1Q.2) use this concept :
polynomial: a0xn+a1xn-1+............+an = 0
sum of roots taking r at a time= (-1)r coefficient at xn-r / coefficient of xn
1) Show that the least value of 2 log 100 a - log a 0.0001, for a > 0 is 4.
2) If the equation x 4-4x 3+ax 2+bx+1=0 has four positive roots, then show that a=6 and b=-4
3) If S = a 1+a 2+a 3+..........+an then show that s/(s-a1) + s/(s-a2) +s/(s-a3) + ...............+ s/(s-a n) > n2 /( n-1).
4) If a+b=1, a>0 , b>0 then prove that (a+1/a) 2 +(b+1/b) 2 ≥ 25/2
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5 Answers
106Q1. 2log100a - loga100-2
= 2log100a + 2 loga100
= 2(log100a + 1log100a)
= 2(x + 1/x)
So the min value is 4 (use AM-GM now)
1Q.4) Use a=sin2x , b=cos2x
now, putting in the equation, u'll get :
(cos4x + sin4x) + (tan4x + cot4x) + 2(tan2x + cot2x) + 6
(now, use AM-GM concept)
1
62Q 3 proved in #3 here: http://www.targetiit.com/iit-jee-forum/posts/a-p-g-p-10650.html
62Q2 was also proved here: http://www.targetiit.com/iit-jee-forum/posts/a-p-g-p-10650.html
