JEE quesn AOD

f ( x ) ≤ c F ( x )

Hence , f ( x ) ≤ [ Minimum value of F ( x ) ] c .............also .

But , F ( x ) ≥ 0 ........And , F ( 0 ) = 0 .

It follows that ,

f ( x ) ≤ 0

But , it is provided that , f ( x ) ≥ 0

Hence , " f ( x ) = 0 " is the only possibility .

The logic is slightly flawed there. All you get is that f(0)=0 from that.

Solution:

Let [0, x_o] be the largest closed interval such that f(x)=0 throughout.

We have for any x>x₀ F(x)>0

Since F'(x) = f(x), the equation is F'(t) ≤ cF(t)

Hence F'(t)/F(t) ≤ c

Hence _x0+ε ∫^x F'(t)/ F(t) dt ≤ c _x0+ε∫^x dt

or F(t) \le e^{c(x-x_0-\epsilon)}F(x_0 + \epsilon )

Continuity of F(t) means that it can be brought arbitrarily close to zero as we please. Hence F(t)=0 for all x which means f(x)=0 for all x

But sir , if

f ( x ) ≤ c F ( x ) ;

would it not be a true statement that " f " must be smaller than the minimum value of " c F " also , which is simply zero ?

:D sorry for misreading the question ;)

Alternative:
Since f(x) is continuous, it means that F(x) is differentiable and so by the Leibniz rule we get
F'(x) =f(x). The given condition, therefore, becomes
F'(x) ≤ c F(x) for all x≥0
i.e. F'(x) - cF(x) ≤ 0
Since e^-cx is a positive quantity for any real x. So multiplication by e^-cx throughout gives
F'(x) e^-cx -c e^-cx F(x) ≤ 0
The LHS is simply the derivative of e^-cx F(x). So we get that for all x≥0,
(e^-cx F(x))' ≤ 0
This implies that for x≥0, the function h(x) = e^-cx F(x) is non-increasing. Hence,
h(x) ≤ h(0)
for any x≥0. But h(0) = F(0) = 0.
Hence, we get
h(x) = e^-cx F(x) ≤0 for all x≥ 0.
Again multiplying by e^cx gives F(x) ≤ 0 for all x≥0.
On the other hand, since f(x) ≥ 0, we have F(x)≥0. And so we must have F(x) ≡ 0 for all x≥0, which in turn imply f(x) = 0 identically for all x≥0.

Remark: The solution remains the same if we replace the point 0 with an arbitrary x₀.