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α + β α + βline BC, x cos + y sin2 2Let point D be (a, b)thenB(β)a cosDα + β2+ b sinC(α)α + β2A(r, 0)= r cos= r cosα − β2α −β2....(3)α −βMultiply (3) by cos2α −β α + β α + β α − βa cos cos + b sin cos2 22 2= r cos 2 α − β2use (1) & (2)a ⎛ 3h ⎞ 3k⎜ −1⎟+ b = r cos 2 α − β...(4)2 ⎝ r ⎠ 2r2square & add (1) & (2)2 24 cos 2 α − β ⎛ 3h ⎞ 9k= ⎜ − 1⎟+...(5)22 ⎝ r ⎠ rfrom (4) & (5)2 21 ⎛ 3h ⎞ 9k a ⎛ 3h ⎞ b3k⎜ − 1⎟+ = ⎜ −1⎟ +24 ⎝ r ⎠ 4r 2r ⎝ r ⎠22rso req. locus is(3x – r) 2 + 9y 2 = 2a(3x – r) + 6.b.y9x 2 + 9y 2 – 6rx + r 2 = 6ax – 2ar + 6.b.yx 2 + y 2 2 2 r 2– (r + a)x – by + +3 3 9⎡ 1 ⎤⎢x− (r + a)3⎥⎣ ⎦=2(r + a)1 ⎤⎢⎡ + y − b3⎥⎣ ⎦222− r − 2ar + b92 +2=a2 +2ar9b92= 0a bIt is a circle and radius is3Since point D is interior of circle so a 2 + b 2 < r 2 ,so radius of this circle is less than r/3.6. Prime factors dividing 10 are 2, 3, 5, 7. As requiredthe number chosen should not be divisible by 2 or 3or 5 or 7. Define, the events asA : divisible by 2B : divisible by 3C : divisible by 5D : divisible by 72A ∪ B ∪ C ∪ D= A + B + C + D – A ∩ B – A ∩ C – A ∩ D– B ∩ C – B ∩ D – C ∩ D + A ∩ B ∩ C+ A ∩ C ∩ D + B ∩ C ∩ D + A ∩ B ∩ D– A ∩ B ∩ C ∩ DP(A ∪ B ∪ C ∪ D)1= [ 1003 + 668 + 401 + 286 – 334 – 2002006– 143 – 133 – 95 – 57 + 66 + 28 + 19 + 47 – 9]1547= 20067. Line BCBb = –i + 3jAaDHCc = 2i + 5j→ r = – î + 3ĵ+ t (3î + 2ĵ)...(1)any pt. D on it = (3t – 1) i + (3 + 2t) jAs HD ⊥ BC,so ((3t – 1 – 1) i + (3 + 2t – 2) j). (3i + 2j) = 03(3t – 2) + 2(2t + 1) = 0413 t – 4 = 0 ⇒ t = 13so point D = – 13i47 j+ 13Now line HD ⇒ → ⎛r = i + 2j + s´ ⎜−⎝= i + 2j + s(–14i + 21j)i1347 j ⎞+ − i − 2j⎟ 13 ⎠⇒ → r = i + 2j + λ(–2i + 3j)Any point A on it = (1 – 2λ)i + (2 + 3λ)jNow as AC ⊥ BHso [ (1 – 2λ – 2)i + (2 + 3λ – 5)j] . [2i – j] = 02 (–1 – 2λ) – (3λ – 3) = 01⇒ –7λ = 2 – 3 ⇒ λ = 75 i 17 jso pt. A = + 7 78. If A is the area of the triangle with sides a, b, and c,then A 2 = s(s – a) (s – b) (s – c);where 2s = a + b + cusing AM – GM inequality for s – a, s – b, s – c, wehave3A 2 ⎧(s− a) + (s − b) + (s − c) ⎫≤ s ⎨⎬⎩ 3 ⎭XtraEdge for IIT-JEE 42 APRIL 2010
A 2 ⎛ 3s − 2s ⎞ s≤ s ⎜ ⎟ =⎝ 3 ⎠ 3A ≤3s 23Let 2s = p,343p 2then A ≤12p 2A max = ,12 3As condition of equality holds iffs – a = s – b = s – c which happen if a = b = cso A max = ; for a = b = cNow again p ≥ 12 3Ap min. = 12 3A;and again equality holds if a = b = c.3at t = 60 min.⎡60⎛ 299 ⎞ ⎤x = 100 ⎢1 − ⎜ ⎟ ⎥ kg⎢⎣⎝ 300 ⎠ ⎥⎦10. Let P (3 cos θ, 3 sin θ)line BC : y = – 3 ; line AC :ABPC9. Let the amount of salt dissolved at any time t is x kg.xSo concentration is 300dx ⎛ 1 x ⎞ ⎛100− x ⎞so = k ⎜ − ⎟⎠ = k ⎜ ⎟dt ⎝ 3 300 ⎝ 300 ⎠dx100 − x= k dt300k– ln (100 – x) = t + C 300at t = 0, x = 0 so C = –ln 100kt100so = ln 100 – ln (100 – x) = ln300100 − xat t = 1 min., x = 1/3k 100so = ln3001100 −3k 100so = ln300 2993k 300so = ln300 299300so k = 300 ln 299300 100so ln . t = ln 299 100 − xso⎛ 300 ⎞⎜ ⎟⎠⎝ 299t100=100 − x⎛ 299 ⎞so 100 – x = 100 . ⎜ ⎟⎠⎝ 300⎡t⎛ 299 ⎞ ⎤so x = 100 ⎢1− ⎜ ⎟ ⎥⎢⎣⎝ 300 ⎠ ⎥⎦tx y cos θ + sin θ = 13 3⎛ 3(1 + sin θ)⎞pt. C ⎜ , − 3 ⎟⎝ cosθ⎠pt. A (0, 3 cosec θ )1 3(1 + sin θ)Area A = . 2 .. ( 3 cosec θ + 3 )2 cosθ223 3(1 + sin θ)6 3(1 + sin θ)==sin θcosθsin 2θ2dA 6 3(2(1 + sin θ)sin 2θcosθ − 2(1 + sin θ)cos 2θ)=2dθsin 2θ12 3(1 + sin θ)(sin 2θcosθ − cos 2θ − sin θcos 2θ)=2sin 2θ12 3(1 + sin θ)(sin(2θ − θ)− cos 2θ)=2sin 2θ212 3(1 + sin θ)(sinθ −1+2sin θ)=2sin 2θ12 3(1 + sin θ)(2sinθ −1)(sinθ + 1)=2sin 2θ212 3(1 + sin θ)(2sin θ −1)=2sin 2θπA min at θ = 6so A min =6⎛ 1 ⎞3⎜1+ ⎟⎝ 2 ⎠ 9= 12 . 4322= 27sq. units.= 0027 Ans.XtraEdge for IIT-JEE 43 APRIL 2010
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