JEE-Mathematics Illustration 14 : Find the equation of plane containing the line of intersection of the plane x + y + z – 6 = 0 and 2x + 3y + 4z + 5 = 0 and passing through (1,1,1). Solution : The equation of the plane through the line of intersection of the given planes is, (x + y + z – 6) + (2x + 3y + 4z + 5) = 0 ......(i) If it is passes through (1,1,1) 3 (1 + 1 + 1 – 6) + (2 + 3 + 4 + 5) =0 14 3 Putting =3/14 in (i); we get (x + y + z – 6) + (2x + 3y + 4z + 5) = 0 14 20x + 23y + 26z – 69 = 0 Ans. 1 7 . PERPENDICULAR DISTANCE OF A POINT FROM THE PLANE : Vector form : If be the plane, then perpendicular distance p, of the point A( ) r .n d a p | a .n d| | n| is d1 d2 . Distance between two parallel planes r .n d1 & r .n d2 | n| Cartesian form : Perpendicular distance p, of the point A(x1, y1, z1) from the plane ax + by + cz + d = 0 is given by p | ax1 by1 cz1 d| (a2 b2 c2 ) Distance between two parallel planes ax + by + cz + d1 = 0 & ax + by + cz + d2 = 0 is d1 d 2 a2 b2 c2 Illustration 15 : Find the perpendicular distance of the point (2, 1, 0) from the plane 2x + y +2z + 5 = 0 Solution : We know that the perpendicular distance of the point (x , y , z ) from the plane 111 ax + by + cz + d = 0 is ax1 by1 cz1 d a2 b2 c2 so required distance = 2 2 1 1 2 0 5 10 Ans. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 22 12 22 3 Illustration 16 : Find the distance between the parallel planes 2x – y + 2z + 3 = 0 and 4x – 2y + 4z + 5 = 0. Solution : Let P(x , y , z ) be any point on 2x – y + 2z + 3 = 0, then 2x – y + 2z + 3 = 0 1 11 11 1 The length of the perpendicular from P(x , y , z ) to 4x – 2y + 4z + 5 = 0 is 1 11 4 x1 2 y1 4 z1 5 2(2 x1 y1 2z1 ) 5 2(3) 5 1 [using (i)] 42 (2)2 42 36 66 Therefore, the distance between the two given parallel planes is 1 Ans. 6 Do yourself - 6 : ( i ) Find the perpendicular distance of the point P(1, 2, 3) from the plane 2x + y + z + 1 = 0. ( i i ) Find the equation of the plane passing through the line of intersection of the planes x + y + z = 5 and 2x + 3y + z + 5 = 0 and passing through the point (0, 0, 0). 64 E
JEE-Mathematics 18. BISECTORS OF ANGLES BETWEEN TWO PLANES : Let the equations of the two planes be ax + by + cz + d = 0 and a1 x + b1 y + c1 z + d1 = 0. Then equations of bisectors of angles between them are given by ax by cz d a1x b1 y c1z d1 (a2 b2 c2 ) (a12 b12 c12 ) ( a ) Equation of bisector of the angle containing origin : First make both constant terms positive. Then positive sign give the bisector of the angle which contains the origin. ( b ) Bisector of acute/obtuse angle : First making both constant terms positive, aa1 + bb1 + cc1 > 0 origin lies in obtuse angle aa1 + bb1 + cc1 < 0 origin lies in acute angle Illustration 17 : Find the equation of the bisector planes of the angles between the planes 2x – y + 2z + 3 = 0 and 3x – 2y + 6z + 8 = 0 and specify the plane which bisects the acute angle and the plane which bisects the obtuse angle. Solution : The two given planes are 2x – y + 2z + 3 = 0 and 3x – 2y + 6z + 8 = 0 where d , d > 0 12 and a a + b b + c c = 6 + 2 + 12 > 0 12 12 12 a 1x b1y c1z d1 = – a 2x b2y c2z d2 (acute angle bisector) a12 b12 2 2 2 2 c 1 a 2 b 2 c 2 and a1x b1y c1z d = a2x b2y c2z d2 (obtuse angle bisector) a12 b12 2 2 b 22 2 c 1 a 2 c 2 2x y 2z 3 3x 2y 6z 8 i.e., = ± 4 14 9 4 36 (14x – 7y + 14z + 21) = ± (9x –6y + 18z + 24) Taking positive sign on the right hand side, we get 5x – y – 4z – 3 = 0 (obtuse angle bisector) and taking negative sign on the right hand side, NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 we get 23x – 13y + 32z + 45 = 0 (acute angle bisector) Ans. 1 9 . POSITION OF T WO POINTS W.R.T. A PL ANE : Two points P(x , y , z ) & Q(x , y , z ) are on the same or opposite sides of a plane ax + by + cz + d = 0 111 2 22 according to ax + by + cz + d & ax + by + cz + d are of same or opposite signs. The plane divides the line 1 11 2 22 joining the points P & Q externally or internally according to P and Q lying on same or opposite sides of the plane. Do yourself - 7 : ( i ) Find the position of the point P(2, –2, 1), Q(3, 0, 1) and R(–12, 1, 8) w.r.t. the plane 2x – 3y + 4z – 7 = 0. ( i i ) Two given planes are –2x + y – 2z + 5 = 0 and 6x – 2y + 3z – 7 = 0. Find (a) equation of plane bisecting the angle between the planes. (b) equation of a plane parallel to the plane bisecting the angle between both the two planes and passing through the point (3, 2, 0). (c) specify which plane is acute angle bisector and which one is obtuse angle bisector. E 65
JEE-Mathematics STRAIGHT LINE 20 . DEFINITION : A straight line in space is characterised by the intersection of two planes which are not parallel and, therefore, the equation of a straight line is present as a solution of the system constituted by the equations of the two planes : a1 x + b1 y + c1 z + d1 = 0; a2 x + b2 y + c2 z + d2 = 0 This form is also known as unsymmetrical form. Some particular straight lines : Straight lines Equation (i) Through the origin y = mx, z = nx (ii) x-axis xyz (iii) y-axis y = 0, z = 0 or 100 (iv) z-axis (v) parallel to x-axis x = 0, z = 0 or xyz (vi) parallel to y-axis 010 (vii) parallel to z-axis x = 0, y = 0 or x y z 0 01 y = p, z = q x = h, z = q x = h, y = p 2 1 . EQUATION OF A STR AIGHT LINE IN SYMMETRICAL FORM : ( a ) One point form : Let A(x1, y1, z1) be a given point on the straight line and , m, n be the d.c’s of the line, then its equation is x x1 y y1 z z1 r (say) m n It should be noted that P(x1 + r, y1 + mr, z1 + nr) is a general point on this line at a distance r from the point A(x1, y1, z1) i.e. AP = r. One should note that for AP = r; , m, n must be d.c.’s not d.r.’s. If a, b, c are direction ratios of the line, then equation of the line is x x1 y y1 z z1 r but here AP r NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 a bc ( b ) Equation of the line through two points A(x1, y1, z1) and B(x2, y2, z2) is x x1 y y1 z z1 x2 x1 y2 y1 z2 z1 x 1 y 2 z 3 Illustration 18 : Find the co-ordinates of those points on the line which is at a distance of 236 3 units from point (1,–2, 3). x 1 y 2 z 3 Solution : Here, .......(i) 236 is the given straight line P(1,–2,3) Let, P = (1,–2,3) on the straight line 3 Units Here direction ratios of line (i) are (2,3,6) Q(2r+1,3r–2,6r+3) 236 E Direction cosines of line (i) are : , , 777 66
JEE-Mathematics Equations of line(i) any may be written as x 1 y 2 z 3 ......(ii) 2/7 3/7 6/7 Co-ordinates of any point on the line (ii) may be taken as 2 3 6 7 r 1, 7 r 2, 7 r 3 2 3 6 Let, Q 7 r 1, 7 r 2, 7 r 3 Given | r| = 3, r = ± 3 13 5 39 1 23 3 Putting the value of r, we have Q , , or Q = , , Ans. 7 7 7 7 7 7 2 2 . ANGLE BETWEEN A LINE AND A PLANE : Let equations of the line and plane be x x1 y y1 z z1 and ax + by + cz + d = 0 respectively and be m n the angle which line makes with the plane. Then ( ) is the angle n NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65between the line and the normal to the plane. linea bm cn So, sin (a2 b2 c2 ) (2 m2 n2 ) Line is parallel to plane if = 0 i.e. if a + bm + cn = 0. Line is perpendicular to the plane if line is parallel to the normal of the plane i.e. if a b c . m n Illustration 19 : Find the angle between the line x 2 y 1 z 3 and the plane 3x + 4y + z + 5 = 0. 3 1 2 Solution : The given line is x 2 y 1 z 3 ....... (i) 3 1 2 and the given plane is 3x + 4y + z + 5 = 0 ....... (ii) If the line (i) makes angle with the plane (ii), then the line (i) will make angle (90° – ) with the normal to the plane (i). Now direction-ratios of line (i) are < 3, –1, –2 > and direction-ratios of normal to plane (ii) are < 3, 4, 1 > cos(90 ) (3)(3) (1)(4 ) (2)(1) sin 9 4 2 3 9 1 4 9 16 1 14 26 14 26 Hence sin 1 3 Ans. 14 26 Do yourself - 8 : ( i ) Find the equation of the line passing through the point (4, 2, 3) and having direction ratios 1, –1, 2 ( i i ) Find the symmetrical form of the line x – y + 2z = 5, 3x + y + z = 6. (iii) Find the angle between the plane 3x + 4y + 5 = 0 and the line x 1 y 2 z 1 . 2 0 1 (iv) Prove that the line x3 y4 z5 is parallel to the plane 4x + 4y – 5z + 2 = 0. 234 E 67
JEE-Mathematics 2 3 . CONDITION IN ORDER THAT THE LINE MAY LIE ON THE GIVEN PLANE : The line x x1 y y1 z z1 will lie on the plane Ax + By + Cz + D = 0 if m n ( a ) A + Bm + Cn = 0 and ( b ) Ax1 + By1 + Cz1 + D = 0 2 4 . IMAGE OF A POINT IN THE PLANE : P In order to find the image of a point P(x1, y1, z1) in a plane ax + by + cz + d = 0, assume it as a mirror. Let Q(x2, y2, z2 ) be the image of the point P(x , y , z ) in the R 1 11 plane, then (a) Line PQ is perpendicular to the plane. Hence equation of PQ is x x1 y y1 z z1 r Q a bc (b) Hence, Q satisfies the equation of line then x2 x1 y2 y1 z2 z1 r. The plane passes through a b c the middle point of line PQ and the middle point satisfies the equation of the plane i.e. a x 2 x1 b y 2 y1 c z 2 z1 d 0 . The co-ordinates of Q can be obtained by solving these 2 2 2 equations. 2 5 . FOOT, LENGTH AND EQUATION OF PERPENDICUL AR FROM A POINT TO A LINE : Let equation of the line be x x1 y y1 z z1 r (say) .......... (i) m n and A ( ) be the point. Any point on the line (i) is P(r + x1, mr + y1, nr + z1) .......... (ii) If it is the foot of the perpendicular, from A on the line, then AP is to the line, so i.e. (r + x1 – ) + m (mr + y1 – ) + n (nr + z1 – ) = 0 A() since r = ( – x1) + ( – y1) m + ( – z1) n L 2 + m2 + n2 = 1 P Putting this value of r in (ii), we get the foot of perpendicular from point A to the line. Length : Since foot of perpendicular P is known, length of perpendicular, AP [(r x1 )2 (mr y1 )2 (nr z1 )2 ] Equation of perpendicular is given by x y z r x1 mr y1 nr z1 Illustration 20 : Find the co-ordinates of the foot of the perpendicular from (1, 1, 1) on the line joining NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 Solution : (5, 4, 4) and (1, 4, 6). Let A (1, 1, 1), B (5, 4, 4) and C (1, 4, 6) be the given points. Let M be the foot of the perpendicular from A on BC. If M divides BC in the ratio : 1, then A(1,1,1) co-ordinates of M are 5 , 4 4 , 6 4 1 1 1 Direction ratios of BC are 1 – 5, 4 – 4, 6 – 4 i.e. –4, 0, 2 D.R.'s of AM are 5 1, 4 4 1, 6 4 1 B 90° C 1 1 1 (5,4,4) M (1,4,6) 4 , 3 3 , 5 3 4, 3 + 3, 5 + 3 1 1 1 Since AM BC 2 (4) + 0(3 + 3) – 1 (5 + 3) = 0 8 – 5 – 3 = 0 = 1 Hence the co-ordinates of M are (3, 4, 5) Ans. 68 E
JEE-Mathematics x 1 y 3 z 2 Illustration 21 : Find the length of perpendicular from P(2, – 3,1) to the line 2 3 1 x 1 y 3 z 2 Solution : Given line is .......(i) 2 3 1 P(2,–3,1) and P(2, – 3,1) Co-ordinates of any point on (i) may be taken as (2r–1,3r+3,–r–2) Let Q = (2r –1, 3r + 3,–r – 2) A 90° B Q Direction ratio's of PQ are : (2r –3, 3r +6, –r –3) Direction ratio's of AB are : (2,3, –1) Since, PQ AB 2 (2r – 3) +3 (3r + 6) – 1 (–r – 3) = 0 15 r=– 14 22 3 13 Q = 7 , 14 , 14 222 3 2 132 531 PQ2 = 2 7 1 14 3 14 14 PQ = 531 units Ans. 14 Do yourself - 9 : ( i ) Find the image of point P(1, 3, 2) in the plane 2x – y + z + 3 = 0 as well as the foot of the perpendicular drawn from the point (1, 3, 2). ( i i ) Find the distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line xy z 2 3 6 (iii) Prove that x 1 y 2 z 5 lies in the plane x + 2y – z = 0. 2 3 4 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 2 6 . EQUATION OF PL ANE CONTAINING TWO INTERSECTING LINES : Let the two lines be x 1 y 1 z 1 .......... (i) 1 m1 n1 .......... (ii) and x 2 y 2 z 2 2 m2 n2 These lines will coplanar if 2 1 2 1 2 1 (It is condition for intersection of two lines) 1 m1 n1 0 2 m2 n2 x 1 y 1 z 1 m1 n1 0 the plane containing the two lines is 1 m2 n2 2 E 69
JEE-Mathematics Illustration 22 : Find the equation of the plane containing the line x 1 y 6 z 1 and parallel to the line 342 x 4 y 1 z 3 . 2 3 5 Solution : Any plane containing the line x 1 y 6 z 1 is 342 a(x – 1) + b(y + 6) + c(z + 1) = 0 ....... (i) where, 3a + 4b + 2c = 0 ....... (ii) Also, it is parallel to the second line and hence, its normal is perpendicular to this line 2a – 3b + 5c = 0 ....... (iii) Solving (ii) & (iii) by cross multiplication, we get a b c k 26 11 17 a = 26k, b = –11k & c = –17k Putting these values in (i), we get 26k(x – 1) – 11k(y + 6) – 17k(z + 1) = 0 26x – 11y – 17z = 109, which is the required equation of the plane. 2 7 . LINE OF GRE ATEST SLOPE : A G-plane Q P Consider two planes G-plane and H-plane. H-plane is treated as a horizontal plane or reference plane. G-plane is a given plane. Let AB be the line of intersection of B H-plane G-plane & H-plane. Line of greatest slope is a line which is contained by G-plane & perpendicular to line of intersection of G-plane & H-plane. Obviously, infinitely many such lines of greatest slopes are contained by G-plane. Generally an additional information is given in problem so that a unique line of greatest slope can be found out. Illustration 23 : Assuming the plane 4x–3y +7z =0 to be horizontal, find the equation of the line of greatest slope through the point (2,1,1) in the plane 2x +y – 5z = 0. Solution : The required line passing through the point P (2,1,1) in the plane 2x + y – 5z = 0 and is having greatest slope, so it must be perpendicular to the line of intersection of the planes NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 2x + y – 5z = 0 ......(i) and 4x – 3y + 7z = 0 ......(ii) Let the d.r'.s of the line of intersection of (i) and (ii) be a, b, c 2a + b – 5c = 0 and 4a – 3b + 7c = 0 {as dr'.s of straight line (a, b,c) is perpendicular to d.r'.s of normal to both the planes} abc 4 17 5 Now let the direction ratio of required line be proportional to , m, n then its equation be x 2 y 1 z 1 m n where 2 + m – 5n = 0 and 4 +17m + 5n = 0 mn so, 3 1 1 x 2 y 1 z 1 Thus the required line is Ans. 3 1 1 E 70
JEE-Mathematics 28 . AREA OF TRIANGLE : To find the area of a triangle in terms of its projections on the co-ordinates planes. Let x, y, z be the projections of the plane area of the triangle on the planes yOz, zOx, xOy respectively. Let , m, n be the direction cosines of the normal to the plane of the triangle. Then the angle between the plane of the triangle and yOz plane is the angle between the normal to the plane of the triangle and the x-axis. x = Similarly y = m ; z = n 2x 2y 2z If A(x1, y1, z1), B(x2, y2, z2), C(x3, y3, z3) be the three vertices of the triangle then 1 y1 z1 1 1 x1 z1 1 1 x1 y1 1 x y2 z2 1 , y 2 x2 z2 1 , z 2 x2 y2 1 2 y3 z3 1 x3 z3 1 x3 y3 1 Do yourself - 10 : (i) Prove that the lines x 1 y 3 z 5 and x 2 y 4 z 6 are coplanar. Find their point of 357 135 intersection. ( i i ) Find the area of the triangle whose vertices are the points (1, 2, 3), (–2, 1, –4), (3, 4, –2). Miscellaneous Illustrations : Illustration 24 : If a variable plane cuts the coordinate axes in A, B and C and is at a constant distance p from the origin, find the locus of the centroid of the tetrahedron OABC. Solution : Let A (a, 0, 0), B (0, b, 0) and C (0, 0, c) Equation of plane ABC is x y z =1 z a b c C(0, 0, c) Now p = length of perpendicular from O to plane (i) O B(0, b, 0) y NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 = 1 or p2 = 1 A(a, 0, 0) x 1 1 1 12 12 1 2 a2 b2 c2 a b c Let G( ) be the centroid of the tetrahedron OABC, then a bc a 0 0 0 a = 4, = 4,= 4 4 4 or, a = 4, b = 4, c = 4 Putting these values of a, b, c in equation (ii), we get p2 = 1 16 or 1 1 1 16 2 2 2 2 p2 1 1 2 2 locus of ( ) is x–2 + y–2 + z–2 = 16 p–2 Ans. E 71
JEE-Mathematics Illustration 25 : Through a point P(h, k, ) a plane is drawn at right angles to OP to meet the coordinate axes in p5 A, B and C. If OP = p, show that the area of ABC is 2| hk| . Solution : OP = h2 k2 2 = p Direction cosines of OP are h k, , h2 k2 2 h2 k2 2 h2 k2 2 Since OP is normal to the plane, therefore, equation of the plane will be, h k y+ h2 k2 2 x+ h2 k2 2 z= h2 k2 2 h2 k2 2 or, hx + ky + z = h2 + k2 + 2 = p2 A p2 , 0, , B p2 , , C p2 h 0 0, k 0 0, 0, Now area of ABC, 2 = A2 + A2 + A2 xy yz zx Now A = area of projection of ABC on xy-plane = area of AOB xy p2 0 1 h 1 p4 = Mod of 1 0 p2 1 = 2 hk 2 k 0 01 Similarly, A = 1 p4 and 1 p4 yz A= 2 k zx 2 h 2 = 1 p8 1 p8 1 p8 p10 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 4 h2k2 4 k22 4 h22 4h2k22 p5 Ans. or = 2| hk| Illustration 26 : Find the locus of a point, the sum of squares of whose distances from the planes : x – z = 0, x – 2y + z = 0 and x + y + z = 0 is 36 Solution : Given planes are x – z = 0, x – 2y + z = 0 and, x + y + z =0 Let the point whose locus is required be P(,,). According to question | |2 | 2 |2 | |2 = 36 26 3 or 3(2+2–2)+2 +42 +2 – 4 – 4 + 2 +2(2 +2 +2 + 2+2+2) =36 × 6 or 62 + 62 + 62 = 36 × 6 or 2 + 2 + 2 = 36 Hence, the required equation of locus is x2 + y2 + z2 = 36 Ans. 72 E
JEE-Mathematics Illustration 27 : Direction ratios of normal to the plane which passes through the point (1, 0, 0) and (0, 1, 0) which makes angle /4 with x + y = 3 are - (A) 1, 1, 2 (B) 2 , 1, 1 (C) 1, 2 , 1 (D) 1, 1, 2 Ans. (D) Solution : The plane by intercept form is x y z 1 11c d.r.’s of normal are 1, 1, 1 and of given plane are 1, 1, 0. c 1 1 .1 1 .1 0 cos c 4 1 1 1 1 1 0 c2 12 1 1 2 4 c 2 c2 21 2 2 c2 d.r.’s are 1, 1, 2 ANSWERS FOR DO YOURSELF 1 : (i) 2 21 (i ii ) 8x + 2y + 24z ± 2k2 + 9 = 0 (iv) 2, 8 , 5 & 1, 10 , 6 3 3 ( v ) ( a ) 7 : 8, externally ( b ) 2 : 3 internally 2 : ( i ) 3, 2, 5 ( i i ) 3, 4, 3 ( i i i ) 2, 3, 4 & 2 , 3 , 4 29 29 29 3 : (i) (iii) 2, 1, 2 2 3 33 4 : (i) (a) 3 , 4 ,1 (b) 7 , 0, 0 , 0, 7 , 0 & 0, 0, 7 (c) x y z 1 52 52 2 3 4 5 7/3 7/4 7/5 (d) 3x 4y z 7 & 3 ˆi 4 ˆj 1 kˆ 7 52 52 2 52 r. 2 52 2 52 5 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 ( i i ) x + y + 2z = 7 5: (ii) c o s 1 1 156 8 (ii) 3x + 4y + 2z = 0 6 : (i) 6 7 : ( i ) P, Q same side & R opposite side ( i i ) ( a ) 4x + y – 5z + 14 = 0 & 32x – 13y + 23z – 56 = 0 ( b ) 4x + y – 5z – 14 = 0 & 32x – 13y + 23z – 70 = 0 ( c ) 4x + y – 5z + 14 = 0 (acute angle bisector) & 32x – 13y + 23z – 56 = 0 (obtuse angle bisector) 8 : (i) x 4 y 2 z 3 (ii) x 11 / 4 y 9 / 4 z 0 (iii) sin 1 6 1 1 2 3 5 4 5 5 9 : (i) 5 , 13 , 2 & 1 , 11 , 4 (ii) 1 3 3 3 3 3 3 10: (i) 1 , 1 , 3 1218 2 2 2 (ii) E 2 73
JEE-Mathematics CHECK YOUR GRASP EXERCISE - 01 SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . The plane XOZ divides the join of (1, –1, 5) and (2, 3, 4) in the ratio : 1, then is - (A) –3 (B) –1/3 (C) 3 (D) 1/3 2 . Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the area of triangles ABC, ACD and ADB be 3, 4 and 5 sq. units respectively. Then the area of the triangle BCD, is - (A) 5 2 (B) 5 5 5 (C) (D) 2 2 3 . Which one of the following statement is INCORRECT ? (A) If n . a = 0, n . b = 0 and n . c = 0 for some non zero vector n , then [a b c] = 0 (B) there exist a vector having direction angles = 30° and = 45° (C) locus of point in space for which x = 3 and y = 4 is a line parallel to the z-axis whose distance from the z-axis is 5 (D) In a regular tetrahedron OABC where 'O' is the origin, the vector OA + OB + OC is perpendicular to the plane ABC. 4 . Consider the following 5 statements (I) There exists a plane containing the points (1, 2, 3) and (2, 3, 4) and perpendicular to the vector = ˆi + ˆj – kˆ V1 (II) There exist no plane containing the point (1, 0, 0); (0, 1, 0); (0, 0, 1) and (1, 1, 1) (III) If a plane with normal vector N is perpendicular to a vector V then N · V = 0 (IV) If two planes are perpendicular then every line in one plane is perpendicular to every line on the other plane (v) Let P1 and P2 are two perpendicular planes. If a third plane P3 is perpendicular to P1 then it must be either parallel or perpendicular or at an angle of 45° to P2. Choose the correct alternative. (A) exactly one is false (B) exactly 2 are false (C) exactly 3 are false (D) exactly four are false 5. Let L be the line = 2 ˆi + ˆj – kˆ + ( ˆi + 2 kˆ ) and let L be the line = 3 ˆi + ˆj + µ( ˆi + ˆj – kˆ ). NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 1 2 r1 r2 Let be the plane which contains the line L1 and is parallel to L2. The distance of the plane from the origin is - (A) 2 / 7 (B) 1/7 (C) 6 (D) none of these 6 . The intercept made by the plane r . n = q on the x-axis is - (A) q (B) ˆi. (C) (ˆi. q q (D) n n) ˆi. q | n| n 7 . If from the point P(f, g, h) perpendiculars PL, PM be drawn to yz and zx planes then the equation to the plane OLM is - xyz xyz (A) +g + =0 (B) + – = 0 f h f gh xy z xyz (C) –g + =0 (D) – +g + =0 f h f h 74 E
JEE-Mathematics 8 . The line which contains all points (x, y, z) which are of the form (x, y, z) = (2, –2, 5) + (1, –3, 2) intersects the plane 2x – 3y + 4z = 163 at P and intersects the YZ plane at Q. If the distance PQ is a b , where a, b N and a > 3 then (a + b) equals - (A) 23 (B) 95 (C) 27 (D) none of these 9 . A plane passes through the point P(4, 0, 0) and Q(0, 0, 4) and is parallel to the y-axis. The distance of the plane from the origin is - (A) 2 (B) 4 (C) 2 (D) 2 2 10. The distance between the parallel planes given by the equations, . (2 ˆi – 2 ˆj + kˆ ) + 3 = 0 and r . (4 ˆi – 4 ˆj + 2 kˆ ) + 5 = 0 is - r (A) 1/2 (B) 1/3 (C) 1/4 (D) 1/6 1 1 . If the plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(k) with x-axis, then k is equal to - 3 2 2 (D) 1 (A) (B) 7 (C) 2 3 1 2 . A variable plane forms a tetrahedron of constant volume 64K3 with the coordinate planes and the origin, then locus of the centroid of the tetrahedron is - (A) x3 + y3 + z3 = 6K2 (B) xyz = 6k3 (C) x2 + y2 + z2 = 4K2 (D) x–2 + y–2 + z–2 = 4k–2 1 3 . The expression in the vector form for the point r1 of intersection of the plane r · n = d and the perpendicular line r = r0 + t n where t is a parameter given by - d nr20 ·n r0n·2n (A) r1 = r0 + n (B) r1 = r0 – n r1 r0 r0 ·n d r1 r0 r0 · n n (C) = – n n (D) = + n 1 4 . The equation of the plane containing the line x 1 y 3 z 2 and the point (0, 7, –7) is - 3 2 1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 (A) x + y + z = 1 (B) x + y + z = 2 (C) x + y + z = 0 (D) none of these SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 15. Consider the plane r .n1 d1 and r .n2 d2 , then which of the follwoing are true - (A) they are perpendicular if n1 .n2 0 | n1 . n2 (B) angle between them is c o s 1 n1 || n2 | (C) normal form of the equation of plane are d1 & d2 r .n1 | n1 | r .n2 | n2| (D) none of these 16. The equation of the plane which contains the lines ˆi 2ˆj kˆ (ˆi 2ˆj kˆ) and ˆi 2ˆj kˆ µ(ˆi ˆj 3kˆ) r r must be - (B) 7(x – 1) – 4(y – 2) – (z + 1) = 0 (A) r.(7ˆi 4ˆj kˆ) 0 (D) r.(ˆi ˆj 3kˆ) 0 (C) r.(ˆi 2ˆj kˆ) 0 E 75
JEE-Mathematics 17. The plane containing the lines and - r a ta ' r a ' sa (A) must be parallel to (B) must be the perpendicular to aa' aa' (C) must be [r, a, a '] = 0 (D) (r a).(a a ') 0 1 8 . The points A(5, –1, 1), B(7, –4, 7), C(1, –6, 10) and D(–1, –3, 4) are the vertices of a - (A) parallelogram (B) rectangle (C) rhombus (D) square 1 9 . If P , P , P denotes the perpendicular distances of the plane 2x – 3y + 4z + 2 = 0 from the parallel planes 123 2x – 3y + 4z + 6 = 0, 4x – 6y + 8z + 3 = 0 and 2x – 3y + 4z – 6 = 0 respectively, then - (A) P + 8P – P = 0 (B) P = 16P 1 23 32 (C) 8P = P (D) P + 2P + 3P = 29 21 123 CHECK YOUR GRASP ANSWER KEY EXERCISE-1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 Que. 1 2 3 4 5 6 7 8 9 10 Ans. D A B D A A B A D D Que. 11 12 13 14 15 16 17 18 19 Ans. B B A C A,B A,B B,C,D A,C A,B,C,D 76 E
EXERCISE - 02 JEE-Mathematics BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1. If the line = 2 ˆi – ˆj + 3 kˆ + ( ˆi + ˆj + 2 kˆ ) makes angles , , with xy, yz and zx planes respectively then r which one of the following are not possible ? (A) sin2 + sin2 + sin2 = 2 and cos2 + cos2 + cos2 = 1 (B) tan2 + tan2 + tan2 = 7 and cot2 + cot2 + cot2 = 5/3 (C) sin2 + sin2 + sin2 = 1 and cos2 + cos2 + cos2 = 2 (D) sec2 + sec2 + sec2 = 10 and cosec2 + cosec2 + cosec2 = 14/3 2 . A plane meets the coordinate axes in A, B, C such that the centroid of the triangle ABC is the point (1, r, r2). The plane passes through the point (4, –8, 15) if r is equal to - (A) –3 (B) 3 (C) 5 (D) –5 3 . Indicate the correct order statements - x4 y6 z6 x 1 y 2 z 3 (A) The lines 3 = 1 = 1 and 1 = 2 = 2 are orthogonal (B) The planes 3x – 2y – 4z = 3 and the plane x – y – z = 3 are orthogonal. (C) The function f(x) = n(e–2 + ex) is monotonic increasing x R. (D) If g is the inverse of the function, f(x) = n(e–2 + ex) then g(x) = n(ex – e–2) x 1 y 1 4 . The coordinates of a point on the line 2 = 3 = z at a distance 4 14 from the point (1, –1, 0) are- (A) (9, –13, 4) (B) ( 8 14 +1, –12 14 –1, 4 14 ) (C) (–7, 11, –4) (D) (– 8 14 +1, 12 14 –1, –4 14 ) x9 y 4 z5 5 . Let 6x + 4y – 5z = 4, x – 5y + 2z = 12 and = = be two lines then- 2 1 1 (B) the angle between them must be cos–1 5 (A) the angle between them must be 3 6 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 (C) the plane containing them must be x + y – z = 0 (D) they are non-coplanar 6. x 1 y 1 z 3 x y z 1 are - The lines == and = = 2 1 1 2 1 (A) coplanar for all (B) coplanar for = 19/3 (C) if coplanar then intersect at 1 , 2 , 4 (D) intersect at 1 , 1 , 1 5 5 5 2 2 7. If two pairs of opposite edges of a tetrahedron are perpendicular then - 8. (A) the third is also perpendicular (B) the third pair is inclined at 60° E (C) the third pair is inclined at 45° (D) (B), (C) are false The equation of a plane bisecting the angle between the plane 2x – y + 2z + 3 = 0 and 3x – 2y + 6z + 8 = 0 is - (A) 5x – y – 4z – 45 = 0 (B) 5x – y – 4z – 3 = 0 (C) 23x – 13y + 32z + 45 = 0 (D) 23x – 13y + 32z + 5 = 0 77
JEE-Mathematics 9. A non-zero vector is parallel to the line of intersection of the plane determined by the vectors ˆi , ˆi + ˆj and the a plane determined by the vectors ˆi – ˆj , ˆi – kˆ ,. The possible angle between and ˆi – 2 ˆj + 2 kˆ is - a (A) /3 (B) /4 (C) /6 (D) 3/4 1 0 . If 1, m1, n1 and 2, m2, n2 are DCs of the two lines inclined to each other at an angle , then the DCs of the bisector of the angle between these lines are- (A) 1 2 , m1 m 2 , n1 n 2 (B) 1 2 , m1 m 2 , n1 n 2 2 sin / 2 2 sin / 2 2 sin / 2 2 cos / 2 2 cos / 2 2 cos / 2 (C) 1 2 , m1 m2 , n1 n 2 (D) 1 2 , m1 m2 , n1 n 2 2 sin / 2 2 sin / 2 2 sin / 2 2 cos / 2 2 cos / 2 2 cos / 2 11. Points that lie on the lines bisecting the angle between the lines x 2 y 3 z6 and x 2 y 3 z6 236 362 are - (A) (7, 12, 14) (B) (0, –3, 14) (C) (1, 0, 10) (D) (–3, –6, –2) BRAIN TEASERS ANSWER KEY EXERCISE -2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 Que. 1 2 3 4 5 6 7 8 9 10 Ans. A,B,D B,C C,D A,C A,C B,C A,D B,C B,D B,C Que. 11 Ans. A,B,C,D 78 E
JEE-Mathematics EXERCISE - 03 MISCELLANEOUS TYPE QUESTIONS TRUE / FALSE 1 . If the plane xbc + yac + zab = abc cuts x, y & z-axis in A, B & C respectively then area of ABC is a2b2 b2c2 c2a2 . r a + b r · n = d is 2 cos–1 b · n 2. The angle between the line = and plane – bn 3 . The perpendicular distance of the plane r · nˆ = d, from the origin is d where d > 0 4 . If A(1, 2, –1), B(2, 6, 2) and C(, –2, –4) are collinear then value of is 0. 5 . The projection of line segment on the axes of reference are 3, 4 and 12 respectively. The length of such a line segment is 13 MATCH THE COLUMN Following questions 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 statement in Column-I can have correct matching with ONE statement in Column-II. 1 . Match the following pair of planes with their lines of intersections : Column-I Column-II (A) x + y = 0 = y + z x 2 y 2007 z 2004 (p) 0 = 1 = 1 (B) x = 2, y = 3 (q) x 2 y z 1 (C) x = 2, y + z = 3 == (D) x = 2, x + y + z = 3 0 1 1 (r) x = –y = z (s) x 2 = y 3 = z 0 01 2.NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 Consider three planes E P 2x + y + z = 1 1 P x – y + z = 2 2 P x – y + 3z = 5 3 The three planes intersects each other at point P on XOY plane and at point Q on YOZ plane. O is the origin. Column-I Column-II (A) The value of is (B) The length of projection of PQ on x-axis is (p) 1 (C) If the co-ordinates of point R situated at a minimum (q) 2 distance from point 'O' on the line PQ are (a, b, c), then value of 7a + 14b + 14c is (r) 4 a (s) 3 (D) If the area of POQ is , then value of a – b is b 79
JEE-Mathematics 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 statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. 3 . Consider the following four pairs of lines in column–I and match them with one or more entries in column–II Column-I Column-II (A) L : x=1 + t, y=t, z=2–5t (p) non coplanar lines 1 (q) lines lie in a unique plane L: r = (2, 1,–3) + (2,2,–10) 2 (r) infinite planes containing both the lines x 1 y 3 z 2 (B) L : == (s) lines are not intersecting 12 2 1 L2 : x2 y 6 z2 1 1 3 (C) L : x = – 6t, y=1 + 9t, z=–3t 1 L : x=1 +2s, y=4–3s, z=s 2 x y 1 z 2 (D) L : = = 11 2 3 x 3 y 2 z 1 L: = = 2 4 3 2 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. b 1. Statement - I : If a plane contains point A( a ) and is parallel bto vcec. tors and c , then its vector equation + & µ are parameters and is r = a + b µ c , where Because Statement - II : If three vectors are co-planar, then any one can be expressed as the linear combination of other two. (A) A (B) B (C) C (D) D NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 2 . Statement - I : If ax + by + cz = a2 b2 c2 be a plane and (x1, y1, z1) and (x2, y2, z2) be two points on this plane then a(x1 – x2) + b(y1 – y2) + c(z1 – z2) = 0. Because Statement - II : If two vectors p1 ˆi + p2 ˆj + p3 kˆ and q1 ˆi + q2 ˆj + p3 kˆ are orthogonal then p1q1 + p2q2 + p3q3 = 0. (A) A (B) B (C) C (D) D 3 . Statement - I : If the lines x x1 = y y1 = z z1 and x x2 = y y2 = z z2 are coplanar then a1 b1 c1 a2 b2 c2 x1 y1 z1 x2 y2 z2 a1 b1 c1 = a1 b1 c1 a2 b2 c2 a2 b2 c2 Because Statement - II : If the two lines are coplanar then shortest distance between them is zero. (A) A (B) B (C) C (D) D 80 E
JEE-Mathematics 4 . Statement - I : ABCDA B C D is a cube of edge 1 unit. P and Q are the mid points of the edges B A , 111 1 11 and B C respectively. Then the distance of the vertex D from the plane PBQ is 8 11 3. Because Statement - II : Perpendicular distance of point (x1, y1, z1) from the plane ax + by + cz + d = 0 is given by ax1 by1 cz1 d . a2 b2 c2 (A) A (B) B (C) C (D) D 5 . Statement - I : If 2a + 3b + 6c = 14, where a, b & c R, then the minimum value of a2 + b2 + c2 is 4. Because Statement - II : The perpendicular distance of the plane px + qy + rz = 1 from origin is 1 . p2 q2 r2 (A) A (B) B (C) C (D) D 6 . Consider following two planes P1 [r p a b] 0 P2 [r p c d] 0 such that | (a b) (c d)| 0 & let x be any vector in space. Statement-I : x.{(a b) (c d)} 0 x.{t1a t2b} 0, t1, t2 R Because Statement-II : x.{t1a t2 b} 0 t1, t2 R x.{(a b) (c d)} 0 . (A) A (B) B (C) C (D) D 7. Consider planes P1 : ˆi ).{(ˆi ˆj kˆ) (ˆi 2 kˆ)} 0 and P2 : ( (2ˆi ˆj kˆ)).{(ˆi 2 kˆ) (2ˆi ˆj 3kˆ)} 0 (r r NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 and line L : 5ˆi (ˆi ˆj kˆ) r Statement-I : P1 & P2 are parallel planes. Because Statement-II : L is parallel to both P1 & P2. (A) A (B) B (C) C (D) D COMPREHENSION BASED QUESTIONS Comprehension # 1 : Given four points A(2, 1, 0); B(1, 0, 1); C(3, 0, 1) and D(0, 0, 2). The point D lies on a line L orthogonal to the plane determined by the point A, B, and C On the basis of above information, answer the following questions : 1 . Equation of the plane ABC is - (A) x + y + z – 3 = 0 (B) y + z – 1 = 0 (C) x + z – 1 = 0 (D) 2y + z – 1 = 0 2 . Equation of the line L is - (A) = 2 kˆ + ( ˆi + kˆ ) (B) = 2 kˆ + (2 ˆj + kˆ ) r r (C) = 2 kˆ + ( ˆj + kˆ ) (D) none of these r E 81
JEE-Mathematics 3 . Perpendicular distance of D from the plane ABC, is - (A) 2 1 (C) 2 1 Comprehension # 2 : (B) (D) 2 2 If a line passes through P (x , y , z ) and having Dr’s a, b, c, then the equation of line is x x1 = y y1 = z z1 111 a b c and equation of plane perpendicular to it and passing through P is a(x – x1) + b(y – y1) + c(z – z1) = 0. Further equation of plane through the intersection of the two planes a x + b y + c z + d = 0 and a x + b y + c z + d = 0 is 1 111 2 222 (a1x + b1y + c1z + d1) + k(a2x + b2y + c2z + d2) = 0 On the basis of above information, answer the following questions : 1 . The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line x y z3 2 = 3 = 4 is - 1 21 1 29 1 13 2 (A) 5 (B) 5 (C) 5 (D) 5 The equation of the plane through (0, 2, 4) and containing the line x 3 y 1 z 2 2. = = is - 3 4 2 (A) x – 2y + 4z – 12 = 0 (B) 5x + y + 9z – 38 = 0 (C) 10x – 12y – 9z + 60 = 0 (D) 7x + 5y – 3z + 2 = 0 3 . The plane x – y – z = 2 is rotated through 90° about its line of intersection with the plane x + 2y + z = 2. Then equation of this plane in new position is - (A) 5x + 4y + z – 10 = 0 (B) 4x + 5y + 3z = 0 (C) 2x + y + 2z = 9 (D) 3x + 4y – 5z = 9 Comprehension # 3 : Consider a triangular pyramid ABCD the position vectors of whose angular point are A(3, 0, 1); B(–1, 4, 1); C(5, 2, 3) and D(0, –5, 4). Let G be the point of intersection of the medians of the triangle BCD. On t he basis of abovei nformat ion, a nswer t he fol low i ng que st ions : 1 . The length of the vector AG is- (A) 17 51 51 59 (B) (C) (D) 3 9 4 2 . Area of the triangle ABC in sq. units is- (A) 24 (B) 8 6 (C) 4 6 (D) none of these 3 . The length of the perpendicular from the vertex D on the opposite face is - NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 14 2 3 (D) none of these (A) (B) (C) (D) x + y – 2z = 1 6 6 6 (C) 2x + y – 2z = 4 4 . Equation of the plane ABC is - (A) x + y + 2z = 5 (B) x – y – 2z = 1 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-03 True / False 1. F 2. T 3. T 4. T 5. F Match the Column 1 . (A) (r); (B) (s); (C) (p); (D) (q) 2 . (A) (r); (B) (p); (C) (q); (D) (s) 3 . (A) (r); (B) (q); (C) (q,s); (D) (p,s) Assertion & Reason 1. C 2. A 3. C 4. D 5. A 6. D 7. B Comprehension Based Questions Comprehension # 1 : 1 . B 2 . C 3 . D Comprehension # 2 : 1 . B 2 . C 3 . A Comprehension # 3 : 1 . B 2 . C 3 . A 4 . D 82 E
EXERCISE - 04 [A] JEE-Mathematics CONCEPTUAL SUBJECTIVE EXERCISE 1 . Find the angle between the two straight lines whose direction cosines , m, n are given by 2 + 2m – n = 0 and mn + n + m = 0. 2 . A variable plane is at a constant distance p from the origin and meets the coordinate axes in points A, B and C respectively. Through these points, planes are drawn parallel to the coordinates planes. Find the locus of their point of intersection. 3 . P is any point on the plane x + my + nz = p. A point Q taken on the line OP (where O is the origin) such that OP.OQ = p2. Show that the locus of Q is p(x + my + nz) = x2 + y2 + z2. 4 . The plane x + my = 0 is rotated about its line of intersection with the plane z = 0 through an angle . Prove that the equation to the plane in new position is x + my ±z 2 m 2 tan = 0 x3 y 3 z 5 . Find the equations of the two lines through the origin which intersect the line = = at an 2 11 angle of 3 x 1 y 2 z 3 6 . Find the equation of the line which is reflection of the line 9 = 1 = 3 in the plane 3x – 3y + 10z = 26 7 . Find the point where the line of intersection of the planes x – 2y + z = 1 and x + 2y – 2z = 5, intersects the plane 2x + 2y + z + 6 = 0 8 . Find the foot and hence the length of the perpendicular from the point (5, 7, 3) to the line x 15 y 29 5 z 3 = 8 = 5 . Also find the equation of the plane in which the perpendicular and the given straight line lie. x 1 y 2 z 9 . Find the equation of the plane containing the straight line 2 = 3 = 5 are perpendicular to the plane x – y + z + 2 = 0 x 1 y z x 3 y z 2 1 0 . Find the equation of the plane containing the line = 3 = 2 and parallel to the line =5 = . 2 2 4 Find also the S.D. between two lines. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-04(A) 1111 xy z xyz 1 . = 90° 2 . x2 + y2 + z2 = p2 5 . = = or = = x 4 y 1 z 7 1 2 1 1 1 2 6. 9 = 1 = 3 7 . (1, – 2, –4) 8. (9, 13, 15) ; 14; 9x – 4y – z = 14 9 . 2x + 3y +z + 4 = 0 1 0 . x – 2y + 2z – 1 = 0; 2 units E 83
JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE EXERCISE - 04 [B] 1 . Through a point P(f, g, h), a plane is drawn at right angles to OP where 'O' is the origin, to meet the coordinate axes in A, B, C. Prove that the area of the triangle ABC is r5 where OP = r. 2fgh 2 . Find the equations to the line which can be drawn from the point (2, –1, 3) perpendicular to the lines x 1 y 2 z 3 x4 y z3 2 = 3 = 4 and 4 = 5 = 3 at right angles. 3 . The position vectors of the four angular points of a tetrahedron OABC are (0, 0, 0); (0, 0, 2); (0, 4, 0) and (6, 0, 0) respectively. A point P inside the tetrahedron is at the same distance 'r' from the four plane faces of the tetrahedron. Find the values of 'r'. x 6 y 10 z 14 4 . The line 5 = 3 = 8 is the hypotenuse of an isosceles right angled triangle whose opposite vertex is (7, 2, 4). Find the equation of the remaining sides. 5 . If two straight lines having direction cosines , m, n satisfy a + bm + cn = 0 and fmn + gn + hm = 0 are f gh perpendicular, then show that a + b + c = 0. 6 . Find the equations to the line of greatest slope through the point (7, 2, –1) in the plane x – 2y + 3z = 0 assuming that the axes are so placed that the plane 2x + 3y – 4z = 0 is horizontal. 7 . Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the area of triangles ABC, ACD and ADB be denoted by x, y and z sq. units respectively. Find the area of the triangle BCD. BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-04(B) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 x2 y 1 z3 2 x 7 y 2 z 4 x 7 y 2 z 4 2 . 11 = 10 = 2 3. 3 4. = = ; = 3 = 3 6 2 2 6 6. x 7 y 2 z 1 7 . (x2 y2 z2 ) = = 22 5 4 84 E
EXERCISE - 05 [A] JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS x 1 y 2 z 3 x 1 y 5 z 6 1 . If the line 3 = 2k = 2 and 3k = 1 = 5 are perpendicular to each other then k = [AIEEE-2002] 5 7 –7 –10 (1) 7 (2) 5 (3) 10 (4) 7 2 . The angle between the lines, whose direction ratios are 1, 1, 2 and 3 –1, – 3 –1, 4, is- [AIEEE-2002] (1) 45° (2) 30° (3) 60° (4) 90° 3 . The acute angle between the planes 2x – y + z = 6 and x + y + 2z = 3 is- [AIEEE-2002] (1) 30° (2) 45° (3) 60° (4) 75° 4 . The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x2 + y2 + z2 + 4x – 2y – 6z = 155 is- [AIEEE-2003] (1) 39 (2) 26 4 (4) 13 (3) 11 13 x 2 y 3 z 4 x 1 y 4 z 5 5 . The lines = = and = = are coplanar if- [AIEEE-2003] 1 1 k k21 (1) k = 3 or –3 (2) k = 0 or –1 (3) k = 1 or –1 (4) k = 0 or –3 6 . The radius of the circle in which the sphere x2 + y2 + z2 + 2x – 2y – 4z – 19 = 0 is cut by the plane x + 2y + 2z + 7 = 0 is- [AIEEE-2003] (1) 4 (2) 1 (3) 2 (4) 3 7 . A tetrahedron has vertices at O(0, 0, 0), A(1, 2, 1), B(2, 1, 3) and C(–1, 1, 2). Then the angle between the faces OAB and ABC will be- [AIEEE-2003] (1) 90° 19 17 (4) 30° (2) cos–1 35 (3) cos–1 31 8 . The two lines x = ay + b, z = cy + d and x = a'y + b', z = c'y + d' will be perpendicular, if and only if- [AIEEE-2003] (1) aa' + cc' + 1 = 0 (2) aa' + bb' + cc' + 1 = 0 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 (3) aa' + bb' + cc' = 0 (4) (a + a') (b + b') + (c + c') = 0 9 . A line makes the same angle , with each of the x and z axis. If the angle , which it makes with y-axis, is such that sin2 = 3sin2, then cos2 equals- [AIEEE-2004] (1) 2/3 (2) 1/5 (3) 3/5 (4) 2/5 1 0 . Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is- [AIEEE-2004] (1) 3/2 (2) 5/2 (3) 7/2 (4) 9/2 11 . A line with direction cosines proportional to 2, 1, 2 meets each of the lines x = y + a = z and x + a = 2y = 2z. The coordinates of each of the points on intersection are given by- [AIEEE-2004] (1) (3a, 3a, 3a), (a, a, a) (2) (3a, 2a, 3a), (a, a, a) (3) (3a, 2a, 3a), (a, a, 2a) (4) (2a, 3a, 3a), (2a, a, a) x – and t and t 12. If the straight lines = 1 + s, y = –3 s, z=1 + s x = , y = 1 + t, z = 2 – t, with parameters s 2 respectively are coplanar then equals- [AIEEE-2004] (1) –2 (2) –1 1 (4) 0 (3) – 2 E 85
JEE-Mathematics 1 3 . The intersection of the spheres x2 + y2 + z2 + 7x – 2y – z = 13 and x2 + y2 + z2 – 3x + 3y + 4z = 8 is the same as the intersection of one of the sphere and the plane- [AIEEE-2004] (1) x – y – z = 1 (2) x – 2y – z = 1 (3) x – y – 2z = 1 (4) 2x – y – z = 1 x 1 y 1 z 2 1 1 4 . if the angle between the line == and the plane 2x – y + z + 4 = 0 is such that sin = 3 122 the value of is- [AIEEE-2005] 5 –3 3 –4 (1) 3 (2) 5 (3) (4) 3 4 1 5 . The angle between the lines 2x = 3y = –z and 6x = –y = –4z is- [AIEEE-2005] (1) 0° (2) 90° (3) 45° (4) 30° 1 6 . If the plane 2ax – 3ay + 4az + 6 = 0 passes through the midpoint of the line joining the centres of the spheres x2 + y2 + z2 + 6x – 8y – 2z = 13 and x2 + y2 + z2 – 10x + 4y – 2z = 8 then a equals- [AIEEE-2005] (1) –1 (2) 1 (3) –2 (4) 2 17. The distance between the line 2 ˆi – 2 ˆj + 3 kˆ + ( ˆi – ˆj + 4 kˆ ) and the plane .( ˆi + 5 ˆj + kˆ ) = 5 is- r= r [AIEEE-2005] 10 10 3 10 (1) 9 (2) 3 3 (3) 10 (4) 3 1 8 . The plane x + 2y – z = 4 cuts the sphere x2 + y2 + z2 – x + z – 2 = 0 in a circle of radius- [AIEEE-2005] (1) 3 (2) 1 (3) 2 (4) 2 1 9 . The two lines x = ay + b, z = cy + d; and x = a'y + b; z = c'y + d' are perpendicular to each other if- [AIEEE-2006] (1) aa' + cc' = 1 (2) a + c = –1 (3) a + c = 1 (4) aa' + cc' = –1 a' c' a' c' [AIEEE-2006] 2 0 . The image of the point (–1, 3, 4) in the plane x – 2y = 0 is - (1) (15, 11, 4) (2) 17 , 19 ,1 (3) (8, 4, 4) (4) None of these 3 3 2 1 . Let L be the line of intersection of the planes 2x + 3y + z = 1 and x + 3y + 2z = 2. If L makes an angle with the positive x-axis, then cos equals - [AIEEE-2007] (1) 1/ 3 (2) 1/2 (3) 1 (4) 1/ 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 2 2 . If a line makes an angle of with the positive directions of each of x-axis and y-axis, then the angle that the line 4 makes with the positive direction of the z-axis is- [AIEEE-2007] (1) /6 (2) /3 (3) /4 (4) /2 2 3 . If (2, 3, 5) is one end of a diameter of the sphere x2 + y2 + z2 – 6x – 12y – 2z + 20 = 0, then the coordinates of the other end of the diameter are- [AIEEE-2007] (1) (4, 9, –3) (2) (4, –3, 3) (3) (4, 3, 5) (4) (4, 3, –3) 24. The line passing through the points (5, 1, a) and (3, b, 1) crosses the yz-plane at the point 0, 17 , 13 Then- 2 2 . [AIEEE-2008] (1) a = 2, b = 8 (2) a = 4, b = 6 (3) a = 6, b = 4 (4) a = 8, b = 2 x 1 y 2 z 3 x 2 y 3 z 1 2 5 . If the straight lines = = and = = intersect at a point, then the integer k is k 2 3 3 k 2 equal to- [AIEEE-2008] (1) –5 (2) 5 (3) 2 (4) –2 86 E
JEE-Mathematics 26. Let the line x 2 y 1 z 2 lie in the plane x + 3y – z + = 0. Then (, ) equals : 3 5 2 [AIEEE-2009] (1) (5, – 15) (2) (–5, 5) (3) (6, –17) (4) (–6, 7) 2 7 . The projections of a vector on the three coordinate axis are 6, –3, 2 respectively. The direction cosines of the vector are :- [AIEEE-2009] 6 3 2 6 3 2 6 3 , 2 (1) , , (2) ,, (3) 6, –3, 2 (4) , 7 7 7 7 77 555 2 8 . A line AB in three-dimensional space makes angle 45º and 120º with the positive x-axis and the positive y-axis respectively. If AB makes an acute angle with the positive z-axis, then equals :- [AIEEE-2010] (1) 30° (2) 45° (3) 60° (4) 75° 2 9 . Statement–1 : The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x – y + z = 5. [AIEEE-2010] Statement–2 : The plane x – y + z = 5 bisects the line segment joining A(3, 1, 6) and B(1, 3, 4). (1) Statement–1 is true, Statement–2 is true; Statement–2 is a correct explanation for Statement–1. (2) Statement–1 is true, Statement–2 is true; Statement–2 is not a correct explanation for Statement–1. (3) Statement–1 is true, Statement–2 is false. (4) Statement–1 is false, Statement–2 is true. 30. If the angle between the line x = y 1 z 3 and the plane x + 2y + 3z = 4 is cos–1 5 , then equals - 2 14 [AIEEE-2011] 2 5 2 3 (1) (2) (3) (4) 5 3 3 2 x y 1 z 2 3 1 . Statement-1: The point A(1, 0, 7) is the mirror image of the point B(1, 6, 3) in the line : 12 3 Statement-2: The line : x y 1 z 2 bisects the line segment joining A (1, 0, 7) and B(1, 6, 3). 12 3 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 [AIEEE-2011] (1) Statement-1 is true, Statement-2 is false. (2) Statement-1 is false, Statement-2 is true (3) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1 (4) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. x y2 z3 32. The length of the perpendicular drawn from the point (3, –1, 11) to the line is : 2 3 4 [AIEEE-2011] (1) 66 (2) 29 (3) 33 (4) 53 33 . The distance of the point (1, –5, 9) from the plane x – y + z = 5 measured along a straight line x = y = z is: [AIEEE-2011] (1) 3 5 (2) 10 3 (3) 5 3 (4) 3 10 3 4 . An equation of a plane parallel to the plane x – 2y + 2z – 5 = 0 and at a unit distance from the origin is : [AIEEE-2012] (1) x – 2y + 2z + 5 = 0 (2) x – 2y + 2z – 3 = 0 (3) x – 2y + 2z + 1 = 0 (4) x – 2y + 2z – 1 = 0 E 87
JEE-Mathematics 35. If the lines x 1 y 1 z 1 and x3 yk z intersect, then k is equal to :- [AIEEE-2012] 234 1 21 (1) 0 (2) – 1 2 9 (3) 9 (4) 2 3 6 . Distance between two parallel planes 2x + y + 2z = 8 and 4x + 2y + 4z + 5 = 0 is :- [JEE-MAIN 2013] 3 5 7 9 (1) (2) (3) (4) 2 2 2 2 3 7 . If the lines x 2 y 3 z 4 and x 1 y 4 z 5 are coplanar, then k can have : [JEE-MAIN 2013] 1 1 k k21 (1) any value (2) exactly one value (3) exactly two values (4) exactly three values. PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 Que. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Ans 4 3 3 4 4 4 2 1 3 3 2 1 4 1 2 Que. 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 Ans 3 2 2 4 4 1 4 1 3 1 4 1 3 2 Que. 31 32 33 34 35 36 37 Ans 4 4 2 2 4 3 3 E 88
EXERCISE - 05 [B] JEE-Mathematics JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS 1 . ( a ) Find the equation of the plane passing through the points (2,1,0),(5,0,1) and (4,1,1) ( b ) If P is the point (2, 1, 6) then find the point Q such that PQ is perpendicular to the plane in (a) and the mid point of PQ lies on it. [JEE 03, 4M out of 60] 2. If the lines x 1 y 1 z 1 and x3 yk z are intersecting each other then ‘k’ is - 234 1 21 2 9 (C) 1 3 (A) 9 (B) 2 (D) 2 [JEE 04 (screening)] 3 . T is a parallelopiped in which A, B, C and D are vertices of one face. And the face just above it has correspond- ing vertices A' , B', C', D'. T is now compressed to S with face ABCD remaining same and A', B', C', D' shifted to A'', B'', C'' and D'' in S. The volume of S is reduced to 90% of T. Prove that locus of A'' is a plane. [JEE 04 (Mains) 2M] 4 . A plane is parallel to two lines whose direction ratios (1, 0, –1) & (–1, 1, 0) and it contains the point (1, 1, 1). If it cuts the coordinate axes at A, B, C. then find the volume of tetrahedron OABC, where O is the origin. [JEE 04 (Mains) 2M] 5 . P1 and P2 are planes passing through origin. L1 and L2 are two line on P1 and P2 respectively such that their intersection is origin. Show that there exists points A, B, C, whose permutation A', B', C' can be chosen such that (i) A is on L1, B on P1 but not on L1 and C not on P1 (ii) A' is on L2, B' on P2 but not on L2 and C' not on P2. [JEE 04 (Mains) 4M] 6 . A variable plane at a distance of 1 unit from the origin cut the coordinate axis at A, B & C. If centroid of 111 triangle ABC is D(x, y, z) satisfy the relation x 2 y 2 z2 k , then value of k is - [JEE 05 (screening) 3M] (A) 3 (B) 1 (C) 1/3 (D) 9 7 . Find the equation of the plane containing the line 2x – y + z – 3 = 0, 3x + y + z = 5 and at a distance of 1 [JEE 05 (Mains) 2M] 6 from the point (2, 1, –1) 8 . A plane passes through (1, –2, 1) and is perpendicular to two planes 2x – 2y + z = 0 and x – y + 2z = 4. The distance of the plane from the point (1, 2, 3) is - [JEE 06, 3M] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 (A) 0 (B) 1 (C) 2 (D) 2 2 9 . Match the following Column-I [JEE 06, 6M] Column-II (A) Two rays in the first quadrant x + y = |a| and ax – y = 1 (p) 2 intersects each other in the interval a (a , ), the value of 0 a is 0 (B) Point ( ) lies on the plane x + y + z = 2. (q) 4/3 Let ˆi ˆj kˆ, kˆ (kˆ 0 , then = a a) 10 10 (C) (1 y2 )dy (y2 1)dy (r) 1 x dx 1 x dx 01 0 1 (D) If sinAsinBsinC + cosAcosB = 1, then the value of sinC = (s) 1 E 89
JEE-Mathematics 1 0 . Match the following [JEE 06, 6M] Column-I Column-II (p) 0 (A) tan 1 1 t , then tant = i 1 2i2 (B) Sides a, b, c of a triangle ABC are in A.P. (q) 1 and cos 1 b a , cos 2 b, cos 3 a c b , c ac then tan2 1 tan2 3 = 5 22 (r) (C) A line is perpendicular to x + 2y + 2z = 0 and passes 3 through (0, 1, 0).The perpendicular distance of this (s) 2/3 line from the origin is 1 1 . Consider the following linear equations [JEE 2007, 6M] ax + by + cz = 0; bx + cy + az = 0; cx + ay + bz = 0 Column-I Column-II (p) the equations represent planes meeting only at (A) a + b + c 0 and a2 + b2 + c2 = ab + bc + ca a single point (q) the equations represent the line x = y = z (B) a + b + c 0 and a2 + b2 + c2 ab + bc + ca (r) the equations represent identical planes. (C) a + b + c 0 and a2 + b2 + c2 (s) the equations represent the whole of the three ab + bc + ca dimensional space (D) a + b + c = 0 and a2 + b2 + c2 = ab + bc + ca 1 2 . Consider the planes 3x – 6y – 2z = 15 and 2x + y – 2z = 5. [JEE 2007, 3M] Statement-1 : The parametric equations of the line of intersection of the given planes are x = 3 + 14t, y = 1 + 2t, z = 15t. because Statement-2 : The vector 14ˆi 2ˆj 15kˆ is parallel to the line of intersection of given planes. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 (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. 1 3 . Consider three planes [JEE 2008 (4M, –1M)] P1 : x – y + z = 1 P2 : x + y – z = –1 P3 : x – 3y + 3z = 2 Let L1, L2, L3 be the lines of intersection of the planes P2 and P3, P3 and P1, and P1 and P2, respectively. Statement-1 : At least two of the lines L1, L2 and L3 are non-parallel. because Statement-2 : The three planes do not have a common point. (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. 90 E
JEE-Mathematics Paragraph for Question 14 to 16 Consider the lines L1 : x 1 y 2 z 1 L2 : x2 y2 z3 , 123 312 1 4 . The unit vector perpendicular to both L1 and L2 is :- [JEE 2008 (4M, –1M)] 7ˆi 7ˆj kˆ ˆi 7ˆj 7kˆ ˆi 7ˆj 5kˆ ˆi 7ˆj 5kˆ (D) (A) (B) (C) 99 99 53 53 [JEE 2008 (4M, –1M)] 1 5 . The shortest distance between L1 and L2 is :- 17 (D) 5 3 (A) 0 17 41 (B) 3 (C) 5 3 1 6 . The distance of the point (1, 1, 1) from the plane passing through the point (–1, –2, –1) and whose normal is perpendicular to both the lines L1 and L2 is :- [JEE 2008 (4M, –1M)] 2 7 13 23 (A) 75 (B) 75 (C) 75 (D) 75 1 7 . A line with positive direction cosines passes through the point P (2, –1, 2) and makes equal angles with the coordinate axes. The line meets the plane 2x + y + z = 9 at point Q. The length of the line segment PQ equals [JEE 2009, 3M, –1M] (A) 1 (B) 2 (C) 3 (D) 2 18. Let P(3, 2, 6) be a point in space and Q be a point on the line (ˆi ˆj 2kˆ) µ(3ˆi ˆj 5 kˆ) . Then the value r [JEE 2009, 3M, –1M] of µ for which the vector PQ is parallel to the plane x – 4y + 3z = 1 is :- 1 1 1 1 (A) (B) – (C) (D) – 4 4 8 8 1 9 . Match the statements/ expressions given in Column I with the values given in Column II [JEE 2009, 8M] Column-I Column-II NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 (A) The number of solutions of the equation xesinx – cosx = 0 (P) 1 (Q) 2 in the interval 0, (R) 3 2 (B) Value(s) of k for which the planes kx + 4y + z = 0, 4x + ky + 2z = 0 and 2x + 2y + z = 0 intersect in a straight line (C) Value(s) of k for which x 1 x 2 + x 1 + x 2 = 4k (S) 4 has integer solution(s) (T) 5 (D) If y' = y +1 and y(0) =1, then value(s) of y (ln 2) 20. Equation of the plane containing the straight line xyz and perpendicular to the plane containing the 234 xyz xyz straight lines and is [JEE 10, 3M, –1M] 342 423 (A) x + 2y – 2z = 0 (B) 3x + 2y – 2z = 0 (D) 5x + 2y – 4z = 0 (C) x – 2y + z = 0 E 91
JEE-Mathematics 2 1 . If the distance of the point P(1,–2,1) from the plane x + 2y – 2z = , where > 0, is 5, then the foot of the perpendicular from P to the plane is- [JEE 10, 5M, –2M] (A) 8 , 4 , 7 (B) 4 , 4 , 1 (C) 1 , 2 , 10 (D) 2 , 1 , 5 3 3 3 3 3 3 3 3 3 3 3 2 22. If the distance between the plane Ax – 2y + z = d and the plane containing the lines x 1 y 2 z 3 and 234 x 2 y 3 z4 is 6 , then |d| is [JEE 10, 3M] 345 [JEE 10, 8M] 2 3 . Match the statements in Column-I with the values in Column-II. Column-I Column-II (A) A line from the origin meets the lines (p) –4 x 2 y 1 z 1 x 8 y 3 z 1 and at P and Q 3 1 2 1 2 1 1 respectively. If length PQ = d, then d2 is (B) The values of x satisfying (q) 0 (r) 4 tan 1 (x 3) tan 1 (x 3) sin 1 3 are 5 (s) 5 (t) 6 (C) Non-zero vectors and satisfy 0 , a, b c a.b bc ba (b a).(b c) 0 and 2 . If a b 4c, then the possible values of are (D) Let f be the function on [–,] given by f(0)= 9 and f(x) = sin 9x sin x for x 0. 2 2 2 The value of f ( x )dx is 2 4 . ( a ) The point P is the intersection of the straight line joining the points Q(2,3,5) and R(1,–1,4) w i t h NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 the plane 5x – 4y – z = 1. If S is the foot of the perpendicular drawn from the point T(2,1,4) to QR, then the length of the line segment PS is - 1 (B) 2 (C) 2 (D) 2 2 (A) 2 ( b ) The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x – y + z = 3 and at a distance 2 from the point (3, 1, –1) is 3 (A) 5x – 11y + z = 17 (B) 2x y 3 2 1 (C) x + y + z = 3 (D) x 2y 1 2 (c) If the straight lines x 1 y 1 z and x 1 y 1 z are coplanar, then the plane(s) containing 2 k2 5 2k these two lines is(are) (A) y + 2z = –1 (B) y + z = –1 (C) y – z = –1 (D) y – 2z = –1 [JEE 2012, 3+3+4] 92 E
JEE-Mathematics 2 5 . Perpendiculars are drawn from points on the line x 2 y 1 z to the plane x + y + z = 3. The feet 2 1 3 of perpendiculars lie on the line [JEE-Advanced 2013, 2] (A) x y 1 z 2 (B) x y 1 z 2 (C) x y 1 z 2 (D) x y 1 z 2 5 8 13 2 3 5 4 3 7 2 7 5 2 6 . A line passing through the origin is perpendicular to the lines 1 : 3 t ˆi 1 2t ˆj 4 2t kˆ, t 2 : 3 2sˆi 3 2s ˆj 2 s kˆ, s Then , the coordinate(s) of the point(s) on at a distance of 17 from the point of intersection of and 2 is(are) - [JEE-Advanced 2013, 4, (–1)] 1 (A) 7 , 7 , 5 (B) (–1,–1,0) (C) (1,1,1) (D) 7 , 7 , 8 3 3 3 9 9 9 27. Two lines yz yz are coplanar. Then can take value(s) L1 : x 5, 3 2 and L 2 : x , 1 2 (A) 1 (B) 2 (C) 3 [JEE-Advanced 2013, 3, (–1)] (D) 4 28. Consider the lines x 1 y z 3 x 4 y 3 z 3 and the planes P :7x+y + 2z = 3, L1 : 2 1 ,L2 : 1 1 1 1 2 P2 : 3x + 5y – 6z = 4. Let ax + by + cz = d be the equation of the plane passing through the point of intersection of lines L and L and perpendicular to planes P and P . 12 12 Match List-I with List-II and select the correct answer using the code given below the lists. List-I List-II P. a = 1. 13 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#10\\ENG\\03-3D-COORDINATE GEO.p65 Q. b = 2. –3 3. 1 R. c = 4. –2 S. d = Codes : [JEE-Advanced 2013, 3, (–1)] P QR S (A) 3 2 4 1 (B) 1 3 4 2 (C) 3 2 1 4 (D) 2 4 1 3 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-05 1 . (a) x + y – 2z = 3 ; (b) (6, 5, –2) 2. B 9 6. D 4 . cubic unit 2 7 . 2x –y + z – 3 = 0, 62x + 29y + 19z – 105 = 0 8. D 9. (A)(s) ; (B)(p) ; (C)(q,r) ; (D)(s) 1 0 .(A)(q) ; (B)(s) ; (C)(r) 11. (A)(r) ; (B)(q) ; (C)(p) ; (D)(s) 1 2 . D 13. D 1 4 .B 1 5 . D 1 6 . C 17. C 1 8 . A 1 9 . (A) (P) ; (B) (Q, S) ; (C) (Q, R, S, T) ; (D) (R) 2 0 .C 2 1 . A 2 2 . 6 2 3 . (A) (t), (B) (p,r), (C) (q,s), (D) (r) 2 4 . (a) A; (b) A; (c) B,C 2 5 .D 26. B,D 27. A,D 28. A E 93
JEE-Mathematics AREA UNDER THE CURVE 1. AREA UNDER THE CURVES : y y=ƒ(x) ( a ) Area bounded by the curve y = f(x), the x-axis and the ordinates at x = a and b x O x=a dx x=b x = b is given by A = y dx , where y = ƒ (x) lies above the x-axis a y bx a and b > a. Here vertical strip of thickness dx is considered at distance x. O ( b ) If y = ƒ (x) lies completely below the x-axis then A is negative and we consider b the magnitude only, i.e. A = y dx a cb y x x=b ( c ) If curve crosses the x-axis at x = c, then A = y dx ydx x=a Oc ac ( d ) Sometimes integration w.r.t. y is very useful (horizontal strip) : y Area bounded by the curve, y-axis and the two abscissae at y=b dy b y=a Ox y = a & y = b is written as A xdy . a Note : If the curve is symmetric and suppose it has 'n' symmetric portions, then total area = n (Area of one symmetric portion). Illustration 1 : Solution : Find the area bounded by y = sec2 x, x = , x = & x-axis Illustration 2 : 63 / 3 ydx = / 3 sec2 xdx = [tan x ] / 3 = tan 3– 1 = 2 sq.units. / 6 – tan = 3 3 /6 /6 Area bounded = 36 Find the area in the first quadrant bounded by y = 4x2, x = 0, y = 1 and y = 4. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 4 4y 1 2 y3 / 2 4 Y Solution : = 2 3 Required area x dy = dy = y=4 1 12 1 x=0 = 1 [43/2 – 1] = 1 [8 – 1] y=1 33 O 7 X = = 2 sq.units. 33 Illustration 3 : Find the area bounded by the curve y = sin2x, x-axis and the lines x = and x 3 44 /2 3/4 cos 2 x /2 cos 2 x 3/ 4 2 2 = sin 2xdx sin 2xdx /4 /2 Solution : Required = area E / 4 / 2 /2 3/4 /4 = 1 [1 0] 1 (0 (1)) = 1 sq. unit 22 1
JEE-Mathematics Do yourself - 1 : ( i ) Find the area bounded by y = x2 + 2 above x-axis between x = 2 & x = 3. ( i i ) Using integration, find the area of the curve y 1 x2 with co-ordinate axes bounded in first quadrant. (i i i ) Find the area bounded by the curve y = 2cosx and the x-axis from x = 0 to x = 2. ( i v ) Find the area bounded by the curve y = x|x|, x-axis and the ordinates x = 1 and x=1. 2 2. AREA ENCLOSED BETWEEN TWO CURVES : y y1=ƒ(x) ( a ) Area bounded by two curves y = ƒ (x) & y = g(x) O x1 y2=g(x) x2 x such that ƒ (x) > g(x) is y x1=ƒ(y) x2 y2 A (y1 y2 )dy x2=g(y) x1 x x2 y1 A [ƒ(x) g(x)]dx y y2=g(x) x1 ( b ) In case horizontal strip is taken we have y2 A (x1 x2 )dy y1 y2 A [ƒ(y ) g(y )]dy y1 ( c ) If the curves y = ƒ (x) and y = g(x) intersect at x = c, then required area 12 cb b A = (g(x) ƒ(x))dx (ƒ(x) g(x))dx = ƒ(x) g(x) dx y1=ƒ(x) x=a c ac a x x=b Note : Required area must have all the boundaries indicated in the problem. Illustration 4 : Find the area bounded by the curve y = (x – 1) (x – 2) (x – 3) lying between the ordinates x = 0 and Solution : x = 3 and x-axis NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 To determine the sign, we follow the usual rule of change of sign. y = +ve for x > 3 Y y = –ve for 2 < x < 3 y = +ve for 1 < x < 2 C y = –ve for x < 1. B DF X O1 23 3 123 E 0| y| dx = 0| y| dx + 1| y| dx + 2| y| dx A 1 23 (0, –6) = 0 – y dx + 1 y dx + 2 –y dx 1 11 Now let F(x) = (x – 1) (x – 2) (x – 3) dx = (x3 – 6x2 + 11x – 6) dx = x4 – 2x3 + x2 – 6x. 42 99 F(0) = 0, F(1) = – , F(2) = –2, F(3) = – . 44 3 Hence required Area = – [F(1) – F(0)] + [F(2) – F(1)] – [F(3) – F(2)] = 2 4 sq.units. 2E
JEE-Mathematics Illustration 5 : Compute the area of the figure bounded by the straight lines x = 0, x = 2 and the curves Solution : y = 2x, y = 2x – x2. Figure is self-explanatory y = 2x, (x – 1)2 = – (y – 1) R(2,4) 2 y=2x x=2 The required area = 0 (y1 y2 ) dx y=2x–x2 M(2,0) where y = 2x and y = 2x – x2 = 2 (2x 2x x2 )dx Q 12 0 (0,1) = 2x x2 1 x3 2 = 4 4 8 – 1 = 3 –4 sq.units. O 3 ln 2 3 ln 2 ln 2 3 ln 2 0 Illustration 6 : Compute the area of the figure bounded by the parabolas x = – 2y2, x = 1 – 3y2. Solution : Solving the equations x = –2y2, x = 1 – 3y2, we find that ordinates of the points of intersection of the two curves as y = – 1, y = 1. (–2, 1) Y x=1–3y2 12 x=–2y2 1 P1 The points are (–2, –1) and (–2, 1). –2 –1 P2 1 O The required area 1 1 [(1 3y2 ) (–2y2 )]dy 0 (x1 x 2 ) dy 0 2 = 2 X 1 (1 y2 )dy = y3 1 4 =2 2 y = 3 sq.units. 0 3 (–2, –1) –1 0 Do yourself - 2 : ( i ) Find the area bounded by y x and y = x. ( i i ) Find the area bounded by the curves x = y2 and x = 3 – 2y2. 1 (i i i ) Find the area of the region bounded by the curves x = , x = 2, y = logx and y = 2x. 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 3 . CURVE TRACING : The following procedure is to be applied in sketching the graph of a function y = f(x) which in turn will be extremely useful to quickly and correctly evaluate the area under the curves. ( a ) Symmetry : The symmetry of the curve is judged as follows : ( i ) If all the powers of y in the equation are even then the curve is symmetrical about the axis of x. ( i i ) If all the powers of x are even, the curve is symmetrical about the axis of y. (i i i ) If powers of x & y both are even, the curve is symmetrical about the axis of x as well as y. (iv) If the equation of the curve remains unchanged on interchanging x and y, then the curve is symmetrical about y = x. ( v ) If on interchanging the signs of x & y both, the equation of the curve is unaltered then there is symmetry in opposite quadrants. ( b ) Find dy/dx & equate it to zero to find the points on the curve where you have horizontal tangents. ( c ) Find the points where the curve crosses the xaxis & also the yaxis. ( d ) Examine if possible the intervals when f(x) is increasing or decreasing. Examine what happens to ‘y’ when x or . Illustration 7 : Find the area of a loop as well as the whole area of the curve a2y2 = x2 (a2 – x2). Solution : The curve is symmetrical about both the axes. It cuts x-axis at (0, 0), (–a, 0), (a, 0) E Area of a loop = 2 a y dx = 2 a x a x2 dx Y a 0 0 =– 1 a a2 x2 (–2x)dx = – 1 2 (a2 x 2 )3 / 2 a 2 a2 O a a 3 0 3 0 X' A' AX (–a,0) (a,0) Total area = 2 × 2 a2 = 4 a2 sq.units. 33 3
JEE-Mathematics Illustration 8 : Find the whole area induded between the curve x2y2 = a2(y2 – x2) and its asymptotes. Solution : (i) The curve is symmetric about both the axes (even powers of x & y) (ii) Asymptotes are x = ± a a y A 4 ydx 0 a ax x=–a dx x=a x 4 dx 0 a2 x2 a 4a a2 x2 0 = 4a2 Illustration 9 : Find the area bounded by the curve xy2 = 4a2(2a–x) and its asymptote. Solution : (i) The curve is symmetrical about the x-axis as it contains even powers of y. (ii) It passes through (2a,0). (iii) Its asymptote is x = 0, i.e., y-axis. 2a 2a 2a x y A 2 ydx 2 2a dx 0 0x Put x = 2a sin2 (2a,0) x /2 A = 16a2 cos2 d 0 = 4a2 4 . IMPORTANT POINTS : ( a ) Since area remains invariant even if the co-ordinate axes are shifted, hence shifting of origin in many cases proves to be very convenient in computing the area. Illustration 10 : Find the area enclosed by |x – 1| + |y + 1| = 1. (0,1) 2 Solution : Shift the origin to (1, –1). (1,0) X=x–1 Y=y+1 (–1,0) (0,–1) |X| + |Y| = 1 Area = 2 2 = 2 sq. units Illustration 11 : Find the area of the region common to the circle x2 + y2 + 4x + 6y – 3 = 0 and the parabola NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 x2 + 4x = 6y + 14. Solution : Circle is x2 + y2 + 4x + 6y – 3 = 0 x (x + 2)2 + (y + 3)2 = 16 Shifting origin to (–2,–3). (–23,2) (0,4) X2 + Y2 = 16 (23,2) equation of parabola (x + 2)2 = 6(y + 3) X2 = 6Y Solving circle & parabola, we get X = ± 2 3 Hence they intersect at 2 3,2 & 2 3,2 4 6Y dY 2 16 Y 2 dY A 2 0 2 2 6 Y 3 / 2 2 1 Y 16 Y2 16 sin 1 Y 4 4 3 16 2 2 3 3 0 4 2 sq. units 3 2 4 E
JEE-Mathematics Do yourself : 3 ( i ) Find the area inside the circle x2–2x + y2 – 4y + 1 = 0 and outside the ellipse x2–2x+4y2–16y+13= 0 t dx t dy If the equation of the curve is in parametric form, then A y .dt or x .dt , where are (b) t dt t dt values corresponding to values of x and & are values corresponding to values of y. Illustration 12 : Find the area bounded by x-axis and the curve given by x = asint, y = acost for 0 t . y dx .dt = a cos t(a cos t)dt = a2 a2 sin 2t a2 a2 Solution : Area = (1 cos 2 t)dt t = 0 dt 20 2 2 0 2 2 0 Alternatively, dy a2 a2 sin 2t a 2 = x .dt = = = t a sin t(a sin t)dt (cos 2t 1)dt Area 0 dt 0 20 22 02 Illustration 13 : Find the area of the figure bounded by one arc of the cycloid x = a(t – sint), y = a(1 – cost) and the x-axis. Solution : To find the points where an arc cuts x-axis a(1 – cost) = 0 t = 0, dx a2 3 sin 2t a2 3 3 a 2 t 2 sin t 2 2 Area = y dt a2 (1 cos t)2 dt 2 40 0 dt 0 Do yourself - 4 : ( i ) Find the area of the loop of the curve : (a) x = 3t2, y = 3t – t3 (b) x = t2 – 1, y = t3 – t NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 ( c ) If y = ƒ (x) is a monotonic function in (a, b), then the area bounded by the ordinates at x = a, x = b, y = ƒ (x) and y = ƒ (c) [where c (a, b)] is minimum when c a b . 2 Proof : Let the function y = ƒ (x) be monotonically increasing. y y=ƒ(x) cb y=ƒ(c) O x=a x=c x=b x Required area A = [ƒ(c) ƒ(x)]dx [ƒ(x) ƒ(c)]dx ac For minimum area, dA 0 dc [ƒ '(c).c ƒ(c) ƒ '(c)a ƒ(c)] [ƒ(c) ƒ '(c).b ƒ '(c).c ƒ(c)] 0 ƒ '( c ) c a b 0 2 ab (ƒ '(c) 0) c= 2 E5
JEE-Mathematics Illustration 14 : Find the value of 'a' for which area bounded by x = 1, x=2, y=6x2 and y=ƒ(a) is minimum. Solution : Let b = ƒ (a). y=6x2 y a2 a 2 y=ƒ(a) (b 6x2 )dx (6 x 2 bx 2x3 2x3 bx A b)dx 1a 1a = 8a3 – 18a2 + 18 x=a x x=1 x=2 For minimum area dA 0 da 24a2 – 36a = 0 a = 1.5 Alternatively, y = 6x2 dy 12x dx Hence y = ƒ (x) is monotonically increasing. Hence bounded area is minimum when a = 1 2 = 1.5 2 Do yourself - 5 : (i ) Find the value of 'a' (0 < a < ) for which the area bounded by the curve ƒ (x) = sin3x + sinx, 2 y = ƒ (a) between x = 0 & x = is minimum. ( d ) The area bounded by a curve & an axis is equal to the area bounded by the inverse of that curve & the other axis, i.e., the area bounded by y = ƒ (x) and x-axis (say) is equal to the area bounded by y = ƒ–1(x) and y-axis. Illustration 15 : If y = g(x) is the inverse of a bijective mapping ƒ : R R, ƒ (x) = 6x5 + 4x3 + 2x, find the area Solution : bounded by g(x), the x-axis and the ordinate at x = 12. y y=x ƒ (x) = 12 E(0,12) y=f(x) D 6x5 + 4x3 + 2x = 12 x = 1 12 1 A(0,1) y=g(x) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 B g(x)dx = area of rectangle OEDF – f(x)dx 00 1 0 F(1,0) x C(12,0) = 1 × 12 – (6x5 4 x3 2x)dx = 12 – 3 = 9 sq. units. 0 Do yourself - 6 : ( i ) Find the area bounded by the inverse of bijective function ƒ (x) = 4x3 + 6x, the x-axis and the ordinates x = 0 & x = 44. 5 . USEFUL RESULTS : ( a ) Whole area of the ellipse, x2/a2 + y2/b2 = 1 is ab sq.units. ( b ) Area enclosed between the parabolas y2 = 4 ax & x2 = 4 by is 16ab/3 sq.units. ( c ) Area included between the parabola y2 = 4 ax & the line y = mx is 8 a2/3 m3 sq.units. ( d ) The area of the region bounded by one arch of sin ax (or cosbx) and x-axis is 2/a sq.units. 1b (e) Average value of a function y = f (x) over an interval a x b is defined as : y(av) = f(x)dx . ba a 6 E
JEE-Mathematics Miscellaneous Illustration : Illustration 16 : Find the smaller of the areas bounded by the parabola 4y2 – 3x – 8y + 7 = 0 and the ellipse Solution : x2 + 4y2 – 2x – 8y + 1 = 0. II (1, 3 / 2 ) C is 4(y2 – 2y) = 3x – 7 L 1 I or 4(y – 1)2 = 3x – 3 = 3 (x – 1) .....(i) AB Above is parabola with vertex at (1, 1) P C is (x2 – 2x) + 4 (y2 – 2y) = –1 O N (2, 0) 2 or (x – 1)2 + 4(y – 1)2 = –1 + 1 + 4 M or (x 1)2 (y 1)2 = 1 ......(ii) (1 - 3 / 2) 22 12 Above represents an ellipse with centre at (1, 1). Shift the origin to (1, 1) and this will not affect the magnitude of required area but will make the calculation simpler. Thus the two curves are 4Y2 = 3X and X2 Y2 = 1 22 1 They meet at 1, 3 2 Required area = 2(A + B) = 2 Y1dX Y2dX = 2 3 1 X dX 2 4 X2 = 3 2 sq.units. 2 2 dX 6 3 0 1 Illustration 17 : Find the area bounded by the regions y x , x > – y & curve x2 + y2 = 2. Solution : Common region is given by the diagram x=– y If area of region OAB = then area of OCD = C A y= x D NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 Because y = x & x = – y –2 O B will bound same area with x & y axes respectively. 12 y = x y2 = x x = – y x2 = y and hence both the curves are symmetric with respect to the line y = x Area of first quadrant OBC = r 2 = ( r = 2 ) 4 2 Area of region OCA = – 2 Area of shaded region = ( – ) + = sq.units. 2 2 Illustration 18 : Find the equation of line passing through the origin & dividing the curvilinear triangle with vertex at the origin, bounded by the curves y = 2x – x2, y = 0 & x = 1 in two parts of equal areas. E7
JEE-Mathematics Solution : Area of region OBA = 1 (2 x x2 )dx 0 = x 2 x3 1 = 2 A 3 0 3 C(1, y) B(1, 0) 21 =A +A A = 3 11 1 3 Let pt. C has coordinates (1, y) 1 1 O 2 3 Area of OCB = × 1 × y = 2 y= 3 C has coordinates 1, 2 3 Line OC has slope m = 2 0 = 2 3 3 10 Equation of line OC is y = mx y = 2 x. 3 Illustration 19 : Find the area bounded by the curves x2 + y2 = 4, x2 = – 2 y and the line x = y, below x-axis. Solution : Let C is x2 + y2 = 4, P is y = – x2 and L is y = x. 2 y=x We have above three curves. Solving P and C we get the points O A(– 2 , – 2 ), B( 2 , – 2 ) Also the line y = x passes through A(– 2 , – 2 ) A B Required area = shaded + dotted (– 2, – 2 ) ( 2,– 2) = 0 2 (y3 y 1 )dx 2 (y2 y1 )dx – 0 = 0 x dx 0 2 x2 dx 2 4 x2 dx 2 – 2 2 = x2 0 – 1 x3 2 – x 4 x2 4 sin 1 x 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 2 – 2 3 0 2 2 2 0 2 |A| = 3 16 sq.units. 6 Illustration 20 : In the adjacent figure, graphs of two functions y = f(x) y and y = sin x are given. y = sin x intersects y = f(x) A1 y=f(x) A A2 at A(a f(a)); B (, 0) and C(2, 0). A (i = 1, 2, 3) is y=sinx i Oa BC the area bounded by the curves y = f(x) and y=sinx 2 y = sinx between x = 0 and x = a (i = 1), between A3 x = a and x = (i = 2), between x = and x = 2 (i = 3). If A = 1 – sina + (a – 1) cosa, determine the 1 function f(x). Hence determine 'a' and A . Also 1 calculate A and A . 23 8 E
JEE-Mathematics a Solution : From the figure it is clear that A = (sin x f(x)) dx = 1 – sin a + (a – 1) cosa 1 differentiate w.r.t. a 0 sina – f(a) = – cos a + cos a – (a – 1) sina sina – f(a) = – asina + sina f(a) = a sina f(x) = xsinx The points where f(x) & sinx intersect are x sinx = sinx sinx = 0 or x = 1 a = 1 (0 < a < ) 1 A = (sin x x sin x) dx = 1 – sin1 sq.units 1 0 A= (f(x) sin x) dx (x sin x sin x) dx = – 1 – sin1) sq.units 2 11 2 A= (sin x x sin x) dx = (3 – 2) sq.units 3 Illustration 21 : The area bounded by y = x2 + 1 and the tangents to it drawn from the origin is :- (A) 8/3 sq. units (B) 1/3 sq. units (C) 2/3 sq. units (D) none of these Solution : The parabola is even function & let the equation of tangent is y= mx Now we calculate the point of intersection of parabola & tangent mx = x2 + 1 x2 – mx + 1 = 0 D = 0 m2 – 4 = 0 m = ± 2 C A (1,2) Two tangents are possible y = 2x & y = –2x B Intersection of y = x2 + 1 & y =2x is x = 1 & y = 2 O (0, 0) 1 1 1) 2x dx 1 = = y2 y1 dx = (x2 3 0 0 Area of shaded region OAB sq. units NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 1 2 Area of total shaded region = 2 3 3 sq. units Illustration 22 : Determination of unknown parameter : Consider the two curves C : y = 1 + cos x & C : y = 1 + cos (x – ) for 0, ; x [0, ] . Find 1 2 2 the value of , for which the area of the figure bounded by the curves C ,C & x = 0 is same as that 12 of the figure bounded by C, y = 1 & x = . 2 Solution : 1 + cos x = 1 + cos(x – ) x 2 C1 2 /2 A= (1 cos x) (1 cos(x ))dx C1 1 0 O /2 sin x sin(x ) /2 2 sin sin 02 E9
JEE-Mathematics 1 + cos(x – ) = 1 x = + 2 A2 1 cos x 1dx sin x 2 2 = |sin – 1| = 1 – sin A1 A2 2 sin sin 1 sin 2 Ans. 3 ANSWERS FOR DO YOURSELF NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 1 : (i) 25 (i i i ) 8 sq. units. 7 3 sq. units ( i i ) sq. units. ( i v ) sq. units 2 : (i) 3 : (i) 4 24 4 : (i) 1 4 25 3 5 : (i) 6 sq. units (ii) 4 sq. units (iii) log 2 sq. units 6 : (i) (b) log 2 2 2 2 sq. units ( a ) 72 3 sq. units 8 5 15 sq. units 8 60 sq. units. 10 E
EXERCISE - 01 JEE-Mathematics CHECK YOUR GRASP SELECT THE CORRECT ALTERNATIVE (ONLY ONE CORRECT ANSWER) 1 . The area of the region bounded by the curves y = | x – 2| , x = 1 , x = 3 and the x-axis is - (A) 3 (B) 2 (C) 1 (D) 4 2 . The area enclosed between the curve y = log (x + e) and the coordinate axes is - e (A) 4 (B) 3 (C) 2 (D) 1 3 . The area of the figure bounded by the curves y = nx & y = (nx)2 is - (A) e + 1 (B) e – 1 (C) 3 – e (D) 1 4 . Suppose y = f(x) and y = g(x) are two functions whose grahps intersect at three points (0, 4), (2, 2) and (4, 0) with f(x) > g(x) for 0 < x < 2 and f(x) < g(x) for 2 < x < 4. 44 If [f(x) g(x)]dx 10 and [g(x) f(x)]dx 5 , the area between two curves for 0 < x < 2, is - 02 (A) 5 (B) 10 (C) 15 (D) 20 5 . The area bounded by the curves y = – x and x = – y where x, y 0 (A) cannot be determined (B) is 1/3 (C) is 2/3 (D) is same as that of the figure bounded by the curves y = x ; x 0 and x = y ; y 0 6 . The area of the closed figure bounded by y = x, y = –x & the tangent to the curve y = x2 5 at the point (3, 2) is - (A) 5 (B) 2 5 (C) 10 5 (D) 2 7 . The area of the region(s) enclosed by the curves y = x2 and y = | x| is - (A) 1/3 (B) 2/3 (C) 1/6 (D) 1 8 . The area bounded by the curve y = xe–x ; xy = 0 and x = c, where c is the x-coordinate of the curve's inflection point, is - (A) 1 – 3e–2 (B) 1 – 2e–2 (C) 1 – e–2 (D) 1 3 3 9 . The area enclosed by the curves y = cos x, y = 1 + sin2x and x = as x varies from 0 to , is - 22 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 3 3 3 3 (A) – 2 (B) (C) 2 + (D) 1 + 2 2 2 2 1 0 . The area enclosed by the curve y2 + x4 = x2 is - 2 4 8 10 (A) (B) (C) (D) 3 3 3 3 1 1 . Consider two curves C : y = 1 and C : y = nx on the xy plane. Let D denotes the region surrounded by 1x 2 1 C , C and the line x = 1 and D denotes the region surrounded by C , C and the line x = a. If D = D then the 12 2 12 12 value of 'a' - e (B) e (C) e – 1 (D) 2(e – 1) (A) 2 1 2 . The area of the region for which 0 < y < 3 – 2x – x2 & x > 0 is - 3 3 1 3 (A) (3 2x x2 )dx (B) (3 2x x2 )dx (C) (3 2x x2 )dx (D) (3 2x x2 )dx 1 0 0 1 1 3 . The area bounded by the curves y = x(1 – nx) and positive x-axis between x = e–1 and x = e is - E 11
JEE-Mathematics e2 4e2 e2 5e 2 4e2 e2 5e2 e 2 (A) 5 (B) 4 (C) 5 (D) 4 1 4 . The curve f(x) = Ax2 + Bx + C passes through the point (1, 3) and line 4x + y = 8 is tangent to it at the point (2, 0). The area enclosed by y = f(x), the tangent line and the y-axis is - (A) 4/3 (B) 8/3 (C) 16/3 (D) 32/3 1 5 . Let y = g(x) be the inverse of a bijective mapping f : R Rf(x) = 3x3 + 2x. The area bounded by graph of g(x), the x-axis and the ordinate at x = 5 is - 5 7 9 13 (A) (B) (C) (D) 4 4 4 4 dy 1 6 . A function y = f(x) satisfies the differential equation, – y = cos x – sin x, with initial condition that y is dx bounded when x . The area enclosed by y = f(x), y = cosx and the y-axis in the 1st quadrant is- (A) 2 –1 (B) 2 (C) 1 1 (D) 2 SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1 7 . Let 'a' be a positive constant number. Consider two curves C : y = ex, C : y = ea – x. Let S be the area of the 12 part surrounding by C , C and the y-axis, then - 12 (A) Lim S 1 S1 a (B) Lim aa 0 2 4 (C) Range of S is [0,) (D) S(a) is neither odd nor even 1 8 . Area enclosed by the curve y = sinx between x =2n to x = 2(n+1) is- 2 /2 (D) 4 (A) sin x dx (B) 2 sin x dx (C) 4 sin x dx 0 0 0 1 9 . If (a, 0) & (b,0) [a,b > 0] are the points where the curve y = sin2x – 3 sinx cuts the positive x-axis first & second time, A & B are the areas bounded by the curve & positive x-axis between x=0 to x=a and x = a to x=b respectively, then - (A) 4A + 8 cosa = 7 (B) AB 1 (C) 4A + 4B + 14cosb = 0 (D) B – A = 4cos a 16 2 0 . For which of the following values of m, is the area of the region bounded by the curve y = x – x2 and the line y = mx equals to 9/2 ? (A) –4 (B) –2 (C) 2 (D) 4 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 2 1 . Let f(x) = |x|– 2 and g(x) = |f(x)|. Now area bounded by x-axis and f(x) is A and area bounded by x-axis and g(x) is A then – 12 (A) A = 3 (B) A = A (C) A = 4 (D) A + A = 8 1 12 2 12 CHECK YOUR GRASP ANSWER KEY EXERCISE-1 Que. 1 2 3 4 5 6 7 8 9 10 Ans. C D C C B A B A C B Que. 11 12 13 14 15 16 17 18 19 20 Ans. B C B B D A A,B,C,D B,C,D A,B,C,D B,D Que. 21 Ans. B,C,D 12 E
JEE-Mathematics EXERCISE - 02 BRAIN TEASERS SELECT THE CORRECT ALTERNATIVES (ONE OR MORE THAN ONE CORRECT ANSWERS) 1. 1 x2 are two curves lying in the XY plane. Then - If C y = and C y = 1 1 x2 2 2 (A) area bounded by curve C and y = 0 is 1 1 (B) area bounded by C and C is 1 2 23 (D) area bounded by curve C and x-axis is (C) area bounded by C and C is 1 – 12 2 12 2 . Area enclosed by the curves y = nx ; y = n|x| ; y = |nx| and y = |n|x|| is equal to - (A) 2 (B) 4 (C) 8 (D) cannot be determined 3 . y = f(x) is a function which satisfies- (i) f(0) = 0 (ii) f''(x) = f'(x) and (iii) f'(0) = 1 then the area bounded by the graph of y = f(x), the lines x = 0, x – 1 = 0 and y + 1 = 0, is - (A) e (B) e – 2 (C) e – 1 (D) e + 1 4 . Let T be the triangle with vertices (0, 0), (0, c2) and (c, c2) and let R be the region between y = cx and y = x2 where c > 0 then - c3 c3 Area(T) 3 Area(T) 3 (A) Area (R) = (B) Area of R = Lim Lim (C) c0 Area(R ) (D) c0 Area(R ) 2 6 3 5 . Suppose g(x) = 2x + 1 and h(x) = 4x2 + 4x + 5 and h(x) = (fog)(x). The area enclosed by the graph of the function y = f(x) and the pair of tangents drawn to it from the origin, is - (A) 8/3 (B) 16/3 (C) 32/3 (D) none 6 . Let f(x) = x2 + 6x + 1 and R denote the set of points (x, y) in the coordinate plane such that f(x) + f(y) 0 and f(x) – f(y) 0. The area of R is equal to - (A) 16 (B) 12 (C) 8 (D) 4 x1 7 . The value of 'a' (a > 0) for which the area bounded by the curves y = 6 x2 , y = 0, x = a and x = 2a has the least value, is - (A) 2 (B) 2 (C) 21/3 (D) 1 8 . Consider the following regions in the plane : R = {(x, y) : 0 x 1 and 0 y 1} and R = {(x, y) : x2 + y2 4/3} 12 of can a 3 b The area the region R R be expressed as , where a and b are integers, then - 1 2 9 (A) a = 3 (B) a = 1 (C) b = 1 (D) b = 3 1 9. The area of the region of the plane bounded by (| x| ,| y| ) 1 & xy is - 2 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 (A) less then 4n 3 15 (C) 2 + 2n2 (D) 3 + n2 (B) 4 1 0 . The line y = mx bisects the area enclosed by the curve y = 1 + 4x – x2 & the lines x = 0, x = 3 & y = 0. Then 2 the value of m is - 13 6 3 (D) 4 (A) (B) (C) 6 13 2 1 1 . Area of the region enclosed between the curves x = y2 – 1 and x = |y| 1 y2 is - (A) 1 (B) 4/3 (C) 2/3 (D) 2 1 2 . If the tangent to the curve y = 1 – x2 at x = , where 0 < < 1, meets the axes at P and Q. As varies, the minimum value of the area of the triangle OPQ is k times the area bounded by the axes and the part of the curve for which 0 < x < 1, then k is equal to - 2 75 25 2 (A) (B) (C) (D) 3 16 18 3 BRAIN TEASERS ANSWER KEY EXERCISE-2 Que. 1 2 3 4 5 6 7 8 9 10 11 12 Ans. A,B B C A,C B C D A,C A,D A D A E 13
JEE-Mathematics EXERCISE - 03 MISCELLANEOUS TYPE QUESTIONS MATCH THE COLUMN Following questions 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 statement in Column-I can have correct matching with ONE statement in Column-II. 1 . Let f(x) = |x|, g(x) = |x – 1| and h(x) = |x + 1|. Column-I Column-II (A) Area bounded by min (f(x), g(x)) and x-axis is (B) Area bounded by min (f(x), h(x)) and x-axis is 1 (C) Area bounded by min ((f(x), g(x), h(x)) and x-axis is (p) sq. unit 1 8 (D) Area bounded by min (f(x), g(x), h(x)) and y = 2 is 1 (q) sq. unit 4 1 (r) sq. unit 2 3 (s) 4 sq. unit 2. Column I Column – II (A) 1 (B) The area bounded by the curve x = 3y2 – 9 and the lines (p) 4 x = 0, y = 0 and y = 1 in square units is equal to (C) 8 (D) If a curve ƒ (x) = a x bx , (ƒ (x) 0 x [0, 9]) passes through (q) 5 the point (1, 2) and the area bounded by the curve, line x = 4 (r) and x-axis is 8 square unit, then 2a + b is equal to (s) The area enclosed between the curves y = sin2x and y = cos2x in the interval 0 x in square units in equal to The area bounded by the curve y2 = 16x and line y = mx is 2 3 square units, then m is equal to 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. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 (D) Statement-I is false, Statement-II is true. 1 . Statement-I : The area of the curve y = sin2x from 0 to will be more than that of curve y = sinx from 0 to . Because Statement-II : t2 > t if t R – [0, 1]. (A) A (B) B (C) C (D) D 2 . Statement-I : The area bounded by the curves y = x2 – 3 and y = kx + 2 is the least, if k = 0. Because Statement-II : The area bounded by the curves y = x2 – 3 and y = kx + 2 is k2 20 . (A) A (B) B (C) C (D) D 3 . Consider the two curves y = x – x2 and y = x2 , ( > 0). Statement-I : The area bounded between the curves is maximum when = 1. Because 2 Statement-II : The area bounded between the curves is (1 2 )2 square units. (A) A (B) B (C) C (D) D 14 E
JEE-Mathematics COMPREHENSION BASED QUESTIONS Comprehension # 1 11 Consider two curves y and y = 4(x 1) . 'a' is a number such that a > 2 & the reciprocal of the area x2 of the figure bounded by the curves, the line x = 2 & x = a is a itself. 'b' is a number such that 1 < b < 2 & the area bounded by the two curves & the lines x = b & x = 2 is equal to 1 1 . b On the basis of above information, answer the following questions : 1 . The value of na – nb is - (A) positive integer (B) negative integer p (D) irrational number (C) rational number of the form q , where p, q are co-prime & q > 1. n(a 1) 0 2. If A = n(b 1) , then A–1 is - 0 A (B) A (C) 4A A (A) – (D) 4 4 3 . If z is a complex number such that z = n(a – 1) + in(b – 1) then arg(z) is - 3 3 (A) (B) (C) (D) 4 4 4 4 Comprehension # 2 Five curves defined as follows : C1 : |x + y| 1 C2 : |x – y| 1 1 C3 : |x| 2 C : |y| 1 4 2 C5 : 3x2 + 3y2 = 1 On the basis of above information, answer the following questions : 1 . The area bounded by C1 and C2 which does not contain the area bounded by C5, is - (D) 2 (A) 2 – (B) 2 – (C) 2 – NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 4 6 3 2 . That part of area of curve C5 which does not contain points satisfying C3 and C4, is - 1 1 (D) 2 1 (A) (B) 1 (C) 9 3 32 3 36 3 . That part of area which is bounded by C1 and C2 but not bounded by C3 and C4, is - (A) 1 1 1 (D) none of these (B) (C) 2 3 MISCELLANEOUS TYPE QUESTION ANSWER KEY EXERCISE-3 Match the Column 2 . (A) (r), (B) (s), (C) (p), (D) (q) 1. (A) (q), (B) (q), (C) (r), (D) (s) 3. C Assertion & Reason 3. A 1. D 2. C 3. C Comprehension Based Quesions Comprehension # 1 : 1. A 2. D Comprehension # 2 : 1. C 2. D E 15
JEE-Mathematics EXERCISE - 04 [A] CONCEPTUAL SUBJECTIVE EXERCISE 1 . Find the area bounded by y = cos–1x, y = sin–1x and y-axis. 2 . Find the value of c for which the area of the figure bounded by the curves y = sin 2x, the straight lines x = /6, x = c & the abscissa axis is equal to 1/2. 3 . Find the area of the region {(x , y) : 0 y x² + 1, 0 y x + 1, 0 x 2}. 4. A figure is bounded by the curves y = 2 sin x , y = 0, x=2 & x = 4. At what angles to the positive 4 x-axis straight lines must be drawn through (4 , 0) so that these lines divide the figure into three parts of the same size. 5 . Find the area bounded by the curves y = 1 x2 and y = x3 x. Also find the ratio in which the y-axis divideds this area. 6 . Find the area bounded on the right by the line x + y = 2, on the left by the parabola y = x2 and below by the x- axis. 7 . The tangent to the parabola y = x2 has been drawn so that the abscissa x of the point of tangency belongs 0 to the interval [1, 2]. Find x for which the triangle bounded by the tangent, x-axis & the straight line y = x20 0 has the greatest area. 8. Find the area of the region bounded by the curves , y = log x , y = sin4 x & x = 0. e 9 . For what value of 'a' is the area bounded by the curve y = a2x2 + ax + 1 and the straight lines y = 0, x = 0 & x = 1 the least ? 1 0 . The line 3x + 2y = 13 divides the area enclosed by the curve, 9x2 + 4y2 18x 16y 11 = 0 in two parts. Find the ratio of the larger area to the smaller area. 1 1 . Find the values of m (m > 0) for which the area bounded by the line y = mx + 2 and the curve x = 2y y2 is , (i) 9/2 square units & (ii) minimum. Also find the minimum area. 1 2 . Let f(x) = Maximum {x2, (1 – x)2, 2x(1 – x)}, where 0 x 1. Determine the area of the region bounded by the curves y = f(x), x-axis, x = 0 & x = 1. 1 3 . If the area enclosed by the parabolas y = a – x2 and y = x2 is 18 2 sq. units. Find the value of 'a'. 1 4 . Find the area enclosed between the curves : y loge (x e) , x loge (1 / y) & the x-axis. 1 5 . Find the value(s) of the parameter 'a' (a > 0) for each of which the area of the figure bounded by the straight line y a2 ax & the parabola y x2 2ax 3a2 is the greatest. NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 1 a4 1 a4 1 6 . Find the positive value of 'a' for which the parabola y = x2 + 1 bisects the area of the rectangle with vertices (0, 0), (a, 0), (0, a2 + 1) and (a, a2 + 1). 1 7 . Find the area bounded by y = x + sinx and its inverse between x = 0 and x = 2. CONCEPTUAL SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(A) 1 . 4 . tan 1 2 2 ; tan 1 4 2 2 2 sq.units 2. c = 6 or 3 3 . 23/6 sq. units 3 3 5. sq.units ; 1 6. 5/6 sq. units 7 . x = 2, A(x ) = 8 8 . 11 sq. units 9 . a = 3/4 2 1 00 8 3 2 1 1 . (i) m = 1, (ii) m = ; A = 4/3 sq.units 12. 17 / 27 1 3 . a = 9 1 4 . 2 sq. units 10. 2 min 15. a = 31/4 1 6 . 3 17. 8 16 E
EXERCISE - 04 [B] JEE-Mathematics BRAIN STORMING SUBJECTIVE EXERCISE 1 . For what value of 'a' is the area of the figure bounded by 1 1 , x = 2 & x = a equal to n 4 ? y= , y= x 2x 1 5 y 2 . Let C & C be two curves passing through the origin as shown in the figure. C2 12 C A curve C is said to \"bisect the area\" the region between C & C , if for each point 12 A C1 P of C, the two shaded regions A & B shown in the figure have equal areas. B Determine the upper curve C , given that the bisecting curve C has the equation O x 2 y = x2 & that the lower curve C has the equation y = x2/2. 1 3. Let A be the area bounded by the curve y = (tan x)n & the lines x = 0, y = 0 & x = /4. Prove that for n n>2, A + A = 1/(n 1) & deduce that 1/(2n + 2) < A< 1/(2n 2). n n2 n 4 . Consider a square with vertices at (1, 1), (–1, 1), (–1, –1) & (1, –1). Let S be the region consisting of all points inside the square which are nearer to the origin than to any edge. Sketch the region S & find its area. 5 . In what ratio does the x-axis divide the area of the region bounded by the parabolas y = 4x – x2 and y = x2 – x ? 6 . A polynomial function f(x) satisfies the condition f(x + 1) = f(x) + 2x + 1. Find f(x) if f(0) = 1. Find also the equations of the pair of tangents from the origin on the curve y = f(x) and compute the area enclosed by the curve and the pair of tangents. 7 . Consider the curve y = xn where n > 1 in the 1st quadrant. If the area bounded by the curve, the x-axis and the tangent line to the graph of y = xn at the point (1, 1) is maximum then find the value of n. 8 . Consider the collection of all curve of the form y = a – bx2 that pass through the point (2, 1), where a and b are positive constants. Determine the value of a and b that will minimise the area of the region bounded by y = a – bx2 and x-axis. Also find the minimum area. 9 . Show that the area bounded by the curve y = nx c , the x-axis and the vertical line through the maximum x point of the curve is independent of the constant c. Also find the area. 10. Let f(x) be a continuous function given by f(x) 2 x for | x| 1 . Find the area of the region in the for | x| 1 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 x 2 ax b third quadrant bounded by the curves, x = –2y2 and y = f(x) lying on the left of the line 8x + 1= 0 [JEE 99, 10M (out of 200)] BRAIN STORMING SUBJECTIVE EXERCISE ANSWER KEY EXERCISE-4(B) 2 1 1. 2 . (16/9) x2 4 . 3 16 2 20 sq. units a = 8 or 6 21 5. 5 8. 4 : 121 2 7. 2 1 E 6 . f(x) = x2 + 1; y = ± 2x ; A = sq. units 3 b = 1/8, A minimum = 4 3 sq. units 9 . 1/2 sq. units 10. 257/192 ; a = 2 ; b = –1 17
JEE-Mathematics JEE-[MAIN] : PREVIOUS YEAR QUESTIONS EXERCISE - 05 [A] 1 . If the area bounded by the x-axis, curve y = f(x) and the lines x = 1, x = b is equal to b2 1 – 2 for all b > 1, then f(x) is- [AIEEE-2002] (1) (x 1) (2) (x 1) (3) (x2 1) x (4) 1 x2 2 . The area of the region bounded by the curves y = |x – 1| and y = 3 – |x| is - [AIEEE-2003] (1) 6 sq. units (2) 2 sq. units (3) 3 sq. units (4) 4 sq. units 3 . The area of the region bounded by the curves y = |x – 2|, x = 1, x = 3 and the x-axis is- [ A IE E E -2 0 0 4 ] (1) 1 (2) 2 (3) 3 (4) 4 4. Area of the greatest rectangle that can be inscribed in the ellipse x2 y2 =1 is - + [AIEEE-2005] a2 b2 (1) 2ab (2) ab (3) ab a (4) b 5 . The area enclosed between the curve y = loge(x+e) and the cooordinate axes is- [AIEEE-2005] (1) 1 (2) 2 (3) 3 (4) 4 6 . The parabolas y2 = 4x and x2 = 4y divide the square region bounded by the lines x = 4, y = 4 and the coordinate axes. If S1, S2, S3 are respectively the areas of these parts numbered from top to bottom; then S1 : S2 : S3 is - [AIEEE-2005] (1) 1 : 2 : 1 (2) 1 : 2 : 3 (3) 2 : 1 : 2 (4) 1 : 1 : 1 7 . Let f(x) be a non-negative continuous function such that the area bounded by the curve y= f(x), x-axis and the ordinates x = and x = > is sin cos 2 . Then is - [AIEEE-2005] 4 4 4 f 2 (1) 2 1 (2) 2 1 (3) 1 2 (4) 1 2 4 4 4 4 8 . The area enclosed between the curves y2 = x and y = |x| is- [AIEEE-2007] NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 2 (2) 1 1 1 (1) 3 (3) 6 (4) 3 9 . The area of the plane region bounded by the curves x + 2y2 = 0 and x + 3y2 = 1 is equal to- [ A IE E E -2 0 0 8 ] 5 1 2 4 (1) 3 (2) 3 (3) 3 (4) 3 1 0 . The area of the region bounded by the parabola (y – 2)2 = x – 1, the tangent to the parabola at the point (2, 3) and the x–axis is :- [AIEEE-2009] (1) 9 (2) 12 (3) 3 (4) 6 3 1 1 . The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and x = is - 2 [AIEEE-2010] (1) 4 2 – 2 (2) 4 2 + 2 (3) 4 2 – 1 (4) 4 2 + 1 18 E
JEE-Mathematics 1 2 . The area of the region enclosed by the curves y = x, x = e, y = 1 and the positive x-axis is - [AIEEE-2011] x 3 5 1 (4) 1 square units (1) square units (2) square units (3) square units [AIEEE-2011] 2 2 2 1 3 . The area bounded by the curves y2 = 4x and x2=4y is:= (1) 0 32 16 8 (2) 3 (3) 3 (4) 3 1 4 . The area bounded between the parabolas x2 y and x2 = 9y, and the straight line y = 2 is : [AIEEE-2012] 4 (1) 10 2 (2) 20 2 10 2 20 2 (3) (4) 3 3 1 5 . The area (in square units) bounded by the curves y x , 2y – x + 3 = 0, x-axis and lying in the first quadrant is : [JEE (Main)-2013] (1) 9 (2) 36 (3) 18 27 (4) 4 NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 PREVIOUS YEARS QUESTIONS ANSWER KEY EXERCISE-5 [A] Qu e. 1 2 3 4 5 6 7 8 9 10 1 1 1 2 13 1 4 15 Ans 4 4 1 1 1 4 4 3 4 1 1 1 3 4 1 E 19
JEE-Mathematics EXERCISE - 05 [B] JEE-[ADVANCED] : PREVIOUS YEAR QUESTIONS 1 . The triangle formed by the tangent to the curve f(x) = x2 + bx – b at the point ( 1 ,1) and the coordinates axes, lies in the first quadrant. If its area is 2, then the value of b is - [JEE 2001] (A) –1 (B) 3 (C) –3 (D) 1 2 . The area bounded by the curves y = |x| – 1 and y = –|x|+ 1 is - [JEE 2002 (Screening), 3M] (A) 1 (B) 2 (C) 2 2 (D) 4 3 . Find the area of the region bounded by the curves y = x2 , y = |2 – x2| and y = 2 , which lies to the right of the line x = 1. [JEE 2002, (Mains) 5M out of 60] x2 y2 4 . ( a ) The area of the quadrilateral formed by the tangents at the end points of latus recta to the ellipse 1 , 95 is - (A) 27/4 sq. units (B) 9 sq. units (C) 27 sq. units (D) 27/2 sq. units ( b ) The area bounded by the curves y x , 2y + 3 = x and x-axis in the 1st quadrant is - (A) 18 (B) 27/4 (C) 36 (D) 9 [JEE 2003 (Screening), 3 + 3M] 5 . ( a ) The area bounded by the angle bisectors of the lines x2 – y2 + 2y = 1 and the line x + y = 3, is - (A) 2 (B) 3 (C) 4 (D) 6 ( b ) The area enclosed between the curves y = ax2 and x = ay2 (a > 0) is 1 sq. unit, then the value of a is - 1 1 (C) 1 1 (A) (B) (D) 3 3 2 [JEE 2004 (Screening), 3+3M] 6 . The area bounded by the parabolas y = (x + 1)2 and y = (x – 1)2 and the line y = 1/4 is - (A) 4 sq. units (B) 1/6 sq. units (C) 4/3 sq. units (D) 1/3 sq. units 7 . Find the area bounded by the curves x2 = y, x2 = –y and y2 = 4x – 3. [JEE 2005 (Screening), 3M] [JEE 2005, (Mains), 4M] ML OP L O L O4a2 4a 1 f(1) NODE6\\E\\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Maths\\Unit#08\\Eng\\01. Area Under the curve.p65 3a2 3a MM PP MM PP MM PP8 . If 4b2 4b 1 f(1) = 3b2 3b , f(x) is a quadratic function and its maximum value occurs at a point MN PQ NM PQ NMM PPQ4c2 4c 1 f(2) 3c2 3c V. A is a point of intersection of y = f(x) with x-axis and point B is such that chord AB subtends a right angle at V. Find the area enclosed by f(x) and chord AB. [JEE 2005 (Mains) 6M out of 60] 9 . Match the following - /2 (p) 1 (q) 0 (A) (sin x )cos x (cos x cot x n(sin x)sin x )dx 0 (B) Area bounded by – 4y2 = x and x – 1 = –5y2 (C) Cosine of the angle of intersection of curves (r) 6 n 2 y = 3x–1 ln x and y = xx – 1 is dy 6 (s) 4/3 (D) Let dx x y , where y (0) = 0, then the value of [JEE 2006, 6M] y when x + y = 6 is E 20
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