Practicing Success
Let $f^{\prime}(x)=\frac{192 x^3}{2+\sin ^4 \pi x}$ for all $x \in R$ with $f\left(\frac{1}{2}\right)=0$. If $m \leq \int\limits_{1 / 2}^1 f(x) d x \leq M$, then the possible values of $m$ and $M$ are |
$m=13, M=24$ $m=\frac{1}{4}, M=\frac{1}{2}$ $m=-11, M=0$ $m=1, M=12$ |
$m=1, M=12$ |
We have, $f'(x)=\frac{192 x^3}{2+\sin ^4 \pi x}$ for all $x \in R$ $\Rightarrow \frac{192}{3} x^3 \leq f'(x) \leq \frac{192}{2} x^3$ for all $x \in\left[\frac{1}{2}, 1\right]$ $\Rightarrow 64 \int\limits_{1 / 2}^x x^3 d x \leq \int\limits_{1 / 2}^x f'(x) d x \leq 96 \int\limits_{1 / 2}^x x^3 d x$ for all $x \in\left(\frac{1}{2}, 1\right)$ $\Rightarrow 16 x^4-1 \leq f(x)-f\left(\frac{1}{2}\right) \leq 24 x^4-\frac{3}{2}$ for all $x \in\left(\frac{1}{2}, 1\right)$ $\Rightarrow 16 x^4-1 \leq f(x) \leq 24 x^4-\frac{3}{2}$ for all $x \in\left(\frac{1}{2}, 1\right)$ $\Rightarrow \int\limits_{1 / 2}^1\left(16 x^4-1\right) d x \leq \int\limits_{1 / 2}^1 f(x) d x \leq \int\limits_{1 / 2}^1\left(24 x^4-\frac{3}{2}\right) d x$ $\Rightarrow 2.6 \leq \int\limits_{1 / 2}^1 f(x) d x \leq 3.9$ Clearly, option (d) is correct. |