Solution - Career Point

20.[C]Let P(h, k) be one of point of contact thenk = sin h ...(1)equation of tangent is y – k = cos h (x – h)which parries through origin∴ k = h cos h ...(2)from (1) & (2)2k+ k 2 = sin 2 h + cos 2 h = 12h⇒ k 2 + k 2 h 2 = h 2 ⇒ x 2 – y 2 = x 2 y 2∴ locus of (h, k) is x 2 – y 2 = x 2 y 224.[A] 1 –4 2= (1 + cos 2 x) 3/2 ⎡⎤⎢− (1 + cos x) + 2⎥ ⎣ 5⎦= 52 (1 + cos 2 x) 3/2 (3 – 2cos 2 x) + cx 2 +3x 44– .......... = cos x∴ I =∫cos x dx = sin x+ c21.[B] x + y = 16, x, y > 0s = x 3 + y 3 = x 3 + (16 – x) 3ds = 3x 2 – 3(16 – x) 2dxds = 0dx⇒ x 2 = 256 – 32x + x 2 ⇒ x = 822.[C]2d s= 6x – 6(16 – x) = 6[2x – 16]2dxat x = 823d s d s= 0 and = 12 ≠ 023dx dxHence there in no minimum exist.1 1 2− −2 2 3b a=c1 1 1− −2b a c1 1 2⇒ + = b a c= c22ab⇒ c =a + c∴ c is H.M. of a & b23.[B] I =∫1+ cos x sin 2x cos 2x dxput 1 + cos 2 x = t 2⇒ –2sin x cos x dx = 2tdt⇒ – sin 2x dx = 2tdt22∴ I = –∫t .(2t dt).(2cos x − 1)= –∫t.2t.(2t − 3) dt2225.[A]πx→226.[A] y =27.[B]∫x−cos t(2 −1)dtπ / 2⎛ 0 ⎞lim ⎜ ⎟2x⎛ π ⎞⎝ 0 ⎠∫⎜ t − ⎟ dt2π / 4⎝2 ⎠−cos x2 −1lim=πx→⎛ π ⎞2 ⎜ x − ⎟.2x⎝ 2 ⎠⎛ 1 ⎞lim ⎜1+2 ⎟⎝ n ⎠n→∞log y =nn→∞r=12limπx→2−cos xln2.sin x=4x− π2462 2 2 2 2n ⎛ 2 ⎞ n ⎛ 3 ⎞ n⎜1+ ⎟ ⎜1⎟2 +2⎜⎝n⎟⎠⎛ n.... ⎜1+⎝ n⎛2∑ ⎟ ⎞⎜rlim log1 + .2⎝ n ⎠n⎛= ∑⎜rlim log→∞1 +nr=1 ⎝ n1=∫2 x log(1 + x2 ) dx02=∫1⇒ y = 4/e22⎟ ⎞⎠log t dt = ( t log t − t) 2 1= 2 log 2 – 1 = log e4required area122⎞⎟⎠⎜⎝2n2n2r2n2 r 1. . n nn⎟⎠y = 2x 2ln2π⎡ 2t= –2 ⎢⎢⎣553t− 33⎤⎥⎥⎦+ c01/√2= – 54 (1 + cos 2 x) 5/2 + 2(1 + cos 2 x) 3/2 + c–1XtraEdge for IIT-JEE 118APRIL 2010