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NARAYANA IIT/PMT ACADEMY JEE ADVANCE : 2013 And for x 2 ,0 3 y 4x … (3) Case (iii) 0 x y 2x x 2 | x 2 2x | y 3x 2 | x 2| For 0 x 2 y 4x … (4) And for, x 2 y 2x 4 … (5) So graph is (PAPER – II) CODE: 0 (0,8) y = 2x+4 y= -2x-4 y= 2x+4 y= -4x (0,8/3) y=4x (-2, 0) (-2/3,0) (0, 0) (2,0) x From graph x 2,0 , are point of minimum, & x 2 is point of maxima 3 43 Let w 3 i n 1 and P : n 1,2,3,... Further H1 z C : Rez 2 2 and 1 H2 z C : Rez 2 , where C is the set of all complex numbers If z1 P H 1, z2 P H 2 and O represents the origin, then z1O z 2 (A) 2 (B) 6 (C) 2 3 (D) 5 6 Ans (C, D) Sol NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS - INDIA 26
NARAYANA IIT/PMT ACADEMY JEE ADVANCE : 2013 (PAPER – II) CODE: 0 2 5 Clearly from the figure, z1Oz 2 or or 3 6 x x 1 44 If 3 4 , then x = 2log3 2 2 (A) (B) 2log3 2 1 2 log2 3 Ans (A, B, C) x x 1 Sol 3 4 Taking log both sides with base 3 x 1 log 3 log 4 3 3 x log 3 x 1 log 4 3 3 x x 1 log3 4 x x log34 log34 log34 x log34 1 log3 4 x log3 4 1 2log3 2 x 2log 2 1 3 x x 1 3 4 Taking log both sides with base 4 log 3 log 4 x x 1 4 4 x log 3 x 1 4 1 x x log 43 x 1 log 43 1 1 2 x or x 1 log 3 2 log 3 4 2 (C) 1 1 log 3 4 2log2 3 (D) 2log 3 1 2 NARAYANA GROUP OF EDUCATIONAL INSTITUTIONS - INDIA 27
Page 1 and 2: NARAYANA IIT/PMT ACADEMY JEE ADVANC
Page 3 and 4: Ans Sol: 9 (C) 5R A, C 2 an o Z 4.5
Page 9 and 10: (A) 2 BQR Ans (C or B) Sol e B 2 R
Page 15 and 16: (B) 9 1 8 2 4Be 1 p 4Be 1D NARAYANA
Page 19 and 20: (C) (D) O O O V HOH 2 C NARAYANA II
Page 23 and 24: 1 0 3 0 3 NARAYANA IIT/PMT ACADEMY
Page 25: NARAYANA IIT/PMT ACADEMY JEE ADVANC
Page 29 and 30: Sol n Lt 1 0 n n n Lt Lt Lt 1 n 1 n
Page 31 and 32: Ans Sol (C) f ' x f x ,0 x (C) For

NARAYANA IIT/PMT ACADEMY - Narayanaicc.com

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