Solution - Career Point

⎯ Ca(OH)⎯⎯⎯2∆→ OC 6 H COOOH⎯⎯⎯5(Baeyer−Villigeroxidation)(Y)O⎯⎯ →O(Z)14. [1]Molecular weight of starch = 162n∴ moles of C 6 H 12 O 6 to be produced = 1 mol3000∴ moles of ATP required = = 100 mol3015. [7]0P − P0 = x1 = mole fraction of solutePx 1 = 0.0125⎛ 1 ⎞ 1000⎜ −1⎟ =⎝ x1⎠ m×18∴ m = 0.7016. [5]Eq. of metal = Eq. of hydrogen0.1 43.9= ⇒ X = 251/ X 11200Now eq. of metal = eq. of KMnO 40.151/ Y − 2= 0 .1×58.81000∴ Y = 517. [7]For the first bulb p 1 v 1 = n 1 RT& second bulb p 2 v 2 = n 2 RTor p 1 v 1 + p 2 v 2 = (n 1 + n 2 ) RT …(1)Suppose equilibrium pressure at each bulb = p atmThen p(v 1 + v 2 ) = (n 1 + n 2 )RT …(2)From eq. (1) & (2)p 1 v 1 + p 2 v 2 = p(v 1 + v 2 )9 × 5 + 6 × 10 = p × 15p = 105/15 = 718. [2]••OF FXeXeOOFO••FXeO 3XeOF 419. [6]Each has two geometrical isomers and two opticalisomers (shown by cis-isomer).1. [A] We haveMATHEMATICSn2n4 4∑ 2r −1)= ∑r−∑r= 1r= 1 r=1n( (2r) = f(2n) – 16f(n)2. [D] Using De Moivre's theorem2i /f r (α) = e α r22i / re α i / r…. e α2(iα/ r )(1+2+...... + r)= e2(iα/ r )[r(r+1)/ 2] (iα/ 2)(1+1/ r)= e= e∴ lim f n ( π)= lim iπ/2(1+1/ n)en→∞n→∞= e iπ/2 ⎛ π ⎞ ⎛ π ⎞= cos ⎜ ⎟ + i sin ⎜ ⎟ = i⎝ 2 ⎠ ⎝ 2 ⎠α + βt3. [D] z = ⇒ ( γ + δt) z = α + βtγ + δt⇒ (δz – β)t = α – γzα − γz⇒ t =[Q αδ – βγ ≠ 0 ]δz− β4. [C]α − γz α − γzAs t is real, =δz− β δz− β⇒ (α – γz)( δ z – β ) = ( α –4γ z )(δz – β)⇒ ( γ δ – γ δ )z z +(γ β – α δ)z + (α δ – β γ ) z= (α β – α β) ...(1)Since δγ is real, δγ =δγ or γ δ – δ γ = 0Therefore (1) can be written as a z + a z = c ...(2)where a = i(α δ – β γ ) and c = i( α β – α β )Note that a ≠ 0 for if a = 0 thenα γ γ γaδ – β γ = 0 ⇒ = = [Q is real]β δ δ δ⇒ αδ – βγ = 0,which is against hypothesis.Also, note that c = i( α β – α β ) is a purely realnumber.α + βtThus, z = represents a straight line.γ + δtdy = (x – 1)(x – 2) 2 sodx2d y= (x – 2) × (3x – 4).2dxd yThe points of inflection are given by = 02dxSo x = 2, x = 4/3 are points of inflection.2XtraEdge for IIT-JEE 96APRIL 2010