SATHEE: Definite Integration Question 49

Definite Integration Question 49

Question 49

Let $f^{'} (x) = \frac{192 x^{3}}{2 + \sin^{4} π x}$ 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) $m = \frac{1}{4}, M = \frac{1}{2}$

(d) $m = 1, M = 12$

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Solution:

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

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

$\begin{aligned} \int_{1 / 2}^{x} \frac{192 x^{3}}{3} d x \leq \int_{1 / 2}^{x} f^{'} (x) d x \leq \int_{1 / 2}^{x} \frac{192 x^{3}}{2} d x \Rightarrow & \frac{192}{12} x^{4} - \frac{1}{16} \leq f (x) - f (0) \leq 24 x^{4} - \frac{3}{2} \Rightarrow & 16 x^{4} - 1 \leq f (x) \leq 24 x^{4} - \frac{3}{2} \end{aligned}$

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

$\begin{aligned} \int_{1 / 2}^{1} (16 x^{4} - 1) d x \leq \int_{1 / 2}^{1} f (x) d x \leq \int_{1 / 2}^{1} 24 x^{4} - \frac{3}{2} d x \Rightarrow & \frac{16 x^{5}}{5} - x \leq \int_{1 / 2}^{1} f (x) d x \leq \frac{24 x^{5}}{5} - \frac{3}{2} x_{1 / 2}^{1} \Rightarrow & \frac{11}{5} + \frac{2}{5} \leq \int_{1 / 2}^{1} f (x) d x \leq \frac{33}{10} + \frac{6}{10} \Rightarrow & 2.6 \leq \int_{1 / 2}^{1} f (x) d x \leq 3.9 \end{aligned}$

$^{*})$ None of the option is correct.

Definite Integration Question 49

Question 49

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Definite Integration Question 49

Question 49

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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