6is the ans to the 1st one 2) ???
f is a real valued infinitely differentiable function.f(0) = 0 and f"(x) >0 for all real x
then f(x)/x is
1) increasing on (0,∞) 2) increasing on (-∞,∞)
3)decreasing on (0,∞) 3)decreasing on (-∞,∞)
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12 Answers
212)There exists a real valued function f satisfying f(x2)- f2(x)≥1/4. The function can't be a
1) one to one function 2)Constant function
3)increasing function 4)Continuous function
21thanks everyone for ur replies
@kunl, yep urri8, try posting the solution if u have time, its not much lengthy ;)
@Vivek, 1st one is correct from ur side,post the soln pls :P for teh 2nd one, see the hint by prophet sir.
211)for +ve x
p(x) = f(x)/x
p'(x) = {xf'(x) - f(x)}/ x2
[f(x) - f(0)]/x = f'(x1)
f(x)/x < f'(x) as f''(x)>0 so f(x) is increasing
or f(x)/x2<f'(x)/x
which gives p(x) >0 for +ve x
the actual problem was to prove this only..I was the stupid person who created those options (effect of JEE weather ,i must say:D)
1u gave options and all used those options to answer rather than sending solution[3][3]