May 2011 - Career Point

3. Find the sum of the terms of G.P. a + ar + ar 2 + ..... + ∞where a is the value of x for which the function7 + 2x log e 25 – 5 x – 1 – 5 2–x has the greatest value andx t dtr is the Ltx→∫ 02x tan( π + x20 )aSol. S =1− r,To get the greatest value f´(x) = 2log e 25 – 5 x – 1 log 5+ 5 2–x log 5f ´(x) = 4 log e 5 – 5 x–1 log e 5 + 5.5 1 – x log e 5⇒ f ´(x) = 0 put 5 x – 1 t(> 0)t 2 – 4t – 5 = 0 ⇒ t = 5 ⇒ 5 x –1 = 5 ⇒ x = 2to evaluate r :x 2t dt 1r = Ltx→∫=0 02x tan( π + x)π1 2πsince a = 2, r = ⇒ sum of G.P. = π π −14. Let [x] stands for the greatest integer function find23 x + sin xthe derivative of f (x) = ( x + [ x + 1]) , where itexists in (1, 1.5). Indicate the point(s) where it doesnot exist. Give reason(s) for your conclusion.Sol. The greatest integer [x 3 + 1] takes jump from 2 to 3 at3 2 and again from 3 to 4 at 3 3 in [1, 1.5] andtherefore it is discontinuous at these two points. As aresult the given function is discontinuous at 3 2 andhence not differentiable.To find the derivative at other points we write :in (1, 3 2 ), f (x) = ( x + 2)2x + sin x−12x + sin x⇒ f ´(x) = ( x + 2){x 2 + sin x + (x + 2) (2x + cos x) log (x + 2)}2x + sin xin ( 3 2 , 3 3 ), f (x) = ( x + 3) ,2x + sin x−1f ´(x) = ( x + 3) {x 2 + sin x+ (2x + cos x) (x + 3) × log e (x + 3)}2x + sin xin ( 3 5 , 1.5), f (x) = ( x + 4) ,2x + sin x−1f ´(x) = ( x + 4) , {x 2 + sin x + (2x + cos x)(x + 4) × log e (x + 4)}5. For three unit vectors â , bˆ and ĉ not all collineargiven that aˆ× cˆ= cˆ× bˆand b ˆ× aˆ= aˆ×c ˆ . Show thatcosα + cos β + cos γ = –3/2, where α, β and γ are theangles between â and bˆ , bˆ and ĉ and ĉ and ârespectively.Sol. â × ĉ = cˆ× bˆ⇒ ( â + bˆ ) × ĉ = → 0⇒ ĉ is collinear with â + bˆ ⇒ â + bˆ = λ ĉ forsame λ ∈ RSimilarly bˆ + ĉ = µ â for some scalar uNow â + bˆ = λ ĉ ⇒ â + bˆ + ĉ = (λ + 1) → cSimilarly ⇒ â + bˆ + ĉ = (µ + 1) âHence (λ + 1) ĉ = (µ + 1) â ,either λ + 1 = µ + 1 = 0 or ĉ is collinear with â .But ĉ can not be collinear to â other wise cˆ × aˆ= 0⇒ cˆ × bˆ= 0⇒ bˆ is collinear to with ĉ⇒ â bˆ and ĉ are collinear.Hence ĉ is not collinear to â⇒ λ + 1 = µ + 1 = 0⇒ λ ± µ = –1Hence bˆ + ĉ = µ â⇒ â + bˆ + ĉ = → 0⇒ ( â + bˆ + ĉ ) . ( â + bˆ + ĉ ) = 0⇒ 1 + 1 + 1 + 2 ( â . bˆ + bˆ . ĉ + ĉ . â ) = 03⇒ â . bˆ + bˆ . ĉ + ĉ . â = – 2⇒ cos α + cos β + cos γ = – 236. If a, b, c and n are positive integers such thata + b + c = n, show that(a a b b c c ) 1/n + (a b b c c a ) 1/n + (a c b a c b ) 1/n ≤ n.Sol. Since a, b, c are integers, from A.M. – G.M.inequality we can write( a + a + .... atimes)+ ( b + b + ... btimes)+ ( c + c + ... ctimes)a + b + c1/a +b+c≥ [(a.a....a times)(b.b....b times)(c.c...c times)]1a.a + b.b + c.c a b c⇒≥ ( a b c ) a+b+ca + b + c1c.a + a.b + b.c c a cSimilarly,≥ ( a b c ) a+b+cc + a + b1b.a + c.b + a.c b c aand≥ ( a b c ) b+c+ab + c + aAdding these three inequalities, we get2 2 2a + b + c + 2ab+ 2bc+ 2ca≥ (a a b b c c ) 1/na + b + c+ (a c b a c b ) 1/n + (a b b c c a ) 1/n2( a + b + c)where LHS =a + b + c= a + b + c Hence provedXtraEdge for IIT-JEE 41 MAY 2011