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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