1first one trying....
Paper 2
Let x1,x2,....,xn be a sequence of integers such that -1 ≤ xi ≤ 2 for all i belongs to {1,2,...n} ;
x1 +x2 + x3 + ....+ xn = 19
x12+x22 +...+xn2 = 99
Let m,M be the least and greatest values of x13 + x23 +...+xn3 then
(A) m=17
(B) m=19
(C)M/m = 7
(D) M/m = 6
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10 Answers
4Let f: R+ ---> R+ be a differentiable function satisfying f(xy) = f(x)/y + f(y)/x
for all x,y ε R+ also
f(!) = 0 , f ' (1) = 1 . If M be the greatest value of f(x) then find the value of [M+e] .
(where [.] denotes the greatest integer function)
3f(x+h)-f(x)/h = f'(x)
[f(x(1+h/x)) -f(x)]/h
= 1/h[f(x)/(1+h/x) + f(1+h/x)/x -f(x) ]
= f(x)[-1/x]/[1+h/x] + [1/x^2]f(1+h/x)/h/x
as h-->0
-f(x)/x + f'(1)/x^2 = f'(x) !!!
1/x^2-f(x)/x = df(x)/dx
1/x^2 = f(x)/x + f'(x)
IF = e^lnx =x
so ∫1/xdx = xf(x)
lnx = xf(x) + c
put x=1
0=c
lnx/x= f(x)
where f'(x) = 0 f(x) = 1/x..[fr prev eq]
lnx = 1 or x =e!!!
so maxima = 1/e!!!!!!!
[e+1/e] = 3!!!!!!
341The functional equation is xyf(xy) = xf(x) + yf(y)
Let g(x) = xf(x)
Then g(xy) = g(x) + g(y)
This equation is g(x) = c ln x
Then f(x) = c ln x/x
From f'(1)=1, c=1
rest is same as iitimcomin
1no of -1 be p
no of 0 be q
no of 1 be r
no of 2 be s
thus -p+0+r+2s=19................(1)
p+r +4s=99...............(2)
and k= x13+x23+..............xn3=--p+r+8s. ..........(3)
adding (1) and (3) we get k=19+6s now s≥0
thus k≥19
thus m=19
adding (1) and (2) we get
6s=118-2r
now r is an integer and r≠1,0 but r≥2 to be divisible by 6
thus minimum value of r is 2
max value of 6s is 118-4=114
thus max value of k=19+114=133
thus M=133
M/m=133/19=7