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⎪⎧∴ tan ⎨⎪⎩∴n∑r=1lim tann→∞⎪⎪⎫−1⎛1 ⎞ (2n + 1) −1tan ⎜ ⎟2 ⎬ ==⎝ 2r ⎠⎪⎭1+(2n + 1).1⎪⎧⎨⎪⎩n∑r=1nn + 1⎪⎫−1⎛1 ⎞tan ⎜ ⎟2 ⎬ =⎝ 2r ⎠lim nn→∞⎭n + 1= 1Numerical Response type questions :12. [3]We have92| z |= (2 + cos θ) 2 + sin 2 θ = 5 + 4 cos θ (1)and3 3 + = 4 + 2 cos θz z (2)Eliminating θ from (1) and (2), we get9 ⎛ 1 1 ⎞2 – 6 ⎜ + ⎟⎠ = – 3| z | ⎝ z z⇒ 3 = 2( z + z) – |z| 213. [6]Clearly x > 0 and x ≠ 1/5log55 − log5xlog 5x (5/x) =log55 + log5xPutting log 5 x = t, then equation becomest 2 1− t+ = 11+t⇔ t 3 + t 2 – 2t = 0⇔ t(t – 1) (t + 2) = 0⇔ t = 0, 1, – 2So integral roots are 1 and 5.14. [2]Let the A.P. be a – d, a, a + d, a + 2d. Note that aand d must be integers. Also as this is an increasinga + 2d is the largest. We havea + 2d = (a – d) 2 + a 2 + (a + d) 2= 3a 2 + 2d 2⇒ 3a 2 – a + 2d 2 – 2d = 0As a is real,1 – 8(d 2 – d) ≥ 0⇒ d 2 – d – 81 ≤ 0⇒2⎛ 1 ⎞ 3⎜d − ⎟ ≤⎝ 2 ⎠ 81 3 1 3⇒ – ≤ d ≤ +2 2 2 2 2 2As d is an integer, d = 0, 1But d ≠ 0, therefore, d = 1.Thus 3a 2 – a = 0 ⇒ a = 0 or a = 1/3.As a is an integer, a = 0.Hence, required number is 2.15. [1] We haveE = 2n + 1 C 1 + 2n + 1 C 2 + … + 2n + 1 C n – 2n + 1 C 0– 2n + 1 C 1 – … – 2n + 1 C n[using n C r = n C n – r ]= – 2n + 1 C 0 = – 1.∴ |E| = 116. [3] The given system of equations will have a nontrivialsolution ifa + 2t b c∆ = b c + 2t a = 0cab + 2tClearly ∆ is a cubic polynomial in t and has 3 roots.17. [3] None that it is not given that f is a differentiablefunction We havef (4 + h) − f (4)f ′(4) = limh→0h2f( 4 + h ) )= limh→0(4 + h)= limh→0h⎡8⎢1+= lim⎣h→018. [2]dy2x + 2y = 0 dx3/ 2h3 h2 4− f (22)− 8 8[(1 + h / 4)= limh→0h3/ 2−1]⎤ ⎡3⎤+ ...... −1⎥8⎢h + .....⎦ 8 ⎥= lim⎣ ⎦hh→0hdy x⇒ = –dx y⇒dydx(1,3)1= –3Therefore, the equation of the tangent at (1,3 ) isy – 3 = (–1/ 3 ) (x – 1)and the point of intersection of this tangent with thex-axis is (4, 0). The equation of the normal at(1, 3 ) is y – 3 = 3 (x – 1), and the point ofintersection of this normal with the x-axis is (0, 0).Hence the required area is1 . 4 3 = 2 3219. [4] The two curves represent parabolas withvertices at (0, 0) and (3, 0). They intersect at (1, 1)and (1, –1), so the required area isarea of OPMQO = 2 (area of OPMO).XtraEdge for IIT-JEE 98APRIL 2010