Solution - Career Point

MATHEMATICAL CHALLENGESSOLUTION FOR THIS ISSUE (SET # 12)a −xna − x −1x n1. I n = ∫ e x dx0 = ( e − x x n ) a 0+ n ∫ e0 dx= – e –a a n + nI n – 1I n = –e –a a n + n[–e –a a n–1 + (n–1)I n–2 ]= –e –a [a n + na n–1 +n(n – 1)a n–2 + n(n – 1)(n – 2)a n–3+ ..... + n(n – 1)..... 2a] + n I 0=⎡⎪⎧n n−12 ⎤−xa −aa a a ⎪⎫n ⎢( − e )0− e ⎨ + + .... + + a⎬⎥⎢⎣⎪⎩ n n −12 ⎪⎭ ⎥⎦I n =⎡ ⎪⎧⎪⎫⎤⎢ − −2na a an 1 e ⎨1+ a + + ..... + ⎬⎥⎢⎣⎪⎩ 2 n ⎪⎭ ⎥⎦Now, I = ∫ ∞ −e x x n =2. Mid pt. M of BC = (0, –1)0C(–2, 1) 60ºM45ºLt I n = na→∞B(2, –3)−3−1slope of BC = = –12 + 2so slope of altitude of ∆PBC is = 1.length BC = 16 + 16 = 4 2Now altitude PM = 4 2 sin 60ºP= 4 2 .A(6, 5)3 = 2 62eq n . of PM line isx − 0 y + 1 = = r (as its slope is 1)1 12 2r rx = & y = – 12 2for req. pt. P take r = 2 6 .⎛⎞So pt. P ≡ ⎜2 6 2 6, −1⎟= ( 2 3, 2 3 − 1)⎝ 2 2 ⎠n3. S n = ∑r=0n1n=1r . n C r = n . 2 n–1so S = ∑S = ∑ nnn=1n.2−1S = 1 + 2.2 1 + 3.2 2 + 4.2 3 + .... + n . 2 n–12S = 2 + 2.2 2 + 3.2 3 + ..... + (n – 1). 2 n–1 + n.2 n(1 – 2)S = (1 + 2 + 2 2 + 2 3 + .... + 2 n–1 ) – n . 2 n2 n 1= 1 . – n . 2n2 −−1S = n . 2 n – 2 n + 1 = (n – 1) 2 n + 1Now A.M. ≥ G. M.n n nnC1+ 2. C2+ 3. C3+ .... + n. Cn≥1+2 + 3 + ..... + n⎧ n⎨ C1.⎩n 2 n 3 n n 1+2+... + n( C ) .( C ) ......( C )⎫23n(n+1)2n⎬⎭⎛ n−1⎞⎜ n.2 ⎟⎜ n(n + 1) ⎟ ≥ n C 1 ( n C 2 ) 2 ..... ( n C n ) n⎜ ⎟2⎝ ⎠⎛nso n C 1 . ( n C 2 ) 2 ..... ( n C n ) n 2 ⎞≤ ⎜ ⎟n 1⎝ + ⎠n+1C24. Let P be a point represented by 1.so as |z 1 – 1| = |z 2 – 1| = |z 3 – 1| so P is thez1 + z2+ z3circumcentre of ABC. Its centroid is3If the ∆ABC is equilateral then circumcentre = centroidsoz1 + z2+ z31 =3so z 1 + z 2 + z 3 = 3Now if z 1 + z 2 + z 3 = 3 then centroid of ∆ABC is 1which is point P and P is already the circumcentre of∆ABC. So now if they are same then ∆ABC isequilateral.5. let the centroid of ∆ABC be (h, k) then3h = r cos α + r cos β + r3hα + β α −β– 1 = 2 cos cosr2 2...(1)&3kr= sin α + sin βα −β α + β= 2 cos sin2 2...(2)1XtraEdge for IIT-JEE 41 APRIL 2010