Discussion Forum

Let f:[0,1]→ R (the set of all real numbers) be a function .Suppose  the function f is twice differentiable

 f(0)=f(1) =0 and satisfies

 $f '' (x)-2 f'(x)+ f(x) \geq e^{x}, x \in [0,1]$

 Which of the following is true for 0<x <1 

Answer:

Explanation:

 Concept involved It is based on the concept of converting into total differential equation (i.e, completing the equation into differential ). So , as to check the function to be increasing or decreasing

 Here   $f''(x) -2 f'(x) + f(x) \geq e^{x}$

$\Rightarrow $    $f''(x)e^{-x} - f'(x)e^{-x} - f'(x)e^{-x}+f(x)e^{-x} \geq0$

$\Rightarrow \frac{d}{dx}(f'(x)e^{-x})-\frac{d}{dx}(f(x)e^{-x})\geq1$

  $\Rightarrow \frac{d}{dx}(f'(x)e^{-x}-f(x)e^{-x})\geq1$

$\Rightarrow \frac{d^{2}}{dx^{2}}(e^{-x}f(x))\geq1$ for all x $\in$ [0,1]

 $\therefore$    $\phi(x) =e^{-x}f(x) $  is concave $n\phi$

 f(0)=f(1)=0

 $\Rightarrow$    $\phi(0)=0=\phi(1)$

 $\Rightarrow  $     $\phi(x)<0$

$\Rightarrow $       $e^{-x}f(x)<0$
$\therefore$     $ f(x) <0$