341-5/3 is not a solution as {x}≥0
1st one is JEE 2010 quesn.....
Lest s ={ 1 , 2 , 3 , 4 } . What r the total no. of unordered pairs of disjoint subsets of S ?
2. Let {x } and [x ] denote the fractional and integral part of a real no. x respectively.
solve 4 {x} = x + [x]
3. Let R = (5 * root 5 + 11) and f = R - [R] , where [ x ] denotes the greatest integer funcn . Prove that Rf = 4 ^ (2n+1).
plzz give the soln. or hints only!!
-
UP 0 DOWN 0 0 9
9 Answers
12)4{x}={x}+2[x]
=>3{x}=2[x]
=> -1<2[x]/3<1
[x]= -1,0,1
and cresponding {x}= -2/3,0,2/3
hence x=-5/3,0,5/3.
6thanx archana
actually i was also gettin three answers but the book in which i saw it has given only 5/3 as the ans..
nevertheless i think the method is correct!!!!
1yes sorry......{x} lies in the interval [0,1)
3)Let R=I+f=(5√5+11)2n+1
f'=(5√5-11)2n+1
R-f'=I+f-f'=2[2n+1C1 (5√5)2n *11+2n+1C2(5√5)2n-1 112......]
=even integer
I+f-f'=an even integer.
hence f-f'=0 as 0≤f-f'<1 f=f'
so,Rf=Rf'=(5√5+11)2n+1 * (5√5-11)2n+1=42n+!.
341The answer depends on whether the union of the pair is S or not. Here I take it that the union of the two sets need not be S
Let us generalize the problem to the set {1,2,...,n}
Consider first the ordered pairs of disjoint subsets L and R.
Every element either belongs to L or to R or to N= S- \left(L\bigcup{R}\right)
Hence each pair of disjoint subsets corresponds bijectively with a sequence (more accurately a word) a_1a_2...a_n where each letter is L, R or N.
Among these the sequence \underbrace{NN...N}_{\text{n places}} is not admissible
Hence the number of ordered pairs is 3^n-1
Since we are looking for unordered pairs, the number of such pairs is
\frac{3^n-1}{2}