Discussion Forum

Let  $f'(x)=\frac{192x^{3}}{2+\sin^{4}\pi x}$ for all x ε R with  $f(\frac{1}{2})=0$   If   $m\leq\int_{1/2}^{1} f(x) dx\leq M,$ , then the possible values of m and M are

Answer:

Explanation:

(b)

             Here,   $f'(x)=\frac{192x^{3}}{2+\sin^{4}\pi x}$

$\therefore$             $\frac{192x^{3}}{3}\leq f'(x)\leq \frac{192x^{3}}{2}$

 On integrating between the limits   $\frac{1}{2}$   to x, we get

  $\int_{1/2}^{x} \frac{192x^{3}}{3}dx\leq\int_{1/2}^{x} f'(x) dx\leq\int_{1/2}^{x}\frac{192x^{3}}{2} dx$

              $\Rightarrow \frac{192}{12}\left(x^{4}-\frac{1}{16}\right)\leq f(x)-f(0)$ $\leq24x^{4}-\frac{3}{2}$

               $\Rightarrow 16x^{4}-1\leq f(x) \leq24x^{4}-\frac{3}{2}$

 Again integrating  between the limits   $\frac{1}{2}$   to 1, we get

      $\int_{1/2}^{1} (16x^{4}-1)dx\leq\int_{1/2}^{1} f(x) dx\leq\int_{1/2}^{1} \left(24x^{4}-\frac{3}{2}\right)dx$

   $\Rightarrow$            $\left[\frac{16x^{5}}{5}-x\right]_{1/2}^1\leq \int_{1/2}^{1} f(x) dx$

                                                                                        $\leq\left[\frac{24x^{5}}{5}-\frac{3}{2}x\right]_{1/2}^1$

     $\Rightarrow$    $\left(\frac{11}{2}+\frac{2}{5}\right)\leq\int_{1/2}^{1}f(x) dx\leq\left(\frac{33}{10}+\frac{6}{10}\right)$

     $\Rightarrow$    $2.6\leq  \int_{1/2}^{1}  f(x) dx\leq3.9$