SATHEE: Indefinite Integration 1 Question 40

Indefinite Integration 1 Question 40

41. Statement I The value of the integral

(2013 Main)

$\int_{π / 6}^{π / 3} \frac{d x}{1 + \sqrt{\tan x}} is equal to π / 6 .$

Statement II $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$

(a) Statement I is correct; Statement II is correct; Statement II is a correct explanation for Statement I

(b) Statement I is correct; Statement II is correct; Statement II is not a correct explanation for Statement I

(d) Statement I is incorrect; Statement II is correct

Passage Based Questions

Passage I

Let $F : R \to R$ be a thrice differentiable function. Suppose that $F (1) = 0, F (3) = - 4$ and $F^{'} (x) < 0$ for all $x \in (1, 3)$ . Let $f (x) = x F (x)$ for all $x \in R$ .

(2015 Adv.)

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Solution:

Let $I = \int_{π / 6}^{π / 3} \frac{d x}{1 + \sqrt{\tan x}}$

$\begin{aligned} ∴ & I & = \int_{π / 6}^{π / 3} \frac{d x}{1 + \sqrt{\tan \frac{π}{2} - x}} \\ = \int_{π / 6}^{π / 3} \frac{d x}{1 + \sqrt{\cot x}} \\ \Rightarrow & I = \int_{π / 6}^{π / 3} \frac{\sqrt{\tan x} d x}{1 + \sqrt{\tan x}} \end{aligned}$

On adding Eqs. (i) and (ii), we get

$\begin{aligned} 2 I & = \int_{π / 6}^{π / 3} d x \\ \Rightarrow & 2 I & = [x]_{Π / 6}^{π / 3} d x \\ \Rightarrow & I & = \frac{1}{2} \frac{π}{3} - \frac{π}{6} = \frac{π}{12} \end{aligned}$

Statement I is false.

But $\int_{a}^{b} f (x) d x = \int_{a}^{b} f (a + b - x) d x$ is a true statement by property of definite integrals.

Indefinite Integration 1 Question 40

41. Statement I The value of the integral

Passage Based Questions

Passage I

Solution:

Engineering Preparation

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Indefinite Integration 1 Question 40

41. Statement I The value of the integral

Passage Based Questions

Passage I

Solution:

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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