CAREER PINT TARGET IIT JEE CEMISTRY, MATEMATICS & PYSICS RS -- I -A INTS & SLUTIN CEMISTRY Section I n +. [B] C n n + n + nc + (n + ) V 7 n + (n + ) / 7 n VC 4 n 4 alkane is C 6 a.[a] P + (v b) RT V at low pessue 'b' is neglected a a P + V RT PV + RT V V a PV RT VRT.[A] 4.[C] PhMgB Ph C C MgB MeMgB (B*) (C*) (D) C C C C C C C Ph C C C C C C C C C MgB Ph C C MgB Me 5. [A, B, C] 6. [A, B] 7. [A, B, C] 8. [B] (A*) / Ph C C Me Ph C C C 9. [A, D] Ph C C C Section - II. A P,T; B R,S; C Q,; D S. A P,Q,T; B Q,T; C S,; D Q,T. [] Section - III W i t E 96500.6 i 60 60 i 0.54amp 08 96500 eo in cuent.60.54.06 % eo.06.6 00 % Ph C C C CAREER PINT, CP Towe, Road No., IPIA, Kota (Raj.), Ph: 0744-040000 Page #
400. [4] Fo initial solution V n s 9 760 05. afte dilution V n S 08 760 V 4V. [4] K b (X ) 0 0 fo conjugate acid base pai K a (X) K 0 4 b (X ) K a (X) 0 4 [ ] [ ] acid salt [Salt] p log ka + log [Acid] log 0 4 4 4. [] Fo tautomeism α hydogen is equied eample of etended conjugation 5.[7] b, c, d, e, f, h, i, ae stonge acid than C 6.[] 7. [] 8. [].[A] MATEMATICS Section I n solving two cuves A 4 + + 4A A 4 (4A + ) + 0 Let t A t (4A + ) t + 0 Let t α, β ae the oots of equation Q α. β > 0 & α + β > 0 & D > 0 so α & β ae eal and distinct fo all A R + So α, β fo A R + ± α, ± β fo A R +. π π.[b] Q f π so g(π). Q f(g()) f (g()). g () g () f (g()).[d] g (π) f (g(π)) π f Q α < < β so if f() + λ then f() < 0 So λ <. 7 7 4.[C] tan 4 + cot 4 4 sin y 4 sin y π π 5π sin y, y ±, ±, ±,. π π But y ±, ±. & coesponding to each value of y thee ae two values of so total numbe of points ae 8. 5.[A,B,C,D] (A) Q g(f(c + )) g(f(c )) g(f(c)) g(b) (B) Diichilet function Q [0, ] f() Q [0,] (C) Q f() is continuous so f() f( + ) f( ) Case I : + is ational then f( + ) g( + ) Case II : If + is iational then f( + ) f() which is equal to g() Q g() g( + ) g( + ) f( + ), So f() g() R (D) Fom Rolle's theoem 6.[A,B] f() / f () / Q f () 0 so is the citical point Q f (0) is not define so 0 is point of etemum. Q f () / is negative fo all R {0} so gaph of y f() is concave down fo all R {0}. But at 0, f () is not define. 7.[A,B,C,D] nπ (A) f() is not define fo but g() is. (B) fo > ; f() while g() (C) f() is define fo but g() is not define (D) f() sin cos + cos sin sin CAREER PINT, CP Towe, Road No., IPIA, Kota (Raj.), Ph: 0744-040000 Page #
8.[B,C,D] 9.[A,C,D] + cos cos 4 cos 4 cos cos + 0 cos ±, nπ ± 6 π, nπ n I Let 00 a. b. 5 c. 7 d. 00 00 00 a + + +.. 97 00 00 00 b + + +. 48 00 00 c + 5 +. 4 5 00 00 d + 7 +. 6 49 Let 50 p. q. 5. 7 s. p 47, q,, s 8. 00 97 48 4 6 So 00..5.7... C 50 94 44 4 6 50 50..5.7 So α, β 4, γ δ 0. Section II. A R, B S, C P, D Q (A) Let α, α, α α + α + α a.() α + α + α b.() α 6.() on solving (), () & () a b 6 ( 4) y (B) Q + 4 9 4 + cosθ & y sinθ y + 5 + 4 cosθ 4 9 So Ma. 9 & Min. (C) Let f() + log f log ( + /) Apply DL'hospital Rule + log f 0 f() g() log g log log Apply DL'hospital CAREER PINT, CP Towe, Road No., IPIA, Kota (Raj.), Ph: 0744-040000 Page # 0 g() (D) 4 e y y + 5 sin y (0) 0 4 + 4e y (y + y ) y + 0 sin cos put 0 4 + 0 y + 0 y 4. A P, B Q, C P, D T π π (A) α π, θ π, cos cosθ π θ, tan tanθ θ π cot cotθ θ π, sec (secθ) π θ, cosec cosecθ π θ. f(α) 0 π π (B) α, 0 θ, π So f(α) π π π π (C) Q α 0, 0 < α < π So θ 0, So f(α) 0 (D) f(α) sin sinθ cos cosθ + tan tanθ cot cotθ + sec secθ cosec cosecθ
whee θ π α f(α) tan tanθ cot cotθ f(α) θ + θ + π π + θ π + π α π α Section - III L ne point common to both of above is (,, 0) and since line is paallel to ( î ĵ + kˆ ) ( î + ĵ kˆ ) So equation of line can be witten as î + ĵ + 0 kˆ + λ( î + ĵ +4 kˆ ) So any point lie on the line is ( +λ, + λ, 4λ). This should also lie on plane so ( + λ) + ( + λ) (4λ) 7 so.[8] β α (0, 4) L 4.[] λ So P (α, β, γ) (,, 4) θ α + β tan + tan tan () tan () π θ 4 Equation of L is y + 4 y + 4 0 Let the equation of cicle is ( ) + (y + ) + y + y + 0 + 4 6 + 6 0 So sum of adii is 8.[] n(s) numbe of ways of distibuting 9 diffeent tiles among thee pesons 9. ( ). n(a) numbe of ways of distibuting 9 tiles in thee goups consist of ( odd even) ( odd even) ( odd) then these thee goups to thee pesons. 4 5 ( ). ( ). m. So m + n 7. n 4.[7] Q n solving y + z 0 & + y z 5 0 A C B Coodinate of C (0, ) A (, 0) B (, 0) Whee + a and Equation of family of cicle passing though point A &B is (y 0) (y 0) + ( ) ( ) + λy 0 Q C(0, ) lies on it so λ 0 So equation of cicum cicle of ABC is + y a + 0 so coodinates of D (0, ) 5.[] Q ω So z + ω + z z ω ω + Let ω + iy ω + y 0 + 0 5,0 A B C CAREER PINT, CP Towe, Road No., IPIA, Kota (Raj.), Ph: 0744-040000 Page # 4
5 4 9 ω ma C A + BC + 5 4 ω min A B AB (m + M) 9 5 5 + 6.[5] Q A 0 & A A so A I So p q 0 & l m 8 n 5 So l ±, m ±, h ± 4 So set S is {,,,, 4, 4, 0} So λ sum of poduct of numbes of set S taken two at a time λ () + ( ) + () + ( ) + (4) + ( 4) + (0) ( + + 4 4 + 0).[C] So b 0 a, 0, 0, So total si pais ae possible of (a, b) (, 0) (, 0) (0, ) (, ) (0, ) (, ). PYSICS v b + 40 ĵ v T 40î Section I vbt vb vt 4(ĵ + î) N y λ 9 7.[7] Q The equation of cuve is y & the equation of pependicula lines be y m and y. So P (m, m ) & Q, m m m So m p m & m q m Equation of tangent at P is y Equation of tangent at Q is y Thei point of intesection m m m + m h m & k m So 4k h So 4y a 4 & b 8.[6] aω + b (aω + b) (aω + b) a ab + b 0 a b ± 4 b + m m W S E.[A] f L µ N µf and f L µ N µf.[a] 4.[A] f T N 5 kg N 50N f m g l ( m + m) v Conceptual. 5.[A,B] v 0 ω v and v 0 + ω v f T N 5 kg F 50N 6.[A,D] Potential of conducto potential at cente v q + v ind. chages 7.[B,C,D]7Ω is shot cicuited. I Ω V R eq. Ω I 6A I 9 6 6 4A I I 6Ω I 6 4 A CAREER PINT, CP Towe, Road No., IPIA, Kota (Raj.), Ph: 0744-040000 Page # 5
8.[A,D] Conceptual. 9.[A,B] (A) δ i + e A fo i e fo minimum deviation δ min i A (B) n A + δ sin A sin (C) δ i + e A min Section - II. A P,R,S ; B Q ; C P,Q,S ; D Q,R Use PV n constant as given and PV nrt R and C C V if PV constant also dq du + dw. A P,Q ; B R,S ; C P,S ; D Q,R Conceptual. Section III dna dnb.[] λna, λna λnb dt dt N B maimum dn B 0 dt λ N λ N o A Bma λ N B N ma λ λ + o NB ma Me, λ λ.[] We have, 0. W + K ma,.(i) and 0. Z W + K ma, (ii) h h Also λ de-boglie p mk A.[4] λ K. K 5.5K (iii) λ K Also, 0. Z enegy coesponding to longest wavelength of the Lyman seies.6 Z fom equations (i), (ii) and (iii), W ev mv m0c o m v mm 0 c o mv h λ mv 0 mm c 4.[] The deviation poduced by small angled pism, δ ( µ ) α (.5 )4º º (always) Deviation caused by mio δ 80º i 80º 45º 90º Net deviation poduced by system δ + δ º + 90º 9º This is moe than 90º, geate is angle of incidence on the mio, smalle is the deviation. If β is the angle of otation of mio in clockwise diection to incease angle of incidence then deviation poduced by the mio will be 80º (45º + β) 90º β Total deviation poduced 90º β + º 9º β given, 9º β 90º β º. 5.[5] Accoding to the question, shift 5 (finge width) ( µ ) td 5λD d d 5λ 5000 t 50, 000 Å µ.5 6.[] The induced emf in loop ABFG AdB 0 0 V dt The induced emf in loop BCD and DEF AdB 0V dt CAREER PINT, CP Towe, Road No., IPIA, Kota (Raj.), Ph: 0744-040000 Page # 6
Ω Ω 0V y z Ω y-z -y Ω -z Ω 0V 0V 4Ω 0 (y z) + ( y) y 0 () 0 ( y) ( z) 0 () 7.[] 0Ω i i V A B i 0Ω 0Ω i i 4V 0 i 0 i () 0 i + 0 i 4 () i + i i () 8.[9] Let be the elongation in the sping when it etuns back K 0 K + mg( + 0) n solving, we get K mg ± (mg K 0 ) and also lowe block will bounce up if K mg mg So, 0 0 9 cm K CAREER PINT, CP Towe, Road No., IPIA, Kota (Raj.), Ph: 0744-040000 Page # 7