11yeah i agree its option B
Q1)The value of ∫a1[x]f'(x)dx,a>1,where [x] denotes the greatest integer not exceding x is?
Q2) If I1=∫01 2x2dx ;I2=∫012x3dx ;I3=∫12 2x2dx; I4=∫122x3dx then
a)I1>I2
b)I2>I1
c)I3>I4
d)I3=I4
Pls explain the above sols also
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32 Answers
1yes the correct ans for 1st is b - post no.20
where are the options for q.1 ??
i cannot c them [12]
1as we hv done in finding limit as a sum u hv to consider all inbetween pts nd if u consider thm thn this cn't be true it is nt equal even in first two cases its jst x2>x3
21@moon....
vivek is correct....
it cant be optn D inque 1
its I4>I3
coz u countover entire interval 1,2 then obviously x3 > x2
1and vivek_aero thr is one mistake in ur sol.
as u hv taken x2≥x3
thn in the limit from 1 to 2 it shd be x2≤x3 thn d option shd also be correct
bt we hv to consider 0<x<1 for I1 and I2 and 1<x<2 for I3 and I4
62f(2)-f(1)
+2f(3)-2f(2)
+3f(4)-3f(3)
......
......
......
[a]f(a)-[a]f([a])
=[a]f(a)-{f(1)+f(2)+....f([a])}
hence option b
21yeah subhash found mist.,....
it gets beter with da optns u know.....
guyz n galz take a = 3.3 and u'll get the answer
11vivek and tapanmast why have u not applied the limits for f(x) as well
in q1
1options of 1st ques:
a) af(a) - {f(1)+f(2)+................+f([a])}
b) [a]f(a) - {f(1)+f(2)+................+f([a])}
c) [a]f([a]) - {f(1)+f(2)+................+f(a)}
d) af([a]) - {f(1)+f(2)+................+f(a)}
21IT CANT BE JUS f(x)
has to depend on a;
I am gettin : [a]*([a]-1)*f(x)/2
1wait i'll put the options there too.............coz ans is nt like u hv given
1Since x2≥x3
I1>I2
for the second part I = 1∫2 f'(x) + 2∫32* f'(x) + .....
= ([a]-1)([a])*f(x)/2
1I1>I2
for x→[0,1]
x2≥x3
for the first qn do you have any other info on f(x)?
1hey don't start guessing work................correct ans is 'a'..............tell hw