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1 INDIA Sec: Sr. IIT-IZ JEE-Main Date: Time: 09:00 AM to 1:00 PM IRTM-06 Max.Marks: 360 KEY SHEET MATHS PHYSICS CHEMISTRY f x ln x ln e x SLUTINS MATHS 1 1 e x ln x ln x e x f ' x ln e x e xln e x xln x e x xln e x 0on 0, since 1 e f x decreases on 0,. The given curve is 3 y 3x 1y
2 dy dy dx dx y 3y 6x _Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s For vertical tangents y For y x 4 3, x dy 1 dx y 0 Fory, x ve not possible Re q. 4 / 3,3, 3. f x x bx cx d,0 b c 4. 3 f ' x 3x bx c Discriminant b c b c f ' x 0 x R x f is strictly increasing x R 5. Sec: Sr.IIT-IZ Page
3 -09-18_Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s 6. The given function is f () x x 3, 3 x 1 /3 x, 1 x The graph of y f x is as shown in the figure.from graph clearly there is one local maximum ( at x 1) and one local minima (at x=0) total number of local maxima of minima = 7. Sec: Sr.IIT-IZ Page 3
4 _Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s ' ' 9. let maximum of f occurs at C. so f c 0 APPLY LMVT for f x ' ' f c f c 0 0 '' f 0, c This implies f ' ' (1) similarly apply LMVT FR f x N c,4 0 5c ' To get f 4 54 c Adding (1) & () f ' ' 0 f () 10. y ax bx c pass through P, Q, S, so a b c 1 for area to be maximum tangent at R s, t should be parallel to QS Then R 1 7, f x 0 has strictly increasing d dx 1. clearly f x cos x 0 Sec: Sr.IIT-IZ Page 4
5 so g x f x.cos x is non increasing _Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s And has roots n 1, n Z g is non-increasing g 0 f x 0 5 f CNCEPTUAL Sec: Sr.IIT-IZ Page 5
6 _Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s Sec: Sr.IIT-IZ Page 6
7 -09-18_Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s 0. Given that 1. dv d 4 dt dt 3 dr 4 r 50 dt dr 50 1 cm / min dt cm / min r 50. Sec: Sr.IIT-IZ Page 7
8 -09-18_Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s Sec: Sr.IIT-IZ Page 8
9 -09-18_Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s 5. dy dx 3 Let y x px q 3x p For 0 3x p 0 x d y dx d y dx dy dx 6 x x d y ve and dx p p x x 3 3 ve p p y has minimum at x and maxima at x p 3 Let f x x 14x 16x 30x f ' x 7x 70x 48x 30 0, x R f is an increasing function on R Also lim f x and lim f x x x The curve y f x crosses x- axis only once x 0 f has exactly one real root 7. Sec: Sr.IIT-IZ Page 9
10 -09-18_Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s 8. Since tangent is parallel to x axis, dy x 3 dx x y 3 Equation of tangent is y-3=0(x-)=>y= CNCEPTUAL 31)Key: 3 PHYSICS SLUTINS. Sec: Sr.IIT-IZ Page 10
11 -09-18_Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s 3)Ans :1 Solution : Let a is the side of square. Torque of current = MB sin 90 = MB cos Mass of each side m Aa Torque of gravity Now, both torques should be same i. e., ia B cos Aga sin ia B cos a mg sin mg a sin mga sin Aga sin cot Ag ib 33). = B 0 origin. 34). (1), so a graph between and B will be a parabola symmetric about axis and passing through the Sec: Sr.IIT-IZ Page 11
12 d d = 5 nib cos 7 m _Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s 5 nib sin 7 m 35. () = 10 nib sin 7 m f = m r mg = m r 5 nib 7 m = 5 nib r 7g m 36) 4 BAB 0l 37)(1) BA B(sin BC 45 sin 0)( k) 4a l 0 ˆ 4 a 0l So net B ˆ Z 4B( Ak) a 0l similarly B( y j) a ˆ 0 or B( j k) l ˆ ˆ a d i 38)() Bnet 0 i1 i3 4d 4d d i d r 5 d r )(4) M IS )() Due to AB : Magnetic field at 1 Due to BCD : Magnetic field at, Due to DE : Magnetic field at, Sec: Sr.IIT-IZ Page 1
13 41)(4) By moment equilibrium about C, _Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s 4)(4)Velocity component along x direction after time t is v 4v v v x x qe t m )(4) d dm B B and dm are in the same direction. 44) (3) CNCEPTUAL 45) 1 Small element having dn turns Sec: Sr.IIT-IZ Page 13
14 db 0 r dn i b a r a b a r b a _Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s 0Ni dr 0Ni b dr 0Ni db, B db ln b / a 46) 3 Q dq R dq i f rdr Q rdr R Q i rdr R 0i 0Q db dr r R R 0Q 0Q B dr R 0 R R 0Q B R 47) 1. CNCEPTUAL 48) 4 a 4 a Ka 3 di j. da i di j. da Kr rdr Kr dr B. dl i B dl i 0ka Br 4 0ka B 4r ) 3. CNCEPTUAL 50) 1 Magnetic field at P is B, perpendicular to op in the direction So B B sin i B cos j Sec: Sr.IIT-IZ Page 14
15 0i B r x sin y / r cos r 0 y i x j 0i 1 B. y i x j r x y _Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s 51) 4 B Then nir R 0 r 0niR B 3 r 1 B r 3 3/ 5) 1 i i a a 0 0 Bx j, B y i B B B x y 53)3 B dl. 0Inet I x b Bx 0 I c b 0I c x Bx c b 0I c x B x c b 54) r x r x 1 r r1 k Sec: Sr.IIT-IZ Page 15
16 B B ir 1 0 0ir r r 1 x x 3/ 3/ _Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s 3 r x 1 x 3/ 3 r 1 1 x r r x 3/ )1 B. dl 0 I Bl l 0 0 B 56) B B B p 1 B p 0 1 r1 r i i 5 T 57).4 CNCEPTUAL 58).3 CNCEPTUAL 59) 60)1 or The electron will be refocused after traveling a distance = pitch of helix pitch Sec: Sr.IIT-IZ Page 16
17 -09-18_Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s moles for glucose and 5 moles for fructose mole of fructose requires (4) conceptual 64. (3) conceptual 65. (3)conceptual CHEMISTRY pb AC 0.5 moles of 4 66.Solution:() Sucrose is composed of -D- glucose & - D fructose. CHH H H H H H H H H HHC CHH H H H H H D glucose D fructose Sucrose 67. Solution:(1) 68. Solution:(3) It is ionic so it exists in crystalline form. Because it is ionic so it soluble in water. Because acidic and basic groups are present in the same molecule, so internal salt formation occurs to form dipolar ion NH 3 + CH C 69. Solution:(3) C C NK Cl CH C C H 5 KCl C C N CH C C H 5 Potassium phthalimide (X) (Y) HCH / C H5H C C Phthalic acid (Z) H H HN CH CH Glycine 70. Solution:(1) The ph at whichh Anion = Cation is called isoelectric point. At isolectric point, - amino acids to not migrate when field is applied. Sec: Sr.IIT-IZ Page 17
18 -09-18_Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s ) S 3H is strongly acidic and donates H + to weakly basic arylamino group. ArCH is not acidic enough to transfer H + to the arylamino group. 3) In p- H 3N + C 6H 4S 3, H 3N + is acidic enough to transfer H + to bases to give the soluble anion p H N C 6H 4S 3 S 3 is too feebly basic and cannot accept H + from acids. 4) - H N + group increases acidity, because of its electron withdrawing inductive effect. 71. Me Me H H Me Me Me Me Me H Me Me 7. Sol.(3) H (CH ) 4 H + H N NH (CH ) 6 H C (CH ) 4 C N (CH ) 6 NH Nylong (1) conceptual 74.( 1) Sol. Polymers whose repeating units are derived from two or more types of monomers units, are called copolymer. nly Buna-S is obtained by the copolymerization of butadiene and styrene monomers. ther polymers ), 3) and 4) are homopolymers, because they consists of only one type of monomers. 75. Myosin is fibrous protein with -pleated structure 76 (3)conceptual 77. (1)conceptual Sec: Sr.IIT-IZ Page 18
19 -09-18_Sr.IIT-IZ_JEE-MAIN_IRTM-06_Key&Sol s 78. ()conceptual 79. (3)conceptual 80. () 81. (4) Tagalose is an aldoketose 8. (4) There is no CH group or NH group. Regents 1 and 8 will fall to give test. 83.() MnNumber average 30 0, , , ,000 Mwweight average , () conceptual 85. ()conceptual 86 ()conceptual 87. ()conceptual 88. (4)conceptual 89 (3)conceptual 90. (1)conceptual Sec: Sr.IIT-IZ Page 19
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