Let f^{prime}(x)=frac{192 x^3}{2+sin ^4 pi x} | vedclass

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(D) Given $f^{\prime}(x) = \frac{192 x^3}{2+\sin^4 \pi x}$. Since $0 \leq \sin^4 \pi x \leq 1$,we have $2 \leq 2+\sin^4 \pi x \leq 3$.Thus,$\frac{192

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$m=1, M=12$

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(D) Given $f^{\prime}(x) = \frac{192 x^3}{2+\sin^4 \pi x}$. Since $0 \leq \sin^4 \pi x \leq 1$,we have $2 \leq 2+\sin^4 \pi x \leq 3$.Thus,$\frac{192 x^3}{3} \leq f^{\prime}(x) \leq \frac{192 x^3}{2}$,which simplifies to $64 x^3 \leq f^{\prime}(x) \leq 96 x^3$.Integrating from $\frac{1}{2}$ to $x$ with $f(\frac{1}{2}) = 0$:$\int_{1/2}^x 64 t^3 dt \leq f(x) \leq \int_{1/2}^x 96 t^3 dt$$16(x^4 - (\frac{1}{2})^4) \leq f(x) \leq 24(x^4 - (\frac{1}{2})^4)$$16x^4 - 1 \leq f(x) \leq 24x^4 - 1.5$.Now,integrate from $\frac{1}{2}$ to $1$:$\int_{1/2}^1 (16x^4 - 1) dx \leq \int_{1/2}^1 f(x) dx \leq \int_{1/2}^1 (24x^4 - 1.5) dx$$[\frac{16x^5}{5} - x]_{1/2}^1 \leq \int_{1/2}^1 f(x) dx \leq [\frac{24x^5}{5} - 1.5x]_{1/2}^1$$(\frac{16}{5} - 1) - (\frac{16}{5 \cdot 32} - \frac{1}{2}) \leq \int_{1/2}^1 f(x) dx \leq (4.8 - 1.5) - (\frac{24}{5 \cdot 32} - 0.75)$$1.1 - (-0.4) \leq \int_{1/2}^1 f(x) dx \leq 3.3 - (-0.6) = 3.9$.Since $1 < 1.5 \leq \int_{1/2}^1 f(x) dx \leq 3.9 < 12$,the values $m=1$ and $M=12$ satisfy the inequality.

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 $m \leq \int_{1 / 2}^1 f(x) d x \leq M$,then the possible values of $m$ and $M$ are

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