41- Question Report (41)

Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 77. Let S = 0 is the locus of centre of a 77. S= 0,  variable circle which intersect the circle x2+ y2 – 4x – 6y = 0  (4,6)  x2 + y2 – 4x – 6y = 0 orthogonally at (4, 6). If    P,S =0  P is a variable point of S = 0, then least    O P    value of OP is (where O is origin) - (O )- (1) 13 (2) 2 13 (1) 13 (2) 2 13 (3) 10 (4) 13 (3) 10 (4) 13 78. Number of integral values of a for which 78. 'a'         smaller root of quadratic equation x2 – 2ax – 4 + a2 = 0 1  x2 – 2ax – 4 + a2 = 0 is smaller than 1 and 6 bigger root is greater than 6 is - (1) 0 (2) 1 (1) 0 (2) 1 (3) 2 (4) infinite values (3) 2 (4)  79. Let a circle S = 0 touches both the circles 79. S=0 x2+y2=400  x2 + y2 = 400 and x2 + y2 – 10x – 24y + 120 = 0 x2 + y2 – 10x – 24y + 120 = 0  externally and also touches x-axis. The x-  S=0  radius of circle S = 0 is -  (1) 200 (2) 33 (3) 120 (4) 240 (1) 200 (2) 33 (3) 120 (4) 240 80. Length of latus rectum of the parabola 80.  9x2 + 16y2 + 24xy – 4x + 3y = 0  9x2 + 16y2 + 24xy – 4x + 3y = 0 is - - 1 1 1 (4) 1 1 1 1 (4) 1 (1) 20 (2) 4 (3) 5 (1) 20 (2) 4 (3) 5 81. P and Q are two points on the parabola 81. P  Q y2 = 8x S y2 = 8x and S is its focus. PS and QS meet the curve again in T and R respectively. If PSQS T PQ passes through a fixed point (–2, 3), RPQ (–2, 3) then TR also passes through a fixed point  TR    whose coordinates are  (1) (2, –3) (2) (3, – 2) (1) (2, –3) (2) (3, – 2) (3) (–2, 3) (4) (–3, 2) (3) (–2, 3) (4) (–3, 2)  1001CT103516012 H-31/34

Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2 82. Slope of common tangents of parabola 82.  (x – 1)2 = 4(y – 2)  (x – 1)2 = 4(y – 2) and ellipse (x –1)2 + (y  2)2  1 (x – 1)2 + (y  2)2  1  2 2 are m1 and m2, then m12  m22 is equal to - m1m2 m12 m22  (1) 2 (2) 3 (3) 4 (4) 6 (1) 2 (2) 3 (3) 4 (4) 6 83. If x, y, z  R+ are such that z > y > x > 1, 83. x, y, z  R+ z> y > x > 1, 5 10 logyx + logxy = 5 logzy + logyz = 10  logyx + logxy = 2 and logzy + logyz = 3 , 2 3 then logxz is equal to - logxz  (1) 2 (2) 3 (3) 6 (4) 12 (1) 2 (2) 3 (3) 6 (4) 12 84. The equation of normal to the curve 84.  y3 + 2xy + x3 = (x – 1)3 (1, –1)  y3 + 2xy + x3 = (x – 1)3 at point (1, –1) is - - (1) 5x + y = 4 (2) 5x – y = 6 (1) 5x + y = 4 (2) 5x – y = 6 (3) x + 5y + 4 = 0 (4) x – 5y = 6 (3) x + 5y + 4 = 0 (4) x – 5y = 6 x2 x2 85. Let ƒ(x) =  (t 1)(t  4)(t  9)dt , then 85. ƒ(x) =  (t 1)(t  4)(t  9)dt  00 (1) ƒ''(x) = 0 have 4 distinct positive solutions (1) ƒ''(x) = 0  (2) ƒ'''(x) = 0 have 2 distinct positive solutions (2) ƒ'''(x) = 0  (3) ƒ'''(x) = 0 have 3 distinct positive solutions (3) ƒ'''(x) = 0  (4) ƒ(x) have 6 critical points. (4) ƒ(x)  86. Let ƒ(x) = a sin( x  b) x0 86. x 6x7  x  1 is differentiable a sin( x  b) x0 x0 6x7  x  1 x0 for all real x. If a  R and b [0,2] , ƒ(x) =    then number of ordered pair(s) of (a, b) a  R b[0,2](a,b)  is -  (1) 1 (2) 2 (1) 1 (2) 2 (3) 4 (4) more than 4 (3) 4 (4) 4  H-32/34 1001CT103516012

Leader Course/Phase-III, IV & V/18-12-2016/Paper-2 87. Number of solutions of the equation 87. [0,5] 2tan–1(cos2x) = tan–1(2cosec2x) in [0,5] is 2tan–1(cos2x) = tan–1(2cosec2x)    m, then - m - (1) m < 1 (2) m  {2, 3, 4} (1) m < 1 (2) m  {2, 3, 4} (3) m = 5 (4) m > 5 (3) m = 5 (4) m > 5 88. Let ƒ(x) = min{sin–1x, cos–1x}, then area 88. ƒ(x) = {sin–1x, cos–1x} ƒ(x) bounded by ƒ(x) and x-axis is - x-- (1) 2 (2) 2  1 (1) 2 (2) 2  1 1 (4) 2  1 1 (4) 2  1 (3) 2 (3) 2 89. Let S = 0 is an ellipse whose vartices are 89.  S=0,    the extremities of minor axis of the ellipse x2 y2 a2  b2 x2 y2 E:  1, a  b  a2 b2 E:   1, a  b . If S = 0 passes through S = 0, E  the foci of E, then its eccentricity is (Ee ) (considering the eccentricity of E as e) 1  2e2 1 1  2e2 1 (1) 1  e2 (2) 1  e2 (1) 1  e2 (2) 1  e2 1  2e2 e2 1  2e2 e2 (3) 1  e2 (4) 1  e2 (3) 1  e2 (4) 1  e2 x2 y2 90. Let L is distance between two parallel 90.  b2 1, a > b  a2 x2 y2 normals of a2  b2 1 , a > b, then LL- maximum value of L is - (1) 2a (2) 2b (1) 2a (2) 2b (3) a + b (4) 2(a – b) (3) a + b (4) 2(a – b)  1001CT103516012 H-33/34

Target : JEE (Main + Advanced) 2017/18-12-2016/Paper-2   H-34/34 1001CT103516012