Discussion Forum

 Let f:R→ R be a continuous  odd function, which vanishes exactly  at one point   $f(1)=\frac{1}{2}$ . Suppose that $F(x)=\int_{-1}^{x} f(t) dt $    for all x ε [-1,2] and    $G(x)=\int_{-1}^{x}t|f\left\{ f(t)\right\}|dt$ for all  x ε [-1,2] If   $\lim_{x \rightarrow 1}\frac{F(x)}{G(x)}=\frac{1}{14}$ , then the value of    $f(\frac{1}{2})$  is

Answer:

Explanation:

Here,   $\lim_{x \rightarrow1}\frac{F(x)}{G(x)}=\frac{1}{14}$

     $\Rightarrow$      $\lim_{x \rightarrow1}\frac{F^{'}(x)}{G^{'}(x)}=\frac{1}{14}$

                                                   [ Using  L' Hospital's rule ]...........(i)

As                  $F(x)=\int_{-1}^{x} f(t)dt$

 $\Rightarrow$   F'(x)= f(x)     .........(ii)

and      $G(x)=\int_{-1}^{x}t|f\left\{ f(t)\right\}|dt$

$\Rightarrow$            $G'(x)=x|f\left\{f(x)\right\}|$            ...........(iii)

    $\therefore$      $\lim_{x \rightarrow 1}\frac{F(x)}{G(x)}=\lim_{x \rightarrow 1}\frac{F'(x)}{G'(x)}=\lim_{x \rightarrow 1}\frac{f(x)}{x|f\left\{f(x)\right\}|}$

                              =   $\frac{f(1)}{1|f\left\{f(1)\right\}|}=\frac{1/2}{|f(1/2)|}$  ........(iv)

 Given,     $\lim_{x \rightarrow 1}\frac{F(x)}{G(x)}=\frac{1}{14}$

  $\therefore$    $\frac{\frac{1}{2}}{|f\left(\frac{1}{2}\right)|}=\frac{1}{14}\Rightarrow |f\left(\frac{1}{2}\right)|=7$