Solution - Career Point
Solution - Career Point Solution - Career Point
(ii)OHH + CH 3OH⊕PorClPClClOCH 3OCH 3ClClCl(b) (i)C 2 H 5alk. KMnO 4Ethyl benzene(ii)CH 2 CH 3BrOHMg +Br 2/hvCOOHBenzoic acidCH 3 CHCOOH2-Phenylpropionic acidCHBrCH 3NaOH–H 2OMgEtherH 2O/H +H⊕COONaNaOH+CaOSodium benzoate+ Na 2 CaO 3BenzeneCH 3 CH – MgBrCO 2CH 3 CHCOOMgBr10. Interpret the non-linear shape of H 2 S molecule andnon-planar shape of PCl 3 using valence shell electronpair repulsion (VSEPR) theory. (Atomic numbers :H = 1, P = 15, S = 16, Cl = 17.) [IIT-1998]Sol. In H 2 S, no. of hybrid orbitals = 21 (6 + 2 – 0 + 0) = 4Hence here sulphur is sp 3 hybridised, so16S = 1s 2 , 2s 2 2p 6 2 2 1 1, 3 s 3px3py3pz14424343sp hybridisationSorH HH HDue to repulsion between lp - lp; the geometry ofH 2 S is distorted from tetrahedral to V-shape.In PCl 3 , no. of hybrid orbitals = 21 [5 + 3 – 0 + 0] = 4Hence, here P shows sp 3 -hybridisation15P = 1s 2 , 2s 2 2p 6 2 1 1 1, 3 s 3px3py3pz14424343sp hybridisationSThus due to repulsion between lp – bp, geometry isdistorted from tetrahedral to pyramidal.TMATHEMATICS11. 7 relatives of a man comprises 4 ladies and3 gentlemen; his wife has also 7 relatives; 3 of themare ladies and 4 gentlemen. In how many ways canthey invite a dinner party 3 ladies and 3 gentlemen sothat there are 3 of man’s relative and 3 of the wife’srelatives ?[IIT-1985]Sol. The possible cases are :Case I : A man invites 3 ladies and women invites 3gentlemen⇒ 4 C 3 . 4 C 3 = 16Case II : A man invites (2 ladies, 1 gentleman) andwomen invites (2 gentlemen, 1 lady)⇒ ( 4 C 2 . 3 C 1 ).( 3 C 1 . 4 C 2 ) = 324Case III : A man invites (1 lady, 2 gentlemen) andwomen invites (2 ladies, 1 gentleman)⇒ ( 4 C 1 . 3 C 2 ).( 3 C 2 . 4 C 1 ) = 144Case IV : A man invites (3 gentlemen) and womeninvites (3 ladies)⇒ 3 C 3 . 3 C 3 = 1∴ Total number of ways = 16 + 324 + 144 + 1 = 48512. Let n be a positive integer and(1 + x + x 2 ) n = a 0 + a 1 x + ..... + a 2n x 2nShow that a 2 0 – a 2 1 + ...... + a 2 2n = a n . [IIT-1994]Sol. (1 + x + x 2 ) n = a 0 + a 1 x + .... + a 2n x 2n ...(1)Replacing x by –1/x. we obtainn⎛ 1 1 ⎞ a⎜1− +2⎟ = a 0 – 1a +2a3 a2n– +...+ ..(2)⎝ x x ⎠ x2 32nx x xNow, a 2 0 – a 2 1 + a 2 2 – a 2 3 + ... + a 2 2n = coefficient ofthe term independent of x in[a 0 + a 1 x + a 2 x 2 + ... + a 2n x 2n ]⎡ a1a2a2n ⎤⎢a0− + −...+22n⎥⎣ x x x ⎦= coefficient of the term independent of x inn(1 + x + x 2 ) n ⎛ 1 1 ⎞⎜1−+2⎟⎝ x x ⎠nBut, R.H.S. = (1 + x + x 2 ) n ⎛ 1 1 ⎞⎜1−+2⎟⎝ x x ⎠2 n 2( 1+x + x ) (x − x +=2nx1)nXtraEdge for IIT-JEE 12 APRIL 2010
2 2 2 n[( x + 1) − x ]=2nx2 41+2x + x=2n2 4 n( 1+2x + x )=2n2( − x )xxThus, a 2 0 – a 2 1 + a 2 2 – a 2 23 + ... a 2n= coefficient of the term independent of x in12n (1 + x 2 + x 4 ) nx= coefficient of x 2n in (1 + x 2 + x 4 ) n= coefficient of t n in (1 + t + t 2 ) n = a n13. Solve for x the following equation :log (2x + 3) (6x 2 + 23x + 21)= 4 – log (3x + 7) (4x 2 + 12x + 9) [IIT-1987]Sol. log (2x + 3) (6x 2 + 23x + 21)= 4 – log (3x + 7) (4x 2 + 12x + 9)⇒ log (2x + 3) (2x + 3) . (3x + 7) = 4 – log (3x + 7) (2x + 3) 2⇒ 1 + log (2x + 3) )(3x + 7) = 4 – 2log (3x + 7) (2x + 3)Put log (2x + 3) (3x + 7) = y⇒ y + y2 – 3 = 0 ⇒ y 2 – 3y + 2 = 0⇒ (y – 1) (y – 2) = 0⇒ y = 1 or y = 2⇒ log (2x + 3) (3x + 7) = 1or log (2x + 3) (3x + 7) = 2⇒ 3x + 7 = 2x + 3 or (3x + 7) = (2x + 3) 2⇒ x = – 4 or 3x + 7 = 4x 2 + 12x + 94x 2 + 9x + 2 = 04x 2 + 8x + x + 2 = 0(4x + 1) (x + 2) = 0x = – 2, –1/4∴ x = – 2, –4, –1/4But, log exists only when, 6x 2 + 23x + 21 > 0,4x 2 + 12x + 9 > 0,2x + 3 > 0 and 3x + 7 > 0⇒ x > –3/2∴ x = –1/4 is the only solution.14. Let f[(x + y)/2] = {f(x) + f(y)} / 2 for all real x and y,if f´(0) exists and equals –1 and f(0) = 1, find f(2).[ΙΙΤ−1992]⎛ x + y ⎞ f (x) + f (y)Sol. f ⎜ ⎟ =⎝ 2 ⎠ 2∀ x, y ∈ R (given)Putting y = 0, we get⎛ x ⎞ f (x) + f (0)f⎜⎟ =⎝ 2 ⎠ 21= [1 + f(x)] [Q f(0) = 1]2⇒ 2f(x/2) = f(x) + 1⇒ f(x) = 2f(x/2) – 1 ∀ x, y ∈ R ...(1)Since f´(0) = –1, we getf (0 + h) − f (0)f (h) −1⇒ lim = – 1 ⇒ limh → 0 hh → 0 hnNow, let x ∈ R then applying formula ofdifferentiability.⎛ 2x + 2h ⎞f ⎜ ⎟ − f (x)f (x + h) − f (x)2f´(x) = lim = lim⎝ ⎠h → 0 hh→0hf (2x) + f (2h)− f (x)= lim2h→0h1 ⎧ ⎛ 2x ⎞ ⎛ 2h ⎞ ⎫⎨2f⎜ ⎟ −1+2f ⎜ ⎟ −1⎬− f (x)2 2 2= lim⎩ ⎝ ⎠ ⎝ ⎠ ⎭h → 0h[using equation (1)]1{2f (x) –1+2f (h) −1}− f (x)= lim 2h→0hf (h) −1= lim = –1h → 0 hTherefore f´(x) = – ∀ x ∈ R⇒∫f´(x)dx =∫−1 dx⇒ f(x) = – x + k where k is a constant.But f(0) = 1, therefore f(0) = – 0 + k⇒ 1 = k⇒ f(x) = 1 – x ∀ x ∈ R⇒ f(2) = – 115. If (a + bx)e y/x = x, then prove thatx 3 2d y ⎛ dy ⎞= ⎜ x − y⎟ [IIT-1983]2dx ⎝ dx ⎠Sol. (a + bx).e y/x = x ...(1)Differentiating both sides, we get⎧ dy ⎫⎪x− y ⎪(a + bx).e y/x .dx⎨ ⎬ + be y/x = 12⎪ x ⎪⎩⎭⎧ dy ⎫⎨x− y⎬dx⇒ x.⎩ ⎭+ be b/x = 1, (using (1))2xdy yor – + be y/x = 1,dx xAgain differentiation both sides,dy⎧ dy ⎫2 x − yd y–dx⎪x− y ⎪+ be y/x .dx22⎨ ⎬ = 0 ..(2)2dx x⎪ x ⎪⎩⎭dy2 x − yd yfrom (2), –dx ⎛ dy y ⎞22⎜ − ⎟ = 0dx x ⎝ dx x ⎠⇒ x 3 2d y ⎛ dy ⎞= ⎜ x − y⎟ 2dx ⎝ dx ⎠22XtraEdge for IIT-JEE 13 APRIL 2010
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(ii)OHH + CH 3OH⊕PorClPClClOCH 3OCH 3ClClCl(b) (i)C 2 H 5alk. KMnO 4Ethyl benzene(ii)CH 2 CH 3BrOHMg +Br 2/hvCOOHBenzoic acidCH 3 CHCOOH2-Phenylpropionic acidCHBrCH 3NaOH–H 2OMgEtherH 2O/H +H⊕COONaNaOH+CaOSodium benzoate+ Na 2 CaO 3BenzeneCH 3 CH – MgBrCO 2CH 3 CHCOOMgBr10. Interpret the non-linear shape of H 2 S molecule andnon-planar shape of PCl 3 using valence shell electronpair repulsion (VSEPR) theory. (Atomic numbers :H = 1, P = 15, S = 16, Cl = 17.) [IIT-1998]Sol. In H 2 S, no. of hybrid orbitals = 21 (6 + 2 – 0 + 0) = 4Hence here sulphur is sp 3 hybridised, so16S = 1s 2 , 2s 2 2p 6 2 2 1 1, 3 s 3px3py3pz14424343sp hybridisationSorH HH HDue to repulsion between lp - lp; the geometry ofH 2 S is distorted from tetrahedral to V-shape.In PCl 3 , no. of hybrid orbitals = 21 [5 + 3 – 0 + 0] = 4Hence, here P shows sp 3 -hybridisation15P = 1s 2 , 2s 2 2p 6 2 1 1 1, 3 s 3px3py3pz14424343sp hybridisationSThus due to repulsion between lp – bp, geometry isdistorted from tetrahedral to pyramidal.TMATHEMATICS11. 7 relatives of a man comprises 4 ladies and3 gentlemen; his wife has also 7 relatives; 3 of themare ladies and 4 gentlemen. In how many ways canthey invite a dinner party 3 ladies and 3 gentlemen sothat there are 3 of man’s relative and 3 of the wife’srelatives ?[IIT-1985]Sol. The possible cases are :Case I : A man invites 3 ladies and women invites 3gentlemen⇒ 4 C 3 . 4 C 3 = 16Case II : A man invites (2 ladies, 1 gentleman) andwomen invites (2 gentlemen, 1 lady)⇒ ( 4 C 2 . 3 C 1 ).( 3 C 1 . 4 C 2 ) = 324Case III : A man invites (1 lady, 2 gentlemen) andwomen invites (2 ladies, 1 gentleman)⇒ ( 4 C 1 . 3 C 2 ).( 3 C 2 . 4 C 1 ) = 144Case IV : A man invites (3 gentlemen) and womeninvites (3 ladies)⇒ 3 C 3 . 3 C 3 = 1∴ Total number of ways = 16 + 324 + 144 + 1 = 48512. Let n be a positive integer and(1 + x + x 2 ) n = a 0 + a 1 x + ..... + a 2n x 2nShow that a 2 0 – a 2 1 + ...... + a 2 2n = a n . [IIT-1994]Sol. (1 + x + x 2 ) n = a 0 + a 1 x + .... + a 2n x 2n ...(1)Replacing x by –1/x. we obtainn⎛ 1 1 ⎞ a⎜1− +2⎟ = a 0 – 1a +2a3 a2n– +...+ ..(2)⎝ x x ⎠ x2 32nx x xNow, a 2 0 – a 2 1 + a 2 2 – a 2 3 + ... + a 2 2n = coefficient ofthe term independent of x in[a 0 + a 1 x + a 2 x 2 + ... + a 2n x 2n ]⎡ a1a2a2n ⎤⎢a0− + −...+22n⎥⎣ x x x ⎦= coefficient of the term independent of x inn(1 + x + x 2 ) n ⎛ 1 1 ⎞⎜1−+2⎟⎝ x x ⎠nBut, R.H.S. = (1 + x + x 2 ) n ⎛ 1 1 ⎞⎜1−+2⎟⎝ x x ⎠2 n 2( 1+x + x ) (x − x +=2nx1)nXtraEdge for IIT-JEE 12 APRIL 2010
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