JEE-Mathematics EXERCISE - 01 CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . The lines 2x – 3y = 5 and 3x – 4y = 7 are diameters of a circle of area 154 sq. units. The equation of the circle is - (A) x2 + y2 – 2x – 2y = 47 (B) x2 + y2 – 2x –2y = 62 (C) x2 + y2 – 2x + 2y = 47 (D) x2 + y2 – 2x + 2y = 62 2 . If a be the radius of a circle which touches x-axis at the origin, then its equation is - (A) x2 + y2 + ax = 0 (B) x2 + y2 ± 2ya = 0 (C) x2 + y2 ± 2xa = 0 (D) x2 + y2 + ya = 0 3 . The equation of the circle which touches the axis of y at the origin and passes through (3,4) is - (A) 4(x2 + y2) – 25x = 0 (B) 3(x2 + y2) – 25x = 0 (C) 2(x2 + y2) – 3x = 0 (D) 4(x2 + y2) – 25x + 10 = 0 4 . The equation of the circle passing through (3,6) and whose centre is (2,–1) is - (A) x2 + y2 – 4x + 2y = 45 (B) x2 + y2 – 4x – 2y + 45 = 0 (C) x2 + y2 + 4x – 2y = 45 (D) x2 + y2 – 4x + 2y + 45 = 0 5 . The equation to the circle whose radius is 4 and which touches the negative x-axis at a distance 3 units from the origin is - (A) x2 + y2 – 6x + 8y – 9 = 0 (B) x2 + y2 ± 6x – 8y + 9 = 0 (C) x2 + y2 + 6x ± 8y + 9 = 0 (D) x2 + y2 ± 6x – 8y – 9 = 0 6 . The equation of a circle which passes through the three points (3,0) (1,–6),(4,–1) is - (A) 2x2 + 2y2 + 5x –11y + 3 = 0 (B) x2 + y2 – 5x +11y – 3 = 0 (C) x2 + y2 + 5x –11y + 3 = 0 (D) 2x2 + 2y2 – 5x +11y – 3 = 0 7 . y 3 x c1 & y 3 x c 2 are two parallel tangents of a circle of radius 2 units, then c1 c2 is equal to - (A) 8 (B) 4 (C) 2 (D) 1 8 . Number of different circles that can be drawn touching 3 lines, no two of which are parallel and they are neither coincident nor concurrent, are - (A) 1 (B) 2 (C) 3 (D) 4 9 . B and C are fixed points having co-ordinates (3, 0) and (–3, 0) respectively. If the vertical angle BAC is 90°, then the locus of the centroid of the ABC has the equation - (A) x2 + y2 = 1 (B) x2 + y2 = 2 (C) 9(x2 + y2) = 1 (D) 9(x2 + y2) = 4 1 0 . If a circle of constant radius 3k passes through the origin ‘O’ and meets co-ordinate axes at A and B then the Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 locus of the centroid of the triangle OAB is - (A) x2 + y2 = (2k)2 (B) x2 + y2 = (3k)2 (C) x2 + y2 = (4k)2 (D) x2 + y2 = (6k)2 1 1 . The area of an equilateral triangle inscribed in the circle x2 + y2 – 2x = 0 is : 33 33 33 (D) none (A) (B) (C) 2 4 8 1 2 . The length of intercept on y-axis, by a circle whose diameter is the line joining the points (–4,3) and (12, –1) is - (A) 3 2 (B) 13 (C) 4 13 (D) none of these 1 3 . The gradient of the tangent line at the point (a cos , a sin ) to the circle x2 + y2 = a2, is - (A) tan ( – ) (B) tan (C) cot (D) – cot 1 4 . x + my + n = 0 is a tangent line to the circle x2 + y2 = r2 , if - (A) 2 + m2 = n2 r2 (B) 2 + m2 = n2 + r2 (C) n2 = r2(2 + m2) (D) none of these 66 E
JEE-Mathematics 1 5 . Line 3x + 4y = 25 touches the circle x2 + y2 = 25 at the point - (A) (4, 3) (B) (3, 4) (C) (–3, –4) (D) none of these 1 6 . The equations of the tangents drawn from the point (0,1) to the circle x2 + y2 – 2x + 4y = 0 are - (A) 2x – y + 1 = 0, x + 2y – 2 = 0 (B) 2x – y – 1 = 0, x + 2y – 2 = 0 (C) 2x – y + 1 = 0, x + 2y + 2 = 0 (D) 2x – y – 1 = 0, x + 2y + 2 = 0 1 7 . The greatest distance of the point P(10,7) from the circle x2 + y2 – 4x – 2y – 20 = 0 is - (A) 5 (B) 15 (C) 10 (D) None of these 3 3 18. The equation of the normal to the circle x2 + y2 = 9 at the point , is - 2 2 (A) x – y 2 (B) x + y = 0 (C) x – y = 0 (D) none of these 3 1 9 . The parametric coordinates of any point on the circle x2 + y2 – 4x – 4y = 0 are- (A) (–2 + 2cos, –2 + 2 sin) (B) (2 + 2cos, 2 + 2 sin) (C) (2 + 2 2 cos, 2 + 2 2 sin) (D) (–2 + 2 2 cos, –2 + 2 2 sin) 2 0 . The length of the tangent drawn from the point (2,3) to the circles 2(x2 + y2) – 7x + 9y – 11 = 0 - (A) 18 (B) 14 (C) 14 (D) 28 2 1 . A pair of tangents are drawn from the origin to the circle x2 + y2 + 20(x + y) + 20 = 0. The equation of the pair of tangents is - (A) x2 + y2 + 5xy = 0 (B) x2 + y2 + 10xy = 0 (C) 2x2 + 2y2 + 5xy = 0 (D) 2x2 + 2y2 – 5xy = 0 2 2 . Tangents are drawn from (4, 4) to the circle x2 + y2 – 2x – 2y – 7 = 0 to meet the circle at A and B. The length of the chord AB is - (A) 2 3 (B) 3 2 (C) 2 6 (D) 6 2 2 3 . The angle between the two tangents from the origin to the circle (x –7)2 + (y + 1)2 = 25 equals - (D) none (A) (B) (C) 2 3 4 2 4 . Pair of tangents are drawn from every point on the line 3x + 4y = 12 on the circle x2 + y2 = 4. Their variable chord of contact always passes through a fixed point whose co-ordinates are - Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 (A) 4 , 3 (B) 3 , 3 (C) (1, 1) (D) 1, 4 3 4 4 4 3 25. The locus of the mid-points of the chords of the circle x2 + y2 – 2x – 4y – 11 = 0 which subtend 60° at 26. the centre is - 27. 28. (A) x2 + y2 – 4x – 2y – 7 = 0 (B) x2 + y2 + 4x + 2y – 7 = 0 E (C) x2 + y2 – 2x – 4y – 7 = 0 (D) x2 + y2 + 2x + 4y + 7 = 0 The locus of the centres of the circles such that the point (2,3) is the mid point of the chord 5x + 2y = 16 is - (A) 2x – 5y + 11 = 0 (B) 2x + 5y – 11 = 0 (C) 2x + 5y + 11 = 0 (D) none The locus of the centre of a circle which touches externally the circle, x2 + y2 – 6x – 6y + 14 = 0 and also touches the y-axis is given by the equation - (A) x2 – 6x – 10y + 14 = 0 (B) x2 – 10x – 6y + 14 = 0 (C) y2 – 6x – 10y + 14 = 0 (D) y2 – 10x – 6y + 14 = 0 The equation of the circle having the lines y2 – 2y + 4x – 2xy = 0 as its normals & passing through the point (2,1) is - (A) x2 + y2 – 2x – 4y + 3 = 0 (B) x2 + y2 – 2x + 4y – 5 = 0 (C) x2 + y2 + 2x + 4y –13 = 0 (D) none 67
JEE-Mathematics 2 9 . A circle is drawn touching the x-axis and centre at the point which is the reflection of (a, b) in the line y – x = 0. The equation of the circle is - (A) x2 + y2 – 2bx – 2ay + a2 = 0 (B) x2 + y2 – 2bx – 2ay + b2 = 0 (C) x2 + y2 – 2ax – 2by + b2 = 0 (D) x2 + y2 – 2ax – 2by + a2 = 0 3 0 . The length of the common chord of circles x2 + y2 – 6x – 16 = 0 and x2 + y2 – 8y – 9 = 0 is - (A) 10 3 (B) 5 3 (C) 5 3 / 2 (D) none of these 3 1 . The number of common tangents of the circles x2 + y2 – 2x – 1 = 0 and x2 + y2 – 2y – 7 = 0 - (A) 1 (B) 3 (C) 2 (D) 4 3 2 . If the circle x2 + y2 = 9 touches the circle x2 + y2 + 6y + c = 0, then c is equal to - (A) –27 (B) 36 (C) –36 (D) 27 3 3 . If the two circles, x2 + y2 + 2g1x + 2f1y = 0 and x2 + y2 + 2g2x + 2f2y = 0 touches each other, then - (A) f1g1 = f2g2 (B) f1 = f2 (C) f1f2 = g1g2 (D) none g1 g2 3 4 . The tangent from the point of intersection of the lines 2x – 3y + 1 = 0 and 3x – 2y – 1 = 0 to the circle x2 + y2 + 2x – 4y = 0 is - (A) x + 2y = 0, x – 2y + 1 = 0 (B) 2x – y – 1 = 0 (C) y = x, y = 3x – 2 (D) 2x + y + 1 = 0 3 5 . The locus of the centers of the circles which cut the circles x2 + y2 + 4x – 6y + 9 = 0 and x2 + y2 – 5x + 4y – 2 = 0 orthogonally is - (A) 9x + 10y – 7 = 0 (B) x – y + 2 = 0 (C) 9x – 10y + 11 = 0 (D) 9x + 10y + 7 = 0 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 36. Equation x x1 y y1 r, may represents - cos sin (A) Equation of straight line, if is constant and r is variable. (B) Equation of a circle, if r is constant & is variable. (C) A straight line passing through a fixed point & having a known slope. (D) A circle with a known centre and given radius. 3 7 . If r represent the distance of a point from origin & is the angle made by line joining origin to that point from Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 line x-axis, then r = |cos| represents - 1 (B) two circles centred at 1 , 0 & 1 , 0 (A) two circles of radii each. 2 2 2 (C) two circles touching each other at the origin. (D) pair of straight line 3 8 . If the circle C : x2 + y2 = 16 intersects another circle C of radius 5 in such a manner that the common chord 12 3 is of maximum length 8 has a slope equal to , then coordinates of centre of C are - 42 (A) 9 , 12 (B) 9 , 12 (C) 9 , 12 (D) 9 , 12 5 5 5 5 5 5 5 5 3 9 . For the equation x2 + y2 + 2x + 4 = 0 which of the following can be true - (A) It represents a real circle for all R. (B) It represents a real circle for || > 2. (C) The radical axis of any two circles of the family is the y-axis. (D) The radical axis of any two circles of the family is the x-axis. E 68
JEE-Mathematics 4 0 . If y = c is a tangent to the circle x2 + y2 – 2x + 2y – 2 = 0, then the value of c can be - (A) 1 (B) 3 (C) –1 (D) –3 4 1 . For the circles S1 x2 + y2 – 4x – 6y – 12 = 0 and S2 x2 + y2 + 6x + 4y – 12 = 0 and the line L x y 0 (A) L is common tangent of S1 and S2 (B) L is common chord of S1 and S2 (C) L is radical axis of S1 and S2 (D) L is perpendicular to the line joining the centre of S1 & S2 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 CHECK YOUR GRASP ANSWER KEY EXERCISE-1 Que. 1 2 3 4 5 6 7 8 9 10 Ans. C B B A C D A D A A Que. 11 12 13 14 15 16 17 18 19 20 Ans. B C D C B A B C C C Que. 21 22 23 24 25 26 27 28 29 30 Ans. C B A D C A D A B B Que. 31 32 33 34 35 36 37 38 39 40 Ans. A A B B C A,B,C,D A,B,C A,B B,C A,D Que. 41 Ans. B,C,D E 69
JEE-Mathematics EXERCISE - 02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1. If 1 , b, 1 , c, 1 & d, 1 are four distinct points on a circle of radius 4 units then, abcd = a, b c d a (A) 4 (B) 1/4 (C) 1 (D) 16 2 . What is the length of shortest path by which one can go from (–2, 0) to (2, 0) without entering the interior of circle, x2 + y2 = 1 ? (A) 23 (B) 3 2 (C) 2 3 (D) none of these 3 3 3 . Three equal circles each of radius r touch one another. The radius of the circle touching all the three given circles internally is - 2 3 2 3 (A) (2 3 )r (B) r (C) r (D) (2 3 )r 3 3 4 . If a2 + b2 = 1, m2 + n2 = 1, then which of the following is true for all values of m, n, a, b - (A) |am + bn| 1 (B) |am – bn| 1 (C) |am + bn| 1 (D) |am – bn| 1 5 . Circles are drawn touching the co-ordinate axis and having radius 2, then - (A) centre of these circles lie on the pair of lines y2 – x2 = 0 (B) centre of these circles lie only on the line y = x (C) Area of the quadrilateral whose vertices are centre of these circles is 16 sq.unit (D) Area of the circle touching these four circles internally is 4(3 2 2 ) 6 . The distance between the chords of contact of tangents to the circle x2 + y2 + 2gx + 2fy + c = 0 from the origin and from the point (g,f) is - (A) g2 f2 g2 f2 c g2 f2 c g2 f2 c (B) (C) (D) 2 2 g2 f2 2 g2 f2 7 . x2 + y2 + 6x = 0 and x2 + y2 – 2x = 0 are two circles, then - (A) They touch each other externally (B) They touch each other internally (C) Area of triangle formed by their common tangents is 33 sq. units. (D) Their common tangents do not form any triangle. 8 . Tangents are draw n to the circle x2 + y2 = 1 at the poi nt s where it is met by the circle s, x2 + y2 – ( + 6)x + (8 – 2)y – 3 = 0, being the variable. The locus of the point of intersection of these Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 tangents is - (A) 2x – y + 10 = 0 (B) x + 2y – 10 = 0 (C) x – 2y + 10 = 0 (D) 2x + y – 10 = 0 9 . 3 circle of radii 1, 2 and 3 and centres at A, B and C respectively, touch each other. Another circle whose centre is P touches all these 3 circles externally and has radius r. Also PAB & PAC - (A) cos 3r (B) cos 2r (C) r 6 (D) r 6 3(1 r) 2(1 r) 23 23 1 0 . Slope of tangent to the circle (x – r)2 + y2 = r2 at the point (x, y) lying on the circle is - x rx y2 x2 y2 x2 (A) y r (B) y (C) 2xy (D) 2xy 1 1 . The circle passing through the distinct points (1,t) , (t,1) & (t,t) for all values of ‘t’, passes through the point - (A) (–1, –1) (B) (–1, 1) (C) (1, –1) (D) (1,1) 1 2 . AB is a diameter of a circle. CD is a chord parallel to AB and 2CD = AB. The tangent at B meets the line AC produced at E then AE is equal to - (A) AB (B) 2AB (C) 2 2AB (D) 2AB 70 E
JEE-Mathematics 1 3 . The locus of the mid points of the chords of the circle x2 + y2 – ax – by = 0 which subtend a right angle at a , b is - 2 2 (A) ax + by = 0 (B) ax + by = a2 + b2 (C) x2 y2 a2 b2 0 (D) x2 y2 a2 b2 0 ax by ax by 8 8 1 4 . A variable circle is drawn to touch the x-axis at the origin. The locus of the pole of the straight line x + my + n = 0 w.r.t. the variable circle has the equation - (A) x(my – n) – y2 = 0 (B) x(my + n) – y2 = 0 (C) x(my – n) + y2 = 0 (D) none 1 5 . (6,0) , (0,6) and (7,7) are the vertices of a triangle. The circle inscribed in the triangle has the equation - (A) x2 + y2 – 9x + 9y + 36 = 0 (B) x2 + y2 – 9x – 9y + 36 = 0 (C) x2 + y2 + 9x – 9y + 36 = 0 (D) x2 + y2 – 9x – 9y – 36 = 0 1 6 . Number of points (x, y) having integral coordinates satisfying the condition x2 + y2 < 25 is - (A) 69 (B) 80 (C) 81 (D) 77 17 . The centre(s) of the circle(s) passing through the points (0, 0), (1, 0) and touching the circle x2 + y2 = 9 is/are - (A) 3 , 1 (B) 1 , 3 (C) 1 , 21 / 2 (D) 1 , 21 / 2 2 2 2 2 2 2 1 8 . The equation(s) of the tangent at the point (0, 0) to the circle, making intercepts of length 2a and 2b units on the co-ordinate axes, is (are) - (A) ax + by = 0 (B) ax – by = 0 (C) x = y (D) bx + ay = 0 1 9 . Tangents are drawn to the circle x2 + y2 = 50 from a point 'P' lying on the x-axis. These tangents meet the y-axis at points 'P1' and 'P2'. Possible co-ordinates of 'P' so that area of triangle PP1P2 is minimum is/are - (A) (10, 0) (B) (10 2, 0) (C) (–10, 0) (D) (10 2, 0) 2 0 . The tangents drawn from the origin to the circle x2 + y2 – 2rx – 2hy + h2 = 0 are perpendicular if - (A) h = r (B) h = –r (C) r2 + h2 = 1 (D) r2 + h2 = 2 2 1 . The common chord of two intersecting circles C1 and C2 can be seen from their centres at the angles of 90° & 60° respectively. If the distance between their centres is equal to 3 + 1 then the radii of C1 and C2 are - (A) 3 and 3 (B) 2 and 2 2 (C) 2 and 2 (D) 2 2 and 4 2 2 . In a right triangle ABC, right angled at A, on the leg AC as diameter, a semicircle is described. The chord Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 joining A with the point of intersection D of the hypotenuse and the semicircle, then the length AC equals to - AB.AD AB . AD (C) AB . AD AB.AD (A) (B) AB AD (D) AB2 AD2 AB2 AD2 2 3 . A circle touches a straight line x + my + n = 0 and cuts the circle x2 + y2 = 9 orthogonally. The locus of centres of such circles is - (B) (x + my – n)2 = (2 + m2) (x2 + y2 – 9) (A) (x + my + n)2 = (2 + m2) (x2 + y2 – 9) (C) (x + my + n)2 = (2 + m2) (x2 + y2 + 9) (D) none of these BRAIN TEASERS ANSWER KEY EXERCISE-2 Que. 1 2 3 4 5 6 7 8 9 10 Ans. C C Que. 11 12 B A,D A,C,D C A,C A A,B,C B,C Ans. D D Que. 21 22 13 14 15 16 17 18 19 20 Ans. C D C A B A C,D A,B A,C A,B E 23 A 71
JEE-Mathematics MISCELLANEOUS TYPE QUESTIONS EXERCISE - 03 MATCH THE COLUMN Following question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given s t a t e m e n t i n C o l u m n - I c a n h a v e c o r r e c t m a t c h i n g w i t h O N E O R M O R E s t a t e m e n t ( s ) i n C o l u m n - I I . 1 . Column-I Column-II (A) If point of intersection and number of common tangents of two (p) µ – = 3 circles x2 + y2 – 2x – 6y + 9 = 0 and x2 + y2 + 6x – 2y + 1 = 0 (q) µ + = 5 (r) µ – = 4 are and µ respectively, then (s) µ + = 4 (B) If point of intersection and number of tangents of two circles x2 + y2 – 6x = 0 and x2 + y2 + 2x = 0 are and µ respectively, then (C) If the straight line y = mx m I touches or lies outside the circle x2 + y2 – 20y + 90 = 0 and the maximum and minimum values of |m| are µ & respectively then (D) If two circle x2 + y2 + px + py – 7 = 0 and x2 + y2 – 10x + 2py + 1 = 0 cut orthogonally and the value of p are & µ respectively then ASSERTION & REASON These questions contains, Statement-I (assertion) and Statement-II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I. (C) Statement-I is true, Statement-II is false. (D) Statement-I is false, Statement-II is true. 1 . Consider two circles C1 x2 + y2 + 2x + 2y – 6 = 0 & C2 x2 + y2 + 2x + 2y – 2 = 0. Statement-I : Two tangents are drawn from a point on the circle C1 to the circle C2, then tangents always p er p end i c u l ar. Because Statement-II : C1 is the director circle of C2. (A) A (B) B (C) C (D) D 2 . Statement-I : The line (x – 3)cos + (y –3)sin = 1 touches a circle (x – 3)2 + (y – 3)2 = 1 for all values Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 of . Because Statement-II : x cos + y sin = a is a tangent of circle x2 + y2 = a2 for all values of . (A) A (B) B (C) C (D) D 3 . Consider the circles C1 x2 + y2 – 6x – 4y + 9 = 0 and C2 x2 + y2 – 8x – 6y + 23 = 0. Statement-I : Circle C1 bisects the circumference of the circle C2. Because Statement-II : Centre of C1 lie on C2. (A) A (B) B (C) C (D) D 4 . Statement-I : Circles x2 + y2 = 4 and x2 + y2 – 8x + 7 = 0 intersect each other at two distinct points Because Statement-II : Circles with centres C1 and C2 and radii r1 and r2 intersect at two distinct points, if |C1C2| < r1 + r2 (A) A (B) B (C) C (D) D 72 E
JEE-Mathematics COMPREHENSION BASED QUESTIONS Comprehension # 1 : Let A (–3, 0) and B (3, 0) be two fixed points and P moves on a plane such that PA = nPB (n 0). On the basis of above information, answer the following questions : 1 . If n 1, then locus of a point P is - (A) a straight line (B) a circle (C) a parabola (D) an ellipse 2 . If n = 1, then the locus of a point P is - (A) a straight line (B) a circle (C) a parabola (D) a hyperbola 3 . If 0 < n < 1, then - (A) A lies inside the circle and B lies outside the circle (B) A lies outside the circle and B lies inside the circle (C) both A and B lies on the circle (D) both A and B lies inside the circle 4 . If n > 1, then - (A) A lies inside the circle and B lies outside the circle (B) A lies outside the circle and B lies inside the circle (C) both A and B lies on the circle (D) both A and B lies inside the circle 5 . If locus of P is a circle, then the circle - (A) passes through A and B (B) never passes through A and B (C) passes through A but does not pass through B (D) passes through B but does not pass through A Comprehension # 2 : P is a variable point of the line L = 0. Tangents are drawn to the circle x2 + y2 = 4 from P to touch it at Q and R. The parallelogram PQSR is completed. On the basis of above information, answer the following questions : 1 . If L 2x + y – 6 = 0, then the locus of circumcetre of PQR is - (A) 2x – y = 4 (B) 2x + y = 3 (C) x – 2y = 4 (D) x + 2y = 3 2 . If P (6, 8), then the area of QRS is - (6 )3 / 2 (24 )3 / 2 48 6 192 6 (A) sq. units (B) sq. units (C) sq. units (D) sq. units 25 25 25 25 3 . If P (3, 4), then coordinate of S is - (A) 46 , 63 (B) 51 , 68 (C) 46 , 68 (D) 68 , 51 25 25 25 25 25 25 25 25 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-3 Match the Column 1 . (A)(r, s) ; (B) (s); (C) (p); (D) (q) Assertion & Reason 1. A 2. A 3. B 4. C Comprehension Based Questions Comprehension # 1 : 1. B 2. A 3. A 4. B 5. B Comprehension # 2 : 1. B 2. D 3. B E 73
JEE-Mathematics EXERCISE - 04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . Find the equations of the circles which have the radius 13 & which touch the line 2x 3y + 1 = 0 at (1, 1). 2 . (x , y ) & (x , y ) are the ends of a diameter of a circle such that x & x are the roots of ax² + bx + c = 0 & 11 22 12 y & y are roots of py² + qy + r = 0. Find the equation of the circle, its centre & radius. 12 3 . If the lines a x + b y + c = 0 & a x + b y + c = 0 cut the coordinate axes in concyclic points. Prove that 1 1 1 2 2 2 a a = b b 1 2 1 2. 4 . A (–a, 0) ; B (a, 0) are fixed points. C is a point which divides internally AB in a constant ratio tan. If AC & CB subtend equal angles at P, prove that the equation of the locus of P is x² + y² + 2ax sec2 + a² = 0. 5 . Let A be the centre of the circle x² + y² 2x 4y 20 = 0. Suppose that the tangents at the points B(1 , 7) & D(4 , -2) on the circle meet at the point C. Find the area of the quadrilateral ABCD. 6 . Determine the nature of the quadrilateral formed by four lines 3x + 4y – 5 = 0; 4x – 3y – 5 = 0; 3x + 4y + 5 = 0 and 4x – 3y + 5 = 0. Find the equation of the circle inscribed and circumscribing this quadrilateral. 7 . A variable circle passes through the point A (a, b) & touches the x-axis ; show that the locus of the other end of the diameter through A is (x a)² = 4by. 8 . A circle is drawn with its centre on the line x + y = 2 to touch the line 4x – 3y + 4 = 0 and pass through the point (0, 1). Find its equation. 9 . Obtain the equations of the straight lines passing through the point A(2, 0) & making 45° angle with the tangent at A to the circle (x + 2)² + (y 3)² = 25. Find the equations of the circles each of radius 3 whose centres are on these straight lines at a distance of 5 2 from A. 1 0 . Suppose the equation of the circle which touches both the coordinates axes and passes through the point with abscissa – 2 and ordinate 1 has the equation x2 + y2 + Ax + By + C = 0, find all the possible ordered triplet (A, B, C). 1 1 . The foot of the perpendicular from the origin to a variable tangent of the circle x2 + y2 2x = 0 is N. Find the equation of the locus of N. 1 2 . The line x + my + n = 0 intersects the curve ax2 + 2hxy + by2 = 1 at the point P and Q. The circle on PQ as diameter passes through the origin. Prove that n2(a + b) = 2 + m2. 13 . Find the equation of the circle which passes through the point (1, 1) & which touches the circle x² + y² + 4x 6y 3 = 0 at the point (2, 3) on it. 1 4 . A circle S = 0 is drawn with its centre at (–1, 1) so as to touch the circle x2 + y2 – 4x + 6y – 3 = 0 externally. Find the intercept made by the circle S = 0 on the coordinates axes. 1 5 . Find the equation of the circle which cuts each of the circles x² + y² = 4 , x² + y² 6x 8y + 10 = 0 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 & x² + y² + 2x 4y 2 = 0 at the extremities of a diameter. 1 6 . If the line x sin – y + a sec = 0 touches the circle with radius 'a' and centre at the origin then find the most general values of '' and sum of the values of '' lying in [0, 100]. 1 7 . Let a circle be given by 2x(x – a) + y(2y – b) = 0, (a 0, b 0). Find the condition on a & b if two chords, each bisected by the x-axis, can be drawn to the circle from the point a, b . 2 1 8 . Find the equation of a line with gradient 1 such that the two circles x2 + y2 = 4 and x2 + y2 – 10x – 14y + 65 = 0 intercept equal length on it. 1 9 . Find the equations of straight lines which pass through the intersection of the lines x 2y 5 = 0, 7x + y = 50 & divide the circumference of the circle x² + y² = 100 into two arcs whose lengths are in the ratio 2 : 1. 2 0 . Find the locus of the middle points of portions of the tangents to the circle x2 + y2 = a2 terminated by the coordinate axes. 2 1 . Show that the equation of a straight line meeting the circle x2 + y2 = a2 in two points at equal distances 'd' from a point (x , y ) on its circumference is xx + yy a2 + d2 = 0. 1 1 1 1 2 74 E
JEE-Mathematics 2 2 . A point moving around circle (x + 4)2 + (y + 2)2 = 25 with centre C broke away from it either at the point A or point B on the circle and move along a tangent to the circle passing through the point D(3, –3). Find the following : (a) Equation of the tangents at A and B. (b) Coordinates of the points A and B. (c) Angle ADB and the maximum and minimum distances of the point D from the circle. (d) Area of quadrilateral ADBC and the DAB. (e) Equation of the circle circumscribing the DAB and also the intercepts made by the this circle on the coordinates axes. 2 3 . Show that the equation x2 + y2 2x 2 y 8 = 0 represents, for different values of , a system of circles passing through two fixed points A, B on the x axis, and find the equation of that circle of the system the tangents to which at A & B meet on the line x + 2y + 5 = 0. 2 4 . Through a fixed point (h, k) secants are drawn to the circle x2 + y2 = r2. Show that the locus of the mid-points of the secants intercepted by the circle is x2 + y2 = hx + ky. 2 5 . A triangle has two of its sides along the coordinate axes, its third side touches the circle x² + y² 2ax 2ay + a² = 0. Prove that the locus of the circumcentre of the triangle is : a² 2a (x + y) + 2xy = 0. 2 6 . Find the equations to the four common tangents to the circles x² + y² = 25 and (x 12)² + y² = 9. 2 7 . Show that the locus of the centres of a circle which cuts two given circles orthogonally is a straight line & hence deduce the locus of the centre of the circles which cut the circles x² + y² + 4x 6y + 9 = 0 & x² + y² 5x + 4y + 2 = 0 orthogonally. Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1 . x² + y² 6x + 4y = 0 or x² + y² + 2x 8y + 4 = 0 2. x² + y² + b x + q y + c r = 0 a p a p 5. 75 sq.units 6 . square of side 2; x2 + y2 = 1; x2 + y2 = 2 8 . x2 + y2 – 2x – 2y + 1 = 0 or x2 + y2 – 42x + 38y – 39 = 0 9. x 7y = 2, 7x + y = 14; (x 1)2 + (y 7)2 = 32; (x 3)2 + (y + 7)2 = 32; (x 9)2 + (y 1)2 = 32; (x + 5)2 + (y + 1)2 = 32 10. x2 + y2 + 10x – 10y + 25 = 0 or x2 + y2 + 2x – 2y + 1 = 0, (10, –10, 25) (2, –2, 1) 11. (x² + y2 x)2 = x2 + y2 1 3 . x² + y² + x 6y + 3 = 0 1 4 . zero, zero 15. x² + y² 4x 6y 4 = 0 16. = n, 5050 17. a² > 2b2 18. 2x – 2y – 3 = 0 19. 4x 3y 25 = 0 or 3x + 4y 25 = 0 20. a2(x2 + y2) = 4x2y2 22. (a) 3x – 4y = 21; 4x + 3y = 3; (b) A(0, 1) and B(–1, –6); (c) 90°, 5( 2 1) units;(d) 12.5 sq. units; (e) x2 + y2 + x + 5y – 6=0, x intercept 5; y intercept 7 23. x2 + y2 2x 6y 8 = 0 26. 2x 5 y 15 = 0, 2x + 5 y 15 = 0, x 35 y 30 = 0, x + 35 y 30 = 0 27. 9x 10y + 7 = 0 E 75
JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE EXERCISE - 04 [B] 1 . Find the equation of the circle inscribed in a triangle formed by the lines 3x + 4y = 12 ; 5x + 12y = 4 & 8y = 15x + 10 without finding the vertices of the triangle. 2 . Consider a curve ax2 + 2 hxy + by2 = 1 and a point P not on the curve. A line is drawn from the point P intersects the curve at points Q & R. If the product PQ · PR is independent of the slope of the line, then show that the curve is a circle. 3 . Find the equation of a circle which is co-axial with circles 2x² + 2y² 2x + 6y 3 = 0 & x² + y² + 4x + 2y + 1 = 0. It is given that the centre of the circle to be determined lies on the radical axis of these two circles. 4 . If 32 + 6 + 1 – 6m2 = 0, then find the equation of the circle for which x + my + 1 = 0 is a tangent. 5 . Circle are drawn which are orthogonal to both the circles S x2 + y2 – 16 = 0 and S' x2 + y2 – 8x – 12y + 16 = 0. If tangents are drawn from the centre of the variable circles to S. Then find the locus of the mid point of the chord of contact of these tangents. 6 . Show that the locus of the point the tangents from which to the circle x² + y² a² = 0 include a constant angle is (x² + y² 2a²)² tan² = 4a² (x² + y² a²). 7 . Find the locus of the mid point of the chord of a circle x² + y² = 4 such that the segment intercepted by the chord on the curve x² 2x 2y = 0 subtends a right angle at the origin. 8 . Prove that the length of the common chord of the two circles x² + y² = a² and (x c)² + y² = b² is 1 (a b c) (a b c) (a b c) (a b c) , where a, b, c > 0. c 9 . Find the equation of the circles passing through the point (2, 8), touching the lines 4x 3y 24 = 0 & 4x + 3y 42 = 0 & having x coordinate of the centre of the circle less than or equal to 8. 10. Lines 5x + 12y 10 = 0 & 5x 12y 40 = 0 touch a circle C of diameter 6. If the centre of C lies in 1 1 the first quadrant, find the equation of the circle C which is concentric with C & cuts intercepts of length 8 21 on these lines. 1 1 . A circle touches the line y = x at a point P such that OP = 4 2 , where O is the origin. The circle contains the point (–10,2) in its interior and the length of its chord on the line x + y = 0 is 6 2 . Determine the equation of the circle. [JEE 1990] 1 2 . Find the intervals of values of 'a' for which the line y + x = 0 bisects two chords drawn from a point 1 2a , 1 2a Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 2 2 2a y = 0. to the circle 2x2 + 2y2 – 1 2a x – 1 [JEE 1996] 1 3 . Find the equations of the circles passing through (–4, 3) and touching the lines x + y = 2 and x – y = 2. 1 4 . P is a variable point on the circle with centre at C . CA & CB are perpendiculars from C on x-axis & y-axis respectively. Show that the locus of the centroid of the triangle PAB is a circle with centre at the centroid of the triangle CAB & radius equal to one third of the radius of the given circle. BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 1 . x2 + y2 2x 2y + 1 = 0 3. 4x2 + 4y2 + 6x + 10y 1 = 0 4 . x2 + y2 – 6x + 3 = 0 5 . x2 + y2 – 4x – 6y = 0 7 . x² + y² 2x 2y = 0 9 . centre (2,3), r = 5; centre 182 , 3 , r = 205 1 0 . x2 + y2 10x 4y + 4 = 0 11. (x – 9)2 + (y – 1)2 = 50 9 9 12. a (–, –2) (2,) 1 3 . x2 + y2 + 2(10 ± 54 )x + 55 ± 8 54 = 0 76 E
JEE-Mathematics EXERCISE - 05 [A] JEE-[MAIN] : PREVIOUS YEAR QUESTIONS 1 . The square of the length of tangent from (3, –4) on the circle x2 + y2 – 4x – 6y + 3 = 0 [AIEEE-2002] (1) 20 (2) 30 (3) 40 (4) 50 2 . Radical axis of the circles x2 + y2 + 6x – 2y –9 = 0 and x2 + y2 – 2x + 9y – 11 = 0 is- [AIEEE-2002] (1) 8x – 11y + 2 = 0 (2) 8x + 11y + 2 = 0 (3) 8x + 11y – 2 = 0 (4) 8x – 11y – 2 = 0 3 . If the two circles (x – 1)2 + (y – 3)2 = r2 and x2 + y2 – 8x + 2y + 8 = 0 intersect in two distinct points, then- [AIEEE-2003] (1) r > 2 (2) 2 < r < 8 (3) r < 2 (4) r = 2 4 . The lines 2x – 3y = 5 and 3x – 4y = 7 are diameters of a circle having area as 154 sq. units. Then the equation of the circle is- [AIEEE-2003] (1) x2 + y2 – 2x + 2y = 62 (2) x2 + y2 + 2x – 2y = 62 (3) x2 + y2 + 2x – 2y = 47 (4) x2 + y2 – 2x + 2y = 47 5 . If a circle passes through the point (a, b) and cuts the circle x2 + y2 = 4 orthogonally, then the locus of its centre is- [AIEEE-2004] (1) 2ax + 2by + (a2 + b2 + 4) = 0 (2) 2ax + 2by – (a2 + b2 + 4) = 0 (3) 2ax – 2by + (a2 + b2 + 4) = 0 (4) 2ax – 2by – (a2 + b2 + 4) = 0 6 . A variable circle passes through the fixed point A(p, q) and touches x-axis. The locus of the other end of the diameter through A is- [AIEEE-2004] (1) (x – p)2 = 4qy (2) (x – q)2 = 4py (3) (y – p)2 = 4qx (4) (y – q)2 = 4px 7 . If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along diameters of a circle of circumference 10, then the equation of the circle is- [AIEEE-2004] (1) x2 + y2 – 2x + 2y – 23 = 0 (2) x2 + y2 – 2x – 2y – 23 = 0 (3) x2 + y2 + 2x + 2y – 23 = 0 (4) x2 + y2 + 2x – 2y – 23 = 0 8 . The intercept on the line y = x by the circle x2 + y2 – 2x = 0 is AB. Equation of the circle on AB as a diameter is- [A IE EE -2 00 4] (1) x2 + y2 – x – y = 0 (2) x2 + y2 – x + y = 0 (3) x2 + y2 + x + y = 0 (4) x2 + y2 + x – y = 0 9 . If the circles x2 + y2 + 2ax + cy + a = 0 and x2 + y2 – 3ax + dy – 1 = 0 intersect in two distinct point P and Q then the line 5x + by – a = 0 passes through P and Q for- [AIEEE-2005] (1) exactly one value of a (2) no value of a (3) infinitely many values of a (4) exactly two values of a 1 0 . A circle touches the x-axis and also touches the circle with centre at (0, 3) and radius 2. The locus of the centre Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 of the circle is- [AIEEE-2005] (1) an ellipse (2) a circle (3) a hyperbola (4) a parabola 1 1 . If a circle passes through the point (a, b) and cuts the circle x2 + y2 = p2 orthogonally, then the equation of the locus of its centre is- [AIEEE-2005] (1) x2 + y2 – 3ax – 4by + (a2 + b2 – p2) = 0 (2) 2ax + 2by – (a2 – b2 + p2) = 0 (3) x2 + y2 – 2ax – 3by + (a2 – b2 – p2) = 0 (4) 2ax + 2by – (a2 + b2 + p2) = 0 1 2 . If the pair of lines ax2 + 2(a + b)xy + by2 = 0 lie along diameters of a circle and divide the circle into four sectors such that the area of one of the sectors is thrice the area of another sector then- [AIEEE-2005] (1) 3a2 – 10ab + 3b2 = 0 (2) 3a2 – 2ab + 3b2 = 0 (3) 3a2 + 10ab + 3b2 = 0 (4) 3a2 + 2ab + 3b2 = 0 1 3 . If the lines 3x – 4y – 7 = 0 and 2x – 3y – 5 = 0 are two diameters of a circle of area 49 square units, the equation of the circle is- [AIEEE-2006] (1) x2 + y2 + 2x – 2y – 62 = 0 (2) x2 + y2 – 2x + 2y – 62 = 0 (3) x2 + y2 – 2x + 2y – 47 = 0 (4) x2 + y2 + 2x – 2y – 47 = 0 E 77
JEE-Mathematics 1 4 . Let C be the circle with centre (0, 0) and radius 3 units. The equation of the locus of the mid points of the chords 2 [AIEEE-2006, IIT-1996] of the circle C that subtend an angle of at its centre is - 3 (1) x2 + y2 = 1 27 9 3 (2) x2 + y2 = (3) x2 + y2 = (4) x2 + y2 = 4 4 2 1 5 . Consider a family of circles which are passing through the point (–1, 1) and are tangent to x-axis. If (h, k) are the co-ordinates of the centre of the circles, then the set of values of k is given by the interval- [AIEEE-2007] (1) 0 < k < 1/2 (2) k 1/2 (3) –1/2 k 1/2 (4) k 1/2 1 6 . The point diametrically opposite to the point (1, 0) on the circle x2 + y2 + 2x + 4y – 3 = 0 is- [AIEEE-2008] (1) (3, –4) (2) (–3, 4) (3) (–3, –4) (4) (3, 4) 1 7 . Three distinct points A, B and C are given in the 2–dimensional coordinate plane such that the ratio of the 1 distance of any one of them from the point (1, 0) to the distance from the point (–1, 0) is equal to 3 . Then the circumcentre of the triangle ABC is at the point :- [AIEEE-2009] (1) 5 , 0 (2) 5 , 0 (3) (0, 0) (4) 5 , 0 2 3 4 1 8 . If P and Q are the points of intersection of the circles x2 + y2 + 3x + 7y + 2p – 5 = 0 and x2 + y2 + 2x + 2y – p2 = 0, then there is a circle passing through P, Q and (1, 1) for :- [ A IE E E -2 0 0 9 ] (1) All except two values of p (2) Exactly one value of p (3) All values of p (4) All except one value of p 1 9 . For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A false statement among the following is :- [AIEEE-2010] (1) There is a regular polygon with r 1 r 1 (2) There is a regular polygon with R 2 R2 r2 (4) There is a regular polygon with r 3 (3) There is a regular polygon with R2 R3 2 0 . The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if :- [AIEEE-2010] (1) – 85 < m < – 35 (2) – 35 < m < 15 (3) 15 < m < 65 (4) 35 < m < 85 2 1 . The two circles x2 + y2 = ax and x2 + y2 = c2 (c>0) touch each other if :- [AIEEE-2011] (1) a = 2c (2) |a| = 2c (3) 2|a| = c (4) |a| = c 2 2 . The equation of the circle passing through the points (1, 0) and (0, 1) and having the smallest radius is: Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 (1) x2 + y2 + x + y – 2 = 0 (2) x2 + y2 – 2x – 2y + 1 = 0 [AIEEE-2011] (3) x2 + y2 – x – y = 0 (4) x2 + y2 + 2x + 2y – 7 = 0 2 3 . The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes through the point (2, 3) is : [AIEEE-2012] (1) 5/3 (2) 10/3 (3) 3/5 (4) 6/5 2 4 . The circle passing through (1, – 2) and touching the axis of x at (3, 0) also passes through the point : [JEE(Main)-2013] (1) (–5, 2) (2) (2, –5) (3) (5, –2) (4) (–2, 5) PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans 3 1 2 4 2 1 1 1 2 4 4 4 3 3 2 Que. 16 17 18 19 20 21 22 23 24 Ans 3 4 4 3 2 4 3 2 3 E 78
JEE-Mathematics EXERCISE - 05 [B] JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS 1 . Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle then 2r equals (A) PQ RS PQ RS 2PQ RS PQ 2 RS 2 (B) (C) PQ RS (D) 2 2 [JEE 2001 (Screening) 1] 2 . Let 2x2 + y2 – 3xy = 0 be the equation of a pair of tangents drawn from the origin 'O' to a circle of radius 3 with centre in the first quadrant. If A is one of the points of contact, find the length of OA. [JEE 2001 (Mains), 5] 3 . Find the equation of the circle which passes through the points of intersection of circles x2 + y2 – 2x – 6y + 6 = 0 and x2 + y2 + 2x – 6y + 6 = 0 and intersects the circle x2 + y2 + 4x + 6y + 4 = 0 orthogonally. [REE 2001 (Mains), 3] 4 . Tangents TP and TQ are drawn from a point T to the circle x2 + y2 = a2. If the point T lies on the line px + qy = r, find the locus of centre of the circumcircle of triangle TPQ. [REE 2001 (Mains), 5] 5 . If the tangent at the point P on the circle x2 + y2 + 6x + 6y = 2 meets the straight line 5x – 2y + 6 = 0 at a point Q on the y-axis, then the length of PQ is (A) 4 (B) 2 5 (C) 5 (D) 3 5 [JEE 2002 (Scr), 3] 6 . If a > 2b > 0 then the positive value of m for which y = mx – b 1 m2 is a common tangent to x2 + y2 = b2 and (x – a)2 + y2 = b2 is 2b a2 4b2 2b b (A) (B) (C) a 2b (D) a 2b a2 4b2 2b [JEE 2002 (Scr), 3] 7 . The radius of the circle, having centre at (2, 1), whose one of the chord is a diameter of the circle x2 + y2 – 2x – 6y + 6 = 0 (A) 1 (B) 2 (C) 3 (D) 3 [JEE 2004 (Scr)] 8 . Line 2x + 3y + 1 = 0 is a tangent to a circle at (1, –1). This circle is orthogonal to a circle which is drawn having diameter as a line segment with end points (0, –1) and (– 2, 3). Find equation of circle. [JEE 2004, 4] 9 . A circle is given by x2 + (y – 1)2 = 1, another circle C touches it externally and also the x-axis, then the locus of Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 its centre is [JEE 2005 (Scr)] (A) {(x, y) : x2 = 4y} {(x, y) : y 0} (B) {(x, y) : x2 + (y – 1)2 = 4} {x, y) : y 0} (C) {(x, y) : x2 = y} {(0, y) : y 0} (D) {(x, y) : x2 = 4y} {(0, y) : y 0} 10 . Let ABCD be a quadrilateral with area 18, with side AB parallel to the side CD and AB = 2CD. Let AD be perpendicular to AB and CD. If a circle is drawn inside the quadrilateral ABCD touching all the sides, then its radius is [JEE 2007, 3] (A) 3 (B) 2 (C) 3/2 (D) 1 1 1 . Tangents are drawn from the point (17, 7) to the circle x2 + y2 = 169. Statement-1 : The tangents are mutually perpendicular. because Statement-2 : The locus of the points from which mutually perpendicular tangents can be drawn to the given circle is x2 + y2 = 338. (A) Statement-1 is true, statement-2 is true; statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true; statement-2 is NOT a correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true. [JEE 2007, 3] E 79
JEE-Mathematics 1 2 . Consider the two curves C1 : y2 = 4x ; C2 : x2 + y2 – 6x + 1 = 0. Then, [JEE 2008, 3] (A) C1 and C2 touch each other only at one point (B) C1 and C2 touch each other exactly at two points (C) C1 and C2 intersect (but do not touch) at exactly two points (D) C1 and C2 neither intersect nor touch each other 1 3 . Consider, L1 : 2x + 3y + p – 3 = 0 ; L2 : 2x + 3y + p + 3 = 0, where p is a real number, and C : x2 + y2 + 6x – 10y + 30 = 0. Statement-1 : If line L1 is a chord of circle C, then line L2 is not always a diameter of circle C. and Statement-2 : If line L1 is a diameter of circle C, then line L2 is not a chord of circle C. (A) Statement-1 is True, Statement-2 is True; statement-2 is a correct explanation for statement-1 (B) Statement-1 is True, Statement-2 is True; statement-2 is NOT a correct explanation for statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True [JEE 2008, 3] 14. Comprehension (3 questions together): A circle C of radius 1 is inscribed in an equilateral triangle PQR. The points of contact of C with the sides PQ, QR, RP are D, E, F respectively. The line PQ is given by the equation 3 x + y – 6 = 0 and the point D is 3 3 3 2 , 2 . Further, it is given that the origin and the centre of C are on the same side of the line PQ. (i) The equation of circle C is (A) (x – 2 3 )2 + (y – 1)2 = 1 (B) (x – 2 3 )2 + (y + 1 )2 = 1 2 (C) (x – 3 )2 + (y + 1)2 = 1 (D) (x – 3 )2 + (y – 1)2 = 1 (ii) Points E and F are given by 3, 3 , 3,0 3, 1 3, 0 2 2 2 2 , (A) (B) 3 3 3 1 (D) 3 , 3 , 3, 1 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 (C) 2 , 2 , 2 , 2 2 2 2 2 (iii) Equations of the sides RP, RQ are 1 22 (B) y = x, y = 0 (A) y = x + 1, y = – x – 1 3 33 33 (D) y = 3 x, y = 0 [JEE 2008, 4+4+4] (C) y = x + 1, y = – x – 1 22 1 5 . Tangent s draw n from the poi nt P(l, 8) to the circle x2 + y2 – 6x – 4y – 11 = 0 touch the circle at the points A and B. The equation of the circumcircle of the triangle PAB is (A) x2 + y2 + 4x – 6y + 19 = 0 (B) x2 + y2 – 4x – 10y + 19 = 0 (C) x2 + y2 – 2x + 6y – 29 = 0 (D) x2 + y2 – 6x – 4y + 19 = 0 [JEE 2009, 3] 1 6 . The centres of two circles C1 and C2 each of unit radius are at a distance of 6 units from each other. Let P be the mid point of the line segment joining the centres of C1 and C2 and C be a circle touching circles C1 and C2 externally. If a common tangent to C1 and C passing through P is also a common tangent to C2 and C, then the radius of the circle C is [JEE 2009, 4] 80 E
JEE-Mathematics 1 7 . Two parallel chords of a circle of radius 2 are at a distance 3 1 apart. If the chords subtend at the 2 [JEE 10, 3] center, angles of and , where k > 0, then the value of [k] is kk [Note : [k] denotes the largest integer less than or equal to k] 1 8 . The circle passing through the point (–1,0) and touching the y-axis at (0, 2) also passes through the point - (A) 3 ,0 (B) 5 ,2 (C) 3 , 5 (D) (–4,0) 2 2 2 2 [JEE 2011, 3, –1] 1 9 . The straight line 2x – 3y = 1 divides the circular region x2 + y2 6 into two parts. If S 2, 3 , 5 , 3 , 1 , 1 , 1 ,1 , 4 2 4 4 4 8 4 then the number of point(s) in S lying inside the smaller part is [JEE 2011, 4] 2 0 . The locus of the mid-point of the chord of contact of tangents drawn from points lying on the straight line 4x – 5y = 20 to the circle x2 + y2 = 9 is- [JEE 2012, 3, –1] (A) 20(x2 + y2) – 36x + 45y = 0 (B) 20(x2 + y2) + 36x – 45y = 0 (C) 36(x2 + y2) – 20x + 45y = 0 (D) 36(x2 + y2) + 20x – 45y = 0 Paragraph for Question 21 and 22 A tangent PT is drawn to the circle x2 + y2 = 4 at the point P 3, 1 . A straight line L, perpendicular to PT is a tangent to the circle (x – 3)2 + y2 = 1. [JEE 2012, 3, –1] 2 1 . A common tangent of the two circles is (A) x = 4 (B) y = 2 (C) x 3y 4 (D) x 2 2y 6 [JEE 2012, 3, –1] 2 2 . A possible equation of L is (A) x 3y 1 (B) x 3 y 1 (C) x 3y 1 (D) x 3y 5 2 3 . Circle(s) touching x-axis at a distance 3 from the origin and having an intercept of length 2 7 or y-axis is (are) [JEE(Advanced) 2013, 3, (–1)] (A) x2 + y2 – 6x + 8y + 9 = 0 (B) x2 + y2 – 6x + 7y + 9 = 0 (C) x2 + y2 – 6x – 8y + 9 = 0 (D) x2 + y2 – 6x – 7y + 9 = 0 Node6\\E_NODE6 (E)\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#05\\Eng\\02 CIRCLE.p65 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [B] 1. A 2 . OA = 3(3 + 10 ) 3 . x2 + y2 + 14x – 6y + 6 = 0; 4 . 2px + 2qy = r 5. C 10. B 6. A 7. C 8 . 2x2 + 2y2 – 10x – 5y + 1 = 0 9. D 16. 8 11. A 12. B 13. C 14. (i) D, (ii) A, (iii) D 1 5 . B 2 3 . A,C 17. 3 18. D 19. 2 20. A 21. D 22. A E 81
JEE-Mathematics COMPLEX NUMBER 1. DEFINITION : Complex numbers are defined as expressions of the form a + ib where a , b R & i = 1 . It is denoted by z i.e. z = a + ib. ‘a’ is called real part of z (Re z) and ‘b’ is called imaginary part of z (Im z). Every Complex Number Can Be Regarded As Purely real Purely imaginary Imaginary if b = 0 if a = 0 if b 0 Note : (i) The set R of real numbers is a proper subset of the Complex Numbers. Hence the Complex Number system is N W I Q R C. (ii) Zero is both purely real as well as purely imaginary but not imaginary. (iii) i = 1 is called the imaginary unit. Also i² = l ; i3 = i ; i4 = 1 etc. In general i4n = 1, i4n+1 = i, i4n+2 = –1, i4n+3 =–i, where n I (iv) a b = a b only if atleast one of either a or b is non- negative. Illustration 1 : The value of i57 + 1/i125 is :- (A) 0 (B) –2i (C) 2i (D) 2 Ans. (A) Solution : 1 i57 + 1/i125 = i56. i + i124 .i 14 1 i =i4 31 i4 i 1i = i i i i2 i i 0 2.NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 ARGAND DIAGRAM : 3. Master Argand had done a systematic study on complex numbers and represented Imaginary axis E every complex number z = x + iy as a set of ordered pair (x, y) on a plane called y P(z) complex plane (Argand Diagram) containing two perpendicular axes. Horizontal axis is known as Real axis & vertical axis is known as Imaginary axis. x All complex numbers lying on the real axis are called as purely real Real axis O z = x + iy Re(z)=x, Im(z)=y and those lying on imaginary axis as purely imaginary. ALGEBR AIC OPER ATIONS : Fundamental operations with complex numbers : ( a ) Addition (a + bi) + (c + di) = (a + c) + (b + d)i (b ) Subtraction (a + bi) – (c + di) = (a – c) + (b – d)i ( c ) Multiplication (a + bi) (c + di) = (ac – bd) + (ad + bc)i ( d ) Division a bi a bi . c di ac bd bc ad i c di c di c di c2 d2 c2 d2 27
JEE-Mathematics Note : (i) The algebraic operations on complex numbers are similar to those on real numbers treating i as a polynomial. (ii) Inequalities in complex numbers (non-real) are not defined. There is no validity if we say that complex number (non-real) is positive or negative. e.g. z > 0, 4 + 2i < 2 + 4i are meaningless. (iii) In real numbers, if a2 + b2 = 0, then a = 0 = b but in complex numbers, z 2 + z 2 = 0 does not imply 12 z = z = 0. 12 3 2i sin Illustration 2 : 1 2i sin will be purely imaginary, if = (A) 2n , n I (B) n , n I (C) n , n I (D) none of these 3 3 3 Ans. (C) Solution : 3 2i sin will be purely imaginary, if the real part vanishes, i.e., 1 2i sin (3 2i sin ) (1 2i sin ) 3 4 sin2 i 8 sin = (1 2i sin ) (1 2i sin ) 1 4 sin2 3 4 sin2 0 3 – 4 sin2 = 0 (only if be real) 1 4 sin2 2 2 3 sin2 = 3 = sin 2 = n ± , n I 3 Do yourself - 1 : (i) Determine least positive value of n for which 1 i n 1 1 i 5 ( i i ) Find the value of the sum (in in2 ) , where i = 1 . n 1 3 . EQUALITY IN COMPLEX NUMBER : NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 Two complex numbers z = a + ib & z = a + ib are equal if and only if their real & imaginary parts 1 1 1 2 2 2 are respectively equal. Illustration 3 : The values of x and y satisfying the equation (1 i)x 2i (2 3i)y i i are 3i 3i (A) x = –1, y =3 (B) x = 3, y = –1 (C) x = 0, y = 1 (D) x = 1, y = 0 Solution : (1 i)x 2i (2 3i)y i i (4 + 2i) x + (9 – 7i) y – 3i – 3 = 10i 3i 3i Ans.(B) Equating real and imaginary parts, we get 2x – 7y = 13 and 4x + 9y = 3. Hence x = 3 and y = –1. E 28
JEE-Mathematics Illustration 4 : Find the square root of 7 + 24 i. Solution : Let 7 24i = a + ib Squaring a2 – b2 + 2iab = 7 + 24i Compare real & imaginary parts a2 – b2 = 7 & 2ab = 24 By solving these two equations We get a = ±4 , b = ±3 7 24i = ±(4 + 3i) Illustr ation 5 : If x 5 2 4 , find the value of x4 + 9x3 + 35x2 – x + 4. Solution : We have , x = –5 + 2 4 x + 5 = 4i (x + 5)2 = 16i2 x2 + 10x + 25 = –16 x2 + 10x + 41 = 0 Now, x4 + 9x3 + 35x2 – x + 4 x2(x2 + 10x + 41) – x(x2 + 10x + 41) + 4(x2 + 10x + 41) – 160 x2(0) – x(0) + 4(0) – 160 –160 Ans. Do yourself - 2 : ( i ) Find the value of x3 + 7x2 – x + 16, where x = 1 + 2i. c i b 2c ( i i ) If a + ib = c i , where c is a real number, then prove that : a2 + b2 = 1 and a c2 1 . (i i i ) Find square root of –15 – 8i 4 . THREE IMPORTANT TERMS : CONJUGATE/MODULUS/ARGUMENT : ( a ) CONJUGATE COMPLEX : If z = a + ib then its conjugate complex is obtained by changing the sign of its imaginary part & is denoted by z . i.e. z = a ib. Note that : (i) z + z = 2 Re(z) (ii) z z = 2i Im(z) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 (iii) z z = a² + b², which is purely real (iv) If z is purely real, then z – z = 0 Im (v) If z is purely imaginary, then z + z = 0 –z z (vi) If z lies in the 1st quadrant, then z lies in the 4th quadrant and Re z lies in the 2nd quadrant. –z z (b) Modulus : If P denotes complex number z = x + iy, then the length OP is called modulus of complex number z. It is denoted by |z|. OP = |z| = x2 y2 Geometrically z represents the distance of point P from origin. ( z 0) z if z 0 Note : Unlike real numbers, z = z if z 0 is not correct. E 29
JEE-Mathematics (c) Argument or Amplitude : Imaginary P(x, y) If P denotes complex number z = x + iy and if OP makes an angle axis with real axis, then is called one of the arguments of z. |z| y Real axis = tan1 (angle made by OP with positive real axis) O x Note : (i) Argument of a complex number is a many valued function. If is the argument of a complex number, then 2n + ; n I will also be the argument of that complex number. Any two arguments of a complex number differ by 2n (ii) The unique value of such that < is called Amplitude (principal value of the argument). (iii) Principal argument of a complex number z = x + iy can be found out using method given below : (a) Find = tan 1 y such that 0, . Im x 2 (b) Use given figure to find out the principal argument according Re as the point lies in respective quadrant. (iv) Unless otherwise stated, amp z implies principal value of the argument. y NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 (v) The unique value of = tan1 x such that 0 2 is called least positive argument. (vi) If z = 0, arg(z) is not defined (vii) If z is real & negative, arg(z) = . (viii) If z is real & positive, arg(z) = 0 (ix) If , z lies on the positive side of imaginary axis. 2 (x) If , z lies on the negative side of imaginary axis. 2 By specifying the modulus & argument a complex number is defined completely. Argument impart direction & modulus impart distance from origin. For the complex number 0 + 0i the argument is not defined and this is the only complex number which is given by its modulus only. Illustration 6 : Find the modulus, argument, principal value of argument, least positive argument of complex numbers (a) 1 + i 3 (b) –1 + i 3 (c) 1 – i 3 (d) –1 – i 3 Solution : (a) For z = 1 + i 3 y (1, 3 ) | z| 12 ( 3 )2 2 arg (z) = 2n + , n I 60° 3 3 1 x Least positive argument is 3 If the point is lying in first or second quadrant then amp(z) is taken in anticlockwise direction. In this case amp(z) = 3 30 E
JEE-Mathematics (b) For z = –1 + i 3 (–1, 3 ) y |z| = 2 arg (z) = 2n + 2 , n I 3 3 60° 120° Least positive argument = 2 1 x 3 amp(z) = 2 3 (c) For z = 1 – i 3 y |z| = 2 1 x –/3 3 5/3 arg (z) = 2n – , n I 3 Least positive argument = 5 (1,– 3 ) 3 If the point lies in third or fourth quadrant then consider amp(z) in clockwise direction. In this case amp(z) = – 3 (d) For z = –1 – i 3 y |z| = 2 4/3 arg (z) = 2n – 2 , n I 1 –2/3 x 3 60° 3 Least positive argument = 4 3 (–1,– 3 ) amp(z) = – 2 3 Illustration 7 : Find modulus and argument for z = 1 – sin + i cos , (0,2) Solution : | z| (1 sin )2 (cos )2 2 2 sin 2 cos sin 22 Case (i) For 0, , z will lie in I quadrant. 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 amp (z) = tan 1 cos amp (z) = tan1 cos2 sin2 tan 1 cos sin 1 sin 22 22 cos sin 2 cos sin 2 2 22 arg z tan 1 tan 4 2 Since , 4 2 4 2 amp (z) = , | z| 2 cos sin 4 2 2 2 Case (ii) at : z 0 0i 2 |z| = 0 amp (z) is not defined. E 31
JEE-Mathematics Case (iii) For , 3 , z will lie in IV quadrant 2 2 Case (iv) Case (v) so amp (z) = –tan–1 tan 2 4 Since , 2 4 2 amp (z) = 3 , |z| = 2 sin cos 2 4 4 2 2 2 at 3 : z = 2 + 0i 2 |z| = 2 amp (z) = 0 For 3 ,2 2 z will lie in I quadrant arg (z) = tan–1tan 2 4 Since , 5 2 4 4 arg z = = 3 , |z| = 2 sin cos 24 24 2 2 Do yourself - 3 : Find the modulus and amplitude of following complex numbers : ( i ) 2 2 3i ( i i ) 3 i (i i i ) –2i 1 2i 2 6 3i (iv) 1 3i (v) 5 3i 5 . REPRESENTATION OF A COMPLEX NUMBER IN VARIOUS FORMS : (a) Cartesian Form (Geometrical Representation) : Every complex number z = x + i y can be represented by a point on the cartesian plane known as NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 complex plane by the ordered pair (x , y). There exists a one-one correspondence between the points of the plane and the members of the set of complex numbers. Imaginary P(x, y) axis For z = x + iy; | z| x2 y2 ; z x iy and tan 1 y |z| x Note : Real axis O (i) Distance between the two complex numbers z & z is given by |z – z |. 12 12 (ii) |z – z | = r, represents a circle, whose centre is z and radius is r. 00 Illustration 8 : Find the locus of : Solution : (a) |z – 1|2 + |z + 1|2 = 4 (b) Re(z2) = 0 (a) Let z = x + iy (|x + iy – 1|)2 + (|x + iy + 1|)2 = 4 (x – 1)2 + y2 + (x + 1)2 + y2 = 4 x2 – 2x + 1 + y2 + x2 + 2x + 1 + y2 = 4 x2 + y2 = 1 Above represents a circle on complex plane with center at origin and radius unity. 32 E
JEE-Mathematics (b) Let z = x + iy z2 = x2 – y2 + 2xyi Re(z2) = 0 x2 – y2 = 0 y = ± x Thus Re(z2) = 0 represents a pair of straight lines passing through origin. Illustration 9 : If z is a complex number such that z2 = ( z )2 , then Solution : (A) z is purely real (B) z is purely imaginary (C) either z is purely real or purely imaginary (D) none of these Let z = x + iy, then its conjugate z x iy Given that z2 = ( z )2 x2 – y2 + 2ixy = x2 – y2 – 2ixy 4ixy = 0 If x 0 then y = 0 and if y 0 then x = 0. Ans. (C) Illustration 10 : Among the complex number z which satisfies |z – 25i| 15, find the complex numbers z having (a) least positive argument (b) maximum positive argument (c) least modulus (d) maximum modulus Solution : The complex numbers z satisfying the condition |z – 25i| 15 are represented by the points inside and on the circle of radius 15 and centre at the point C(0, 25). The complex number having least positive argument and maximum positive arguments in this region are the points of contact of tangents drawn from origin to the circle Here = least positive argument and = maximum positive argument NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 In OCP, OP OC 2 CP 2 252 152 20 D40i Taonriggeinnt from OP 20 4 C 25i and sin Q P OC 25 5 E tan 4 tan 1 4 3 3 O N Thus, complex number at P has modulus 20 and argument tan 1 4 3 zp 20 cos i sin 20 3 i 4 5 5 zp 12 16i Similarly z = –12 + 16i Q From the figure, E is the point with least modulus and D is the point with maximum modulus. Hence, zE OE OC EC 25i 15i 10i and zD OD OC CD 25i 15i 40i Do yourself - 4 : ( i ) Find the distance between two complex numbers z = 2 + 3i & z = 7–9i on the complex plane. 12 ( i i ) Find the locus of |z – 2 – 3i| = 1. (i i i ) If z is a complex number, then z2 + z 2 2 represents - (A) a circle (B) a straight line (C) a hyperbola (D) an ellipse E 33
JEE-Mathematics ( c ) Trigonometric / Polar Representation : z = r (cos + i sin ) where z = r ; arg z = ; z = r (cos i sin ) Note : cos + i sin is also written as CiS Euler's formula : The formula eix = cosx + i sin x is called Euler's formula. It was introduced by Euler in 1748, and is used as a method of expressing complex numbers. Also cos x = eix e ix & sin x = eix e ix are known as Euler's identities. 2 2i (d) Exponential Representation : Let z be a complex number such that z = r & arg z = , then z = r.ei Illustration 11 : Express the following complex numbers in polar and exponential form : 1 3i i 1 (i) (ii) cos i sin 1 2i 33 Solution : (i) Let z 1 3i 1 3i 1 2i 1 i 1 2i 1 2i 1 2i | z| (1)2 12 2 tan 1 1 tan 1 4 4 Re(z) < 0 and Im(z) > 0 z lies in second quadrant. = arg (z) = – = – 3 44 Hence Polar form is z = 2 cos 3 i sin 3 4 4 and exponential form is z 2 e3/ 4 (ii) Let z i 1 i 1 2(i 1) cos i sin 1i 3 = 22 33 (1 i 3 ) z 2(i 1) (1 i 3) 3 1 i 3 1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 3) z 2 2 (1 i 3 ) (1 i Re(z) > 0 and Im(z) > 0 z lies in first quadrant. 3 1 22 2 3 1 | z| 2 2(3 1) 2. 4 tan 3 1 tan 5 5 3 1 12 12 Hence Polar form is z 2 cos 5 i sin 5 12 12 and exponential form is z 2e5/12 34 E
JEE-Mathematics Illustration 12 : If xn = cos i s in then x1x2x3......... is equal to - 2n 2n (A) –1 (B) 1 (C) 0 (D) Ans. (A) Solution : x = cos i sin = 1 × i n 2n 2n e 2n x x x .. . . . . . . . 1 2 3 i i i = ei 2 22 2n = e 21 .e 22 e 2n = cos ........ + i sin ........ = –1 2 22 23 2 22 23 as ........ / 2 1 1/2 2 22 23 Do yourself - 5 : Express the following complex number in polar form and exponential form : ( i ) –2 + 2i (i i ) 1 3i (1 7i) ( i v ) (1 – cos + isin), (0) (iii) (2 i)2 6 . IMPORTANT PROPERTIES OF CONJUGATE : ( a ) z + z = 2 Re (z) ( b ) z z = 2 i Im (z) (c) ( z ) = z ( d ) z1 z2 = z1 + z2 ( e ) z1 z2 = z1 z2 ( f ) z1 z2 = z1 . z2 . In general z1z2.........zn z1.z2 .........zn (g) z1 = z1 ; z 0 ( h ) If f( + i) = x + iy f( – i) = x – iy z2 z2 2 7 . IMPORTANT PROPERTIES OF MODULUS : ( a ) z 0 ( b ) z Re (z) (c) z Im (z) ( d ) z = z = z – z (e) z z = z 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 ( f ) z1 z2 = z1.z2 . In general | z1z2.........zn || z1 | .| z2 | .........| zn | (g) z1 = z1 , z 0 z2 z2 2 ( h ) zn = zn, n I ( i ) |z + z |2 = |z |2 + |z |2 + 2 Re z1 z2 12 1 2 (j) |z + z |2 = |z |2 + |z |2 + 2|z ||z | co s( – ), where , are arg(z ), arg (z ) respectively . 1 2 1 12 1 2 2 (k) z1 z2 2 + z1 z2 2 = 2 z1 2 z2 2 ( l ) z1 z2 z1 + z2 z1+ z2 [Triangle Inequality] ( m ) z1 z2 z1 – z2 z1+ z2 [Triangle Inequality] E 35
JEE-Mathematics 8 . IMPORTANT PROPERTIES OF AMPLITUDE : (a) amp (z . z ) = amp z + amp z + 2 k k I 1 2 1 2 (b) amp z1 = amp z amp z + 2 k k I z2 1 2 ( c ) amp(zn) = n amp(z) + 2k ; n,k I where proper value of k must be chosen so that RHS lies in ( , ]. (3 2 Illustration 13 : Find amp z and |z| if z 4i)(1 i)(1 3i) . (1 i)(4 3i)(2i) Solution : amp z = 2 amp (3 4i) amp(1 i) amp(1 3i) a m p (1 i) amp(4 3i) am p (2i ) 2k where k I and k chosen so that amp z lies in (–,]. amp z 2 ta n 1 4 tan 1 3 2k 3 4 3 4 4 2 amp z = 2 ta n 1 4 co t 1 4 + 2k amp z 2 2k 3 3 3 2 3 amp z [at k = –1] 3 4i 1 i 1 3i 2 Ans. 3 Ans. | z| 1 i | | 4 3 i| | Also, | 2 i | 2 3 4i 1 i 1 3i | z| 1 i 4 3i2i 5 2 2 2 |z| = 2 5 2 1 Aliter 2 (3 4i)(1 i) 1 3i 3 i 2 z z 1 i4 3i2i 2 2 3i 1 3i z = 2 2 2 4 Hence |z| = 1, amp(z) = . NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 3 Illustration 14 : If zi 1 , then locus of z is - zi (A) x-axis (B) y-axis (C) x = 1 (D) y = 1 Solution : We have, zi 1 x iy 1 1 zi x iy 1 x iy 1 2 1 x2 y 12 x2 y 12 4 y 0; y 0 , which is x-axis Ans. (A) x iy 1 2 Illustration 15 : If |z + z |2 = |z |2+|z |2 then z1 is - 12 12 z2 (A) zero or purely imaginary (B) purely imaginary (D) none of these (C) purely real 36 E
JEE-Mathematics Solution : Here let z = r cos 1 i sin 1 ,| z1| r1 1 1 z = r cos 2 i sin 2 ,| z2 | r2 22 |(z + z )|2 = r1 cos 1 r2 cos 2 i r1 sin 1 r2 sin 2 2 1 2 = r12 r22 2r1r2 cos(1 2 ) = | z1 |2 | z2 |2 if cos(1 2 ) 0 1 2 2 amp(z ) – amp(z ) = a m p z1 z1 is purely imaginary Ans. (B) 12 z2 z2 2 2 Illustration 16 : z and z are two complex numbers such that z1 2z2 is unimodular (whose modulus is one), while 12 2 z1 z2 z is not unimodular. Find |z |. 21 Solution : Here z1 2z2 = 1 z1 2z2 1 2 z1 z2 2 z1 z2 z1 2z2 2 z1 z2 z1 2z2 2 2 z1 z2 2 z1 2z2 z1 2z2 2 z1 z2 2 z1 z2 z1 2z2 z1 2z2 2 z1 z2 2 z1z2 z1 z1 2z1 z2 2z2 z1 4z2 z2 4 2z1z2 2z1 z2 z1 z1z2 z2 z1 2 4 z2 2 4 z1 2 z2 2 z1 2 z1 2 z2 2 4 z2 2 4 0 z1 2 4 1 z2 2 0 But |z | 1 (given) 2 |z |2 = 4 1 Hence, |z1| = 2. I l l u s t r a t i o n 1 7 : The locus of the complex number z in argand plane satisfying the inequality NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 lo g1 / 2 | z 1| 4 1 where | z 1| 2 is - 3| z 1| 2 3 (A) a circle (B) interior of a circle (C) exterior of a circle (D) none of these Solution : We have, log1 / 2 z 1 4 1 log1/ 2 1 3 z 1 2 2 z 1 4 1 3 z 1 2 2 loga x is a decreasing function if a 1 2 z 1 8 3 z 1 2 as |z – 1| > 2/3 z 1 10 Ans. (C) which is exterior of a circle. (D) 5 + 1 Illustration 18 : If z4 = 2, then the greatest value of z is - z (A) 1 + 2 (B) 2 + 2 (C) 3 + 1 E 37
JEE-Mathematics Solution : We have z z4 4 z4 4 = 2 + 4 z z z z z z 2 2 z 4 z 12 5 z 1 5 z 5 1 Therefore, the greatest value of z is 5 + 1 . Ans. (D) Illustration 19 : Shaded region is given by - C(1+3 3i ) (A) |z + 2| 6, 0 arg(z) 6 z (B) |z + 2| 6, 0 arg(z) 3 A 0 B(4) (C) |z + 2| 6, 0 arg(z) 2 –2 (D) None of these Solution : Note that AB = 6 and 1 + 3 3i = –2 + 3 + 3 3i = –2 + 6 1 3 = –2 + 6 cos i sin 2 2 i 3 3 BAC = 3 Thus, shaded region is given by |z + 2| 6 and 0 arg (z + 2) Ans. (C) 3 Do yourself - 6 : ( i ) The inequality |z – 4| < |z – 2| represents region given by - (A) Re(z) > 0 (B) Re(z) < 0 (C) Re(z) > 3 (D) none (D) e–rsin ( i i ) If z = rei, then the value of |eiz| is equal to - (A) e–rcos (B) ercos (C) ersin 9 . SECTION FORMULA AND COORDINATES OF ORTHOCENTRE, CENTROID, CIRCUMCENTRE, INCENTRE OF A TRIANGLE : If z & z are two complex numbers then the complex number z = nz1 mz2 divides the join of z & z in the 12 m n 12 ratio m : n. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 Note : (i) If a , b , c are three real numbers such that az + bz + cz = 0 ; where a + b + c = 0 and a,b,c are 1 2 3 not all simultaneously zero, then the complex numbers z , z & z are collinear. 1 2 3 (ii) If the vertices A, B, C of a triangle represent the complex numbers z1, z2, z3 respectively, then : • Centroid of the ABC = z1 z2 z3 3 • Orthocentre of the ABC = a sec A z1 b sec B z2 c sec C z3 or z1 tan A z2 tan B z3 tan C tan A tan B tan C a sec A b sec B c sec C • Incentre of the ABC = (az1 bz2 cz3 ) (a b c) • Circumcentre of the ABC = (z1 sin 2A z2 sin 2B z3 sin 2C ) E (sin 2A sin 2B sin 2C) 38
JEE-Mathematics 1 0 . VECTORIAL REPRESENTATION OF A COMPLEX NUMBER : y ( a ) In complex number every point can be represented in terms of P(z) position vector. If the point P represents the complex number z O x y Q(z2) then, OP = z & OP = z ( b ) If P(z ) & Q(z ) be two complex numbers on argand plane then P(z1) 12 represents complex number z – z. 2 1 PQ O x Note : y Q(z1) P(z) r (i) If OP = z = r ei then OQ = z = r ei ( + ) = z. e i. If OP and OQ are r 1 x of unequal magnitude then ^ ^ ei i.e. z1 z e i OQ OP z1 z O (ii) In general, if z , z , z be the three vertices of ABC then y C(z3) 1 2 3 z3 z1 | z3 z1| ei . Here arg z3 z1 . z2 z1 z2 z1 | z2 z1 | A(z1) (iii) Note that the locus of z satisfying arg z z1 is: B(z2) z z2 x Case (a) 0 < < /2 z Locus is major arc of circle as shown excluding z & z z2 z1 12 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 Case (b) 2 Locus is minor arc of circle as shown z2 z1 excluding z & z z 12 (iv) If A, B, C & D are four points representing the complex numbers z , z , z & z then AB CD if z4 z3 is purely real ; 1 2 3 4 z2 z1 AB CD if z4 z3 is purely imaginary. z2 z1 (v) If z , z , z are the vertices of an equilateral triangle where z is its circumcentre then 123 0 (1) z 2 + z 2 + z 2 z z z z z z = 0 (2) z 2 + z 2 + z 2 = 3 z 2 1 2 3 1 2 2 3 3 1 1 2 3 0 E 39
JEE-Mathematics Illustration 20 : Complex numbers z , z , z are the vertices A, B, C respectively of an isosceles right angled Solution : 1 2 3 triangle with right angle at C. Show that (z – z )2 = 2(z – z )(z – z ). 12 1 33 2 In the isosceles triangle ABC, AC = BC and BCAC. It means that AC is rotated through angle /2 to occupy the position BC. Hence we have, z2 z3 e i / 2 i z – z = +i(z – z) z1 z3 2 3 1 3 z 2 2 z12 z 2 B(z2) 2 z 3 2z2z3 3 2z1z3 z12 z 2 2z1z2 2z1z3 2z2z3 2z1z2 2 z 2 2 3 = 2 z1 z3 z3 z2 C(z3) A(z1) z1 z2 2 2 z1 z3 z3 z2 Illustration 21: If the vertices of a square ABCD are z , z , z & z then find z & z in terms of z & z . 123 4 34 12 Solution : Using vector rotation at angle A z3 z1 z3 z1 i A(z1) D(z4) z2 z1 z2 z1 e4 4 z3 z1 AC and z2 z1 AB Also AC = 2 AB B(z2) C(z3) z3 z1 2 z2 z1 z3 z1 2 cos i sin z2 z1 = 4 4 z – z = (z – z ) (1 + i) 3 1 2 1 z = z + (z – z ) (1 + i) 3 1 2 1 Similarly z = z + (1 + i)(z – z ) 4 2 1 2 Illustration 22 : Plot the region represented by arg z 1 2 in the Argand plane. 3 z 1 3 Solution : Let us take arg z 1 = 2 , clearly z lies on the minor arc of NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 z 1 3 the circle passing through (1, 0) and (–1, 0). Similarly, z 1 (–1, 0) (1, 0) z 1 3 arg = means that 'z' is lying on the major arc of the 2/3 circle passing through (1, 0) and (–1, 0). Now if we take any point in the region included between two arcs say P (z ) we get 11 arg z 1 2 3 z 1 3 Thus arg z 1 2 represents the shaded region (excluding points (1, 0) and (–1, 0)) . 3 z 1 3 40 E
JEE-Mathematics Do yourself - 7 : ( i ) A complex number z = 3 + 4i is rotated about another fixed complex number z = 1 + 2i in anticlockwise 1 direction by 45° angle. Find the complex number represented by new position of z in argand plane. (ii) If A, B, C are three points in argand plane representing the complex number z , z , z such that 123 z = z2 z3 , where R , then find the distance of point A from the line joining points B and C. 1 1 (iii) If A(z ), B(z ), C(z ) are vertices of ABC in which ABC = and AB 2 , then find z in 123 4 BC 2 terms of z and z . 13 ( i v ) If a & b are real numbers between 0 and 1 such that the points z = a + i, z = 1 + bi and z = 0 form 12 3 an equilateral triangle then a and b are equal to :- (A) a = b = 1/2 (B) a = b = 2 – 3 (C) a = b = –2 + 3 (D) a = b = 2 1 (v) If arg z 1 , find locus of z. z 1 4 11 . DE'MOIVRE’S THEOREM : The value of (cos + isin)n is cosn + isinn if 'n' is integer & it is one of the values of (cos + isin)n if n is a rational number of the form p/q, where p & q are co-prime. Note : Continued product of the roots of a complex quantity should be determined by using theory of equations. Illustration 23: If cos + cos + cos = 0 and also sin + sin + sin = 0, then prove that (a) cos2 + cos2 + cos2 = sin2 + sin2 + sin2 = 0 (b) sin3 + sin3 + sin3 = 3sin( ) (c) cos3 + cos3 + cos3 = 3cos( ) Solution : Let z = cos + i sin, z = cos + isin & z = cos + isin. 1 2 3 z + z + z = (cos + cos + cos) + i(sin + sin + sin) 1 2 3 = 0 + i . 0 = 0 .......... (i) (a) Also 1 cos i sin 1 cos i sin z1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 11 cos i sin , cos i sin z2 z3 111 z1 z2 z3 = (cos + cos + cos) – i(sin + sin + sin) ........ (ii) = 0 – i . 0 = 0 2 2 Now z12 2 2 z 2 z 3 z1 z2 z3 z1z2 z2z3 z3z1 = 0 – 2z z z 1 1 1 = 0 – 2z z z . 0 = 0 {using (i) and (ii)} 123 z3 z1 z2 1 2 3 or cos i sin 2 (cos i sin )2 cos i sin 2 0 or cos2 + isin2 + cos2 + isin2 + cos2 + isin2 = 0 + i.0 Equating real and imaginary parts on both sides, cos2 + cos2 + cos2 = 0 and sin2 + sin2 + sin2 = 0 E 41
JEE-Mathematics (b) If z + z + z = 0 then z13 z 3 z 3 3z1z2z 3 1 2 3 2 3 (cos + isin)3 + (cos + isin)3 + (cos + isin)3 = 3(cos + isin) (cos + isin) (cos + isin) or cos3 + isin3 + cos3 + isin3 + cos3 + isin3 = 3{cos() + isin()} Equating imaginary parts on both sides, sin3 + sin3 + sin3 = 3sin( ) (c) Equating real parts on both sides, cos3 + cos3 + cos3 = 3cos( ) Do yourself - 8 : (i) If zr cos 2r i sin 2r, r 0,1,3, 4,......... , then z z z z z is equal to - 55 12345 (A) –1 (B) 0 (C) 1 (D) none of these (D) none of these ( i i ) If (x – 1)4 – 16 = 0, then the sum of nonreal complex values of x is - (D) 12 (A) 2 (B) 0 (C) 4 (i i i ) If ( 3 i)n 2n , n Z , then n is a multiple of - (A) 6 (B) 10 (C) 9 1 2 . CUBE ROOT OF UNITY : (a) The cube roots of unity are 1 , 1 i 3 () , 1 i 3 (2 ) . 2 2 ( b ) If is one of the imaginary cube roots of unity then 1 + + ² = 0. In general 1 + r + 2r = 0 ; where r I but is not the multiple of 3 & 1 + r + 2r = 3 if r = 3 ; I ( c ) In polar form the cube roots of unity are : Im 2 2 4 4 1 = cos 0 + i sin 0 ; = cos + i sin , 2 = cos + i sin 33 33 ( d ) The three cube roots of unity when plotted on the argand 2/3 plane constitute the vertices of an equilateral triangle. O 1 Re ( e ) The following factorisation should be remembered : (a, b, c R & is the cube root of unity) 2 a3 b3 = (a b) (a b) (a ²b) ; x2 + x + 1 = (x ) (x 2) ; a3 + b3 = (a + b) (a + b) (a + 2b) ; NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 a3 + b3 + c3 3abc = (a + b + c) (a + b + ²c) (a + ²b + c) Illustr ation 24 : If & are imaginary cube roots of unity then n +n is equal to - 2n 2n 2n 2n (A) 2cos 3 (B) cos 3 (C) 2i sin 3 (D) i sin 3 Solution : cos 2 i sin 2 33 cos 2 i sin 2 33 cos2 i sin 2 n cos2 i sin 2 n 3 3 3 3 n n + = cos 2n i sin 2n cos 2n i sin 2n = 2cos 2n Ans. (A) 3 3 3 3 3 E 42
JEE-Mathematics Illustr ation 25 : If are roots of x3 – 3x2 + 3x + 7 = 0 (and is imaginary cube root of unity), then find the value of 1 1 1 . 1 1 1 Solution : We have x3 – 3x2 + 3x + 7 = 0 (x – 1)3 + 8 = 0 (x – 1)3 = (–2)3 x 13 1 x 1 11 / 3 = 1, , 2 (cube roots of unity) 2 2 x = –1, 1 – 2 1 – 22 Here = –1, = 1 – 2, = 1 – 22 – 1 = –2, – 1 = –2, – 1 = –22 Then 1 1 1 = 2 2 22 = 1 1 2 2 2 2 1 1 1 2 22 2 Therefore 1 1 1 = 32. Ans. 1 1 1 Do yourself - 9 : ( i ) If is an imaginary cube root of unity, then (1 + – 2)2 equals : - (A) (B) –4 (C) 2 (D) 4 ( i i ) If is a non real cube root of unity, then the expression (1 – )(1 – 2)(1 + 4)(1 + 8) is equal to : - (A) 0 (B) 3 (C) 1 (D) 2 1 3 . nth ROOTS OF UNITY : If 1 , 1 , 2 , 3..... n 1 are the n , nth root of unity then : ( a ) They are in G.P. with common ratio ei(2/n) 2 A2() ( b ) Their arguments are in A.P. with common difference n ( c ) The points represented by n, nth roots of unity are located at the vertices of a (2)A3 regular polygon of n sides inscribed in a unit circle having center at origin, 2/n one vertex being on positive real axis. 2/n A1(1) 2/n (d) 1p + p + p +.... + p = 0 if p is not an integral multiple of n An(n–1) 1 n 1 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 = n if p is an integral multiple of n ( e ) (1 1) (1 2)...... (1 n 1) = n (f) (1 + 1) (1 + 2)....... (1 + n ) = 0 if n is even and 1 = 1 if n is odd. ( g ) 1. 1. 2. 3......... n 1 = 1 or 1 according as n is odd or even. I l l u s t r a t i o n 6 sin 2 k cos 2 k 26: Find the value k 1 7 7 6 6 6 6 2k 2k 2k 2 k Solution : sin 7 cos 7 = sin 7 cos 7 1 k 1 k 1 k 1 k0 6 = (Sum of imaginary part of seven seventh roots of unity) k 0 6 – (Sum of real part of seven seventh roots of unity) +1 = 0 – 0 + 1 = 1 k 0 E 43
JEE-Mathematics 1 4 . THE SUM OF THE FOLLOWING SERIES SHOULD BE REMEMBERED : sin n / 2 n 1 ( a ) cos + cos 2 + cos 3 +..... + cos n = sin / 2 cos 2 sin n / 2 n 1 ( b ) sin + sin 2 + sin 3 +..... + sin n = sin / 2 sin 2 Note : If = (2/n) then the sum of the above series vanishes. 1 5 . STRAIGHT LINES & CIRCLES IN TERMS OF COMPLEX NUMBERS : y z x ( a ) amp(z–) = is a ray emanating from the complex point and inclined at an angle to the x axis. O y (a) (b) ( b ) z a = z b is the perpendicular bisector of the segment joining a & b. O x ( c ) The equation of a line joining z & z is given by ; y z1 12 z = z + t (z z ) where t is a parameter. z2 1 1 2 O x z1 ( d ) z = z (1 + it) where t is a real parameter, is a line through the point z & y 11 perpendicular to z . Ox 1 (e) The equation of a line passing through z & z can be expressed in the determinant form as 12 z z1 z1 z1 1 = 0. This is also the condition for three complex numbers to be collinear. z2 z2 1 (f) Complex equation of a straight line through two given points z & z can be written as 12 z z1 z2 z z1 z2 z1 z2 z1 z2 = 0, which on manipulating takes the form as z z r = 0 where r is real and is a non zero complex constant. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 (g) The equation of circle having centre z & radius is : (z) 0 z z0= or z z z z z0 z + z0 z ² = 0 which is of the form z0 0 0 z z z z r = 0 , r is real centre = & radius = r . Circle will be real if r 0 . (h) arg z z2 = ± or (z z ) ( z z ) + (z z ) ( z z ) = 0 z z z1 2 1 2 2 1 z1 z2 this equation represents the circle described on the line segment joining z & z as diameter. 12 (i) Condition for four given points z , z , z & z to be concyclic is, the number z3 z1 . z4 z2 z3 z2 z4 z1 1 2 3 4 is real. Hence the equation of a circle through 3 non collinear points z , z & z can be taken as 12 3 z z2 z3 z1 is real z z2 z3 z1 = z z2 z3 z1 z z1 z3 z2 z z1 z3 z2 z z1 z3 z2 44 E
JEE-Mathematics Miscellaneous Illustration : Illustration 27 : z 2 Solution : If z is a point on the Argand plane such that |z – 1| = 1, then is equal to - z (A) tan (arg z) (B) cot (arg z) (C) i tan (arg z) (D) none of these Since |z – 1| = 1, let z 1 cos i sin Then, z 2 cos i sin 1 2 sin2 2i sin cos 2i sin cos i sin .... (i) 2 22 2 2 2 and z 1 cos i sin 2 cos2 2i sin cos 2 cos cos i sin .... (ii) 2 2 2 2 2 2 From (i) and (ii), we get z 2 i tan i tan arg z arg z from ii Ans. (C) z 2 2 Illustration 28 : Let a be a complex number such that |a| < 1 and z , z , ....... , z be the vertices of a polygon such that z = 1+ a + a2 + .... ak, then show that v1erti2ces of then polygon lie within the circle k 1 1 . z 1a 1a Solution : We have, zk 1a a2 ..... ak 1 a k1 1a 1 a k 1 1 a k 1 1 a 1 1 1a zk 1 a 1 a 1 a zk a 11 z Vertices of the polygon z1, z2 ,....., zn lie within the circle 1a 1a Illustration 29 : If z and z are two complex numbers and C > 0, then prove that Solution : 12 |z + z |2 (1 + C) |z |2 + (1 + C–1) |z |2 2 1 2 1 We have to prove that : |z + z |2 (1 + C) |z |2 + (1 + C–1)|z |2 1 2 1 2 i.e. |z |2 + |z |2 + z1 z2 z1 z2 1 C |z |2 + (1 +C–1)|z |2 12 12 or z1 z2 z1 z2 C z1 2 C 1 z2 2 or C z1 2 1 z2 2 z1 z2 z1 z2 0 (using Re z1 z2 z1 z2 ) C NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 1 2 C or C z1 z2 0 which is always true. Illu str ation 30 : If [/6, /3], i = 1, 2, 3, 4, 5 and z4 cos1 + z3 cos2 + z2 cos3 + z cos4 + cos5 = 2 3 , then 3 show that |z| > 4 Solution : Given that cos 1. z4 cos 2 . z3 cos 3 . z2 cos 4 . z cos 5 2 3 or cos 1 . z4 cos 2 . z3 cos 3 . z2 cos 4 . z cos 5 2 3 2 3 cos 1.z4 cos 2 .z3 cos 3 .z2 cos 4 .z cos 5 i / 6, / 3 13 2 cos i 2 E 45
JEE-Mathematics 2 3 3 z4 3 z3 3 z2 3 3 z 2 2 2 22 3 z 4 z 3 z 2 z 3 z z 2 z 3 z 4 z 5 .......... | z| 3 1| z| 3 – 3|z| < |z| 4|z| > 3 | z| 3 4 2 11 Illustration 31 : If z , z , z are complex numbers such that z1 z2 z3 , show that the points represented by 123 z , z , z lie on a circle passing through the origin. 12 3 Solution : We have, 2 1 1 1 1 1 1 z2 z1 z1 z3 O z1 z2 z1 z3 z1z2 z1z3 z2 z3 z2 z1 z2 z1 z2 arg z 2 z1 z2 z3 z3 z1 z3 z 3 z1 arg z3 z1 a rg z 2 z1 arg z2 or arg z3 = z 3 z1 z3 z2 Thus the sum of a pair of opposite angle of a quadrilateral is 180°. Hence, the points 0, z1, z2 and z3 are the vertices of a cyclic quadrilateral i.e. lie on a circle. Illustration 32 : Two given points P & Q are the reflection points w.r.t. a given straight line P z1 if the given line is the right bisector of the segment PQ. Prove that the two points denoted by the complex numbers z & z will be the reflection 12 points for the straight line z z r 0 if and only if ; z1 z2 r 0 , where r is real and is non zero complex constant. Q z2 Solution : Let P(z ) is the reflection point of Q(z ) then the perpendicular bisector of z & z must be the line NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 12 12 z z r 0 ......... (i) Now perpendicular bisector of z & z is, z z1 z z2 12 or (z – z ) z z1 z z2 z z2 1 z z1 z1 z z1 z1 z z 2 z2 z z2 z2 ( zz cancels on either side) or z2 z1 z z2 z1 z z1 z1 z2 z2 0 ......... (ii) r Comparing (i) & (ii) z2 z1 z2 z1 z1 z1 z2 z2 z2 z1 ........ (iii) z2 z1 ......... (iv) r z1 z1 z2 z2 ......... (v) Multiplying (iii) by z ; (iv) by z2 and adding 1 z1 z2 r 0 Note that we could also multiply (iii) by z & (iv) by z1 & add to get the same result. 2 46 E
JEE-Mathematics Hence z1 z2 r 0 Again, let z1 z2 r 0 is true w.r.t. the line z z r 0 . Subtracting z z1 ( z z2 ) 0 or z z1 z z2 or z z1 z z2 z z2 Hence 'z' lies on the perpendicular bisector of joins of z & z . 12 ANSWERS FOR DO YOURSELF 1 : ( i ) n = 4 (ii) 0 2 : (i ) –17 + 24i (iii) ±(1 – 4i) 3 : 2 ( i i ) | z| 2; amp(z) 5 (i i i ) | z| 2; amp(z) ( i ) |z| = 4; amp(z) = 3 6 2 ( i v ) | z| 1 ; amp(z) 3 ( v ) | z| 2; amp(z) 24 3 4 : ( i ) 13 units (ii) locus is a circle on complex plane with center at (2,3) and radius 1 unit. (i i i ) C 3 3 ; i 3 4 4 ; i 4 4 4 4 3 3 3 5 : (i) 2 2 cos i sin 2 2 e (ii) 2 cos i sin 2e NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 3 3 ; i 3 i 4 4 4 2 2 2 2 2 ; 2 22 (iii) 2 co s i sin 2e (iv) 2 sin cos i sin 2 sin e 6 : (i ) C (ii) D 7 : ( i ) 1 (2 2 2 )i ( i i ) 0 (iii) z2 = z3 + i(z1 – z3) (iv) B ( v ) Locus is all the points on the major arc of circle as shown excluding points 1 & –1. Im z /4 c(0,1) –1 O 1 Re 8 : (i ) C (ii) A (iii) D 9 : (i ) D (ii) B E 47
JEE-Mathematics CHECK YOUR GRASP EXERCISE - 01 SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 13 [JEE 98] 1 . The value of the sum in in1 , where i 1 , equals n 1 (A) i (B) i – 1 (C) –i (D) 0 2 . The sequence S = i + 2i2 + 3i3 + ...... upto 100 terms simplifies to where i = 1 - (A) 50(1 – i) (B) 25i (C) 25(1 + i) (D) 100(1 – i) 3 . Let i 1 . The product of the real part of the roots of z2 – z = 5 – 5i is - (A) –25 (B) –6 (C) –5 (D) 25 11 4 . If z1 = a i , a 0 and z2 = 1 bi , b 0 are such that z1 z2 then - (A) a = 1, b = 1 (B) a = 1, b = –1 (C) a = –1, b = 1 (D) a = –1, b = –1 5 . The inequality |z – 4| < |z – 2| represents the following region - (A) Re(z) > 0 (B) Re(z) < 0 (C) Re(z) > 2 (D) none of these 6 . If (1 + i) ( 1 + 2i) ( 1 + 3i) .... (1 + ni) = + i then 2 . 5 . 10 ... (1 + n2) = (A) – i (B) 2 – 2 (C) 2 + 2 (D) none of these 7 . In the quadratic equation x2 + (p +iq) x + 3i = 0 , p & q are real. If the sum of the squares of the roots is 8 then : (A) p = 3, q = –1 (B) p = –3, q = –1 (C) p = 3, q = 1 or p = –3, q = –1 (D) p = –3, q = 1 8 . The curve represented by Re(z2) = 4 is - (B) an ellipse (A) a parabola (D) a rectangular hyperbola (C) a circle 9 . Real part of e ei is - (A) ecos [cos (sin )] (B) ecos [cos (cos )] (C) esin [sin (cos )] (D) esin [sin (sin )] 1 0 . Let z and are two non-zero complex numbers such that |z| = || and arg z + arg = , then z equal to - (A) (B) – (C) (D) – 1 1 . Number of values of x (real or complex) simultaneously satisfying the system of equations 1 + z + z2 + z3 + ........ + z17 = 0 and 1 + z + z2 + z3 + ......... + z13 = 0 is - NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 (A) 1 (B) 2 (C) 3 (D) 4 1 2 . If |z | = 1, |z | = 2, |z | = 3 and |9z z + 4z z + z z | = 12 then the value of |z + z + z | is equal to- 123 12 13 23 123 (A) 2 (B) 3 (C) 4 (D) 6 1 3 . A point ‘z’ moves on the curve | z – 4 – 3i |= 2 in an argand plane. The maximum and minimum values of |z| are - (A) 2, 1 (B) 6, 5 (C) 4, 3 (D) 7, 3 1 4 . The set of points on the complex plane such that z2 + z + 1 is real and positive (where z = x + iy, x, y R ) is- (A) Complete real axis only (B) Complete real axis or all points on the line 2x + 1 = 0 (C) Complete real axis or a line segment joining points 1 , 3 & 1 , 3 excluding both. 2 2 2 2 (D) Complete real axis or set of points lying inside the rectangle formed by the lines. 2x + 1 = 0 ; 2x – 1 = 0 ; 2y 3 0 & 2y 3 0 E 48
JEE-Mathematics 1 5 . If is an imaginary cube root of unity, then (1 2 )7 equals [JEE 98] (A) 128 (B) –128 (C) 1282 (D) –1282 334 365 1 i 3 1 i 3 1 6 . If i 1 , then 4 5 2 2 3 2 2 is equal to : [JEE 99] (A) 1 i 3 (B) 1 i 3 (C) i 3 (D) i 3 17. The set of points on an Argand diagram which satisfy both |z| 4 & Arg z are lying on - 3 (A) a circle & a line (B) a radius of a circle (C) a sector of a circle (D) an infinite part line 18. If Arg (z – 2 – 3i) = , then the locus of z is - 4 y y (2,3) (2,3) (A) (B) (C) x (D) x (–2,–3) (–2,–3) 1 9 . The origin and the roots of the equation z2 + pz + q = 0 form an equilateral triangle if - (A) p2 = 2q (B) p2 = q (C) p2 = 3q (D) q2 = 3p 2 0 . Points z1 & z2 are adjacent vertices of a regular octagon. The vertex z3 adjacent to z2 (z3 z1) can be represented by - 1 1 (A) z2 2 (1 i)(z1 z2 ) (B) z2 2 (1 i)(z1 z2 ) 1 (D) none of these (C) z2 2 (1 i)(z2 z1 ) 6655 1 i 3 1 i 3 1 i 3 1 i 3 21. is equal to - 2 2 2 2 (A) 1 (B) –1 (C) 2 (D) none of these 2 2 . If z and are two non-zero complex numbers such that |z| = 1, and Arg (z) – Arg() = /2, then z is equal to - NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 (A) 1 (B) –1 (C) i (D) –i SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 23. For two complex numbers z and z : (az1 b z1 )(cz2 d z2 ) (cz1 dz1 )(az2 b z2 ) if (a, b, c, d R) - 12 (A) a c (B) a b (C) | z1 || z2 | (D) arg(z ) = arg(z ) b d d c 12 2 4 . Which of the following, locii of z on the complex plane represents a pair of straight lines ? (A) Re(z2) = 0 (B) Im(z2) = 0 (C) |z| + z = 0 (D) |z – 1| = |z – i| 2 5 . If the complex numbers z , z , z represents vertices of an equilateral triangle such that |z |=|z |=|z |, then 123 123 which of following is correct ? (A) z + z + z 0 (B) Re(z + z + z ) = 0 (C) Im(z + z + z ) = 0 (D) z + z + z = 0 1 2 3 123 123 1 2 3 26. |x 1 2i| 2 0 , then S contains - If S be the set of real values of x satisfying the inequality 1 – log 2 1 2 (A) [–3, –1) (B) (–1, 1] (C) [–2, 2] (D) [–3, 1] E 49
JEE-Mathematics 2 7 . If amp (z z ) = 0 and |z | = |z | = 1, then :- 12 1 2 (A) z + z = 0 (B) z z = 1 (C) z = z2 (D) none of these 12 12 1 2 8 . If the vertices of an equilateral triangle are situated at z =0, z=z , z =z , then which of the following is/are true - 12 (A) |z | = |z | (B) |z – z | = |z | 12 12 1 (C) |z + z | = |z | + |z | (D) | arg z – arg z |= /3 12 1 2 1 2 2 9 . Value(s) of (–i)1/3 is/are - 3 i 3 i 3 i 3 i (A) (B) (C) (D) 2 2 2 2 3 0 . If centre of square ABCD is at z=0. If affix of vertex A is z , centroid of triangle ABC is/are - 1 (A) z1 (cos + i sin ) 3 (B) 4 cos i sin 2 2 (C) z1 (D) z1 3 cos 2 i sin 3 cos 2 i sin 2 2 x 1 2 3 1 . If is an imaginary cube root of unity, then a root of equation x 2 1 = 0, can be :- 2 1 x 2 (A) x = 1 (B) x = (C) x = 2 (D) x = 0 CHECK YOUR GRASP ANSWER KEY EXERCISE-1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 Que. 1 2 3 4 5 6 7 8 9 10 Ans. B A B BD CCD AD Que. 11 12 13 14 15 16 17 18 19 20 Ans. A A DBDCCACB Que. 21 22 23 24 25 26 27 28 29 30 Ans. A D A,D A,B B,C,D A,B B.C A,B,D A,C C,D Que. 31 Ans. D 50 E
EXERCISE - 02 JEE-Mathematics BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1 . On the argand plane, let 2 3z , 2 3z & | z | = 1. Then the correct statement is - (A) moves on the circle, centre at (–2,0) and radius 3 (B) & describe the same locus (C) & move on different circles (D) – moves on a circle concentric with |z|=1 2 . The value of in + i–n, for i 1 and n I is - 2n (1 i)2n (1 i)2n (1 i)2n (1 i)2n 2n 2n 2n (A) (1 i)2n 2n (B) 2n 2n (C) (D) (1 i)2n (1 i)2n 2n (1 i)2n 3 . The common roots of the equations z3 + (1 + i)z2 + (1 + i)z + i = 0, (where i = 1 ) and z1993 + z1994 + 1 = 0 are - (where denotes the complex cube root of unity) (A) 1 (B) (C) 2 (D) 981 4. If xr C iS for 1 r n ; r, n N then - 2r n n n Lim Im n 0 (A) Lim R e xr 1 (B) Lim Re xr 0 (C) Im r 1 x r (D) xr Lim 1 n n n r 1 r 1 n r 1 5 . Let z1, z2 be two complex numbers represented by points on the circle |z1| = 1 and |z2|=2 respectively, then - (A) max|2z1+ z2| = 4 (B) min |z1 – z2| = 1 1 (D) none of these (C) z2 z1 3 6 . If , be any two complex numbers such that 1 , then which of the following may be true - 1 (A) | | 1 (B) | | 1 (C) ei, R (D) ei, R 7 . Let z, z and z + z represent three vertices of ABC, where is cube root unity, then - NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 2 2 (A) centroid of ABC is 3 (z z) (B) orthocenter of ABC is 3 (z z) (C) ABC is an obtuse angled triangle (D) ABC is an acute angled triangle 8 . Which of the following complex numbers lies along the angle bisectors of the line - L1 : z = (1 + 3) + i(1 + 4) L2 : z = (1 + 3) + i(1 – 4) (A) 11 i (B) 11 + 5i 3i (D) 5 – 3i 5 (C) 1– 5 9 . Let z and are two complex numbers such that |z| 1, || 1 and |z + i| = |z – i | = 2, then z equals - (A) 1 or i (B) i or –i (C) 1 or –1 (D) i or –1 1 0 . If g(x) and h(x) are two polynomials such that the polynomial P(x) = g(x3) + xh(x3) is divisible by x2 + x + 1, then - (A) g(1) = h(1) = 0 (B) g(1) = h(1) 0 (C) g(1) = –h(1) (D) g(1) + h(1) = 0 BRAIN TEASERS ANSWER KEY EXERCISE-2 Que. 1 2 3 4 5 6 7 8 9 10 C A,C,D Ans. A,B,D B,D B,C A,D A,B,C A,B,C,D A,C A,C E 51
JEE-Mathematics EXERCISE - 03 MISCELLANEOUS TYPE QUESTIONS MATCH THE COLUMN Following question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given s t a t e m e n t i n C o l u m n - I c a n h a v e c o r r e c t m a t c h i n g w i t h O N E s t a t e m e n t i n C o l u m n - I I . 1 . Column-I Column-I (A) If z be the complex number such that z1 2 (p) 0 z then minimum value of | z| is tan 8 (B) |z| = 1 & z2n+1 0 then zn zn is equal to (q) 3 z2n 1 z2n 1 11 (C) If 8iz3 + 12z2 – 18z + 27 i = 0 then 2|z| = (r) (D) If z1, z2, z3, z4 are the roots of equation (s) 1 4 z4 + z3 + z2 + z + 1 = 0, then (zi + 2) is i 1 Following question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled as A, B, C and D while the statements in Column-II are labelled as p, q, r and s. Any given s t a t e m e n t i n C o l u m n - I c a n h a v e c o r r e c t m a t c h i n g w i t h O N E O R M O R E s t a t e m e n t ( s ) i n C o l u m n - I I . 2 . Match the figure in column-I with corresponding expression - Column-I Column-I z1 z2 (p) z4 z3 z4 z3 = 0 z2 z1 z2 z1 (A) two parallel lines z3 z4 (B) z1 z4 (q) z2 z1 z2 z1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 z4 z3 z4 z3 z1 z2 two perpendicular lines (r) z4 z1 . z2 z3 z4 z1 . z2 z3 (C) z3 z2 z1 z4 z3 z2 z1 z4 z3 z4 z2 (s) z1 + z3 = z2 + z4 z4 a parallelogram (D) z3 z1 z3 z2 52 E
JEE-Mathematics ASSERTION & REASON These questions contains, Statement I (assertion) and Statement II (reason). (A) Statement-I is true, Statement-II is true ; Statement-II is correct explanation for Statement-I. (B) Statement-I is true, Statement-II is true ; Statement-II is NOT a correct explanation for statement-I. (C) Statement-I is true, Statement-II is false. (D) Statement-I is false, Statement-II is true. 1 . Statement-I : There are exactly two complex numbers which satisfy the complex equations |z – 4 – 5i|= 4 and Arg (z – 3 – 4i) = simultaneously. 4 Because Statement-II : A line cuts the circle in atmost two points. (A) A (B) B (C) C (D) D 2. Let z1, z2, z3 satisfy z 2 2 and z0 = 2. Consider least positive arguments wherever required. z 1 Statement–1 : 2 arg z1 z3 arg z1 z0 . z2 z3 z2 z0 and Statement–2 : z1, z2, z3 satisfy |z – z0| = 2. (A) A (B) B (C) C (D) D 3 . Statement-I : If z = i + 2i2 + 3i3 + ............. + 32i32, then z, z , –z & – z forms the vertices of square on argand plane. Because Statement-II : z, z , –z, – z are situated at the same distance from the origin on argand plane. (A) A (B) B (C) C (D) D 4. Statement-I : If z1 = 9 + 5i and z2 = 3 + 5i and if arg z z1 then |z – 6 – 8i| = 3 2 z z2 4 Because Statement-II : If z lies on circle having z1 & z2 as diameter then arg z z1 . z z2 4 (A) A (B) B (C) C (D) D NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 5 . Statement-1 : Let z1, z2, z3 be three complex numbers such that |3z1 + 1|= |3z2 + 1|= |3z3 + 1| and 1 + z1 + z2 + z3 = 0, then z1, z2, z3 will represent vertices of an equilateral triangle on the complex plane. and Statement-2 : z1, z2, z3 represent vertices of an equilateral triangle if z 2 z 2 z32 z1z2 z2z3 z3z1 . 1 2 (A) A (B) B (C) C (D) D COMPREHENSION BASED QUESTIONS Comprehension # 1 : Let z be any complex number. To factorise the expression of the form zn – 1, we consider the equation zn = 1. This equation is solved using De moiver's theorem. Let 1, 1, 2,........ n–1 be the roots of this equation, then zn – 1 = (z – 1)(z – 1)(z – 2)........(z – n–1) This method can be generalised to factorize any expression of the form zn – kn. for z7 + 1 = 6 z C iS 2m example, m 0 7 7 This can be further simplified as z7 + 1 = (z + 1) z 2 2z cos 1 z 2 2z cos 3 1 z 2 2z cos 5 1 ............ (i) 7 7 7 E 53
JEE-Mathematics These factorisations are useful in proving different trigonometric identities e.g. in eqaution (i) if we put z = i, then equation (i) becomes (1 i) (i 1) 2i cos 2i cos 3 2i cos 5 7 7 7 i.e. cos cos 3 cos 5 1 7 7 7 8 On the basis of above information, answer the following questions : 1 . If the expression z5 – 32 can be factorised into linear and quadratic factors over real coefficients as z5 32 z 2 (z2 pz 4)(z2 qz 4) , where p > q, then the value of p2 – 2q - (A) 8 (B) 4 (C) –4 (D) –8 2 . By using the factorisation for z5 + 1, the value of 4 sin cos comes out to be - 10 5 (A) 4 (B) 1/4 (C) 1 (D) –1 3 . If (z2n+1 – 1) = (z – 1)(z2 – p1z + 1)........ (z2 – pnz + 1) where n N & p1, p2 ............. pn are real numbers then p1 + p2 + ........... + pn = (A) –1 (B) 0 (C) tan(/2n) (D) none of these Comprehension # 2 : In the figure |z| = r is circumcircle of ABC.D,E & F are the middle A(za) points of the sides BC, CA & AB respectively, AD produced to meet the circle at L. If CAD = , AD = x, BD = y and altitude of ABC O from A meet the circle |z|= r at M, za, zb & zc are affixes of vertices C(zc) A, B & C respectively. On the basis of above information, answer the following questions : (zb)B D 1 . Area of the ABC is equal to - P L M (A) xy cos ( + C) (B) (x + y) sin (C) xy sin ( + C) 1 (B) zbei(–2B) (C) zbeiB (D) 2 xy sin ( + C) 2 . Affix of M is - (B) 2zbei(2A – 2) (C) zbei(A – ) (D) 2zbeiB (A) 2zbei2B (D) 2zbei(A – ) 3 . Affix of L is - (A) zbei(2A – 2) MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-3 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 Match the Column 1. (A) (s), (B) (p), (C) (q), (D) (r) 2. (A) (q), (B) (p), (C) (q, s), (D) (r) Assertion & Reason 1. D 2. A 3. B 4. C 5. B Comprehension Based Questions Comprehension # 1 : 1. A 2. C 3. A Comprehension # 2 : 1. C 2. B 3. A 54 E
JEE-Mathematics EXERCISE - 04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . Find the modulus, argument and the principal argument of the complex numbers. (a) z = 1 + cos 10 + i sin 10 (b) (tan1 – i)2 9 9 5 12 i 5 12 i (c) z = 5 12 i 5 12 i 2 . Given that x , y R , solve : 4x² + 3xy + (2xy 3x²)i = 4y² (x2/2) + (3xy 2y²)i z1 2z2 3 . Let z1 and z2 be two complex numbers such that 2 z1 z2 = 1 and |z2| 1, find |z1|. 4 . If iz3 + z2 – z + i = 0, then prove that |z|=1. e 2 iA e iC e iB 5 . If A, B and C are the angle of a triangle D = eiC e2iB eiA where i = 1 , then find the value of D. e iB e iA e 2iC 6 . For complex numbers z & , prove that, | z|2 | |2 z z if and only if, z = or z 1 7. Let z , z b e c o mp le x numb ers wi th | z |=|z |= 1, prove th at |z + 1|+ |z + 1|+|z z + 1| 2 1 2 12 1 2 12 8 . Interpret the following locii in z C. (a) 1 < z 2i < 3 (b) Re z 2i 4 (z 2i) iz 2 (c) Arg (z + i) Arg (z i) = /2 (d) Arg (z a) = /3 where a = 3 + 4i. 9. Let A = {a R| the equation (1 + 2i)x3 – 2(3 + i)x2 + (5 – 4i)x + 2a2 = 0} has at least one real root. Find the value of a2 . a A NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 1 0 . ABCD is a rhombus in the Argand plane. If the affixes of the vertices be z1, z2, z3, z4 and taken in anti-clockwise sense and CBA = /3, show that (a) 2z2 = z1(1 + i 3 ) + z3(1 – i 3 ) & (b) 2z4 = z1(1 – i 3 ) + z3(1 + i 3 ) 1 1 . P is a point on the Argand plane. On the circle with OP as diameter two points Q & R are taken such that POQ = QOR = . If 'O' is the origin & P, Q & R are repre sented by the complex numbers Z , Z & Z respectively, show that : Z 2 cos2 = Z . Z cos2. 12 3 2 1 3 1 2 . Let A z1 ; B z2 ; C z3 are three complex numbers denoting the vertices of an acute angled triangle. If the origin ‘O’ is the orthocentre of the triangle, then prove that z1 z2 z1z2 z2 z3 z2 z3 z3 z1 z3 z1 . 1 3 . (a) If is an imaginary cube root of unity then prove that : (b) (1 + 2) (1 2 + 4) (1 4 + 8)..... to 2n factors = 22n. E If is a complex cube root of unity , find the value of ; (1 + ) (1 + 2) (1 + 4) (1 + 8)..... to n factors. 55
JEE-Mathematics 1 4 . If the biquadratic x4 + ax3 + bx2 + cx + d = 0 (a, b, c, d R) has 4 non real roots, two with sum 3 + 4i and the other two with product 13 + i. Find the value of 'b'. 1 5 . If x = 1+ i 3 ; y = 1 i 3 & z = 2 , then prove that xp + yp = zp for every prime p > 3. CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1 . (a) Principal Arg z = 4 ; z = 2 cos 4 ; Arg z = 2 k 4 k I NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 9 9 9 (b) Modulus = sec21, Arg z = 2 n +(2 – ), Principal Arg z = (2 – ) (c) Principal value of Arg z = & z = 3 , Arg z = 2n , n I 22 2 2 Principal value of Arg z = 2 & z = 3 , Arg z = 2n 2 , n I 3K 2 . x = K , y = K R 3. 2 5. –4 2 8 . (a) The region between the concentric circles with centre at (0 , 2) & radii 1 & 3 units 11 (b) region outside or on the circle with centre 2 + 2i and radius 2 (c) semi circle (in the 1st & 4th quadrant) x² + y² = 1 (d) a ray emanating from the point (3 + 4i) directed away from the origin & having equation 3 x y 4 3 3 0 9. 18 1 3 . (b) one if n is even ; ² if n is odd 1 4 . 5 1 56 E
EXERCISE - 04 [B] JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE 1 . (a) Let z = x + iy be a complex number, where x and y are real numbers. Let A and B be the sets defined by A = {z||z| 2} and B = {z|(1 – i)z + (1 + i) z 4}. Find the area of the region A B. 1 (b) For all real numbers x, let the mapping f(x) = , where i = 1 . If there exist real numbers a, b, c x i and d for which f(a), f(b), f(c) and f(d) form a square on the complex plane. Find the area of the square. pq r 2 . If q r p 0 ; where p , q , r are the moduli of non-zero complex numbers u, v, w respectively, prove r pq w w u2 that, arg v = arg v u . 3 . 1 sin ( n x ) a b b For x (0, /2) and sin x = 3 , if n0 3n c then find the value of (a + b + c), where a, b, c are positive integers. (You may use the fact that sin x = eix e ix ) 2i 4 . If z , z are the roots of the equation az2 + bz + c = 0, with a, b, c > 0 ; 2b2 > 4ac > b2 ; z third quadrant ; 12 1 z2 second quadrant in the argand's plane then, show that z1 b2 1 / 2 z2 4ac a rg 2 cos 1 5 . If Zr, r = 1, 2, 3,..... 2m, m N are the roots of the equation Z2m + Z2m–1 + Z2m–2 +....... + Z +1 = 0 2m 1 then prove that r1 Zr 1 = – m 6. If (1 + x)n = C + C x + ......+ C xn (n N), prove that : 0 1 n 1 2 n 1 2n /2 cos n 1 2 n 1 2n /2 sin n (a) C0 + C4 + C8 + .... = 2 4 (b) C1 + C5 + C9 + .... = 2 4 (c) C + C + C + .... = 1 2 n 1 2n /2 cos n (d) C + C + C + .... = 1 2 n 1 2n /2 sin n 2 6 10 4 3 7 11 4 2 2 1 2 n 2 cos n (e) C0 + C3 + C6 + C9 + .... = 3 3 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 x n 2 7 . Prove that : (a) cos x + nC1 cos 2x + nC2 cos 3x +..... + nCn cos (n + 1) x = 2n. cosn 2 . cos 2 x (b) sin x + nC sin 2x + nC sin 3x +..... + nC sin (n + 1) x = 2n. cosn x n 2 . sin x 12 n 2 2 8 . The points A, B, C depict the complex numbers z , z , z respectively on a complex plane & the angle B & C of 123 1 (z2 z3 )2 4(z3 z1 )(z1 z2 ) sin2 the triangle ABC are each equal to ( ) . Show that : 2 2 9 . : 32 10 2q 2q p . Evaluate p 1 (3p 2) q 1 sin 11 i cos 11 10. Let a, b, c be distinct complex numbers such that 1 a b 1 b c 1 c a =k. Find the value of k. BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 1 . (a) – 2 (b) 1/2 3. 41 9 . 48(1 – i) 1 0 . – or – 2 E 57
JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [A] 1 . The inequality |z – 4| < |z – 2| represents the following region [AIEEE-2002] (1) Re(z) > 0 (2) Re(z) < 0 (3) Re(z) > 2 (4) none of these 2 . Let z and are two non-zero complex numbers such that |z| = || and arg z + arg = , then z equal to [AIEEE-2002] (1) (2) – (3) (4) – 3 . Let z and z be two roots of the equation z2 + az + b = 0, z being complex, Further, assume that the origin 12 z3, z1 and z2 form an equilateral triangle. then- [AIEEE-2003] (1) a2 = b (2) a2 = 2b (3) a2 = 3b (4) a2 = 4b 4 . If z and are two non-zero complex numbers such that |z| = 1, and Arg(z) –Arg() = /2, then z is equal to [AIEEE-2003] (1) 1 (2) –1 (3) i (4) –i 5. If 1 + ix = 1 , then [AIEEE-2003] 1 – i (1) x = 4n, where n is any positive integer (2) x = 2n, where n is any positive integer (3) x = 4n + 1, where n is any positive integer (4) x = 2n + 1, where n is any positive integer 6 . Let z, w be complex numbers such that z + i w = 0 and arg zw = . Then arg z equals [AIEEE-2004] (1) /4 (2) /2 (3) 3/4 (4) 5/4 (3) a circle 7 . If |z2 – 1| = |z|2 + 1, then z lies on [AIEEE-2004] (1) the real axis (2) the imaginary axis (4) an ellipse x y [AIEEE-2004] 8 . If z = x – iy and z1/3 = p + iq, then p + q is equal to- (p2 + q2 ) (1) 1 (2) –1 (3) 2 (4) –2 9 . If z and z are two non zero complex numbers such that |z + z |=|z |+|z | then arg z – arg z is equal 12 12 1 2 1 2 to- [AIEEE-2005] (1) – (4) 0 (2) (3) – 2 2 z [AIEEE-2005] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 1 0 . If w = 1 and |w|= 1 then z lies on (4) a straight line z– i 3 [AIEEE-2007] (1) a circle (2) an ellipse (3) a parabola (4) 0 1 1 . If |z + 4| 3, then the maximum value of |z + 1| is- [AIEEE-2008] (1) 4 (2) 10 (3) 6 1 1 2 . The conjugate of a complex number is , then that complex number is- i 1 1 1 1 1 (1) (2) (3) (4) i 1 i 1 i 1 i 1 13. If 4 2 , then the maximum value of |Z| is equal to :- [AIEEE-2009] Z Z (1) 2 (2) 2 + 2 (3) 3 +1 (4) 5 +1 58 E
JEE-Mathematics 1 4 . The number of complex numbers z such that |z – 1| = |z + 1| = |z – i| equals :- [AIEEE-2010] (1) 0 (2) 1 (3) 2 (4) 1 5 . Let , be real and z be a complex number. If z2 + z+ = 0 has two distinct roots on the line Re z = 1, then it is necessary that :- [AIEEE-2011] (1) 1 (2) (1, ) (3) (0,1) (4) (1,0) 1 6 . If (1) is a cube root of unity, and (1 + )7 = A + B. Then (A, B) equals :- [AIEEE-2011] (1) (1, 0) (2) (–1, 1) (3) (0, 1) (4) (1, 1) 1 7 . If z 1 and z2 is real, then the point represented by the complex number z lies : [AIEEE-2012] z 1 (1) on the imaginary axis. (2) either on the real axis or on a circle passing through the origin. (3) on a circle with centre at the origin. (4) either on the real axis or on a circle not passing through the origin. 18. If z is a complex number of unit modulus and argument , then arg 1 z equals [JEE (Main)-2013] z 1 (1) – (2) (3) (4) – 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans 4 4 3 4 1 3 2 4 4 4 3 3 4 2 2 Que. 16 17 18 Ans 4 2 3 59 E
JEE-Mathematics JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [B] 1. (a) If z , z , z are complex numbers such that | z 1 || z 2 | | z 3 | 1 1 1 1 then |z +z + z | is - 1 2 3 z1 z3 1 2 3 z2 (A) equal to 1 (B) less than 1 (C) greater than 3 (D) equal to 3 (b) If arg (z) < 0, then arg (–z) – arg (z) = [JEE 2000 Screening) 1+1M out of 35] (A) (B) – (C) 2 (D) 2 2. (a) The complex numbers z , z and z satisfying z1 z3 1i 3 12 3 z2 z3 2 are the vertices of a triangle which is - (A) of area zero (B) right-angled isosceles (C) equilateral (D) obtuse-angled isosceles (b) Let z and z be nth roots of unity which subtend a right angle at the origin. Then n must be of the form 12 (A) 4k + 1 (B) 4k + 2 (C) 4k + 3 (D) 4k [JEE 2001 (Screening) 1+1M out of 35] 11 1 1 3 . Then the value of the determinant 1 3. (a) Let i 1 2 2 is - [JEE 02 (Screening) 3M] 22 1 2 4 (A) 3 (B) 3( 1) (C) 32 (D) 3(1 ) ( b ) For all complex numbers z , z satisfying |z | = 12 and |z –3–4i| = 5 , the minimum value of |z – z | is 12 1 2 1 2 [JEE 02 (Screening) 3M] (A) 0 (B) 2 (C) 7 (D) 17 ( c ) Let a complex number , 1, be a root of the equation zp+q –zp – zq + 1 =0 where p,q are distinct primes. Show that either 1 + + 2 + ....+ p -1 = 0 or 1 + + 2 + .. + q-1 =0, but not both together. [JEE 02 (Mains) 5M] 4. If | z | = 1 and z 1 (where z –1), then Re (w) equals – [JEE 03 (Screening) 3M] z 1 1 z1 2 . (D) | z 1|2 (A) 0 (B) | z 1|2 (C) z 1 | z 1|2 5. If z and z are two complex numbers such that | z | < 1 and | z | > 1 then show that 1 z1 z2 1 12 12 z1 z2 [JEE 03 (mains) 2M out of 60)] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#11\\Eng\\02.Complex number.p65 1 n and arzr 3 6 . r 1 1 Show that there exists no complex number z such that | z | where |ai| < 2 for i = 1, 2,.......n. [JEE 03 (mains) 2M out of 60)] 7 . The least positive value of ‘n’ for which (1 + 2)n = (1 + 4)n, where is a non real cube root of unity is - (A) 2 (B) 3 (C) 6 (D) 4 [JEE 04 (screening) 3M] 8 . Find the centre and radius formed by all the points represented by z = x + i y satisfying the relation | z | K (K 1) where & are constant complex numbers, given by 1 i2 & 1 i2 | z | [JEE 04 (Mains) (2 out of 60)] 9 . If a, b, c are integers not all equal and is cube root of unity ( 1) then the minimum value of |a + b + c2| is - [JEE 05 (screening) 3M] (A) 0 (B) 1 3 1 (C) (D) 2 2 60 E
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