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In the present case⎡(50+ 1)(2x) ⎤ ⎡(51)(2 / 5) ⎤ ⎡102 ⎤k = ⎢ ⎥ =⎣ 3 + 2x⎢ ⎥ =⎦ ⎣ 3 + 2 / 5⎢ ⎥ = 6⎦ ⎣ 17 ⎦Thus, 6 th term is the largest term.6.[D] We have | z | =7.[D]4 4z − + ≤z z4z − +z4| z |4= 2 +| z |⇒ | z | 2 ≤ 2 | z | + 4 ⇒ (| z | – 1) 2 ≤ 5⇒ | z | – 1 ≤ 5 ⇒ | z | ≤ 5 + 1Also, for z = 5 + 14z −z= 2Therefore, the greatest value of | z | is 5 + 1.Integrating by parts, the given integral is equal to16 16x tan –1 x 1x − 1 −∫dx1 x 4 x x −16= π316= π316= π3161 dx–4 ∫x −1 131 4t(1 + t–4 ∫ t0– ( 3 + 3)= 31 12)dt ( x= 1 + t 2 )16 π – 2 38.[C] The intersection of y – x + 1 = 0 and y + x + 5 = 0is (– 2, –3). Put x = X – 2, y = Y – 3. The givendY Y − Xequation reduces to = . This is adX Y + Xhomogeneous equation, so putting Y = υX, we get2dυ υ +X = − dX υ + 11⎛ υ 1 ⎞ dX⇒ ⎜−− ⎟ dυ =2 2⎝ υ + 1 υ + 1 ⎠ X9.[A]⇒ – 21 log (υ 2 + 1) – tan –1 υ = log | X | + C⇒ log (Y 2 + X 2 ) + 2 tan –1 Y = CX⇒ log ((y + 3) 2 + (x + 2) 2 ) + 2 tan –1 y + 3x + 2= CWe have(1 + x) n – nx – 1= C 0 + C 1 x + C 2 x 2 + … + C n x n – nx – 1= x 2 [C 2 + C 3 x + … + C n x n–2 ][Q C 1 = n, C 0 = 1]Thus, (1 + x) n – nx – 1 is divisible by x 2 .(cos 2A − sin 2A) + 110.[A,B,C,D] y =2(cos 2A − sin 2A) −1± (cos 2A − sin 2A) + 1⇒ y =± (cos 2A − sin 2A) −1which gives us four values of y, say y 1 , y 2 , y 3 andy 4 . We havecos 2A − sin 2A + 1 (1 + cos 2A) − sin 2Ay 1 ==cos 2A − sin 2A −1(cos 2A −1)+ sin 2A22cos A − 2sin A cos A=2− 2sin A + 2sin A cos Acos A(cos A − sin A)== cot Asin A(cos A − sin A)−(cos 2A − sin 2A) + 1 (1 − cos 2A) + sin 2Ay 2 ==− (cos 2A + sin 2A) −1− (1 + cos 2A) − sin 2A22 sin A + 2 sin A cos A== – tan A2− 2 cos A − 2 sin A cos A(cos 2A − sin 2A) + 1 (1 + cos 2A) − sin 2Ay 3 ==− (cos 2A + sin 2A) −1− (1 + cos 2A) − sin 2A22cos A − 2sin A cos A cos A − sin A== –2− 2cos A − 2sin A cos A cos A + sin A1−tan A ⎛ π ⎞ ⎛ π ⎞= – = – tan ⎜ − A ⎟ = – cot ⎜ + A ⎟1+tan A ⎝ 4 ⎠ ⎝ 4 ⎠−(cos 2A − sin 2A) + 1 (1 − cos 2A) + sin 2Ay 4 ==(cos 2A + sin 2A) −1− (1 − cos 2A) + sin 2A22sin A + 2sin A cos A cos A + sin A==2− 2sin A + 2sin A cos A cos A − sin A1+tan A ⎛ π ⎞= = tan ⎜ + A ⎟ .1−tan A ⎝ 4 ⎠11.[B, C, D]Equations of the given circles can be written as(x – 3) 2 + y 2 = 3 2 (1)and (x + 1) 2 + y 2 = 1 2 (2)Equation of any tangent to circle (2) is(x + 1) cos θ + y sin θ = 1 (3)This will be a tangent to circle (1) also if(3 + 1)cosθ −1= ± 3 ⇒ 4 cos θ – 1 = ± 32 2cos θ + sin θThat is, cos θ = 1 or cos θ = – 21 . When cos θ = 1,we have sin θ = 0, and the equation of thecommon tangent (3) becomesx + 1 = 1 or x = 0 (4)When cos θ = –1/2, we have sin θ = ± 3 / 2 , andthe equations of the common tangents are1 3– (x + 1) ± y = 1 ⇒ x – 3 y + 3 = 0 (5)2 2and x + 3 y + 3 = 0 (6)2XtraEdge for IIT-JEE 90APRIL 2010
12.[A, B, C, D]13.[B]2x t − 5t + 4Let F(x) =∫dt0t2 + e⇒ F′(x) =2So from F′(x) = 0, we get x = 0 orHence x = 0, ±2, ±1.x = –4Shaded region is S 0 . Area of S 0= 4 × 2 – 21 π (1) 2 = 8 – π/214.[D] y ∈ [0, 4], x ∈ [–1, 1]m (t) = costlines y = 2x + 0.4 lies inside the region so⇒ t ∈ [0, 1]t 2 + (2t + 0.4) 2 – 1 ≥ 0 ⇒ t ∈ [0.28, 1]15.[B] (Slope) max. = (cos t) max = cos (0.28) and point is(π, 1)16.[C]18.[A]24 2x − 5x + 4.2xx2 + ex 2 5 ± 25 −165 ± 3== = 4, 12 2DCy = +4–1 1x = 4Ay = –4 B1 r1=2 r23= 13( −1)−1(3)− 6= = – 33 −12x + RTy −1= cos (0.28)x − πC SC Sx =17.[D] tangents = y = ±3tan30º = 32•C 2C 160º1(h + 1) 2 + k 2 = (1 + 2) 2 (circle)Column Matching19. [A] → r; [B] → p,r; [C] → s; [D] → r10(A) ∑ 20 C r = 20 C 0 + 20 C 1 +……+ 20 C 10r=0But, 20 C 0 + 20 C 1 +……+ 20 C 20 = 2 20(B) ∑r=Also, 20 C 20 = 1 = 20 C 0 , 20 C 19 = 20 C 1 , 20 C 18 = 20 C 2etc.∴ given sum = ( 20 C 0 + 20 C 1 +……+ 20 C 20 )– ( 20 C 11 +…..+ 20 C 20 )2 20 + 20 C 10 – ( 20 C 10 + 20 C 9 + ……+ 20 C 0 )∴ 2 ( 20 C 0 + 20 C 1 +…..+ 20 C 10 )= 2 20 + 20 C 101000100 C r (x –3) 100–r 2 r= ((x–3) +2) 100 = (x –1) 100 = (1 –x) 100100100∑r=0C r ( −x)= ∑=r100r0r( −1)( −1)r 100C x∴ Coeff. of x 53 = (–1) 53 100 C 53 = – 100 C 53(C) We have(1+ x) 10 = 10 C 0 + 10 C 1 x + 10 C 2 x 2 +……+ 10 C 10 x 10....(1)Also (1–x) 10 = 10 C 0 – 10 C 1 x + 10 C 2 x 2 +…….…..+ 10 C 10 x 10 ....(2)Multiplying, we get(1 –x 2 ) 10 = ( 10 C 0 + 10 C 1 x + 10 C 2 x 2 +……….+10 C 10 x 10 ) × ( 10 C 0 – 10 C 1 x +10 C 2 x 2 +……...+ 10 C 10 x 10)Equating the coefficients of x 10 , we get10 C 5 (–1) 5 = 10 C 10 0 C 10 – 10 C 10 1 C 9 + 10 C 10 2 C 8 +…….+ 10 C 10 10 C 0⇒ – 10 C 5 = ( 10 C 0 ) 2 – ( 10 C 1 ) 2 + ( 10 C 2 ) 2 +………+ ( 10 C 10 ) 2rrXtraEdge for IIT-JEE 91APRIL 2010
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Volume - 5 Issue - 10April, 2010 (M
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Volume-5 Issue-10April, 2010 (Month
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Hospital in Parel and ApolloHospita
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KNOW IIT-JEEBy Previous Exam Questi
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CHEMISTRY6. From the following data
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(ii)OHH + CH 3OH⊕PorClPClClOCH 3O
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Physics Challenging ProblemsSet #12
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8 QuestionsSolutionSet # 11Physics
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8 QuestionsSolutionSet # 12Physics
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PHYSICSSStudents'ForumExpert’s So
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If the body loses this heat in time
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Perfect gas equation :From the kine
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The internal energy of n molecules
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KEY CONCEPTOrganicChemistryFundamen
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KEY CONCEPTPhysicalChemistryFundame
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UNDERSTANDINGOrganic Chemistry1. An
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The starting compound (A) reacts wi
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