Indefinite Integration 1 Question 40

41. Statement I The value of the integral

(2013 Main)

$$ \int _{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}} \text { is equal to } \pi / 6 \text {. } $$

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

(c) Statement I is correct; Statement II is false

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

Passage Based Questions

Passage I

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

(2015 Adv.)

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Solution:

  1. Let $I=\int _{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}}$

$$ \begin{array}{rlrl} \therefore & I & =\int _{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan \frac{\pi}{2}-x}} \\ & =\int _{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\cot x}} \\ \Rightarrow \quad & I=\int _{\pi / 6}^{\pi / 3} \frac{\sqrt{\tan x} d x}{1+\sqrt{\tan x}} \end{array} $$

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

$$ \begin{array}{rlrl} & & 2 I & =\int _{\pi / 6}^{\pi / 3} d x \\ \Rightarrow & 2 I & =[x] _{\Pi / 6}^{\pi / 3} d x \\ \Rightarrow & I & =\frac{1}{2} \frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{12} \end{array} $$

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.



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