q1)let A={x1,x2,...,xm},B={y1,y2,...,yn} and f:A→B,n>=m
1.number of increasing function from A to B:
a)mCn b)nCm c)mn d)nC1+nC2+.....+nCn
2.number of non decreasing functions from A to B
3.number of onto functions from A to A
a)m! b)mm c)m!-1 d)none
q2)multi option correct:
if f(x) is a differentiable function satisfying f(100x)=x+f(100x-100),x belongs to R,if f(100)=1,then f(104)=
a)5049 b) r=1Σ100 r c) r=2Σ100 r d)5050
plz show ur working for both questions
1061. nCm
Assuming that increasing means strictly increasing (i.e. one-one function)
no. of ways in which m elements of B can be selected = nCm
In each case of this selection .. .only one permutation exists where the elements of B are arranged in increasing order (which will correspond to elements of A)
So, no. of increasing functions = nCm
2. is it (d)?? (assuming the options for 2 to be same as 1
1is there any other method for q 2.
this is what i did:
f(10000)=100+f(9900) [putting x=100]
f(9900) = 99+f(9800)
f(9800)= 98+f(9700)
.
.
.
f(200)= 2+f(100)
f(100)= 1
---------------------------------
f(10000)+ f(9900)+ ...+ f(200)+ f(100)=(100+99+98+...+2+1)+ (f(9900)+ ...+ f(200)+ f(100))
therefore f(104)= r=1Σ100 r =5050
thus the correct options are B and D.
if any1 used any other process plz post.
1thanks ashish.
well ashish the options for 2 of q1 were longer and complicated so didnt add them.
for ur info the right ans for 2 of q1 was r=0Σm-1 m-1CrnCm-r
1arka the q was number of onto functions from A -> A
where f(x)≠x for any x
did u deliberately change the q??????
in that case number of onto functions is zero as n>m
only possible when n=m which wud have been m!
but as f(x)≠x we shud derange hence ans none
1the answer(as given by fiitjee for 3) is the derangement of m things:
m![1-(1/1!)+(1/2!)-(1/3!)+....+((-1)m/m!)]
1for q2)
put x=2 we get
f(200)=2+f(100)=2+1
x=3
f(300)=3+f(200)=3+2+1
x=4
f(400)=4+f(300)=4+3+2+1
f(100*n)=n+(n-1)+.....+3+2+1=n(n+1)/2
n is integer
so ans b,d