Pioneer Education The Best Way To Success NTSE | Olympiad | AIPMT | JEE - Mains & Advanced JEE Mains Paper With Solutions PART A – MATHEMATICS Straight Objective Type This section contains 30 multiple choice questions. Each question has 4 choices (a), (b), (c) and (d) for its answer, out of which Only One is correct. 1. It S is the set of distinct values of ‘b’ for which the following system of linear equations x + y + z = 1 x + ay + z = 1 ax + by + z = 0 has no solution, then S is: (a) an empty set (b) an infinite set (c) a finite set containing two or more elements (d) a singleton Ans. (d) Solution: 1 1 1 D = 1 a 1  a12 0  a = 1 a b 1 For a = 1 we have first two planes co-incident x + y + z = 1 ax + by + z = 0 For no solution these two are parallel 1 1 1   a 1,b1 a b 1 2. The following statement pq  [~p q q] is: (a) a tautology (b) equivalent to ~p q (c) equivalent to p ~q (d) a fallacy Ans. (a) Solution: pq[~pqq] pqpqq pq~p~qq pq~p q(~qq) (p q) pqwhich is tautology Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 1 of 40 www.pioneermathematics.com Pioneer Education The Best Way To Success NTSE | Olympiad | AIPMT | JEE - Mains & Advanced 3. If 5(tan2 x – cos2x) = 2 cos 2x + 9, then the value of cos4x is: 3 1 2 7 (a) (b) (c) (d)  5 3 9 9 Ans. (d) Solution: 5(tan2x – cos2x) = 2cos 2x + 9  1  1t2  5t2 2 9  1t2  1t2  5(t4 + t2 – 1) = 2 – 2t2 + 9 + 9t2 5t4 – 2t2 – 16 =0 5t4 – 10t2+ 8t2 – 16 = 0 5t2(t2 – 2) + 8(t2 – 2) =0 (5t2 + 8) (t2 – 2) = 0 tan2x = 2 12 1 cos 2x =  12 3 7 cos 4x = 2 cos2 2x – 1 = – 9 4. For three events A, B and C, P (Exactly one of A or B occurs) = P(Exactly one of B or C occurs) 1 1 = P(Exactly one of C or A occurs) = and P(All the three events occurs simultaneously) = . 4 16 Then the probability that at least one of the events occurs, is: 7 7 7 3 (a) (b) (c) (d) 32 16 64 16 Ans. (b) Solution: 1 P(A) + P(B) – 2PAB 4 1 PBPC2PBC 4 1 PAPC2PAC 4 1 PABC 16 3 1 61 7 P(A) + P(B) + P(C) –P(A  B) – P(B  C) –P(A C) + P(A  B  C) =    8 16 16 16 1 1 1 5. Let be a complex number such that 21z where z  3 . If 1 21 2 3k, then k is 1 2 7 equal to: (a) – z (b) z (c) – 1 (d) 1 Ans. (a) Solution: Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 2 of 40 www.pioneermathematics.com Pioneer Education The Best Way To Success NTSE | Olympiad | AIPMT | JEE - Mains & Advanced z i 3  21 3 i 2 3i1 1i 3   12 0 & 3 1 2 1 1 1 3 1 1  1  2 3k  0  2 3k 1 2  0 2  3243k 1i 3 1i 3 i 3 i 3 k2      3 z 2 2 2 2 6. Let k be an integer such that triangle with vertices (k, – 3k), (5, k) and (–k, 2) has area 28 sq. units. Then the orthocentre of this triangle is at the point:  1  3  3  1 (a) 2, (b) 1, (c) 1, (d) 2,          2  4  4  2 Ans. (d) Solution: k 3k 1 5 k 1  56 k 2 1 k(k – 2) – 5(–3k – 2) – k(–3k – k) =  56 k2 – 2k + 15k + 10+ 3k2 + k2 =  56 5k2 + 13k + 10  56 = 0 5k2 + 13k + 66 = 9 or 5k2 + 13k – 46 = 0 13 169920 No solution or k 10 1333 46 k   k2 or k (which is not an integer) 10 10  vertices A(2, – 6), B(5, 2), C(–2, 2) Equation of altitude from vertex A is x = 2 …(i) Equation of altitude dropped from vertex C is 3x+ 8y – 10 = 0 …(ii) solving both (i) and (ii)  1 orthocentre 2,    2 7. Twenty meters of wire is available for fencing off a flower-bed in the form of a circular sector. Then the maximum area (in sq. m) of the flower-bed, is: (a) 12.5 (b) 10 (c) 25 (d) 30 Ans. (c) Solution: Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 3 of 40 www.pioneermathematics.com Pioneer Education The Best Way To Success NTSE | Olympiad | AIPMT | JEE - Mains & Advanced 202r 2r 20  2rr 20   r r2 r2 202r A   . r10r 360 2 r A = 10r – r2 dA 102r0 r5 dr 10   2 15 1  Maximum area = 25225sq.m. 2 8. The area (in sq. units) of the region {(x, y): x  0, x + y  3, x2  4y and y 1 x } is: 59 3 7 5 (a) (b) (c) (d) 12 2 3 2 Ans. (d) Solution: y1 x y12 x Required area = 11 xdx  23xdx  2x2dx 0 1 0 4 1 2 2  2x3/2   x2   x3  2  5 8 = x  3x     1 62   3   7  12 3  2 12 0 1 0 2  5 2 3 5 = 1  4  1    3  2 3 2 2 9. If the image of the point P(1, –2, 3) in the plane, 2x + 3y – 4z + 22 = 0 measured parallel to the line, x y z   is Q, then PQ is equal to: 1 4 5 (a) 3 5 (b) 2 42 (c) 42 (d) 6 5 Ans. (b) Solution: Let R be the point of intersection of plane and line passing through P and parallel to given line. Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 4 of 40 www.pioneermathematics.com Pioneer Education The Best Way To Success NTSE | Olympiad | AIPMT | JEE - Mains & Advanced So, R is 1,24,35 substituting co-ordinates of R in plane 226 121220220  6 6  1 So, R is (2, 2, 8) Hence PR  11625  42 So, PQ = 2 42  1  6x x  10. If for x0,4, the derivative of tan119x3 is x.gx,then g(x) equals: 9 3x x 3x 3x (a) (b) (c) (d) 19x3 19x3 19x3 19x3 Ans. (a) Solution:   3x x 3x x    Let ytan1 2tan1 3x x   2   1 3x x    dy 1 3  9  9 2 3  x  x gx   dx 19x3 2 19x3  19x3 dy  11. If 2sinx + (y + 1) cos x = 0 and y(0) = 1, then y is equal to:   dx 2 1 2 1 4 (a) (b)  (c)  (d) 3 3 3 3 Ans. (a) Solution: dy y1cosx  dx 2sin x dy cosx    dx y1 2sin x ny1 n2sinxc (y + 1) (2 + sin x) = A; for x= 0, y = 1  A = 4  (y +1) (2 + sin x) = 4  1 for x  y 2 3 12. Let a vertical tower AB have its end A on the level ground. Let C be the mid-point of AB and P be a point on the ground such that AP = 2AB. If BPC,then tan is equal to 6 1 2 4 (a) (b) (c) (d) 7 4 9 9 Ans. (c) Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 5 of 40 www.pioneermathematics.com Pioneer Education The Best Way To Success NTSE | Olympiad | AIPMT | JEE - Mains & Advanced Solution: 1 1 tan , tan  ,tany 2 4 tantan tan 1tan tan 1 y 1 1 14y 4    2 y 2 4y 1 4 4 – y = 2 + 8y 2 y 9  2 3 13. If A = , then adj (3A2 + 12A) is equal to   4 1   72 84 51 63 51 84  72 63 (a) (b) (c) (d)         63 51  84 72 63 72 84 51  Ans. (b) Solution:  2 3 A    4 1  A2 – 3A – 10I = 0 A2 = 3A + 10I 3A2 + 12A = 3(3A + 10I) + 12A = 21A + 30I  42 63 30 0   72 63 21A  30I          84 21   0 30 84 51  51 63 adj 21A  30I    84 72 14. For any three positive real numbers a, b and c, 9(25a2 + b2) + 25(c2 – 3ac) = 15b(3a + c), Then (a) b, c and a are in G.P. (b) b, c and a are in A.P. (c) a, b and c are in A.P. (d) a, b and c are in G.P. Ans. (b) Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 6 of 40 www.pioneermathematics.com Pioneer Education The Best Way To Success NTSE | Olympiad | AIPMT | JEE - Mains & Advanced Solution: 225a2 + 9b2 + 25c2 – 75ac – 45ab – 15bc = 0 (15a)2 + (3b)2 + (5c)2 – (15a) (3b)– (3b) (5c) – (15a) (5c) = 0 1 [(15a – 3b)2 + (3b – 5c)2 + (5c – 15a)]2 = 0 2 15a = 3b, 3b = 5c, 5c = 15a 5a = b, 3b = 5c, c = 3a a b c    1 5 3 a = ,b5,c3 a,c,bare inA.P or b, c, a are in A.P 15. The distance of the point (1, 3, – 7) from the plane passing through the point (1, –1, –1), having normal x1 y2 z4 x2 y1 z7 perpendicular to both the lines   and   , is 1 2 3 2 1 1 20 10 5 10 (a) (b) (c) (d) 74 83 83 74 Ans. (b) Solution: Let the plane be a(x – 1) + b(y + 1) + c(z + 1)= 0 a – 2b + 3c = 0 2a – b – c = 0 a b c   5 7 3 5(x – 1) + 7(y + 1) + 3(z + 1) = 0 5x + 7y + 3z + 5 = 0 P(1, 3, – 7) 521215 10 d  25499 83 16. Let I  tannxdx, n 1. If I I a tan5xbx5C,where C is a constant of integration, then the n 4 6 ordered pair (a, b) is equal to  1  1  1   1  (a)  ,1 (b) ,0 (c) ,1 (d)  ,0          5  5  5   5  Ans. (b) Solution: I  tannxdx  tann2(sec2x1)dx n (tanx)n1 (tan x)n2sec2xdx(tan x)n2dx I n1 n2 (tan x)n1 I I  n n2 n1 put n = 6 Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 7 of 40 www.pioneermathematics.com Pioneer Education The Best Way To Success NTSE | Olympiad | AIPMT | JEE - Mains & Advanced 1 1 I I  tan5xatan5xbx5c  a b0 c0 4 6 5 5 1   a,b ,0   5  1 17. The eccentricity of an ellipse whose centre is at the origin is . If one of its directrices is x = - 4, then 2  3 the equation of the normal to it at 1, is    2 (a) 2y – x = 2 (b) 4x – 2y = 1 (c) 4x + 2y – 7 (d) x + 2y = 4 Ans. (b) Solution: a 4 e a = 4e = 2 b2 e2 = 1 – a2 1 4b2  4 4 b 3 x2 y2  1 4 3 3 equation of normal at (1, ) 2 a2x b2y  a2b2 x y 1 1 4x 3y  43 1 3/2 4x – 2y = 1 18. A hyperbola passes through the point P( 2, 3)and has foci at 2,0.Then the tangent to this hyperbola at P also passes through the point: (a) (3 2,2 3) (b) (2 2,3 3) (c) ( 3, 2) (d) ( 2, 3) Ans. (b) Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 8 of 40 www.pioneermathematics.com Pioneer Education The Best Way To Success NTSE | Olympiad | AIPMT | JEE - Mains & Advanced Solution: Here ae = 2 PF ( 22)23  94 2 (2 21) PF’ = ( 22)23 94 2 (2 21)  PFPF' 2  a = 1  e = 2 b2 e2 1 a2  b2 = 3 y2  Equation of hyperbola x2 1 3 or 3x2 – y2 = 3  equation of tangent at ( 2, 3)will be 3 2x 3y3 hence tangent passes through (2 2,3 3)  1 1 x 19. The function f : R   , defined as f(x) = , is:    2 2 1x2 (a) Invertible (b) injective but not surjective (c) surjective but not injective (d) neither injective nor surjective Ans. (c) Solution: x  1 1 fx ;f:R   ,   1x2  2 2 from the graph of f(x) we can observe that function is many one and onto. Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 9 of 40 www.pioneermathematics.com Pioneer Education The Best Way To Success NTSE | Olympiad | AIPMT | JEE - Mains & Advanced cot xcosx 20. lim equals  2x3 x 2 1 1 1 1 (a) (b) (c) (d) 24 16 8 4 Ans. (b) Solution:  x t 2 tan tsin t lim t0 (2t)3 sint(1cost) 1 lim  t0 8t3cost 16   21. Let a 2i j2kand bi j. Let c be a vector such that ca 3, ab c 3 and the angle between cand ab be 300. Then a. c is equal to 25 1 (a) (b) 2 (c) 5 (d) 8 8 Ans. (b) Solution: a 2ij2k, bij, ca 3    ab c 3, c ab 6 Now |ab||c|sin300 3,|ab||c|6  a b c sin6, a b 21  a 3, b 2 cos1    3 2  4 6 c  . 22 3 2 ca 3 2 c 2 2 Squaring, we get c 2a.c  a 9  a.c  2 2 22. The normal to the curve y(x – 2) (x – 3) = x + 6 at the point where the curve intersects the y-axis passes through the point:  1 1 1 1 1 1 1 1 (a)  , (b) , (c) , (d) ,          2 2 2 2 2 3 2 3 Ans. (b) Solution: y(x – 2) (x – 3) = x + 6 Intersection with y-axis; Pioneer Education| SCO 320, Sector 40–D, Chandigarh +91-9815527721, 0172-4617721 Page 10 of 40 www.pioneermathematics.com