Definite Integration Question 49
Question 49
- Let $f^{\prime}(x)=\frac{192 x^{3}}{2+\sin ^{4} \pi x}$ for all $x \in R$ with $f \frac{1}{2}=0$. If $m \leq \int_{1 / 2}^{1} f(x) d x \leq M$, then the possible values of $m$ and $M$ are
(2015 Adv.)
(a) $m=13, M=24$
(b) $m=\frac{1}{4}, M=\frac{1}{2}$
(c) $m=-11, M=0$
(d) $m=1, M=12$
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Solution:
- Here, $f^{\prime}(x)=\frac{192 x^{3}}{2+\sin ^{4} \pi x} \therefore \frac{192 x^{3}}{3} \leq f^{\prime}(x) \leq \frac{192 x^{3}}{2}$
On integrating between the limits $\frac{1}{2}$ to $x$, we get
$$ \begin{aligned} & \quad \int_{1 / 2}^{x} \frac{192 x^{3}}{3} d x \leq \int_{1 / 2}^{x} f^{\prime}(x) d x \leq \int_{1 / 2}^{x} \frac{192 x^{3}}{2} d x \ \Rightarrow \quad & \frac{192}{12} x^{4}-\frac{1}{16} \leq f(x)-f(0) \leq 24 x^{4}-\frac{3}{2} \ \Rightarrow \quad & 16 x^{4}-1 \leq f(x) \leq 24 x^{4}-\frac{3}{2} \end{aligned} $$
Again integrating between the limits $\frac{1}{2}$ to 1 , we get
$$ \begin{aligned} & \int_{1 / 2}^{1}\left(16 x^{4}-1\right) d x \leq \int_{1 / 2}^{1} f(x) d x \leq \int_{1 / 2}^{1} 24 x^{4}-\frac{3}{2} d x \ \Rightarrow \quad & \frac{16 x^{5}}{5}-x \quad \leq \int_{1 / 2}^{1} f(x) d x \leq \frac{24 x^{5}}{5}-\frac{3}{2} x_{1 / 2}^{1} \ \Rightarrow \quad & \frac{11}{5}+\frac{2}{5} \leq \int_{1 / 2}^{1} f(x) d x \leq \frac{33}{10}+\frac{6}{10} \ \Rightarrow \quad & 2.6 \leq \int_{1 / 2}^{1} f(x) d x \leq 3.9 \end{aligned} $$
$\left.{ }^{*}\right)$ None of the option is correct.
