Solution - Career Point

1/ 21 –2∫2 x 2 dx⇒⇒01 2 – 3 2] 1 /[ x 02 31 2 1 – .2 3 2 2=322=2 2628.[B] y ´´ = (y´ + 3) 1/3 ⇒ (y´´) 3 = (y´ + 3) 23⎛2d y ⎞⇒ ⎜ ⎟⎛ dy ⎞ dy–2dx⎜ ⎟⎠ – 6 – 9 = 0⎝ ⎠ ⎝ dx dx∴ order is 2 & degree = 329.[A] x 18 = y 21 = z 28⇒ 18 log x = 21 log y = 28 log z⇒ log y x = 7/6, log z y = 4/3, log x z = 9/14Now, 3, 3 log y x, 3 log z 7, 7 log x 27 4 9= 3, , 3 × , 7 × 2 3 147 9= 3, , 4,2 2which are in A.P.30.[B] log 2 x + log 2 y ≥ 6Here x > 0, y > 0∴ log 2 xy ≥ 6 ⇒ xy ≥ 2 6⇒ xy ≥ 64Now, A.M ≥ G.M.x + y∴ ≥ (xy) 1/22⇒ x + y ≥ 2(64) 1/2x + y ≥ 2 × 8⇒ x + y ≥ 16∴ (x + y) min = 1631.[A] f(x) = x 3 + x 2 + 10x + sinxf´(x) = 3x 2 + 2x + 10 + cos x32.[B]2⎛ 1 ⎞= 3 ⎜ x + ⎟ +⎝ 3 ⎠⇒ f(x) is strictly increasingAlso x → ∞ ⇒ f(x) → ∞,x → – ∞ ⇒ f(x) → – ∞∴ f(x) has only one real root.229 + cos x > 0 ∀ x3Let roots be (2k – 1) & (2k + 1) k ∈ Nthe Sum of roots : 4k = – abQ a ∈ R + , b < 0 as k ≥ 1We have – b = 4ak ⇒ – b ≥ 4a⇒ |b| ≥ 4a {b < 0 ∴ |b| = –b}33.[B] Let G.P. be a + ar + ar 2 ......G.P is infinite so – 1 < r < 1G.P. is decreasing⇒ r > 0 so 0 < r < 1 and therefore a > 0f´(x) = 3x 2 + 3 > 0⇒ f(x) is strictly increasing function∴ f(x) max on [–2, 3] is f(3) = 27 & f´(0) = 3a∴ = 27 & a – ar = 31−r2 4⇒ r = or Q r < 13 334.[B]∴ r = 32& if r = 32 ; a = 9∴ Sum of first three terms = 9 + 6 +4 = 19If z 1 , z 2 & z 3 vertex of equilateral triangle then2z 1 +2z 2 +2z 3 = z 1 z 2 + z 2 z 3 + z 3 z 1∴ (a + i) 2 + (1 + ib) 2 + 0 = (a + i) (1 + ib) + 0 +0⇒ a 2 – 1 + 2ia + 1 – b 2 + 2ib = a + iba + i – b⇒ a 2 – b 2 + i(2a + 2b) = a – b + i(ab + 1)on comparinga 2 – b 2 = a – b and 2a + 2b = ab + 1⇒ (a – b) (a + b – 1) = 0 & 2a + 2b = ab + 1⇒ a = b or a + b = 1 ....(1)⇒ 2a + 2b = ab + 1 .... (2)Now from (1) take a = b put in (2)2a + 2a = a 2 + 1⇒ a 2 – 4a + 1 = 0 ⇒ a = 2 ± 3Q a < 1 ⇒ a = 2 – 3Q a = 2 – 3 & a = b = 2 – 3It we take a + b = 1 & put in (2) then it becomesab = 0 which not possible because a & b liesbetween 0 and 1n ⎡⎤35.[A] S = ⎢⎪⎧r r⎛ ⎞ ⎛ ⎞ ⎪⎫∑ − r 1 3 7( 1) ⎨ + ⎜ ⎟ + ⎜ ⎟ + .... ⎬⎥n Crr=⎢⎥⎣⎪⎩2 ⎝ 4 ⎠ ⎝ 8r 0⎠ ⎪⎭ ⎦nr 1 n= ∑ ( 1) . Cr rr=02nr ⎛ 3 ⎞ n− + ∑( − 1) . ⎜ ⎟ . Cr⎝ 4 ⎠r=0nrr ⎛ 7 ⎞ n+ ( 1) . ⎜ ⎟ . Cr⎠∑ −r=0⎝ 8nnn⎛ 1 ⎞ ⎛ 3 ⎞ ⎛ 7 ⎞= ⎜1 − ⎟ + ⎜1 − ⎟ + ⎜1 − ⎟ + .....⎝ 2 ⎠ ⎝ 4 ⎠ ⎝ 8 ⎠=1 1 +nn222+13n2+ ..... =21n −r1+ ....XtraEdge for IIT-JEE 119APRIL 2010