2 = 0 (d) x + 7y – 2 = 0 (a) x2 + y2 – 1 = 0 (b) x2 + y2 – x – y = 0 4. e equation of the base of an equilateral triangle (c) x2 + y2 – 2x – 2y = 0 ABC is x + y = 2 and the vertex is (2, –1). e area (d) none of these of the triangle ABC is (sq. units) 10. ecommonchordofthecirclex2+y2+6x+8y–7=0 2 3 3 and a circle passing through the origin, and (a) 6 (b) 6 (c) 8 (d) none of these touching the line y = x, always passes through the 5. e line 3x – 4y + 7 = 0 is rotated through an angle point p/4 in the clockwise direction about the point (–1, 1). (a) (– 1/2, 1/2) (b) (1, 1) (c) (1/2, 1/2) (d) none of these e equation of the line in its new position is (a) 7y + x – 6 = 0 (b) 7y – x – 6 = 0 (c) 7y + x + 6 = 0 (d) 7y – x + 6 = 0 11. ree equal circles each of radius r touch one another. e radius of the circle touching all the 6. e equation of a line through the point (1, 2) three given circles internally is whose distance from the point (3, 1) has the least (a) (2 + 3)r (b) (2 + 3 3) r possible value is (c) (2 − 3 3) r (d) (2 − 3) r (a) x + 2y = 3 (b) y = 2x (c) y = x + 1 (d) x + 2y = 5 Sanjay Singh Mathematics Classes, Chandigarh, Ph : 9888228231, 9216338231 8 MATHEMATICS TODAY | SEPTEMBER ‘17
12. e set of points P(x, y) such that their distance (a) x2 + y2 – 3x + 6y + 3 = 0 (b) x2 + y2 – 3x + 6y – 3 = 0 from (3, 0) is 2 times their distance from (0, 2) (c) x2 + y2 + 3x + 6y – 3 = 0 form a circle (d) x2 + y2 – 3x – 6y – 3 = 0 (a) whose radius is greater than 5 units (b) whose radius is equal to 5 units 20. AB is a chord of the parabola y2 = 4ax with vertex (c) whose radius is less than 5 units at A. BC is drawn perpendicular to AB meeting the (d) none of these axis at C. e projection of BC on the x-axis is 13. If the circles ax2 + ay2 + 2bx + 2cy = 0 and (a) a (b) 2a (c) 4a (d) 8a Ax2 + Ay2 + 2Bx + 2Cy = 0 touch each other, then (a) bC = cB (b) aC = cA 21. If normal at point P on parabola y2 = 4ax, (a > 0) meet it again at Q in such a way that OQ is (c) aB = bA (d) a = b = c of minimum length where O is vertex of parabola, ABC then DOPQ is 14. Let f(x , y) = 0 be the equation of a circle. If (a) a right angled triangle f(0, l) = 0 has equal roots l = 2, 2 and f(l, 0) = 0 (b) obtuse angled triangle (c) acute angled triangle has roots λ= 4 , 5 then the centre of the circle is (d) none of these 5 (a) (2, 29/10) (b) (29/10, 2) 22. If two distinct chords drawn from the point (4, 4) (c) (– 2, 29/10) (d) none of these on the parabola y2 = 4ax are bisected on the line y = mx, then the set of value of m is given by 15. e minimum distance between the circle x2 + y2 = 9 and the curve 2x2 + 10y2 + 6xy =1 is (a) 1− 2 , 1 + 2 (b) R 2 2 (a) 2 2 (b) 2 1 11 (c) 3 − 2 (d) 3 − (c) (0, ∞) (d) (–2, 2) 16. If one end of the diameter of a circle which touches 23. yIf2t=1 and t2 be the ends of a focal chord of the parabola 4ax, then the equation t1x2 + ax + t2 = 0 has x-axis is (3, 4) then the locus of other end of the (a) imaginary roots diameter of the circle is/an (a) parabola (b) hyperbola (b) both roots positive (c) one positive and one negative root (c) ellipse (d) circle (d) both roots negative 17. If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes 24. If the normal at three points (ap2, 2ap), (aq2, 2aq) through the points of intersection of the parabolas and (ar2, 2ar) are concurrent then the common y2 = 4ax and x2 = 4ay, then root of equations px2 + qx + r = 0 and a(b – c)x2 + (a) d2 + (3b – 2c)2 = 0 b(c – a)x + c(a – b) = 0 is (b) d2 + (3b + 2c)2 = 0 (a) p (b) q (c) r (d) 1 (c) d2 + (2b – 3c)2 = 0 (d) d2 + (2b + 3c)2 = 0 25. e second degree equation x2 + 4y2 + 2x + 16y +13 = 0 represent itself as 18. e equation of the tangent to the parabola (a) a parabola (b) a pair of straight line y = (x – 3)2 parallel to the chord joining the points (c) an ellipse (d) a hyperbola (3, 0) and (4, 1) is (a) 2x – 2y + 6 = 0 (b) 2y – 2x + 6 = 0 26. A tangent having a slope − 4 to the ellipse (c) 4y – 4x + 13 = 0 (d) 4x + 4y = 13 intersects 3 axes at x2 y2 19. e abscissa and ordinate of the end points A 18 + 32 =1 the major and minor and B of a focal chord of the parabola y2 = 4x are respectively the roots of x2 – 3x + a = 0 and A and B. If O is the origin, then the area of DOAB is y2 + 6y + b = 0. e equation of the circle with AB (a) 48 sq. units (b) 9 sq. units as diameter is (c) 24 sq. units (d) 16 sq. units 10 MATHEMATICS TODAY | SEPTEMBER ‘17
MATHEMATICS TODAY | SEPTEMBER‘17 11
27. Let P be a variable point on the ellipse x2 + y2 =1 (a) x2 + y2 = 25 a2 b2 16 9 with foci S1 and S2. If A be the area of DPS1S2, then (b) 4x2 + 4y2 – 32x –24y + 75 = 0 the maximum value of A (a) ab sinq (b) abe x2 y2 (c) a sinq (d) b sinq (c) 16 − 9 = 25 28. Maximum distance of any point on the circle (d) none of these (x − 7)2 + (y − 2 30)2 = 16 from the centre of the ellipse 25x2 + 16y2 = 400 is 34. Equation of a common tangents to the curves y2 = 8x and xy = –1 is (a) 3y = 9x + 2 (b) y = 2x + 1 1− 3 3 (c) 2y = x + 8 (d) y = x + 2 (a) 2 (b) 2 35. If the foci of the ellipse 16x2 + 7y2 = 112 and of the (c) 3 (d) none of these y2 x2 144 a2 1 If the eccentricity of the ellipse x2 y2 1 hyperbola − = 25 coincide, then a = a2 + a2 + 29. 2 + 5 = (a) 3 (b) 9 (c) 81 (d) 8 be 1 , then length of latus rectum of ellipse is 36. e point of the curve 3x2 – 4y2 = 72 which is 3 nearest to the line 3x + 2y – 1 = 0 is 18 10 (a) (6, 3) (b) (6, –3) (a) 4 (b) 6 (c) 6 (d) 8 (c) (6, 6) (d) (6, 5) 30. e number of rational points on the ellipse 37. Let PQ be a double ordinate of hyperbola x2 y2 9 + 4 =1 is x2 − y2 =1. If O be the centre of the hyperbola and a2 b2 (a) in nite (b) 4 OPQ is an equilateral triangle then the eccentricity e is (c) 0 (d) 2 (a) > 3 (b) > 2 31. e radius of the circle passing through the points of 2 (d) none of these x2 y2 (c) > 3 intersection of ellipse a2 b2 =1 and x2 – y2 = 0 is + 38. If x = 9 is the chord of contact of the hyperbola x2 – y2 = 9, then the equation of the corresponding (a) ab (b) 2 ab pair of tangents is a2 + b2 a2 + b2 (a) 9x2 – 8y2 + 18x – 9 = 0 (c) a2 − b2 (d) a2 + b2 (b) 9x2 – 8y2 – 18x + 9 = 0 a2 + b2 a2 − b2 (c) 9x2 – 8y2 – 18x – 9 = 0 32. e pair of lines joining origin to the intersection of (d) 9x2 – 8y2 + 18x + 9 = 0 x2 y2 the curve a2 + b2 =1 by the line lx + my + n = 0 39. e equation x2 + (y −1)2 − x2 + (y +1)2 =K will represent a hyperbola for are coincident if (a) K ∈ (0, 2) (b) K ∈ (0, ∞) (a) a2l2 + b2m2 = n2 (b) a2 + b2 = 1 (c) K ∈ (1, ∞) (d) none of these l2 m2 n2 l2 m2 40. A(3, 4), B(0, 0) and C(3, 0) are vertices of DABC. a2 b2 If 'P' is a point inside DABC, such that d(P, BC) < (c) + = n2 (d) none of these min. {d(P, AB), d(P, AC)}. en maximum of 33. e auxiliary circle of a family of ellipse passes d(P, BC) is [d(P, BC) represent distance between through origin and makes intercept of 8 and 6 units P and BC] (a) 1 (b) 1/2 on x-axis and y-axis respectively. If eccentricity of (c) 2 (d) none of these all such family is 1/2 then locus of focus will be 12 MATHEMATICS TODAY | SEPTEMBER ‘17
MATHEMATICS TODAY | SEPTEMBER‘17 13
41. If coordinates of the vertices of a triangle are 4. (b) : Let side AB is x (2, 0), (6, 0) and (1, 5) then distance between its | 2 1− 2 | 1 orthocentre and circumcentre is \\ Length AD = − 2 = 2 (a) 4 (b) 6 In DABD, sin60º = 1 60° 60° (c) 5 (d) none of these 2x 42. If the vertices of a triangle are A(1, 4), B(3, 0) and 3 1 2 C(2, 1) then the length of the median passing 2 2x 3 through C is = ⇒ x= (a) 1 (b) 2 (c) 2 (d) 3 \\ Area of an equilateral triangle 43. e equation to the locus of a point which moves = 3 2 = 3 sq. units 4 3 6 so that its distance from x-axis is always one half its distance from the origin is 5. (a) : As (–1, 1) is a point on 3x – 4y + 7 = 0, the rotation is possible. (a) x2 + 3y2 = 0 (b) x2 – 3y2 = 0 3 (c) 3x2 + y2 = 0 (d) 3x2 – y2 = 0 Slope of the given line = 4 −po1s=it−io71n(x=+1341+)−431 44. A ray of light coming from the point (1, 2) is Slope of the line in its new = − 1 re ected at a point A on the x-axis and then passes 7 through the point (5, 3). e coordinates of the e required equation point A are or 7y + x – 6 = 0. is y (a) 13 , 0 (b) 5 , 0 5 13 6. (b) : Line at least distance from (3, 1) points will (c) (–7, 0) (d) none of these be perpendicular to line joining given two points (1, 2) 45. e locus of the point of intersection of the lines and (3, 1). x + 4y = 2a sinq, x – y = a cosq where q is a variable Slope of the line joining the points (1, 2) and (3, 1) is parameter is 2−1 1 (a) 5x2 + 20y2 = a2 (b) 5x2 + 20y2 = 2a2 m = 1−3 = − 2 (c) 5x2 + 20y2 = 3a2 (d) 5x2 + 20y2 = 4a2 Hence the slope of the required line is 2 and its equation is y – 2 = 2(x – 1). 1. (d) : e perpendicular distance of (1, 3) from the 7. (d) : It passes through a xed point (3, 4) line 3x + 4y = 5 is 2 units while, −6 Slope of line joining (3, 4) and (1, – 2) is −2 = 3 sec2q + 2cosec2q ≥ 3 {as sec2q, cosec2q ≥ 1} \\ Slope of required line = –1/3 2. (d) : Let the common di erence of A.P. is d 1 then b = a + 2d and c = a + 6d, so variable straight Equation is y − 4 = − 3 (x − 3) line will be ax + (a + 2d)y + a + 6d = 0 ⇒ a(x + y + 1) + d(2y + 6) = 0 ⇒ x + 3y – 15 = 0 which always passes through (2, –3). 8. (d) : Given, y = 1 x2 − 2x + 27 3. (b) : We have AB = 10, BC = 5. Let 6 ′(x) ≥ 1 2 By bisector property AD 10 2 f (x) = 6 x − 2x + 27 and f 0 in given interval DC = 5 = 1 6 < x < 9. Hence range of f (x) = [f (6), f(9)]. 1 1 For x ∈[6, 9], y take only two integral values. ⇒ Co-ordinates of D are 3 , 3 . Hence equation of BD is 9. (d) : Equation of required circle is x2 + y2 + 2x + l(x – y) = 0 (1/ 3) − 1 y − 1 = (1/ 3) − 5 (x − 5) or x −7y +2 = 0 whose centre of circle is − 2 + λ , λ 2 2 14 MATHEMATICS TODAY | SEPTEMBER ‘17
and radius of circle is 2 + λ 2 + λ2 = (λ +1)2 +1 Also, r2 = 1 2 4 2 2 + 8sin2 θ + 3sin2θ 1 Radius of circle is minimum when l = –1. = 6 + 3sin 2θ − 4 cos 2θ ⇒ rmax = 1 Hence, required equation of circle is x2 + y2 + x + y = 0 Minimum distance between curve = 2. 10. (c): Let circle touching to the line y = x at (0, 0) point is :+x2y2++y26+x l(x – y) = 0 and 16. (a) : Let other end is (h, k) S2 : Sx12 +8 y –7= 0 3 4 So centre ≡ h + , k + touching x-axis means 2 2 \\ Common chord is S1 – S2 = 0 k + 4 ⇒ l(x – y) – 6x – 8y + 7 = 0 r = 2 ⇒ x = y = 1/2 4 3 32 4 2 11. (b) : Q DDEF is an equilateral So k + = h + − + k + − 4 2 2 2 with side 2r if radius of circumcircle DEF is R1, then Hence locus is x2 – 6x – 16y + 9 = 0 Which is a parabola. Area of DDEF = 3 (2r)2 = 3r 2 4 17. (d) : e two parabolas intersect at (0, 0) and (4a, 4a). e equation of their common chord must be 3r2 = 2r ⋅2r ⋅2r ⇒ R1 = 2r y = x which must be same as given line 4R1 3 2bx + 3cy + 4d = 0 \\ Radius of the circle touching all the three given ⇒ 2b = –3c, d = 0 ⇒ (2b + 3c)2 + d2 = 0 circles = r + R1 = r + 2r = (2 + 3)r 18. (c): y′ = 2(x – 3) = 1 gives the point 7 , 1 and the 3 3 2 4 1 7 12. (a) : e prescribed condition gives required tangent is y − 4 = 1 x − 2 (x – 3)2 + y2 = 2{x2 + (y – 2)2} or 4y – 4x + 13 = 0. i.e., x2 + y2 + 6x – 8y – 1 = 0 which is a circle with centre (–3, 4) and radius = 9 +16 +1 = 26 > 5 19. (b) : t1t2 = –1 as AB is focal chord x2 – 3x +a=0; + = 3 and = 13. (a) : Obviously both circles, xy21 + x2 + b =0 ; x1x2 a pass through the origin and 6y therefore O must be the point y1 + y2 = – 6 and y1y2 = b of contact. 1 \\ Centres P, O, Q are collinear. x1x2 = t12 ⋅ t12 = 1 = a − b , − c , (0, 0) and − B , − C are collinear y1y2 = 2t1 − 2 = −4 = b a a A A t1 ∴ c=C or cB = bC \\ a = 1, b = – 4 bB Equation of circle with AB as diameter is x2 + y2 – 3x + 6y – 3 = 0 14. (b) : Given that circle 20. (c): Let B be (at2, 2at) touches y-axis at (0, 2) point Slope of AB = 2 . and intersects the x-axis at t (4/5, 0) and (5, 0). Equation of BC is Centre: 29 2 10 , y − 2at = − t (x − at 2) 2 15. (b) : Let (r cosq, r sinq) be any point on the curve is meets y = 0 at C whose x-coordinate = 4a + at2 2x2 + 10y2 + 6xy =1 then Also, D = (at2, 0) \\ DC = 4a + at2 – at2 = 4a. 2r2 cos2q + 10r2 sin2q + 6r2sinq cosq = 1 MATHEMATICS TODAY | SEPTEMBER‘17 15
21. (a) : \\ Normal at P(t1) meets at Q(t2) Hence the equation of the tangent is 2 1 1 t2 = − t1 − t1 x⋅ 2 y⋅ 2 =1 i.e. x y 1 3 2 4 2 6 8 + + = |t2 | ≥ 2 2 Hence A = (6, 0), B = (0, 8) For minimum length of OQ, |t2| should be minimum Area of DOAB = 1 × 6 × 8 = 24 sq. units i.e. |t2 | = 2 2 2 27. (b) : Let P(a cosq, b sinq) be the variable point on If t2 = −2 2 ⇒ t1 = 2 x2 y2 the ellipse a2 b2 =1. en, Slope of OQ = 2 = m1 and of OP = 2 = m2 + t2 t1 A = Area of DPS1S2, \\ m1m2 = –1 ⇒ DOPQ is right angled triangle. 1 a cos θ b sin θ 1 1 22. (a) : Any point on the line y = mx can be taken as 2 ae 0 1 2 (t, mt). = 0 1 = b sin θ × 2ae = abe sinq Equation of the chord of parabola with this as mid point −ae ymt – 2(x + t) = m2t2 – 4t Area = abe sinq, which is maximum when q = p/2 It passes through (4, 4) \\ Amax = abe m4m2tt2––22((42m+ t) = m2t2 – 4t 28. (d) : Q Centre of ellipse (0, 0) and centre of circle ⇒ + 1)t + 8 =0 have > For two such chords, we must 0 is (7, 2 30) and radius is 4 ⇒ (2m + 1)2 – 8m2 > 0 D \\ Maximum distance of any point on the circle from the centre of the ellipse 1− 2 1 + 2 4m2 – 4m – 1 < 0 ⇒ 2 < m < 2 = (7 − 0)2 + (2 30 − 0)2 + 4 = 17 23. (c): Product of roots < 0 29. (a) : Clearly a2 + 5 > a2 + 2 1 24. (d) : Given points are co-normal points So (a2 + 2) = (a2 + 5) 1 − 3 ⇒ p+q+r=0 Common root is 1. 3a2 + 6 = 2a2 + 10 ⇒ a2 = 4 x2 y2 a h g 10 1 \\ Equation of ellipse is 6 + 9 =1 25. (c): ∆ = h b f = 0 4 8 = −16 =/ 0 2 6 g f c 1 8 13 \\ Length of latus rectum = × = 4 30. 3 ⇒ does not represent a pair of straight line. x2 y2 Again h2 = 0 and ab = 4 ⇒ h2 < ab (a) : 9 + 4 =1 ⇒ An ellipse. x2 y2 Let any point on ellipse be (3 cosq, 2 sinq) 26. (c): Any point on the ellipse (3 2)2 2)2 =1 Since sinq and cosq can be rational for in nite many + (4 value of θ ∈[0, 2π] . can be taken as (3 2 cos θ, 4 2 sin θ) and the slope of 31. (b) : e circle through the points of intersection of the two curves will have centre at origin. the tangent = b2x 32(3 2 cos θ) 4 cot ... (i) x2 y2 − a2 y = − 18(4 = − 3 θ ... (ii) a2 b2 2 sin θ) Solving x2 – y2 = 0 and + =1 , we get 4 = Given slope of the tangent = − 3 x2 = y2 = a2b2 2ab a2 +b2 a2 +b2 From equations (i) and (ii), we get 2a2b2 a2 +b2 cot θ = 1 ⇒ θ = π erefore radius of circle = 4 16 MATHEMATICS TODAY | SEPTEMBER ‘17
32. (a) : lx + my + n = 0 ⇒ lx + my = 1 36. (b) : For the nearest point on the curve, tangent −n drawn to curve at that point should be parallel to the By eory of Homogenization, we can get the pairs of given line 2 line x2 + y2 = lx + my ∴ 6x1 − 8 y1 dy y1) = 0 a2 b2 −n dx (x1, n2 − l2 x2 + n2 − m2 y2 − 2lmxy = 0 ⇒ dy = 3 ⋅ x1 = − 3 a2 b2 dx 4 y1 2 is represent a pair of coincident lines if ⇒ x1 = –2y1 3x2 – 4y2 = 72 ⇒ 12y12 – 4y12 = 72 n2 n2 which satisfy ⇒ x1 =±6 l2m2 a2 l2 b2 m2 0 ⇒ y12 − − − = = 9 ⇒ y1 = ± 3 n4 n2m2 n2l2 Hence required points are (–6, 3) and (6, –3). a2b2 a2 b2 = + ⇒ a2l2 + b2m2 = n2 37. (c): Let P be (a, b) then PQ = 2b and OP = α2 + β2 33. (b) : Centre is (4, 3) and distance of focus from Since OPQ is an equilateral triangle, OP = PQ centre is ae = 5 a2 + b2 = 4b2 = a2 = 3b2 ⇒ α = ± 3β 2 ∵ (a, b) is situated on the hyperbola \\ Locus is (x – 4)2 + (y – 3)2 = 25 4 ∴ α2 − β2 = 1⇒ 3β2 − β2 =1 ⇒ 3 − 1 = 1 > 0 t(xyd+)=:axtE2+qwu2hatte2iroewn4hoaifc=ha tangent at (at2, 2at) to y2 = 8x is a2 b2 a2 b2 a2 b2 β2 34. 8 i.e. a = 2 the curve xy = –1 at b2 ty = intersects a2 > 1 ⇒ e2 − 1 > 1 ⇒ e2 > 4 ⇒ e> 2 ⇒ 3 3 3 3 x(x + 2t2) = the points given by and twill be −1 clearly t =/ 0 38. (b) : x = 9 meets the hyperbola x2 – y2 = 9 at or x2 + 2t2x + t =0 a tangent to the curve (9, 6 2) and (9, − 6 2) . e equations of the tangents if the roots of this quadratic equation are equal, for to the hyperbola at these points are 3x − 2 2y − 3 = 0 which 4t4 – 4t = 0 ⇒ t = 0 or t = 1 and an equation of a common tangent is y = x + 2. and 3x + 2 2y − 3 = 0 (taebh2e)e==:h4aFy2p×oa−re234rtbbh=2oe3=leal1l9i61py4s24e,−⇒x7ax222+e=1y=162 \\ Joint equation of the two tangents is 35. =1 (3x − 2 2y − 3)(3x + 2 2y − 3) = 0 ⇒ (3x − 3)2 − (2 2 y)2 = 0 ⇒ 9x2 – 8y2 – 18x + 9 = 0. \\ 3 For 4 39. (a) : We have, x2 + (y −1)2 − x2 + (y +1)2 = K ... (i) which is equivalent to |S1P – S2P| = const. 25 25 Where S1 ≡ (0, 1), S2 ≡ (0, –1) and P ≡ (x, y) Now, 2a = K [where 2a is the transverse axis and e is the eccentricity] and 2ae = S1S2 = 2 a2 2 e′2 a2 +b2 144 + 25 144 + a2 Dividing, we have e= K a2 25 12 = = 144 ⇒ e′ = Since, e > 1 for a hyperbola, therefore K < 2 25 Also, K must be a positive quantity. Hence, we have, K ∈ (0, 2). A∴ccoared′i=ng152to×giv1e41n42+coan2dition, (ii) 40. (a) : e required point is point of intersection of internal angle bisectors \\ P(x, y) = incentre of D. 3 12 144 + a2 ⇒ 15 = 144 + a2 = 5 × 12 41. (c): Centroid ≡ 2 + 6 +1 , 0 + 0 + 5 ≡ 3, 5 3 3 3 ⇒ a2 = 225 – 144 = 81 ⇒ a = 9 MATHEMATICS TODAY | SEPTEMBER‘17 17
and for orthocentre equation 43. (b) : Let the point (h, k) of line perpendicular to 1 AB passing through C(1, 5) is Given, |k| = 2 h2 + k2 x=1 ... (i) \\SquaLroincugs4xk22 = 3hy22+= k2 ⇒ h2 – 3k2 = 0 Eq. of line perpendicular to AC – 0 and passing through B(6, 0) 44. (a) : Let, the coordinates of point A is (a, 0) B(6, 0) is y = (x – 6)1/5 INfoAwR–mmAaBke=s manARangle q ⇒ 5y = x – 6 ... (ii) Solving (i) and (ii) we get Orthocentre ≡ (1, –1) with +ve x-axis, then AB makes (p – q), therefore We known cent roid divides or t ho cent re and –mAB = mAR circumcentre in 2 : 1 (internally) \\ Circumcentre is (4, 3) 0−2 0−3 \\ Distance between orthocentre and circumcentre − α −1 = α−5 = (1 − 4)2 + (−1 − 3)2 = 5 ⇒ 2(α − 5) = −3(α −1) ⇒ α = 13 \\ 5 42. (a) : Median CD = (2 − 2)2 + (2 −1)2 = 1 A is 13 , 0 5 45. (d) : We have, x + 4y = 2a sinq ... (i) x – y = a cosq …..(ii) ⇒ 5(xx2++442y0)2y2+=(x4−a2y)2 = a2 [From (i) and (ii)] 18 MATHEMATICS TODAY | SEPTEMBER ‘17
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. *ALOK KUMAR, B.Tech, IIT Kanpur z A number of the form x + iy, where x, y ∈ R and ALGEBRAIC OPERATIONS WITH COMPLEX NUMBERS i = −1, is called a complex number and ‘i’ is tLheetnz,1 = a + ib and z2 = c + id be two complex numbers called iota. AMSudubdlttiritpailcoictniaot(inzo1n(z+(1zz–12z)z22:))(:a: ((aa+ ++ibii)bb)+)(c–(+c(c+id+)id=id) ()=a=c(–a(ab+–dc)c)+)++i(iai((dbb++–bddc))) A complex number is usually denoted by z and the Division (z1/z2()w:heacr+e+iaidbt least one of c and d is non-zero) set of complex numbers is denoted by C. (a + ib) (c id) (ac bd) i(bc − ad) i.e., C = {x + iy : x ∈ R, y ∈ R, i = −1} = (c + id) ⋅ (c − id) = c2 + d2 + c2 + d2 . 5 + 3i, –1 + i, 0 + 4i, 4 + 0i etc. are complex − + numbers. z For any positive real number a, we have z Properties of algebraic operations on complex −a = −1 × a = −1 a = i a : Let zt1h, ezi2r aanldgezb3raairce any three complex numbers then operations satisfy z e property a b = ab is valid only if at numbers least one of a and b is non-negative. following properties : z z1 + z2 = z2 + z1 z Integral powers of iota (i) : Since i = −1 hence z (z1 + z2) + z3 = z1 + (z2 + z3) we have i2 = –1, i3 = –i and i4 = 1. To nd the value z z1z2 = z2z1 of in(n > 4). z (z1z2)z3 = z1(z2z3). Let n = 4q + r where 0 r 3, z z1(z2 + z3) = z1z2 + z1z3 \\ in = i4q + r = (i4)q.(i)r = (1)q (i)r = ir z (z2 + z3)z1 = z2z1 + z3z1. In general we have the following results i4n = 1, i4n+1 = i, i4n + 2 = –1, i4n + 3 = –i, where n is any integer. EQUALITY OF TWO COMPLEX NUMBERS Two complex numbers oz1n=lyxi1f +thieyi1r arneadlza2n=d xi2m+agiyin2 aarrye REALAND IMAGINARY PARTS OFA COMPLEX NUMBER said to be equal if and If x and y are two real numbers, then a number of the parts are separately equal. i.e., z1 = z2 form z = x + iy is called a complex number. Here ‘x’ is Comxxp11l=+exxiyn21ua=mndxb2ey+r1s=idyy2o2n. ot possess the property of order called the real part of z and ‘y’ is known as the imaginary i.e., (a + ib) < (or) > (c + id) is not de ned. For example, part of z. e real part of z is denoted by Re(z) and the the statement (9 + 6i) > (3 + 2i) makes no sense. imaginary part by Im(z). If z = 3 – 4i, then Re(z) = 3 and Im(z) = – 4. A complex number z is purely real if its imaginary part is zero i.e., Im(z) = 0 and purely imaginary if its real part is zero i.e., Re(z) = 0. * Alok Kumar is a winner of INDIAN NAtIoNAl MAtheMAtIcs olyMpIAD (INMo-91). he trains IIt and olympiad aspirants. MATHEMATICS TODAY | SEPTEMBER‘17 19
CONJUGATE OF A COMPLEX NUMBER z |z1z2| = |z1||z2| z Conjugate complex number : If there exists a In general |z1z2z3...zn| = |z1||z2||z3|...|zn| complex number z = a + ib, (a, b) ∈R, then its z z1 = |z1 | , (z2 ≠ 0) conjugate is de ned as z = a – ib. z2 |z2 | Y z | zn | = | z |n,n ∈ N Imaginary P(z) z |z1 ± z2|2 = |z1|2 + |z2|2 ± (z1z2 + z1z2) Imaagxiinsary q or |z1|2 + |z2|2 ± 2Re(z1z2) -q Real O X |z1 + z2|2 = |z1|2 + |z2|2 ⇒ z1 is purely real z2 z Q(z) z |z1 + z2|2 + |z1 – z2|2 = 2{|z1|2 + |z2|2} (Law of parallelogram) Geometrically, the conjugate of z is the re ection or mirror image of z about real axis. VARIOUS REPRESENTATIONS OF A COMPLEX z Properties of conjugate : wIfez,hza1vaendthz2eafroelelxoiwstiinngg complex numbers, then NUMBER results: A complex number can be represented in the following form: z (z ) = z z1 ± z2 = z1 ± z2 z Geometrical representation (Cartesian z z1z2 = z1 z2 representation): ecomplexnumberz=a+ib=(a,b) In general, z1 ⋅ z2 ⋅ z3.....zn = z1 ⋅ z2 ⋅ z3.....zn is represented by a point P whose coordinates are referred to rectangular axes XOX′ and YOY′ which (( ))z z1 = z1 ,[z2 ≠ 0] (z )n = (zn) are called real and imaginary axis respectively. is z2 z2 plane is called argand plane or argand diagram or complex plane or Gaussian plane. z z + z = 2Re(z) = 2Re(z ) = purely real z z − z = 2iIm(z) = purely imaginary Y P(a, b) | z | = a2 + b2 z z1z2 + z1z2 = 2Re(z1z2 ) = 2Re(z1z2 ) Reciprocal of a complex number : For an existing qb non-zero complex number z = a + ib, the reciprocal aM 1 X′ Real axis O X z z z is given by z −1 = = z⋅z = | z |2 . MODULUS OF A COMPLEX NUMBER Y′ Modulus of a complex number z = a + ib is de ned Angle of any complex number with positive direction of x–axis is called amplitude or argument by a positive real number given by |z | = a2 + b2 , of z. where a, b are real numbers. Geometrically |z| i.e., represents the distance of point P from the origin, amp(z) = arg (z) = tan−1 b a i.e. |z| = OP. z Trigonometrical (Polar) representation : In DOPM, If |z | = 1 the corresponding complex number is known as Y Pz let OP = r, then a = r cos q and b = r sin q. Hence z can be expressed as z = r(cos q + i sin q) unimodular complex number. where r = |z| and q = principal value of argument of z. Clearly z lies on a circle of unit For general values of the argument radius having centre (0, 0). q MX O z z = r[cos(2np + q) + isin(2np + q)] Sometimes (cos q + i sin q) is written in short Properties of modulus as cisq. z |z| ≥ 0 ⇒ |z| = 0 if z = 0 and |z| > 0 if z ≠ 0. z Vector representation : If P is the point (a, b) on the argand plane corresponding to the complex z –|z| Re(z) |z| and –|z| Im(z) |z| number z = a + ib. z |z| = |z| = |–z| = |–z|=|zi| z zz = |z|2 ⇒ Purely real 20 MATHEMATICS TODAY | SEPTEMBER ‘17
en OP = aiˆ + bjˆ , \\ |OP | = a2 + b2 = |z | and To nd the square root of a – ib, replace i by –i in the above results. tan−1 b arg(z) = direction of the vector OP = a LOGARITHM OF A COMPLEX NUMBER Let log(x + iy) = loge(reiq) = loger + loge eiq = loge r + iq z Eulerian representation (Exponential form) : Since we have eiq = cosq + i sinq and thus z can be = loge (x2 + y 2 ) + itan−1 y expressed as z = reiq, where |z| = r and q = arg(z). x z Principal value of arg (z) : Y Hence, loge (z) = loge |z| + i ampz The value q of the p(––, +a) a GEOMETRY OF COMPLEX NUMBERS argument, which satis es X′ –((p–,––)a)O (+, +) z Distance formula : e distance the inequality –p < q p (+–,a–) X bQe(tzw2=)eies|nagitvwxeonofpbQyoiPn–Qtsa=P|(xzz21o)–f and Q(z2) is called the principal Pz1|| value of argument, where P(z1) α = tan−1 b (acuteangle) Y′ z Section formula : If R(z) divides the line segment a and principal values of argument z will be a, joiningP(z1)andQ(z2)intheratiom1:m2(m1,m2>0) then p – a, –p + a and –a as the point z lies in the 1st , 2nd , 3rd and 4th quadrants respectively. z For internal division z = m1z2 + m2z1 m1 + m2 z Properties of arguments aaIrrngg((gzz11ezzn22+e))r==aarlaagarr(rggzg((3(zz)z11+1))z+–2..z.aa3r+r.g.g.((zazznr22)g))(=+zna2)rk+gp(z,21k)p+K For external division m1z2 − m2z1 z ∈I z z = m1 − m2 z arg(z2) z Equation of a straight line k∈I z Parametric form : Equation of a straight line joining the point having a xes and is z1 z = tz1 + (1 – t)z2, when t ∈ R z1 z2 z2 = arg z1 – arg z2 + 2kp z arg k∈I z Non-parametric form : Equation of a straight z arg z = 2arg z + 2kp k∈I line joining the points having a xes z1 and z2 z z z1 z arg(zn) = n arg z + 2kp k ∈ I is z1 z1 1 = 0 z2 z2 1 z If arg z2 = q, then arg z1 = 2kp – q, where k ∈ I z1 z2 ⇒ z(z1 – z2) – z(z1 – z2) + z1z2 – z2z1 = 0. z arg z = –arg z = arg 1 z z Condition of Collinearity of three points : ree pozi1nts1A(z1), B(z2) and C(z3) are collinear arg(z – z) = (4k ± 1) π , z not being purely z1 z imaginary 2 z arg(–z) = arg(z) + (2n + 1)p if, z2 z2 1 = 0 z arg(z) – a r g (z) = ±p (If z is purely imaginary) z3 z3 1 z z1z2 + z1z2 = 2|z1| |z2| cos (q1 – q2) or z1 − z2 = z2 − z3 = z1 − z3 where q1 = arg(z1) and q2 = arg(z2) z1 − z2 z2 − z3 z1 − z3 z e general value of arg(z) is 2np – arg(z). z General equation of a straight line : The general equation of a straight line is of the form SQUARE ROOT OF A COMPLEX NUMBER az + az + b = 0, where a is non zero complex Let z = a + ib be a complex number, number and b is any real number. en z + a + i z − a , for b> 0 z Slope of a line : e complex slope of the line a + ib = ± 2 2 az + az + b = 0 is −a = − coeff. of z and real a coeff. of z z + a −i z − a , for b < 0. slope of the line az + az + b = 0 is −i (a + a ) . = ± 2 2 (a − a ) MATHEMATICS TODAY | SEPTEMBER‘17 21
z Length of perpendicular : The length of z Complex number as a rotating arrow in the perpendicular from a point z1 to the line argand plane : Let z = r(cosq + isinq) = reiq ...(i) i eiq be a complex number representing a point P az + az + b = 0 is given by | az1 + az1 + b | or in the argand plane. | az1 + az1 + b | |a|+|a | en OP = |z| = r Y Q(zeif) P(z) 2|a| and∠POX = q f z Equation of a circle : P(z) Now consider complex X′ q X r number O z The equation of a circle whose centre is at point zo1r = zeif = reiq eif having a x z0 and radius r C(z0) z1 = rei(q Y′ + f) is |z – z0| = r {From (i)} CQleinartlyhethaergcaonmdpplelxanneu, mwbheernzO1 rQep=rerseanntds a point z If the centre of the circle is at origin and radius ∠QOX r, then its equation is |z| = r. = q + f. z |z – z0| < r represents interior of a circle, |z – z0| = r Clearly multiplication of z with eif rotates the lie on the circle and |z – z0| > r represents exterior of the circle. vector OP through angle f in anticlockwise sense. Similarly multiplication of z with e–if will rotate z General equation of a circle : The general the vector OP in clockwise sense. equation of the circle is zz + az + az + b = 0 Iafnzd1,Cz2suacnhd tzh3aatrAeCth=e a xes of the points A, (where a is complex number and b ∈ R) with z AB and ∠CAB = q. B centre and radius as – a and |a |2 −b = aa − b C(z3) respectively. B(z2) z Equation of circle in diametric form : If end q points of dPi(azm) beetearnryepproeinsetnotnedthbeycAirc(zle1)thaennd, A(z1) B(z2) and (z – z1)(z – z2) + (z – z2)(z – z1) = 0. erefore, AB = z2 – z1, AC = z3 – z1. which is required equation of circle in en AC will be obtained by rotating AB diametric form. through an angle q in anticlockwise sense, and therefore, ROTATION THEOREM AC = AB eiθ or (z3 – z1) = (z2 – z1)eiq Rotational theorem is used to nd the angle between two or z3 − z1 = eiq intersecting lines. is is also known as coni method. z2 − z1 Let z1, z2 and z3 be the a xes of three points A, B and C Y C(z3) B(z2) z If A, B and C are three points in argand plane respectively taken on argand a such that AC = AB and ∠CAB = q then use the rotation about A to nd eiq, but if AC ≠ AB use plane. A(z1) coni method. en we have AC = z3 – z1 fq X z If four points z1, z2, z3 and z4 are concyclic then and AB = z2 – z1 O (z4 − z1)(z2 − z3) (z4 − z2)(z1 − z3) and let arg AC = arg(z3 – z1) = q and AB = arg(z2 – z1) = f is real Let ∠CAB = a, or (z2 − z3 )(z 4 − z1) π, 0 Q ∠CAB = a = q – f = arg AC – arg AB arg (z1 − z3 )(z 4 − z2) = ± = arg(z3 – z1) – arg(z2 – z1) = arg z3 − z1 TRIANGLE INEQUALITIES angle between AC and AB z2 − z1 In any triangle, sum of any two sides is greater than the or third side and di erence of any two sides is less than the third side. By applying this basic concept to the set of = arg affix of C − affix of A complex numbers we are having the following results. affix of B − affix of A 22 MATHEMATICS TODAY | SEPTEMBER ‘17
z |z1 ± z2| |z1| + |z2| z (sin θ + i cos θ)n = cos n π − θ + i sinn π − θ z |z1 ± z2| ≥ ||z1| – |z2|| 2 2 is theorem is not valid when n is not a rational number STANDARD LOCI IN THE ARGAND PLANE or the complex number is not in the form of cos q + i sin q. If z is a variable point and z1, are two xed points in the argand plane, then z2 ROOTS OF A COMPLEX NUMBER |bzis–eczt1o| r=o|fzt–hez2l|in⇒e sLeogcmuesnotfjzoiinsitnhge perpendicular z z1 and z2. z nth roots of complex number (z1/n) Let z = r(cos q + i sin q) be a complex number. z |z – zL1|o+cu|sz o–fzz2|i=s constant > |z1 – z2| By using De’moivre’s theorem nth roots having ⇒ an ellipse n distinct values of such a complex number are given by z |z – zL1o| +cu|szo–f zz2i|s=th|ez1li–nez2s|egment 2kπ + 2kπ + ⇒ n n joining z1 and z2. z1/n = r1/n cos θ + i sin θ , z |z – zL1|o–cu|zs –ofzz2|i=s a|zs1t–razig2|ht line joining ⇒ z does not lie between z1 and z2. but z1 and z2 where k = 0, 1, 2, ..., (n – 1). z Properties of the roots of z1/n |z – zL1|o–cu|zs –ofzz2| = constant (> 0 but |z1 – z2|) z All roots of z1/n are in geometrical progression z ⇒ is a hyperbola. ≠ with common ratio e2pi/n. z c|zir–clez1w|2it+h |z –anzd2|z22=as|zt1h–e ezx2|t2rem⇒itiLesoocuf sdioafmzeitsera. z Sum of all roots of z1/n is always equal to zero. z1 z Product of all roots of z1/n = (–1)n – 1. z Modulus of all roots of z1/n are equal and each z |z – z1| = k|z – z2|, (k ≠ 1) ⇒ Locus of z is a circle. equal to r1/n or |z|1/n. z arg z − z1 =a( xed) ⇒ Locus of z is a segment z Amplitude of all the roots of z1/n are in A.P. with z − z2 of circle. common di erence 2π . n z arg z − z1 = ±p/2 ⇒ Locus of z is a circle with z − z2 z All roots of z1/n lies on the circumference of a z1 and z2 as the vertices of diameter. circle whose centre is origin and radius equal to |z|1/n. Also these roots divides the circle into n equal parts and forms a polygon of n sides. z e nth roots of unity : e nth roots of unity are z arg z − z1 = 0 or p ⇒ Locus of z is a straight given by the solution set of the equation z − z2 line passing through z1 and z2. xn = 1 = cos 0 + i sin 0 = cos 2kp + i sin 2kp x = [cos 2kp + i sin 2kp]1/n DE’ MOIVRE’S THEOREM cos 2kπ 2kπ z If n is any rational number, then x= n + i sin n , where k = 0, 1, 2, ..., (n – 1). (cos q + i sin q)n = cos nq + i sin nq. z Properties of nth roots of unity z t+Ihf zeqn3=+z(c=.o..sc+oqs1q(+qn)1(ic+wsoihsnqeq2qr3+e1+)(qqci13o,s+siqnq2..,q2. q3++)3..qi....(nsqci)nno+s∈qqi2n)Rsi+.ni(qs1in+qqn2) z If z = r(cos q + i sin q) and n is a positive integer, z Let α = cos 2π + i sin 2π = ei(2 π/n) , the nth roots n n ofunitycan be expressed in the form ofaseriesi.e., 1, a, a2, ... an – 1. Clearly the series is G.P. with common ratio a i.e., ei(2p/n). cos 2kπ + θ + i sin 2kπ + θ z e sum of all n roots of unity is zero i.e., then z1/n = r1/n n n , 1 + a + a2 + ... + an – 1 = 0. where k = 0, 1, 2, 3,...(n – 1). z Product of all n roots of unity is (–1)n–1. Deductions: If n ∈ Q, then z Sum of pth power of n roots of unity 1 + ap + a2p + ... + a(n – 1)p z (cos q – i sin q)n = cos nq – i sin nq 0,when pis not multiple of n = n,when pis a multiple of n z (cos q + i sin q)–n = cos nq – i sin nq z (cos q – i sin q)–n = cos nq + i sin nq MATHEMATICS TODAY | SEPTEMBER‘17 23
z e n, nth roots of unity if represented on a z ||Izzzzf11|z–+||1zzz|z1122||–+221==zz,22||||z|zz22121|||≥22 1 then complex plane locate their positions at the +(+(||z|z|zz1122||||–2+2 –+||zz2222|||z|z))1122||||+z–z22(|(|aaccroroggss((((qzqz1111))––––qqa2a2))rr..gg((zz22))))22.. vertices of a regular polygon of n sides inscribed z in a unit circle having centre at origin, one vertex z on positive real axis. z Cube roots of unity : Cube roots of unity are the z z solution set of the equation x3 – 1 = 0 ⇒ x = (1)1/3 zIf2|za1r|e=co|zn2j|uagnadteacmopm(zp1l)ex+naummpb(ze2r)s=of0e, athchenozth1 earnd z ⇒ x = (cos0 + i sin0)1/3 e area of the triangle whose vertices are z, iz and 2kπ 2kπ z ⇒ x = cos 3 + i sin 3 , where k = 0, 1, 2 z z + iz is 1 | z |2. 2 z erefore roots are 1, cos 2π + i sin 2π , e area of the triangle with vertices z, wz and 3 3 4π 4π z + wz is 3 | z2 |. cos 3 + i sin 3 or 1, e2πi/3, e4πi/3 4 Alternative : x = (1)1/3 ⇒ x3 – 1 = 0 Iafnzd1,zz02b, ez3thbeectihrecuvmerctiecnetsreo,ftahneneqzu21i+lazte22r+alzt23ri=an3gzl20e. ⇒ (x – 1)(x2 + x + 1) = 0 zIItorff21fizazn+11n,,sgzzizld222e2,e,+iszsz.3.a3e..n+.q.bzduzneiz2nlbt0ah=etbeeternhavzielte20sriv.tceiecrnteistcreoosifdoa,fttahreiranenggullea,r polygon then the x = 1, −1 +i 3 , −1 −i 3 2 2 If one of the complex root is , then other root will be 2 or vice-versa. (z1 – z2)2 + (z2 – z3)2 + (z3 – z1)2 = 0 z Properties of cube roots of unity z 1+ + 2=0 z 3=1 or z21 + z22 + z23 = z1z2 + 1z2z3 + z3z1 or 1 + 1 + =0 0,if n isnot a multiple of 3 z 1 + ωn + ω2n = o3f,uifnnitiys,awmhuenltirpelperoefse3nted Irfeszp1e, czzt1i2v−, ezzly32 bezt2h−e za3 xezs3 −ofz1the vertices A, B, C z e cube roots of a triangle ABC, then its orthocentre is on z complex plane, lie on vertices of an equilateral triangle inscribed in a unit circle having centre z a(sec A)z1 + b(sec B)z2 + (c secC)z3 z asec A + bsec B + c secC at origin, one vertex being on positive real axis. Re(iz) = –Im(z), Im(iz) = Re(z). z Cube root of – 1 are –1, – , – 2. tInhfeecthesseusmacroizmly1 p+cloeznx2juinsguaamtreebcaeolrmnsupzml1ebxa.enrd, tzh2enartehesyucahretnhoatt IMPORTANT POINTS z 0 = 0 + i 0, is the identity element for addition. 1 = 1 + i 0 is the identity element for multiplication. pInfrezoc1edasusncadtrizzly21zac2roeinstjwuagoarecteoalmconpmulempxlbenexur.,mtbheerns such that the z The additive inverse of a complex number z they are not z z = a + ib is –z (i.e. – a – ib). z For every non-zero complex number z, the z If and 2 are the complex cube roots of unity, then (a + b 2)(a 2 + b ) = a2 + b2 – ab multiplicative inverse of z is 1. (a + b)(a + b )(a 2 + b2 ) = a3 + b3 z (a + b + c 2)(a + b 2 + c ) = a2 + b2 + c2 – ab – bc – ca (a + b + c)(a + b + c 2)(a + b 2 + c ) = a3 + b3 + z |z| ≥ |Re(z)| ≥ Re(z) and |z| ≥ |Im(z)| ≥ Im(z) z | z | is always a unimodular complex number if z ≠ 0. c3 – 3abc z z aIpfzo1itnh+trsbezae2re+poccoiznl3lti=sn0ezaw1r,.hze2r,eza3+cbon+nce=ct0e,dthbeny relation z z the three |Re(z)| + |Im(z)| 2 |z| z If z + 1 = a, the greatest and least values of |z| are z If z is a complex number, then ez is periodic. z a+ and a− . a2 + 4 a2 + 4 If three complex numbers are in A.P., then they lie respectively 2 2 on a straight line in the complex plane. 24 MATHEMATICS TODAY | SEPTEMBER ‘17
PROBLEMS 9. e complex numbers sin x + i cos 2x and cos x – i sin 2x are conjugate to each other for Single Correct Answer Type 1 1. e value of i1 + 3 + 5 +...+ (2n + 1) is (a) x = np (b) x= n + 2 π (a) i if n is even, – i if n is odd (b) 1 if n is even, – 1 if n is odd (c) x = 0 (d) No value of x (c) 1 if n is odd, – 1 if n is even (d) i if n is even, – 1 if n is odd 10. e number of solutions of the equation z2 + z– = 0 is 1 (a) 1 (b) 2 (c) 3 (d) 4 2. If x+ x = 2 cos θ, then x is equal to z − i (a) cos q + i sin q (b) cos q – i sin q 11. If zz– + i (z ≠ −i) is a purely imaginary number, then z (c) cos q ± i sin q (d) sin q ± i cos q is equal to 3. 3 + 2i sin θ will be real, if q = (a) 0 (b) 1 (b) np + (c) 2 (d) None of these 1 − 2i sin θ π (a) 2np 2 12. e maximum value of |z| where z satis es the (c) np (d) None of these condition z + 2 =2 is [where n is an integer] z 4. e real part of (1 – cos q + 2i sin q)–1 is (a) 3 −1 (b) 3 +1 1 1 (a) 3 + 5cos θ (b) 5 − 3cos θ (c) 3 (d) 2 + 3 (c) 1 (d) 1 13. aIfnzd1|azn1|d=z|2z2b|e. Icfozm1 hpalesxpnousimtivbeerrseaslupcahrtthaantdzz12≠hazs2 3 − 5cos θ 5 + 3cos θ negative imaginary part, then (z1 + z2 ) may be 5. If x + iy = 3 , then x2 + y2 is equal to (z1 − z2 ) 2+ cos θ+ i sin θ (a) Purely imaginary (b) Real and positive (a) 3x – 4 (b) 4x – 3 (c) Real and negative (d) None of these (c) 4x + 3 (d) None of these 14. e product of two complex numbers each of unit 6. If (p + i)2 = µ + iλ, then 2 + l2 is equal to modulus is also a complex number, of 2p −i (a) Unit modulus (p2 + 1)2 (p2 −1)2 (b) Less than unit modulus (a) 4 p2 −1 (b) 4 p2 −1 (c) Greater than unit modulus (d) None of these (p2 −1)2 (p2 + 1)2 (c) 4 p2 +1 (d) 4 p2 +1 15. |I(zfa1)|+z11|z2=+|zz32|+=....+....z.n=| =|zn| = 1, then the value of 7. If z(1 + a) = b + ic and a2 + b2 + c2 = 1, then 1+ iz = (b) |z1| + |z2| +....+ |zn| (b) 1− iz (a) a + ib b − ic 1+ c 1+ a (c) 1 + 1 + ......... + 1 a + ic (c) 1+b (d) None of these z1 z2 zn If a 8. = cos q + i sin q, then 1+ a = (d) None of these 1−a 16. e values of z for which |z + i| = |z – i| are (a) cot q (b) cot θ (a) Any real number 2 (b) Any complex number (c) Any natural number (c) i cot θ (d) i tan θ (d) None of these 2 2 MATHEMATICS TODAY | SEPTEMBER‘17 25
17. Let z be a complex number (not lying on X-axis) of 27. i log x − i is equal to 1 x + i maximum modulus such that z + z =1. en (a) p + 2tan–1 x (b) p – 2tan–1 x (c) –p + 2tan–1x (d) –p – 2tan–1 x (a) Im(z) = 0 (b) Re(z) = 0 (c) amp(z) = p (d) None of these 28. e equation zz– + az– + a–z + b = 0, b ∈ R represents 18. e minimum value of |2z – 1| + |3z – 2| is a circle if (a) 0 (b) 1/2 (c) 1/3 (d) 2/3 (a) |a|2 = b (b) |a|2 > b 19. isIsfucezhq1utahanladtto|zz12+arze2| two non-zero complex numbers (c) |a|2 < b (d) None of these = |z1| + |z2|, then arg (z1) – arg (z2) 29. vcLieerrtctuitcmheesceocnoftmraenpolefextqhuneilutarmtieabrnaeglrlsetr,zita1hn,egnzl2ez.12aLn+edtz22zz+30 be the be the (a) –p (b) − π (c) π (d) 0 (b) –z20 (c) 3z02 z23 = 2 2 (a) z02 (d) –3z02 20. If z– be the conjugate of the complex number z, then which of the following relations is false ? 30. If the vertices of a quadrilateral be A = 1 + 2i, B = –3 + i, C = –2 – 3i and D = 2 – 2i, then the (a) |z| = |z–| (b) z z– = |z–|2 quadrilateral is (c) z1 + z2 = z1 + z2 (d) arg (z) = arg (z–) (a) Parallelogram (b) Rectangle (c) Square (d) Rhombus If for complex numbers and z2, arg(z1/z2) = 0, 21. t(hae)n||zz11| –+z|2z|2i|s equal to z1 |z1| – |z2| 31. If is a complex number satisfying ω + 1 = 2 , ω (b) then maximum distance of from origin is (c) ||z1| – |z2|| (d) 0 (a) 2 + 3 (b) 1+ 2 22. |z1 + z2| = |z1| + |z2| is possible if (c) 1+ 3 (d) None of these (a) z2 = z–1 (b) z2 = 1 Multiple Correct Answer Type z1 (c) arg(z1) = arg(z2) (d) |z1| = |z2| 32. Let z1, z2, z3, ...., zn are the complex numbers such n n 23. If −1+ −3 = reiθ , then q is equal to ∑ ∑that |z1| = |z2| = .... = |zn| =1. If z = k =1 1 zk zk 2π 2π then 3 3 k =1 (a) π (b) − π (c) (d) − (a) z is purely imaginary 3 3 (b) z is real 24. e value of (–i)1/3 is (c) 0 < z n2 1+ 3i 1 − 3i (d) z is a complex number of the form a + ib (a) 2 (b) 2 33. Lx2et+x1a,xx+2 are the roots of the quadratic equation b = 0 where a,b are complex numbers (c) − 3−i (d) 3 3+i and y|a1|,yy+2 are the roots of the quadratic equation 2 2 y2 + |b| = 0. If |x1| = |x2| = 1 then 25. e amplitude of ee−iθ is equal to (d) esin q (a) |y1| = 1 (b) |y2| = 1 (a) sin q (b) –sin q (c) ecos q (c) |y1| ≠ |y2| (d) |y1| = |y2| = 2 26. e real part of (1 – i)–i is 34. If the equation z3 + (3 + i)z2 – 3z – (m + i) = 0 1 (a) e −π/4 cos 2 log 2 where m ∈ R, has atleast one real root then ‘m’ can have the value equal to (a) 1 (b) 2 (c) 3 (d) 5 (b) −e −π/4 sin 1 log 2 2 Lz3et–z1b,zz22,+z33zin–G1.P=. b0e roots of the equation 35. then (c) e π/4 cos 1 log 2 (d) e −π/4 sin 1 log 2 (a) zb2==31 (b) zb2c=an2be –3 2 2 (c) (d) 26 MATHEMATICS TODAY | SEPTEMBER ‘17
36. e complex numbers satisfying the equation (3z + 1) 42. If f(x) = 1 + 2x + 3x2 + 4x3 + 5x4 + 6x5 + 7x6 then (4z + 1)(6z+ 1)(12z + 1) = 2 is /are for a ≠ 1, f (x) + f (ax) + f (a2x) + f (a3x) +....+ f (a6x) = (a) 33 − 5 (b) 33 + 5 (a) 42 (b) 21 (c) 14 (d) 7 24 24 43. sin π sin 2π sin 3π = sin 4π sin 5π sin 6π = (c) −i 23 − 5 (d) −i 23 + 5 7 7 7 7 7 7 24 24 37. If all the three roots of az3 + bz2 + cz + d = 0 have (a) 7 (b) 7 (c) 7 (d) 7 16 8 32 64 negative real parts (a, b, c ∈ R) then (a) ab > 0 (b) bc > 0 Matrix-Match Type (c) ad > 0 (d) bc – ad > 0 44. Match the following. Comprehension Type Column-I Column-II Paragraph for Q. No. 38 to 40 (A) e maximum value of (p) 0 e complex slope M of a line joining two points z1 ||z – | – –|zc–om–p||le(xwchuebree |z| = 5 and , root and z2 in complex plane is de ned as M = z1 − z2 .Its of unity) is z1 − z2 real slope m is tanq, where q is inclination of the line . (B) Tangent drawn to circle (x – 1)2 + (q) 2/3 (y – 1)2 = 5 at a point P meets 38. M = the line 2x + y + 6 = 0 at Q on 1+ im (a) 1 − im the x-axis then the value of (PQ)2 (b) 2m + i 1− m2 2 is 1+ m2 m2 1+ (C) One vertex of an equilateral (r) 3 triangle is at the origin and the (c) 2m + i 1+ m2 other two vertices are given by 1+ m2 m2 2z2 + 2z + k = 0, then k is 1 − (d) 1 − im (s) 6 m−i 45. Match the following. 39. e inclination of a line whose complex slope is – 2 (where is a non real cube root of unity) is Column –I Column –II (a) 2π (b) π (c) π (d) π (A) If z = z1 + i z2 then |z| equals (p) 0 3 3 6 12 z2 + i z1 40. Which of the following is false? (a) |M| = 1 z − 2i (q) (b) M = m is never possible (B) If z = z + 2i be purely 1 (c) m = i M − 1 imaginary then |z| equals M + 1 If |z + 6| = |2z + 3| then |z| (r) 2 (d) M = cis 2q (C) equals Paragraph for Q. No. 41 to 43 cis 2K π (D) Let ≠ 1 be a cube root of (s) 3 7 unity and If a is any of 7th roots of unity, then a = cω2 z= a + bω + bω2 + 6 c + aω + + bω (K = 0 to 6) and ∑ αi = −1(α ≠ 1) and a7 = 1 a i=1 + cω2 41. e equation whose roots are a + a2 + a4 and a3 + a5 + a6 is (a ≠ 1) b + cω + aω2 then |z| equals (a) x2 + x – 2 = 0 (b) x2 + x + 2 = 0 (c) x2 – x + 2 = 0 (d) x2 – x – 2 = 0 (t) 4 MATHEMATICS TODAY | SEPTEMBER‘17 27
Integer Answer Type 3. (c) : We have, 3 + 2i sin θ = (3 + 2i sin θ)(1+ 2i sin θ) 1 − 2i sin θ (1 − 2i sin θ)(1+ 2i sin θ) 46. CzL1e(tz=3l)z,0bz+0e be two complex naumtrbiaenrsg.leA(szu1)c,hB(tzh2a)t, ∠ABC = the vertices of 3lk, πz2 = z0 + leip/4, z3 = + lei7p/11 and = 3 − 4 sin2 θ 8sin θ 22 then the value of k z0 + i + 4 sin2 is 1 + 4 sin2 θ 1 θ Now, since it is real, therefore Im (z) = 0 8sin θ 47. If the argument of (z – a)(z– – b) is equal to that ⇒ 1+ 4 sin2 θ = 0 ⇒ sin q = 0 \\ q = np of ( 3 + i)(1+ 3 i) , where a, b are real numbers. where n = 0, 1, 2, 3, ...... 1+i a circle with then nd Remark : Check for (a), if n = 0, q = 0 the given number 3+i is absolutely real but (c) also satis es this condition and If locus of z is centre 2 in (a) and (c), (c) is most general value of q. a + b. 48. xIf5=a–a11,(46=a6920+,,1aw/33)h,, ear=e4,aaa(55k05=3 a+ar1ek/2–)t,h1th,eaenr=o[|oleti2s20p1o1/5f+atnh2de01l1eq+=uaa2t310i01o011|n], 4. (d) : Let z = 0 {(1 – cosq) + i·2 sinq}–1 2 sin θ −1 sin θ 2 cos θ −1 2 2 2 = + i ⋅ (where [ ] denotes the greatest integer function) is θ −1 1 sin θ − i ⋅ 2cos θ 49. Two lines zi – z–i + 2 = 0 and z(1 + i) + z–(1 – i) + 2 = 0 2 i⋅ sin 2 − i ⋅ 2cos 2 intersect at a point P. ere is a complex number = 2 sin θ θ × θ θ 2 2 2 2 =x+ iy at a distance of 2 units from the point sin + 2 cos which lies on line z(1 + i) + z–(1 – i) + 2 = 0. a sin θ − i ⋅ 2 cos θ P 2 2 Find [|x|] (where [ ] represents greatest integer = function). θ sin2 θ cos2 θ 2sin 2 2 + 4 2 50. Suppose that w is the imaginary (2009)th roots of ∑2008 θ 1 = (a)(2b ) + c where ∴ Re(z) = sin 2 unity. If (22009 – 1) 1 + 3cos2 r=1 2 − wr θ θ a, b, c ∈ N and the least value of (a + b + c) is (2008)K. 2 sin 2 2 e numerical value of K is 3 θ+ SOLUTIONS 5. (b) : If x + iy = 2 + cos i sin θ 1. (c) : Let z = i[1+3+5+....+(2n+1)] 3(2 + cos θ − i sin θ) 6 + 3cos θ − 3i sin θ Clearly series is A.P. with common di erence = 2 (2 + cos θ)2 + sin2 θ cos2 θ + 4 cos θ + sin2 Q nTunm=b2enr −1 tearnmdsTinn+1A=.P2. n=+n1 = = 4 So, of + θ + 1 = 6 + 3 cos θ −3sin θ Now, Sn+1 = n 2+1[2⋅1 + (n +1 −1)2] 5 + 4 cos θ 5 + 4 cos θ + i ⇒ Sn+1 = n + 1[2 + 2n] = (n + 1)2 i.e. z = i(n+1)2 x = 3(2 + cos θ) , y = −3sin θ 2 5 + 4 cos θ 5 + 4 cos θ 9 Now put n = 1, 2, 3, 4, 5, ..... ∴ x2 + y2 = (5 + 4 cosθ)2 [4 + cos2 q + 4 cos q + sin2 q] n = 1, z = i4 = 1, n = 2, z = i6 = –1, n = 3, z = i8 = 1, n = 4, z = i10 = – 1, = 5 + 9 θ = 4 6 + 3 cos θ − 3 = 4x − 3 n = 5, z = i12 = 1, .......... 4 cos 5 + 4 cos θ 2. (c) : x + 1 = 2 cos θ ⇒ x2 – 2x cos q + 1 = 0 6. (d) : We have, x 4 cos2 θ − 4 (p + i)2 (p2 −1+ 2pi)(2p + i) ⇒ x = 2 cos θ ± 2 ⇒ x = cos q i sin q µ + iλ = 2p −i = (2p − i)(2p + i) 28 MATHEMATICS TODAY | SEPTEMBER ‘17
MATHEMATICS TODAY | SEPTEMBER‘17 29
= 2p(p2 − 2) + i(5p2 −1) 9. (d) : sin x + i cos 2x and cos x – i sin 2x are conjugate 4 p2 +1 to each other if sin x = cos x and cos 2x = sin 2x 5π 9π µ2 λ2 4 p2(p2 − 2)2 + (5p2 − 1)2 or tan x = 1 ⇒ x = π , 4 , 4 , ...... ...(i) (4 p2 + 1)2 4 \\ + = π 5π 9π 4 p6 + 6p2 + 9p4 +1 and tan 2x = 1 ⇒ 2x = 4 , 4 , 4 , ........ (4 p2 + 1)2 = ⇒ x = π , 5π , 9π ....... ...(ii) 8 8 8 p4 (4 p2 + 1) + 2 p2(4 p2 + 1) + (4 p2 + 1) There exists no value of x common in (i) and (ii). = (4 p2 + 1)2 erefore there is no value of x for which the given = p4 + 2 p2 + 1 (p2 + 1)2 complex numbers are conjugate. 4 p2 +1 4 p2 +1 = 10. (d) : Let z = x + iy, so, z = x − iy, 7. (a) : Given, z(1 + a) = b + ic ⇒ z = b + ic \\ z2 + z = 0 ⇔ (x2 − y2 + x) + i (2xy − y) = 0 1+a Equating real and imaginary parts, we get x2 – y2 + x = 0 ...(i) 1+ iz 1+ i(b + ic) / (1+ a) 1+ a − c + ib 1 1 − iz = 1 − i(b + ic) / (1+ a) = 1+ a + c − ib and 2xy – y = 0 ⇒ y=0 or x= 2 = (1+ a − c + ib)(1+ a + c + ib) If y = 0, then (i) gives x2 + x = 0 ⇒ x = 0 or x = –1 (1+ a + c)2 + b2 1 y2 1 1 3 = 1+ 2a + a2 − b2 − c2 + 2ib + 2iab If x = 2 , then x2 – y2 + x = 0 ⇒ = 4 + 2 = 4 1+ a2 + c2 + b2 + 2ac + 2(a + c) ⇒ y=± 3 a2 + b2 + c2 + 2a + a2 − b2 − c2 + 2ib(1+ a) 2 = 1+ 1+ 2ac + 2(a + c) Hence, there are four solutions in all. = 2a(a + 1) + 2ib(1+ a) = a + ib 11. (b) : Here, z −i = x + i(y − 1) . x − i(y + 1) 2(1+ a)(1+ c) 1+ c z +i x + i(y + 1) x − i(y + 1) 8. (c) : a = cos q + i sin q = (x2 + y2 −1) + i(−2x) 1+ (1 + cosθ) + i sinθ x2 + (y + 1)2 ∴ 1− a = (1 − cosθ) − i sinθ . a As z − i is purely imaginary, we get On Rationalizing the denominator, we get z +i 1+ a = (1 + cos θ) + i sin θ × (1 − cos θ) + i sin θ x2 + y2 – 1 = 0 ⇒ x2 + y2 = 1 ⇒ zz = 1 1− a (1 − cos θ) − i sin θ (1 − cos θ) + i sin θ 2 2 ⇒ |z|2 – 2 |z| –2 (1+ cos θ)(1− cos θ)+ (1+ cos θ)i sinθ 12. (b) : z + z = 2 ⇒| z | − | z | ≤ 2 0 + (1− cos θ)i sin θ + i2 sin2 θ = (1 − cos θ)2 −(i sin θ)2 |z|≤ 2± 4 + 8 ≤ 1± 3. 2 1 − cos2 θ + i sin θ + i sin θ cos θ Hence max. value of |z| is 1+ 3 + i sin θ − i sin θ cos θ − sin2 θ = 1+ cos2 θ − 2 cos θ + sin2 θ 13. (a) : Let za1n=d a + ib = (a, b) and z2 = c – id = (c, –d) where a >0 d > 0 ...(i) 1 − (cos2 θ + sin2 θ) + 2i sinθ en |z1| = |z2| ⇒ a2 + b2 = c2 + d2 = 1 + (cos2 θ + sin2 θ) − 2cosθ = 2i sinθ 2(1 − cosθ) z1 + z2 (a + ib) + (c − id) Now z1 − z2 = (a + ib) − (c − id) θ θ θ i ⋅ 2 sin 2 cos 2 i cos 2 cot θ 2 = sin2 θ = θ = i = [(a + c) + i(b − d)][(a − c) − i(b + d)] 2 2 [(a − c) + i(b + d)][(a − c) − i(b + d)] 2 sin 30 MATHEMATICS TODAY | SEPTEMBER ‘17
= (a2 + b2 ) − (c2 + d2 ) − 2(ad + bc)i Since |z| = r is maximum, therefore dr = 0 a2 + c2 − 2ac + b2 + d2 + 2bd Di erentiating (i) w.r.t. q, we get dθ = a2 −(ad + bc)i [using (i)] 2r dr − 2 dr − 4 sin 2θ = 0 + b2 − ac + bd dθ r3 dθ \\ (z1 + z2 ) is purely imaginary. Putting dr = 0, we get sin 2q = 0 ⇒ q = 0 or π (z1 − z2 ) dθ 2 (z1 + z2 ) \\ z is purely imaginary. (Q q ≠ 0) (z1 − z2 ) However if ad + bc = 0, then will be equal 18. (c) : Given expression, |2z – 1| + |3z – 2|, minimum to zero. According to the conditions of the equation, value of |2z – 1| is 0 at z = 1 . So value of given we can have ad + bc = 0 2 1 1 Remark : Assume any two complex numbers satisfying expression = 0 + 2 = 2 , minimum value of |3z – 2| is both conditions i.e., z1 ≠ z2 and |z1| = |z2| 0 at 2 . So value of given expression =1313. 0 1 . Let z1 = 2 + i, z2 = 1 – 2i, \\ z1 + z2 3−i z = 3 + = 3 z1 − z2 1+ 3i = = −i So minimum value of given expression is Hence the result. 19. (d) : Let z1 = r1 (cos q1 + i sin q1) and z2 = r2 (cos q2 + i sin q2) 14. (a) 15. (c) : We have |zk| = 1, k = 1, 2, .... n \\ |z1 + z2| = [(r1 cos q1 + r2 cos q2)2 + (r1 sin q1 + r2 sin q2)2]1/2 ⇒ | zk |2 = 1 ⇒ zk zk = 1⇒ zk = 1 = [r12 + r12 + 2r1r2 cos (q1 – q2)]1/2 = [(r1 + r2)2]1/2 zk (Q |z1 + z2| = |z1| + |z2|) erefore, | z1 + z2 + .... + zn | = | z1 + z2 + .... + zn | erefore cos(q1 – q2) = 1 ⇒ q1 – q2 = 0 ⇒ q1 = q2 (Q | z |=|z |) us arg(z1) – arg(z2) = 0 20. (d) : Let z = x + iy, z = x − iy ⇒ | z1 + z2 + ..... + zn | = 1+1 + .... + 1 z1 z2 zn arg(z) = θ = tan−1 y 16. (a) : Let, z = x + iy ...(i) Since x Given |z + i| = |z – i| or |x + iy + i| = |x + iy – i| arg (z ) = θ = tan−1 −y x or |x + i(y + 1)| = |x + i(y – 1)| us arg(z) ≠ arg(z ) ⇒ x2 + (y + 1)2 = x2 + (y −1)2 21. (c) : We have, |z1 – z2|2 = |z1|2 + |z2|2co–s2(q|1z1–| |qz22)| where q1 = arg(z1) and q2 = arg(z2) ⇒ x2 + (y + 1)2 = x2 + (y – 1)2 Since arg(z1) – arg(z2) = 0 \\ |z1 – z2|2 = |z1|2 + |z2|2 –2|z1| |z2| = (|z1| – |z2|)2 ⇒ y2 + 2y + 1 = y2 – 2y + 1 ⇒ 4y = 0 or y = 0 ⇒ |z1 – z2| = ||z1| – |z2|| Hence from (i), we get z = x, where x is any real number. 17. (b) : Let z = r(cos q + i sin q) en z + 1 = 1 ⇒ z + 1 2 = 1 z z 22. (c) : |z1 + z2| = |z1| + |z2| r(cos θ sin θ) 1 (cos sin θ) 2 1 |z1 + z2|2 = |z1|2 + |z2|2 + 2 |z1| |z2| r ⇒ |z1|2 + |z2|2 + 2Re | z1z2 | = |z1|2 + |z2|2 + 2 |z1| |z2| ⇒ + i + θ − i = ⇒ 2Re | z1z2 | = 2|z1| |z2| ⇒ 2|z1| | z2 | cos(q1 – q2) = 2|z1| |z2| 1 2 cos2 1 2 sin2 1 ⇒ cos(q1 – q2) = 1 or q1 – q2 = 0 r r \\ arg(z1) = arg(z2) ⇒ r + θ + r − θ = ⇒ r2 + 1 + 2 cos 2θ = 1 ...(i) r2 MATHEMATICS TODAY | SEPTEMBER‘17 31
23. (c) : Here, −1+ −3 = reiθ [Q log(a + ib) = log(reiq) = log r + iq = log a2 + b2 + i tan−1(b / a) ] ⇒ −1+ i 3 = reiθ = r cos q + ir sin q Equating real and imaginary parts, we get z = log x4 +1− 2x2 + 4x2 tan−1 2x i (x2 +1) −x r cos q = –1 and r sin θ = 3 ⇒ + i 1 2 Hence, tan θ = − 3 ⇒ tan θ = tan 2π . Hence θ = 2π . = log 1 + i(2 tan–1 x) 3 3 ⇒ z = i2 2 tan–1 x = –2tan–1 x = p –2tan–1 x. 24. (c) : Since − 3 − i = − cos π + i sin π 28. (b) : By adding aa on both the sides of 2 6 6 zz + az + az = −b 3 3 we get, (z + a)(z + a) = aa − b − 3 − i cos π sin π ⇒ 2 = − 6 + i 6 = −i ⇒ |z + a|2 = |a|2 – b, {Q zz = | z |2 } 25. (b) : Let z = ee–iq = ecosq – i sin q = ecos q e–i sin q is equation will represent a circle with centre z = –a, z = ecosq [cos(sin q) – i sin(sin q)] if |a|2 – b > 0, i.e. |a|2 > b since |a|2 = b represents z = ecosq cos(sin q) – iecosq sin(sin q) point circle only. amp(z) tan−1 ecos θ sin(sin θ) 29. (c) : Let r be the circumradius of the equilateral − ecos θ cos(sin θ) triangle and the cube root of unity. = A(z1) = tan–1[tan(–sin q)] = – sin q 2π 3 26. (a) : Let z = (1 – i)–i. Taking log on both sides, O′(z0 ) ⇒ log z = –i log(1 – i) = −i log 2 cos π − i sin π B(z2) C(z3) 4 4 LitestvAerBtCicebseAth, eBeaqnudilaCterreaslptreicatnivgelelywwitihthz1c,irzc2uamndcezn3tares ( )= −i log 1 π/4 O′(z0). 2e−i π/4 = −i 2 log 2 + log e −i en the vectors O′A = z1 − z0 = reiθ = −i 1 log 2 − iπ = − i log 2 − π 2 4 2 4 2π i θ+ 3 −i (log 2) = rωeiθ e−π/4 2 O′B = z2 − z0 = re ⇒ z = e 4π Taking real part only, we get i θ + 3 = rω2eiθ 1 O′C = z3 − z0 = re 2 Re(z) = e− π/4 cos log 2 \\ z1 = z0 + reiθ , z2 = z0 + rωeiθ , z3 = z0 + rω2eiθ Squaring and adding 27. (b) : Let z = i log x − i ⇒ z = log x −i z12 + z22 + z32 = 3z02 + 2(1 + ω + ω2 )z0reiθ x + i i x + i + (1 + ω2 + ω4 )r2ei2θ ⇒ z = log x −i × x − i = log x2 −1 − 2ix = 3z02, since 1 + + 2 = 0 = 1 + 2 + 4 i x +i x − i x2 +1 30. (c) : Given the vertices of quadrilateral A(1 + 2i), B(–3 + i), C(–2 – 3i) and D(2 – 2i) ⇒ z = log x2 −1 − i 2x Now, AB = 16 + 1 = 17, BC = 16 + 1 = 17 i x2 +1 x2 + 1 z = log x2 −1 2 −2x 2 −2x CD = 16 + 1 = 17, DA = 16 + 1 = 17 i x2 +1 + x2 +1 x2 −1 ⇒ +i tan−1 AC = 9 + 25 = 34, BD = 25 + 9 = 34 Hence it is a square. 32 MATHEMATICS TODAY | SEPTEMBER ‘17
31. (b) : Since maximum distance of any complex 39. (c) number from origin is given by | | therefore, | ω |= ω + 1 − 1 ω+ 1 1 1 40. (c) : Given M= z1 − z2 = (x1 − x2 ) + i( y1 − y2 ) ωω ≤ ω + ω = 2 + ω z1 − z2 (x1 − x2 ) − i( y1 − y2 ) | | 1+ i tan θ 1+ im ⇒ | |2 –2 | | –1 0 ⇒ | ω |≤ 2 ± 2 2 = 1− i tan θ = 1 − im 2 m2 Hence, max | | is 1+ 2 M = i 1 2m + 1 − m2 32. (b, c) : We have, + m2 1 + z = (z1 + z2 + ... + zn ) 1 + 1 + ... + 1 \\ M = cos 2q + i sin 2q = cis 2q ⇒ |M| = 1 z1 z2 zn Also, −ω2 = 1+i 3 = cis 2θ ⇒θ = π 2 6 = |z1 + z2 + ... + zn|2 |z1|2 + |z2|2 + |z3|2 + ... + |zn|2 = n2 41. (b) 33. (a, b) : x2 + ax + b = 0 ⇒ |a| 2 and |b| = 1 42. (d) − a ± a2 −4 b − a ±i 4− a2 43. (b) : Let a = cis 2π then a = a + a2 + a4, y= 2 = 2 b = a3 + a5 + a6 7 \\ |y| = 1 ⇒ a + b = a + a2 + a3 + a4 + a5 + a6 = –1 and ab = (a + a2 + a4)(a3 + a5 + a6) = 2 34. (a, d) : Let a is a real root then \\ Equation whose roots are a, b is x2 – x(–1) + 2 = 0 a3 + (3 + i)a2 – 3a = m + i f(x) + f(ax) + f(a2x) + f(a3x) + .... + f(a6x) ⇒ a3 + 3a2 – 3a – m = 0 or a2 – 1 = 0 = (1 + 1 + 1 + .... + 1) + 2x (1 + a + a2 + a3 + .... + a6) ⇒ a = 1 or –1 ⇒ m = 1 or 5 + 3x2 (1 + a2 + a4 + .... + a12) + .... + 7x6(1 + a6 + H\\S3o5e,.nzz(ca222e,,==c1)1z–:1, zWb3,e+⇒h32a–vze231=z=1,10z2,⇒z3 in G.P. a12 + a18 + .... + a36) = 7 b=3 Also, x7 −1 = (x – a)(x – a2)(x – a3)(x – a4)(x – a5) x −1 (x – a6) 36. (a, c) : (3z + 1)(4z + 1)(6z + 1)(12z + 1) = 2 ⇒ 8(3z + 1) 6 (4z + 1) 4 (6z + 1) 2 (12z + 1) Putting a = cis 2π of applying x 1 gives 7 =2×8×6×4×2 (24z + 8)(24z + 6)(24z + 4)(24z + 2) = 768 7 = 1 − cis 2π 1 − cis 4π .... 1 − cis 12π Let 24z + 5 = U 7 7 7 (U + 3)(U + 1)(U – 1)(U – 3) = 768 ⇒ (U2 – 9)(U2 – 1) = 768 7 = 2sin π 2sin 2π .... 2 sin 6π ⇒ U4 – 10U2 – 759 = 0 ⇒ U2 = 33 or –23 7 7 7 ⇒ 24z + 5 = ± 33 or ± i 23 π 2π 3π 4π 5π 6π 7 7 7 7 7 7 7 7 64 8 z=± 33 − 5 or ±i 23 − 5 ⇒ sin sin sin = sin sin sin = = 24 24 37. (a, b, c, d) : Let z1 = x1; z2, z3 = x2 ± iy2 44. (A) → (s), (B) → (q), (C) → (r) ⇒ z1 + z2 + z3 = −b (A) ||z – | – |z – ω ||max = | – ω | = 3 a (B) Q (–3, 0) \\ PQ = S1 = 12 b (C) 2z2 + 2z + k = 0 ⇒ x1 + 2x2 = − a < 0 ⇒ ab > 0 Also, z1 z2 z3 = x1 x22 + y22 = − d ⇒ ad > 0 \\ z = −2 ± 4 − 8k a 4 Since 'z' is a complex number 4 – 8k will be negative ( )Also bc − a2 < x1 x22 + y22 ⇒ bc > ad 1 2 38. (c) ⇒ k > MATHEMATICS TODAY | SEPTEMBER‘17 33
\\ Points are (0, 0), −1 , 2k −1 −1 , −1 2k −1 48. (1) : Clearly, a1 = 1 ; a2 = a; a3 = a2; a4 = a3; a5 = a4; 2 2 2 2 where a = ei2p/5 \\ l = a13001 = (a2)1001 = a2002 = a5 × 400 + 2 = a2 Since triangle is equilateral = (a4)669+1/3 = (a3)(669+1/3) = a2008 = a3 \\ 1 (2k − 1) + 1 = (2k − 1) ⇒ k = 2 / 3 = (a5)503+1/2 = (a4)503+1/2 = a2014 = a5·402+4 = a4 4 4 Also sum of 2011th power of roots of unity is 0 So, 1 + a2011 + l2011 + 2011 + v2011 = 0 45. (A) → (q), (B) → (r), (C) → (s), (D) → (q) l2011 + 2011 + v2011 = –(1 + a2011) l2011 + 2011 + v2011 = –(1 + a) (A) | z | = z1 + i z2 = | i || −iz1 + z2 | = 1 × z2 − iz1 =1 |l2011 + 2011 + v2011| = |–(1 + ei2p/5)| z2 + i z1 | z2 + iz1 | z2 − iz1 (B) Here, arg z − 2i = π or − π , So it represents a = |1 + cos 2p/5 + i sin 2p/5| = |2 cos p/5 (cos p/5 + i sin p/5)| z + 2i 2 2 semicircle with diametric ends at points (0, 2) and (0, –2) 5+ 1 5 +1 So, |z| = radius = 2 =2 4 = 2 \\ [|l2011 + 2011 + v2011|] = 1 (C) (x + 6)2 + y2 = (2x + 3)2 + (2y)2 ⇒ x2 + y2 + 12x + 36 = 4x2 + 4y2 + 12x + 9 49. (1) : Solving the equation of the lines we get ⇒ 3x2 + 3y2 = 27 ⇒ x2 + y2 = 9, |z| = 3 z = −z ⇒ z = i |a – i| = 2; a = i ± 2eiq , put it in the equation of the (D) z= 1 aw + bw2 + cw3 + 1 aw2 + bw3 + cw4 second line, we get cos q – sin q = 0 w c + aw + bw2 w2 b + cw + aw2 iπ = 1 × 1 + 1 ×1 = w2 + w = −1 ⇒ | z | = 1 w w2 α = i ± 2e 4 \\ x = ± 2 ⇒ [|x|] = 1 46. (5) : |z1 – z0| = |z2 – z0| = |z3 – z0| = |l| 50. (2) : Let x be the (2009)th root of unity ≠ 1, then x2009 – 1 = (x – 1)(x – w)....(x – w2008) Now, z3 − z0 = ei7π/11 = ei17π/44 Taking log on both sides, we get z2 − z0 eiπ/4 ln (x2009 – 1) = ln (x – 1) + ln (x – w) + ln (x – w2) +....+ ln (x – w2008) 17 π (where S represents z0) \\ Di erentiating both the sides w.r.t. x, we get 44 ⇒ ∠BSC = ∑(2009)x2008 1 2008 1 − r =1 − wr π x2009 −1 = x 1 + x ...(i) 88 ⇒ ∠BAC = 17 Putting x = 2 in equation (i), we get ∑ ( )12008 Similarly z2 − z0 = eiπ/4 ⇒ ∠ACB = π + r =1 1 = 2009 22008 z1 − z0 8 − wr 22009 −1 2 17π 15π ∴ ∠ ABC = π − π − 88 = 22 Multiplying both sides of above equation by (22009 – 1), 8 get 22009 −1 47. (3) : tan−1 x2 (a − b)y = π ( ) ∑we 2008 1 = 2009 ⋅ 22008 − 22009 +1 y2 − (a + b)x + ab 4 r=1 2 − wr \\ + = 22008 (2009 – 2) + 1 = 22008 2007 + 1 = [(a)(2b) + c] ⇒ x2 + y2 – (a + b)x – (a – b)y + ab = 0 \\ a = 2007, b = 2008, c = 1 Centre = 3+ i ⇒ a + b = 3 Hence a + b + c = 4016 2 34 MATHEMATICS TODAY | SEPTEMBER ‘17
CLASS XI Series 5 CBSE Permutations and Combinations | Binomial Theorem PERMUTATIONS AND COMBINATIONS X n factorial = n or n! = n(n − 1)(n − 2) ... 2.1 X nCr = n = C(n,r) = n! (r < n) = n Pr X 0 = 0! = 1 r r! r!(n − r)! X nPr = P(n, r) = n(n – 1)(n – 2) ... (n – r + 1) X nCr = nCn–r = n! (r < n) (n − r)! X nCx = nCy ⇒ x = y or x + y = n X nPn = n(n – 1) ... 2 .1 = n! X nCr + nCr–1 = n+1Cr X Number of permutations of n di erent things taken X nCr = n n−1Cr −1 r at a time when r (i) repetition is allowed = nr (ii) a particular thing is included = r · n–1Pr–1 X No. of combinations of n di erent things taken 'r' at (iii) a particular thing is not included = n–1Pr a time when 'p' particular things are always included X (i) No. of permutations when 'r' particular things = n–pCr–p are together = (n – r + 1)! r! X No. of combinations when 'p' particular things are (ii) If 'r' particular things are identical then no. of always excluded = n–pCr permutations = (n – r + 1)! X No. of ways of selecting zero or more things out of 'n' X No. of ways of arranging 'n' things when 'p' things are identical things = 1 + 1 + 1 + ... upto (n + 1) terms = n + 1 of one kind, 'q' things are of second kind and rest are all di erent = n! X No. of ways of selecting zero or more things out of n p!q! di erent things = nC0 + nC1 + ... + nCn = 2n MATHEMATICS TODAY | SEPTEMBER‘17 35
BINOMIAL THEOREM X (a + x)n = nC0 anx0 + nC1an–.1.x..1++nCnCn 2aa0xn–n2x2 + ...... (ii) rth term from end = rth term from beginning n ∈ I+ in the expansion of (y + x)n X General term in the expansion of (a + b)n is X Some Important Expansions Tr + 1 i.e., (r + 1)th term = nCr an – r . br h (1 + x)n = 1 + nC1 x + nC2 x2 + ..... + nCnxn X Middle term : In the expansion of (a + b)n, the n middle term is given by ∑= nCr xr r=0 h n + 1 th term, if n is even 2 h (1 – x)n = 1 – nC1 x + nC2 x2 – ..... + (–1)n nCnxn th + 1th n + 1 n + 1 n h 2 term and 2 term, if n is odd ∑= (−1)r.nCrxr X Number of terms in the binomial expansion r=0 of (a + x)n = n + 1 h (1 + x)–1 = 1 – x + x2 – x3+ ..... + (–1)rxr + ...... ∞ X In the expansion of (x + y)n h (1 + x)–2 = 1 – 2x + 3x2 – ..... + (–1)r(r + 1) xr + ..... ∞ (i) rth term from end = (n – r + 2)th term from h (1 – x)–1 = 1 + x + x2 + x3 + ..... + xr + ..... ∞ beginning h (1 – x)–2 = 1 + 2x + 3x2 + 4x3 + ..... + (r + 1) xr + ..... ∞ WORK IT OUT the same sequence of answers. What is the maximum number of students in the class for this to be possible? VERY SHORT ANSWER TYPE 10. Find 1. If (n + 2)! = 2550 (n!), nd n. (i) Coe cient of x2 in (1 – 2x + 3x2) (1 – x)14 (ii) Coe cient of x11 in the expansion of 2. Prove that : (2n)! = 2n (n!) [1 3 5 ... (2n–1)] (1 – 2x + 3x2) (1 + x)11. 3. Find the (n+1)th term from the end in the expansion of x − 1 3n . LONG ANSWER TYPE - I x 11 If C(n, r) : C(n, r + 1) = 1 : 2 and C(n, r + 1) : C(n, r + 2) 4. How many 4-letter words, with or without = 2 : 3, determine the values of n and r. meaning, can be formed out of the letters of the word 'LOGARITHMS' if repetition of letters is not 12. Assuming that x is so small that its second and allowed? higher powers may be neglected, simplify the following: (1 − 2x)2/3(4 + 5x)3/2 11 1− x Expand 1 , 0 2n 5. x − y y ≠ If xp occurs in the expansion of x2 1 , prove 13. + x SHORT ANSWER TYPE that its coe cient is (2n)! 6. Find the middle terms in the expansion of 4n − p ! 2n + p ! 3 3 x3 7 3x − 6 . 14. e letters of the word 'OUGHT' are written in all possible orders and these words are written out as 1 in a dictionary. Find the rank of the word 'TOUGH' 7. Find the greatest term in (5 + 2y)17 if y = 2 . in this dictionary. 8. Find r, if 5P(4, r) = 6P(5, r – 1). 15. In how many ways can 9 examination papers be arranged so that the best and the worst papers are 9. For a set of ve true or false questions, no student has never together? written all correct answer and no two students have given 36 MATHEMATICS TODAY | SEPTEMBER ‘17
LONG ANSWER TYPE - II 1 11 11C0x11 1 0 11C1x10 1 y y y 16. In how many ways can 3 prizes be distributed 5. x − = − among 4 boys, when (i) no boy gets more than 1 prize + 11C2 x 9 1 2 − 11C3x8 1 3 + 11C4 x 7 1 4 (ii) a boys may get any number of prizes y y y (iii) no boy gets all the prizes? − 11C5 x 6 1 5 + 11C6x5 1 6 − 11C7 x 4 1 7 17. Find a, b and n in the expansion of (a – b)n if the y y y rst three terms of the expansion are 729, 7290 and 30375 respectively. 18. How many numbers are there between 100 and 11C8x3 1 8 11C9 x 2 1 9 11C10x 1 10 1000 such that at least one of their digits is 7? y y y 19. cIfoea1,ciean2,ts ain3 thane dexpaa4nsbieonaonfy(1f+ouxr)n,cponrosveecuthtiavte + − + a1 + a3 = 2a2 − 11C11 1 11 a1 + a2 a3 + a4 a2 + a3 y 20. If p is nearly equal to q and n > 1, show that = x11 − 11 x10 + 55 x9 − 165 x8 + 330 x7 − 462 x6 1/n y y2 y3 y4 y5 (n + 1) p + (n − 1)q = p . (n − 1) p + (n + 1)q q 1/6. x5 x4 x3 x2 11x 1 Hence, nd the approximate value of 99 + 462 y6 − 330 y7 + 165 y8 − 55 y9 + y10 − y11 101 SOLUTIONS 6. Clearly, the given expansion contains 8 terms 1. (n + 2)! = 2550 (n!) 8 th 8 + 1th ⇒ (n + 2) (n + 1) (n!) = 2550 (n!) Middle terms are 2 2 terms, ⇒ (n + 2) (n + 1) = 2550 ⇒ n2 + 3n – 2548 = 0 \\ and ⇒ (n – 49) (n + 52) = 0 i.e,, 4th and 5th terms ⇒ n = 49 as n = –52 is rejected being n ∈ N. \\ n = 49 = (−1)3 7C3 ⋅ (3x)(7 − 3) x3 3 ==(2n)[2!(n=2n×()2([2nnn)((n2–n–2)–1()21(nn)(––2n24))–.....2.2)4(.2.1n2]]×–[([32(n)2n.–..–411).()32(2n.n2––.313))......33 Now, t4 = t3+1 ⋅ 6 2. . 1] . 1] = 2n (n!) [1 3 5 ... (2n – 1)]. = −35 × 81 × x4 × x9 = −105x13 . = 216 8 4 3. Given expansion is x − 1 3n And, t5 t4 +1 = (−1)4 ⋅ 7C4 ⋅ (3x)(7 − 4) ⋅ x3 x 6 (n + 1)th term from the end in the given expansion x3 x12 35x15 = (3n – n + 1)th i.e., (2n + 1)th term from the beginning = 35 × 27 × × 1296 = 48 . in the same expansion . 1 2n \\ Required term = T2n +1 = 3nC2n x 3n − 2n − x 7. Here, Tr + 1 = 17 −r +1⋅ 2y = 18 − r Q y = 1 Tr r 5 5r 2 = 3n ! ! xn (−1)2n = 3n ! 1 . \\ Tr + 1 > , = or < Tr according as 18 − r >, = or < 1, (2n)!n x 2n (2n)! n ! xn 5r 4. e word 'LOGARITHMS' contains 10 di erent Now, 18 – r = 5r ⇒ 18 = 6r ⇒ r = 3 letters. \\ Number of required words = number of Hence the 3rd and the 4th terms are equal when arrangements of 10 letters, taken 4 at a time 1 = 10P4 = 10 × 9 × 8 × 7 = 5040 y = 2 \\ T3 = T4 = 17C2(5)15 = 17 × 8 × (5)15. MATHEMATICS TODAY | SEPTEMBER‘17 37
8. 5P(4, r) = 6P(5, r – 1) ⇒ r+ 2 = 2 ⇒ 2n − 5r − 8 = 0 ...(ii) 4! 5! −r − 3 ⇒ 5 ⋅ (4 − r)! = 6 ⋅ [5 − (r − 1)]! n 1 5! 6 × 5! Solving (i) and (ii) we get, n = 14, r = 4 − r)! (6 − r)! ⇒ (4 = 12. Using binomial expansion, we have 1 6 (1 − 2x)2/3(4 + 5x)3/2 − r)! − r)(4 ⇒ (4 = (6 − r)(5 − r)! 1− x ⇒ (6 – r) (5 – r) = 6 ⇒ r2 – 11r + 30 = 6 = (1 − 2x)2/3 ⋅ 43/2 1 + 5 x 3/2 ⋅ (1 − x)−1/2 ⇒ (r – 3) (r – 8) = 0 4 ⇒ r = 3 or 8 Since P(n, r) is de ned only when r n, we reject = 8(1 − 2x)2/3 ⋅ 1 + 5 x 3/2 ⋅ (1 − x)−1/2 r = 8. Hence r = 3. 4 9. Clearly, there are 2 ways of answering each of the { } { }= 1+ 2 3 5 5 questions, i.e., true for false. 8 × 3 × (−2x) + ... × 1+ 2 × 4 x + ... \\ Total number of di erent sequences of answers × 1 + − 1 (− x) + .... = 2 × 2 × 2 × 2 × 2 = 32. 2 ere is only one all-correct answer. So, the maximum number of sequences leaving all- 4 15 1 correct answer is (32 – 1) = 31. = 8 × 1 − 3 x 1 + 8 x 1 + 2 x Since di erent students have given di erent sequences of answers, so the maximum possible number of students = 31. [neglecting x2 and higher powers of x] 10. (i) We have, (1 – 2x + 3x2) (1 – x)14 = (1 – 2x + 3x2) =8× 1 − 4 x + 15 x 1 + 1 x [neglecting x2] (\\×ii()1wC–eo1-4heCav1xeci+(e1n1t4–Co2f2xxx2+2 –=3.x.1.24. )Cto(211–5+2txe×)r1m114sC=)1(1+ 3 8 2 3 = 66 1+×111C–9x29 – 2x + 3x2) × = 8 × 1 + 13 x 1 + 1 x = 8 × 1 + 1 x + 13 x (1 +C11oCe1xc+ie1n1tCo2fxx21+1 ... + 1111CC1100x+103+×x1111C) 9 24 2 2 24 \\ = × [neglecting x2] 11 × 10 = 1− 2 × 11 + 3 × 2! = 144 = 8 × 1 + 25 x = 8 + 25 x . 24 3 11. C(n, r) : C(n, r + 1) = 1 : 2 1 2n ⇒ n! : n! = 1 : 2 13. e general term in the expansion of x2 + x is : r !(n − + 1)!(n − r)! (r r − 1)! 2nCr (x2 )2n −r 1 r x n! (r + 1) r ! (n − r − 1) ! 1 Tr +1 = ⋅ r ) (n n! 2 ⇒ r !(n − − r − 1) ! × = = 2nCr x4n − 2r ⋅ x−r = 2nCr x4n − 3r ⇒ r +1 = 1 ⇒ n − 3r − 2 = 0 ...(i) is contains xp if 4n − 3r = p ⇒ r = 4n − p n −r 2 3 Also C(n, r + 1) : C(n, r + 2) = 2 : 3 \\ Co-e cient of xp = 2nCr where r = 4n − p 3 ⇒ n! : n! = 2 : 3 (2n) ! (2n)! (r + 1)!(n − r − 1)! (r + 2)!(n − r − 2)! ! 2n − = 4n − 4n − = 4n − 2n + 3 p 3 p 3 p 3 p ! n! ! ! ⇒ − 1) × (r + 1) ! (n − r (n − r − 2) ! 14. Total number of letters in the word OUGHT is (r + 2)(r + 1)!(n − r − 2)! = 2 5 and all the ve letters are di erent. Alphabetical order n! 3 of letters is G, H, O, T, U. 38 MATHEMATICS TODAY | SEPTEMBER ‘17
Number of words beginning with G = 4! = 24 Dividing (i) by (ii), we get Number of words beginning with H = 4! = 24 Number of words beginning with O = 4! = 24 an = 729 = 1 ⇒ a = 1 ⇒ a = n ...(iv) Number of words beginning with TG = 3! = 6 nC1an −1b 7290 10 nb 10 b 10 Number of words beginning with TH = 3! = 6 Number of words beginning with TOG = 2! = 2 Dividing (ii) by (iii), we get Number of words beginning with TOH = 2! = 2 There will be two words beginning with TOU and nC1an −1b = 7290 TOUGH is rst among them nC2an − 2b2 30375 \\ Rank of word 'TOUGH' in the dictionary ⇒ nan −1b = 7290 = 6 = 24 + 24 + 24 + 6 + 6 + 2 + 2 + 1 = 89th n(n − 1) 30375 25 2 an − 2b2 15. us e number of ways onfuamrrbaenrgionfgp9erpmapuetrasti=on9Ps,9 =9! 2 6 Let rst consider the each − 25 containing the best and the worst papers together. ⇒ 1 × a = ...(v) n b Let us tie these two papers together and consider them as one paper. 2 × n = 6 (using (iv)) ⇒ n=6 n−1 10 25 en, these 8 tpwapoerpsacpaenrsbemararyanbgeedarinra8nPg8e=d8a!mwoanysg. Putting n = 6 in (i) we get, a6 = 729 ⇒ a = 3 A lso, these themselves in 2! i.e., 2 ways. Putting n = 6, a = 3 in (iv), we get b = 5 So, the number of permutations with the worst and the 18. We have to form 3-digit numbers such that at least best papers together = (2 × 8!). one of their digits is 7. Now, \\ e number of arrangements with best and worst (i) 3-digit numbers with 7 at the unit's place: papers never together = [(9!) –(2 × 8!)] = (9 – 2) × 8! = 7 × 8 ! = 282240 e number of ways to ll the hundred's place = 9 16. (i) e rst prize can be given away in 4 ways as [by any digit from 1 to 9]. it may be give to anyone of the 4 boys. So, the second prize can be given away in 3 ways, e number of ways to ll the ten's place = 10 and third prize can be given away to anyone of the [by any digit from 0 to 9]. remaining 2 boys in 2 ways. \\ e of ways in which all the prizes can be e number of ways to ll the unit's place = 1 given away = (4 × 3 × 2) = 24. [by 7 only] (ii) e rst prize can be given away in 4 ways as it may be given to anyone of the 4 boys. \\ Number of such numbers = 9 × 10 × 1 = 90. e second prize can be given away in 4 ways, (ii) 3-digit numbers with 7 at the ten's place but not at because there is no restriction as to the number of prizes a student gets. unit's place : Similarly, the third prize can be given away in 4 e number of ways to ll the hundred's place = 9 ways. Hence, the number of ways in which all the [by any digit from 1 to 9]. prizes can be given away = (4 × 4 × 4) = 64. e number of ways to ll the ten's place = 1 (iii) e number of ways in which a boy gets all the 3 [by 7 only] prizes is 4, as anyone of the 4 boys may get all the e number of ways to ll the unit's place = 9 prizes. \\ e number of ways in which a boy does not [by any digit from 0, 1, 2, 3, 4, 5, 6, 8, 9]. get all the prizes = 64 – 4 = 60 \\ Number of such numbers = 9 × 1 × 9 = 81. MPP-5 CLASS X I ANSWER KEY 17. T1 of (a + b)n = an = 729 ...(i) 1. (b) 2. (c) 3. (d) 4. (c) 5. (c) T2 of (a + b)n = nC1an – 1 b = 7290 ...(ii) T3 of (a + b)n = nC2a n – 2 b2 = 30375 ...(iii) 6. (d) 7. (a,b,c) 8. (b,c) 9 . (a,b,c) 10. (d) 11. (b, c) 12. (a,b,c,d) 13. (a,d) 14. (c) 15. (b) 16. (a) 17. (4) 18. (5) 19. (3 ) 20. (8) MATHEMATICS TODAY | SEPTEMBER‘17 39
(iii)3-digit numbers with 7 at the hundred's place but 1 n +1 h 1 + n −1 h −1 neither at the unit's place nor at ten's place: 2nq 2nq e number of ways to ll the hundred's place = 1. = + [by 7 only]. e number of ways to ll the ten's place = 9. = 1 + n +1 h 1 − n −1 h [by any digit from 0, 1, 2, 3, 4, 5, 6, 8, 9]. 2nq 2nq e number of ways to ll the unit's place = 9 [by any digit from 0, 1, 2, 3, 4, 5, 6, 8, 9]. [expanding the 2nd expression and neglecting h2, h3, etc] \\ Number of such numbers = 1 × 9 × 9 = 81. Hence, the number of required type of numbers = 1 + n +1 h − n −1 h [neglecting h2] = (90 + 81 + 81) = 252. 2nq 2nq = 1 + h . nq 19. (Gr i+ve1n),tha, 1(,r a+2,2)at3h and (ar4+ar3e)ththteercmose cients of the Also, p 1/n = q + h 1/n = 1 + h 1/n = 1 + h . rth, and respectively in q q q nq the expansion of (1 + x)n aaaN243o===w,(((rrra1+++=231)))rtttthhhh coe cient = nnCCrr ...(i) [Expanding and neglecting h2 and higher powers of h] coe cient = nnCCrr – 1 ...(ii) 1/n coe cient = ...(iii) ∴ (n + 1)p + (n − 1)q = p coe cient = + 1 ...(iv) (n − 1)p + (n + 1)q q + 2 Deduction : Taking p = 99, q = 101 and n = 6 in the Now, a2 = n Cr = n! ⋅ (r − 1)!(n − r + 1)! above result, we get a1 nCr −1 !(n − n! 1/6 r r)! = n−r +1 99 = (6 + 1) × 99 + (6 − 1) × 101 = 1198 = 599 . 101 (6 − 1) × 99 + (6 + 1) × 101 1202 601 ∴ a2 + a1 = a2 + 1 = n − r + 1 + 1 = n + 1 r a1 a1 r r MADHYA PRADESH at ∴ a1 = r ...(v) • Telang Book Corner - Bhopal a1 + a2 n +1 Ph: 0755-2741446; Mob: 9827058587, 9425301772 Putting r + 1 in place of r in (v), we get r +1 • Books N Books - Bhopal a2 = n +1 ...(vi) a2 + a3 ...(vii) Ph: 0755-2673602, 2574603; Mob: 9826086858, 9826431103 a3 r+2 Similarly, a3 + a4 = n +1 • UBS Publishers Distributors Pvt Ltd. - Bhopal ∴ a1 + a3 = n r 1 + r+2 Ph: 0755-4203183, 4203193, 2555285; Mob: 9425013150 a1 + a2 a3 + a4 + n +1 • Krishna Pustak Kendra - Gwalior = 2(r + 1) = 2a2 n +1 a2 + a3 Ph: 0751-2625982; Mob: 9425118370, 9425110809, 8989475382 20. Let p = q + h, where h is so small that its square and • New Bhartiya Prakashan - Gwalior Ph: 0751-2625534, 2423004; Mob: 9407079998 higher powers may be neglected. en, • Chelawat Book Depot - Indore Ph: 0731-3027430; Mob: 9926500362, 9479723775 • Shiv Shakti Book Centre - Indore Ph: 0731-2459956, 2459955; Mob: 9425063647 (n + 1)p + (n − 1)q = (n + 1)(q + h) + (n − 1)q • Arun Prakashan - Indore (n − 1)p + (n + 1)q (n − 1)(q + h) + (n + 1)q Ph: 0731-2454372, 4054544; Mob: 9826369234, 9425313294 = 2nq + (n + 1)h = 1 + n +1 h 2nq + (n − 1)h 2nq • Bhaiya Store - Indore Ph: 0731-2531207, 2540166; Mob: 9425318103 1 + • Case Traders - Indore Mob: 9827259249, 9827620395 n−1 h • Chelawat Book Centre - Indore Mob: 9406666147 2nq • Jain Book Palace - Indore Ph: 0371-4054817, 4054839; Mob: 9926636333 • Kitab Kunj - Indore Ph: 0731-4054442; Mob: 9826325258, 9926058881 • Sushil Prakashan - Indore Ph: 0731-2535892; Mob: 9425322330 • New Radhika Book Palace - Jabalpur Mob: 9425411566, 7614083315 • Shankar Competition Book House - Rewa Ph: 07662-403161, 421128; Mob: 9407041342 • M.P. Publishers & Suppliers - Ujjain Ph: 0734-2515680, 2514262; Mob: 9826099912 • O.K. Book Depot - Ujjain Mob: 9425091418 [on dividing num. and denom. by 2nq] 40 MATHEMATICS TODAY | SEPTEMBER ‘17
Class XI This 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. Binomial Theorem Total Marks : 80 Time Taken : 60 Min. Only One Option Correct Type One or More Than One Option(s) Correct Type 1. If O be the sum of odd terms and E that of even 7. If (1 + x –2x2)6 = 1 + a1x + a2x2 + a12x12, then terms in the expansion of (x + a)n, then O2 – E2 = (a) a2 + a4 + a6 + ...... + a12 = 31 (a) (x2 + a2)n (b) (x2 – a2)n (b) a1 + a3 + a5 + ...... + a11 = –32 (c) a1 + a2 + a3 + ...... + a12 = –1 (c) (x – a)2n (d) none of these (d) none of these 2. (115)96–(96)115 is divisible by 8. Which of the following is an even positive integer? (a) ( 3 + 1)2n +1 + ( 3 − 1)2n+1 (a) 15 (b) 17 (b) ( 3 + 1)2n + ( 3 − 1)2n (c) 19 (d) 21 (c) ( 3 + 1)2n+1 − ( 3 − 1)2n+1 3. e number of terms in the expansion of (a + b + c)10 is (d) none of these (a) 11 (b) 21 If (1 + x)n = C0 + C1x ..+....+C+2n(xCn2n++=1..n). +2CnnC–1=nx2nn,–t1h(enn (c) 55 (d) 66 9. (a) C1 + 2C2 + 3C3 + (b) C0 + 2C1 + 3C2 + 4. e coe cient of the term independent of x in the + 2) expansion of (1 + x + 2x3 ) 3 x2 − 1 9 , is (c) C0 + C1 + C2 + ... + Cn = 2n+1 − 1 2 3x 2 3 n +1 n +1 (a) 1 (b) 19 (c) 17 (d) 1 (d) none of these 3 54 54 4 n r−1 m 1i0 20 10. nCr rC p 2 p is equal to i=0 m− ∑∑ ∑5. , p e sum i where q = 0 if p < q, r =1 p=0 is maximum when m is (a) 4n – 3n + 1 (b) 4n – 3n –1 (c) 4n – 3n + 2 (d) 4n –3n (a) 5 (b) 10 1 20 (c) 15 (d) 20 11. In the expansion of 3 4 + 46 6. If C0, C1, C2, ....., Cn denote the coe cients in the expansion of (1 + x)n, then (a) the number of rational terms = 4 C0 + 3·C1 + 5·C2 + ..... + (2n + 1)Cn = (b) the number of irrational terms = 19 (a) n·2n (b) (n – 1)2n (c) the middle term is irrational (c) (n + 1)2n–1 (d) the number of irrational terms = 17 (d) (n + 1)2n MATHEMATICS TODAY | SEPTEMBER‘17 41
12. Coe cient of xn in the expansion of (1 + x)2n is (a) 9n −1 (b) 32n −1 equal to 2 4 (a) P(2n,n) (c) 32n +1 (d) 9n +1 n 4 2 (b) (n +1)(n + 2)(n + 3).....(2n) Matrix Match Type n 16. Match the following : (c) C(2n, n) Column I Column II (d) 1⋅3⋅5.....⋅(2n −1) 2n P. If last digit of the number 1. l+ =6 n 999 is l and last digit of 2l100 13. If n is a positive integer and (3 3 + 5)2n+1 = α + β, is , then Q. If last digit of the number 2. l + = 11 where a is an integer and 0 < b < 1, then 2999 is l and last digit of (a) a is an even integer 3lll is , then (b) (a + b)2 is divisible by 22n+1 3. l– =4 (c) the integer just below (3 3 + 5)2n+1 divisible Let a = 72! −1 is divisible R. (36!)2 by 3 (d) a is divisible by 10 by 10l + , then 4. l + l = 9 Comprehension Type P QR (a) 2 43 ∑Consider (1 + x + x2)2n = 4n where (b) 3 41 (c) 1 43 ar xr , r =0 (d) 2 13 a0, a1, a2, ...., a4n are real numbers and n is a positive Integer Answer Type integer. 17. e remainder when 9103 is divided by 25 is equal n−1 to 18. If (2r + 3)th and (r – 1)th terms in the expansion of ∑14. e value of a2r is r =0 (1 + x)15 have equal coe cients then r is (a) 9n − 2a2n −1 (b) 9n + 2a2n +1 19. If the coe cient of the middle term in the binomial 4 4 expansions of (1 + ax)4 and (1 – ax)6 is same, then –10a equals (c) 9n − 2a2n +1 (d) 9n + 2a2n −1 4 4 20. If n–1C3 + n–1C4 > nC3, then, the least value of n would be n ∑15. e value of a2r−1 is r =1 Keys are published in this issue. Search now! J Check your score! If your score is > 90% EXCELLENT WORK ! You are well prepared to take the challenge of final exam. No. of questions attempted …… 90-75% GOOD WORK ! You can score good in the final exam. No. of questions correct …… 74-60% SATISFACTORY ! You need to score more next time. Marks scored in percentage …… < 60% NOT SATISFACTORY! Revise thoroughly and strengthen your concepts. 42 MATHEMATICS TODAY | SEPTEMBER ‘17
This column is aimed at Class XII 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. *ALOK KUMAR, B.Tech, IIT Kanpur FUNCTIONS Testing for a function by vertical line test : A Functions can be easily de ned with the help of concept relation f : A B is a function or not, it can be mapping. Let X and Y be any two non-empty sets. “A checked by a graph of the relation. If it is possible function from X to Y is a rule or correspondence that to draw a vertical line which cuts the given curve at assigns to each element of set X, one and only one more than one point then the given relation is not element of set Y”. Let the correspondence be ‘f ’ then a function and when this vertical line means line mathematically we write f : X Y where y = f(x), x ∈ X parallel to Y-axis cuts the curve at only one point and y ∈ Y. We say that ‘y’ is the image of ‘x’ under f then it is a function. Figure (iii) and (iv) represents (or x is the pre image of y). a function. YY A mapping f : X Y is said to be a function if each element in the set X has its image in set Y. It is also X′ O X possible that there are few elements in set Y which X′ O X are not the images of any element in set X. Y′ (ii) Y′ (i) Every element in set X should have one and only one image. at means it is impossible to have more than YY one image for a speci c element in set X. Functions can not be multi-valued (A mapping that is multi- valued is called a relation from X and Y) e.g. X′ O X X′ O X Y′ (iii) Y′ (iv) Set X Set Y Set X Set Y a 1a 1 Number of functions : Let X and Y be two nite sets having m and n elements respectively. en b 2b 2 each element of set X can be associated to any one of n elements of set Y. So, total number of functions c3 3 from set X to set Y is nm. Function Function Set X Set Y Set X Set Y Real valued function : If R, be the set of real a a numbers and A, B are subsets of R, then the function b 1 b 1 f : A B is called a real function or real valued c 2 c 2 function. 3 3 Not Function Not Function * Alok Kumar is a winner of INDIAN NAtIoNAl MAtheMAtIcs olyMpIAD (INMo-91). he trains IIt and olympiad aspirants. MATHEMATICS TODAY | SEPTEMBER‘17 43
DOMAIN, CO DOMAIN AND RANGE OF Equal functions : Two function f and g are said to FUNCTION be equal functions, if and only if (i) Domain of f = Domain of g If a function f is de ned from a set A to set B then (ii) Co-domain of f = Co-domain of g (f : A B) set A is called the domain of f and set B (iii) f(x) = g(x) x ∈ their common domain is called the co-domain of f. e set of all f-images of the elements of A is called the range of f. KINDS OF FUNCTION In other words, we can say One-one function (injection) : A function f : A B Domain = All possible values of x for which f(x) is said to be a one-one function or an injection, if exists. di erent elements of A have di erent images in B. Range = For all values of x, all possible values e.g. Let f : A B and g : X Y be two functions of f(x). represented by the following diagrams. Range Af B Xg Y a1 b1 x1 y1 Domain Co-domain a2 b2 x2 y2 a3 b3 x3 y3 Af B Domain = (a, b, c, d) = A a4 b4 x4 y4 a p b5 y5 b c q Co-domain = (p, q, r, s) = B Clearly, f : A B is a one-one function. But d g : X Y is not one-one function because two r Range = (p, q, r) distinct elements and have the same image s under function g. x1 x3 Methods for nding domain and range of function Domain : Expression under even root (i.e., square root, fourth root etc.) ≥ 0. Denominator ≠ 0. Method to check the injectivity of a function If domain of y = f(x) and y = g(x) are and (i) Take two arbitrary elements x, y (say) in the respectively then the domain of D1 D2 domain of f. (i) f(x) g(x) or f(x) g(x) is D1 D2. (ii) Solve f(x) = f(y). If f(x) = f(y) gives x = y only, (ii) f (x) is D1 D2 – {g(x) = 0} g(x) then f : A B is a one-one function (or an ( )(iii) f (x) = D1 {x : f(x) ≥ 0} injection). Otherwise not. Range : Range of y = f(x) is collection of all outputs If function is given in the form of ordered pairs f(x) corresponding to each real number in the and if two ordered pairs do not have same second domain. element then function is one-one. If domain ∈ nite number of points ⇒ range If the graph of the function y = f(x) is given and ∈ set of corresponding f(x) values. each line parallel to x-axis cuts the given curve at maximum one point then function is one-one. e.g. YY If domain ∈ R or R – [some nite points]. en (0, 1) express x in terms of y. From this nd y for x to be de ned (i.e., nd the values of y for which X′ O X X′ O X x exists). Y′ f(x) = ax + b f(x) = ax (0 < a < 1) If domain ∈ a nite interval, nd the least and Y′ greatest value for range using monotonicity. Number of one-one functions (injections) : If Algebra of functions : For functions f : x R; A and B are nite sets having m and n elements g : x R, we have x ∈ X, respectively, then number of one-one functions (cf )(x) = cf(x), where c is a scalar. (f g)(x) = f(x) g(x). MfroamnyA-otnoeBfu=nctni0Pomn, ,:ifiAfnnf<u≥mnmction f : A (fg)(x) = (gf )(x) = f(x) g(x). B is said f f (x) to be a many-one function if two or more elements g (x) = g (x) , g(x) ≠ 0 of set A have the same image in B. 44 MATHEMATICS TODAY | SEPTEMBER‘17
us, f : A B is a many-one function if there Method to nd onto or into function : exist x, y ∈ A such that x ≠ y but f(x) = f(y). (i) Solve f(x) = y by taking x as a function of y In other words, f : A B is a many-one function if it is not a one-one function. i.e., g(y)(say). (ii) Now if g(y) is de ned for each y ∈ co-domain A f B Xg Y a1 bbbbbb154326 x1 y1 and g(y) ∈ domain then f(x) is onto and if any a2 x2 y2 one of the above requirements is not ful lled, a3 x3 y3 then f(x) is into. a4 x4 y4 a5 x5 y5 One-one onto function (bijection) : A function f : A B is a bijection if it is one-one as well as onto. If function is given in the form of set of ordered pairs In other words, a function f : A B is a bijection if and the second element of atleast two ordered pairs (i) It is one-one i.e., f(x) = f(y) ⇒ x = y for all x, are same then function is many-one. If the graph of y = f(x) is given and the line parallel y ∈ A. to x-axis cuts the curve at more than one point then (ii) It is onto i.e., for all y ∈ B, there exists function is many-one. x ∈ A such that f(x) = y. YY AB X′ O f(x) = x2 X X′ O f(x) = |x| X Y′ Y′ a1 b1 a2 b2 a3 b3 a4 b4 Onto function (surjection) : A function f : A B Clearly, f is a bijection since it is both injective as well as surjective. is onto if each element of B has its pre-image in A. Number of one-one onto function (bijection) : If In other words, Range of f = Co-domain of f. e.g. e following arrow-diagram shows onto function. A and B are nite sets and f : A B is a bijection, then A and B have the same number of elements. Af B XgY If A has n elements, then the number of bijection a1 b1 x1 y1 a2 from A to B is the total number of arrangements of a3 b2 x2 x3 y2 n items taken all at a time i.e. n!. b3 x4 y3 Algebraic functions : Functions consisting of nite Number of onto function (surjection) : If A and B number of terms involving powers and roots of the are two sets having m and n elements respectively such that 1 n m, then number of onto functions independent variable and the four fundamental n operations +, –, × and are called algebraic ∑from A to B is (−1)n−r nCrrm. functions. r =1 3 (ii) x +1 , x ≠ 1 Into function : A function f : A B is an into x −1 function if there exists an element in B having no e.g., (i) x 2 + 5x pre-image in A. (iii) 3x4 – 5x + 7 In other words, f : A B is an into function if it is not an onto function e.g., e following arrow- Transcendental function : A function which is not diagram shows into function. algebraic is called a transcendental function. e.g., trigonometric; inverse trigonometric , exponential A fB X gY and logarithmic functions are all transcendental a1 b1 x1 y1 functions. a2 b2 a3 b3 x2 y2 Trigonometric function : A function is said to y3 be a trigonometric function if it involves circular functions (sine, cosine, tangent, cotangent, secant, x3 y4 cosecant) of variable angles. MATHEMATICS TODAY | SEPTEMBER‘17 45
Sine function Cotangent function Y (p/2, 1) Y (–3p/2, 1) (–3p/2, 0) (–p/2, 0) (3p/2, 0) X′ O (p/2, 0) X′ –p O p X (–p/2, –1) Y′ x = –2p x = –p Y′ x = p x = 2p Domain = R ; Range = [–1, 1] (3p/2, –1) Domain = R – {np | n ∈ I}; Range = R Inverse trigonometric functions Cosine function Function Domain Range De nition Y (0, 1) of the function sin–1 x [–1, 1] [–p/2, p/2] y = sin–1 x cos–1 x x = sin y X′ p O p X tan–1 x 22 cot–1 x cosec–1 x [–1, 1] [0, p] y = cos–1 x x = cos y (–p, –1) Y′ (p, –1) Domain = R; Range = [–1, 1] (–∞, ∞) or R (–p/2, p/2) y = tan–1 x x = tan y Tangent function (–∞, ∞) or R (0, p) y = cot–1 x Y x = cot y X′ –3p/2 –p O p X R – (–1, 1) [–p/2, p/2] y = cosec–1 x –p/2 p/2 3p/2 – {0} x = cosecy sec–1 x R – (–1, 1) [0, p] – {p/2} y = sec–1 x x = sec y Y′ Domain = R – {(2n + 1) p/2 |n ∈ I}; Range = R Exponential function : Let a ≠ 1 be a positive real number. en f : R (0, ∞) de ned by f(x) = ax Cosecant function called exponential function. Its domain is R and range is (0, ∞). (–3p/2, 1) Y (p/2, 1) y=1 Y a>1 a<1 Y X′ O y = –X1 (0, 1) f(x) = ax f(x) = ax (0, 1) X′ O X X′ O X (–p/2, –1) (–3p/2, –1) Y′ Y′ Y′ x = p x = 2p x = –2p x = –p Graph of f(x) = ax, when a > 1 Graph of f(x) = ax, when a < 1 Domain = R – {(np|n ∈ I}; Range = (–∞, –1] [1, ∞) Logarithmic function : Let a ≠ 1 be a positive real number. en f : (0, ∞) R de Itnseddobmyafi(nx)is=(l0o,g∞ax) Secant function is called logarithmic function. (–2p, 1) Y (2p, 1) and range is R. (0, 1) YY X′ O X f(x) = logax (1, 0) X′ O (1, 0) X X′ O X (–p, –1) Yx′ (p, –1) f(x) = logax x = –3p/2 x = –p/2 p/2 x= = 3p/2 [1, ∞) Y′ Y′ Domain = R – {(2n + 1) p/2 |n ∈ I}; Range = (–∞, –1] Graphwhofenf(xa)>=1logax, Graphwhofenf(xa)<=1logax, 48 MATHEMATICS TODAY | SEPTEMBER‘17
Explicit and implicit functions : A function is said Y to be explicit if it can be expressed directly in terms 3 of the independent variable otherwise is an implicit. 2 e.g., y = sin–1 x + log x is explicit function, while x2 + y2 = xy and x3y2 = (a – x)2 (b – y)2 are implicit 1 functions. X′ –3 –2 –1 O –11 2 3 X Constant function : Let –2 k be a xed real number. Then a function f(x) Y –3 given by f(x) = k for all X′ k f(x) = k x ∈ R is called a constant Y′ function. e domain of OX Signum function : The Y Y′ function defined by (0, 1) (x) | x | , x ≠ 0 o r X′ O X x (0, –1) f = 0, x = 0 Y′ the constant function f(x) = k is the complete set 1, x > 0 of real numbers and the range of f is the singleton set {k}. The graph of a constant function is a f (x) = 0, x = 0 is called the signum function. straight line parallel to x-axis as shown in gure and it is above or below the x-axis according as k e dom−a1in, xis<R0and the range is the set {–1, 0, 1}. is positive or negative. If k = 0, then the straight line coincides with x-axis. Reciprocal function : e function that associates 1 is Identity function : e Y each non-zero real number x to be reciprocal xand function defined by called the reciprocal function. e domain f(x) = x for all x ∈ R, f(x) = x range of the reciprocal is called the identity X′ OX function are both equal Y f(x) = 1/x function on R. Clearly, Y′ to R – {0} i.e., the set of X′ OX the domain and range of all non-zero real numbers. e graph is the identity function is R. e graph of the identity as shown. Y′ function is a straight line passing through the Power function : A function f : R R de ned by, f(x) = xa, a ∈ R is called a power function. origin and inclined at an angle of 45° with positive direction of x-axis. EVEN AND ODD FUNCTION Even function : If f(–x) = f(x), x ∈ domain then Modulus function : The function defined by f(x) is called even function. e.g., f(x) = ex + e–x, f(x) = x2, f(x) = x sin x, f(x) = cos x, f(x) = x2 cos x, f (x) =| x |= x, when x≥0 is called the modulus all are even functions. −x, when x<0 Odd function : If we put (–x) in place of x in the given function and if f(–x) = –f(x), x ∈ domain function. The domain Y then f(x) is called odd function. e.g., f(x) = ex – ex, f(x) = sin x, f(x) = x3, f(x) = xcos x, f(x) = x2 sin x, of the modulus function f(x) =–x f(x) = x all are odd functions. is the set R of all real X′ OX numbers and the range is Properties of even and odd function (i) e graph of even function is always symmetric the set of all non-negative Y′ with respect to y-axis. The graph of odd real numbers. function is always symmetric with respect to origin. Greatest integer function : Let f(x) = [x] where [x] (ii) e product of two even functions is an even denotes the greatest integer less than or equal to x. function. e domain is R and the range is I. e.g. [1.1] = 1, [2.2] = 2, [– 0.9] = –1, [– 2.1] = – 3 etc. e function f de ned by f(x) = [x] for all x ∈ R, is called the greatest integer function. MATHEMATICS TODAY | SEPTEMBER‘17 49
(iii) e sum and di erence of two even functions f –1 : B A, f –1(b) = a ⇒ f (a) = b is an even function. In terms of ordered pairs inverse function is de ned as f –1 = (b, a) if (a, b) ∈ f. (iv) e sum and di erence of two odd functions For the existence of inverse function, it should be is an odd function. one-one and onto. (v) e product of two odd functions is an even Properties of Inverse function function. (i) Inverse of a bijection is also a bijection and (vi) e product of an even and an odd function is unique. is an odd function. It is not essential (ii) (f –1)–1 = f that every function is even or odd. It is (iii) If f : A B is bijection and g : B A is possible to have some functions which are neither even nor odd function. e.g., wsinehvtseerrAese,aIonAfdafn.BdrIeBesnpaerfecotgiidv=eenlyItB.itaynfdungcotfio=nIsAo. n the (v) If f : A B and g : B C are two bijections, then (vii) f(x)e=suxm2 +ofxe3v, ef(nxa)n=dloodgedxfu, fn(cxt)io=neixs.neither even nor odd function. gof : A C is bijection and (gof )–1 = ( f–1og–1). (iv) fog ≠ gof but if, fog = gof then either f –1 = g or (viii) Zero function f(x) = 0 is the only function which is even and odd both. g –1 = f also, ( fog)(x) = (gof )(x) = (x). PERIODIC FUNCTION PROBLEMS A function is said to be periodic function if its each value is repeated a er a de nite interval. So Single Correct Answer Type a function f(x) will be periodic if a positive real number T exist such that, f(x + T) = f(x), x ∈ 1. If f (x) = log 1 + x , then f 2x is equal to domain. Here the least positive value of T is called 1 − x 1+ x2 the period of the function. (a) [f(x)]2 (b) [f(x)]3 (c) 2f(x) (d) 3f(x) COMPOSITE FUNCTION If f : A B and g : B C are two functions then 2. If f (x) = x −3 , then f [f {f (x)}] equals −1 the composite function of f and g, (a) x x +1 –x x gof : A C will be de ned as (b) (c) x (d) gof(x) = g(f(x)), x ∈ A 2 Properties of composition of function 3. If f (x) = cos(logx), then the value of (i) f is even, g is even ⇒ fog is even function. (ii) f is odd, g is odd ⇒ fog is odd function. f (x)⋅ f (4) − 1 f x + f (4x) (iii) f is even, g is odd ⇒ fog is even function. 2 4 (iv) f is odd, g is even ⇒ fog is even function. (v) Composite of functions is not commutative (a) 1 (b) –1 (c) 0 (d) 1 i.e., fog ≠ gof. (vi) Composite of functions is associative i.e., 4. Let f : R R be de ned by f(x) = 2x + |x|, then (fog)oh = fo(goh) f(2x) + f (–x) – f(x) = (vii)Function gof will exist only when range of (a) 2x (b) 2|x| (c) –2x (d) –2|x| f is the subset of domain of g. (viii) If both f and g are one-one, or onto then 5. If f(x + ay, x – ay) = axy, then f(x, y) is equal to fog and gof are also one-one or onto. (a) xy (b) x2 – a2y2 x2 − y2 x2 − y2 INVERSE FUNCTION (c) 4 (d) a2 If f : A B be a one-one onto (bijection) function, then the mapping f –1 : B A which associates 6. If f(x) = cos[p2]x + cos[–p2]x, then each element b ∈ B with element a ∈ A, such that f(a) = b, is called the inverse function of the (a) f π = 2 (b) f (–p) = 2 function f : A B. 4 (c) f (p) = 1 (d) f π = −1 2 7. If e f (x) = 10 + x , x ∈(−10, 10) and 10 − x 50 MATHEMATICS TODAY | SEPTEMBER‘17
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