SATHEE: Definite Integration Question 19

Definite Integration Question 19

Question 19

Let $a + b = 4$ , where $a < 2$ and let $g (x)$ be a differentiable function. If $\frac{d g}{d x} > 0, \forall x$ prove that $\int_{0}^{a} g (x) d x + \int_{0}^{b} g (x) d x$ increases as $(b - a) \underset{(1997, 5 M)}{increases.}$

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Answer:

Correct Answer: 19. $(n = 3)$

Solution:

Let $t = b - a$ and $a + b = 4$

[given]

$\begin{aligned} \Rightarrow t = 4 - a - a & \Rightarrow t = 4 - 2 a & \Rightarrow a = 2 - \frac{t}{2} & and t = b - (4 - b) & \Rightarrow t = 2 b - 4 & \Rightarrow \frac{t}{2} = b - 2 & \Rightarrow b = 2 + \frac{t}{2} \end{aligned}$

Again, $a < 2$

$\begin{array}{ll} \Rightarrow & 2 - \frac{π}{2} < 2 \Rightarrow & \frac{π}{2} > 0 \Rightarrow t > 0 \end{array}$

[given]

Now, $\int_{0}^{a} g (x) d x + \int_{0}^{b} g (x) d x$

$= \int_{0}^{2 - t / 2} g (x) d x + \int_{0}^{2 + t / 2} g (x) d x$

Let $F (x) = \int_{0}^{2 - t / 2} g (x) d x + \int_{0}^{2 + t / 2} g (x) d x$

For $t > 0, F^{'} (t) = - \frac{1}{2} g 2 - \frac{t}{2} + \frac{1}{2} g 2 + \frac{t}{2}$

[using Leibnitz’s rule]

$= \frac{1}{2} g 2 + \frac{t}{2} - \frac{1}{2} g 2 - \frac{t}{2}$

Again, $\frac{d g}{d x} > 0, \forall x \in R$

[given]

Now, $2 - t / 2 < 2 + t / 2 ∴ t > 0$

We get $g (2 + t / 2) - g (2 - t / 2) > 0, \forall t > 0$

So, $F^{'} (t) > 0, \forall t > 0$

Hence, $F (t)$ increases with $t$ , therefore $F (t)$ increases as $(b - a)$ increases.

Definite Integration Question 19

Question 19

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Definite Integration Question 19

Question 19

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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