Discussion Forum

let   $f:[\frac{1}{2},1]\rightarrow R$ ( the set of all real numbers) be a positive, non -constant and differentiable function such that f'(x)<2f(x)  and   $f(\frac{1}{2})=1$   .Then , the value of   $\int_{1/2}^{1} f(x) dx$ lies in the interval 

Answer:

Explanation:

Concept involved 

 When evere we have linear differential equation containing inequality we should always check for increasing or decreasing 

i.e, for    $\frac{dy}{dx}+Py<0$

   $\Rightarrow$     $\frac{dy}{dx}+Py>0$

 Multiply by integrating factor i.e, $e^{\int_{}^{}Pdx }$  and  convert into total differential equation

 Here, f'(x) <2 f(x) , multiplying by  $e^{-\int_{}^{}2dx }$

 $f'(x).e^{-2x}-2e^{-2x}f(x)<0$

 $\Rightarrow$     $\frac{d}{dx} (f(x).e^{-2x}) <0$

$\therefore$     $\phi(x)=f(x)e^{-2x}$   is decreasing for   $x \in \left[\frac{1}{2},1\right]$

 thus, when  x >1/2

       $\phi(x) < \phi(\frac{1}{2})$

$\Rightarrow$      $e^{-2x} f(x)<e^{-1}.f(\frac{1}{2})$

 $\Rightarrow$      $f(x)<e^{2x-1}.1 given f(\frac{1}{2})=1$

$\Rightarrow$      $0< \int_{1/2}^{1} f(x) dx <\int_{1/2}^{1} e^{2x-1}dx$

$\Rightarrow$     $0< \int_{1/2}^{1} f(x) dx <\left(\frac{e^{2x-1}}{2}\right)^{1}_{1/2}$

$\Rightarrow$        $0< \int_{1/2}^{1} f(x) dx <\frac{e-1}{2}$