Solution - Career Point

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