1q1)For g(x) to be continuous,
[2/a]=[√26/a]
so least integer value of a is 6.
[.]-G.I.F.
1) If f(x) is continous for allx belonging to R and range of f(x) is (2, √26) and g(x) = [ f(x) / a ] is continous for all x belonging to R (where [.] denotes the greatest integral function). Then the least posiitve integral value of a is
(a)2 (b)3 (c)6 (d)5
2) Let f(x) = max {x ,0 } for all x belonging to R. Then f(xy) ≠f(x) . f(y) if,
(a) x>0 , y>o (b)x <0 , y<0 (c) x>0, y<0 (d)x <0, y>0
-
UP 0 DOWN 0 0 2
2 Answers
11For x>0,y>0 ...f(x)=x,f(y)=y...hence given condition is satisfied...
for x<0,y<0. f(x)=f(y)=0. but f(xy) is +ve, as xy is positive.Hence (b) is the right option....
similarly, check the other options......
