
Statement – I: The value of the integral $\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan x}}} }}} $is equal to$\dfrac{\pi }{6}$.
Statement – II: $\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}}} = \int_{\text{a}}^{\text{b}} {{\text{f}}\left( {{\text{a + b - x}}} \right){\text{dx}}} $
$
{\text{A}}{\text{. Statement - I is true, Statement - II is true; Statement - II is not a correct explanation for Statement - I}} \\
{\text{B}}{\text{. Statement - I is true, Statement - II is false}} \\
{\text{C}}{\text{. Statement - I is false, Statement - II is true}} \\
{\text{D}}{\text{. Statement - I is true, Statement - II is true; Statement - II is a correct explanation for Statement - I}} \\
$
Answer
521.1k+ views
Hint: In order to determine the correct answer we compute statements I and II first and then verify which of the options holds true. We solve the given statements using the formula of property of definite integrals, which says$\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}}} = \int_{\text{a}}^{\text{b}} {{\text{f}}\left( {{\text{a + b - x}}} \right){\text{dx}}} $. We refer to the concept of definite integrals to solve and verify the both statements.
Complete step-by-step solution:
Given data,
$\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan x}}} }}} $
$\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}}} = \int_{\text{a}}^{\text{b}} {{\text{f}}\left( {{\text{a + b - x}}} \right){\text{dx}}} $
Let us consider the integral function given in statement I to be ‘X’
$ \Rightarrow {\text{X = }}\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan x}}} }}} {\text{ - - - - }}\left( 1 \right)$
Now according to the property of definite integrals we know that an integral function of the form $\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}}} $can be expressed as$\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {{\text{a + b - x}}} \right){\text{dx}}} $.
I.e., $\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}}} = \int_{\text{a}}^{\text{b}} {{\text{f}}\left( {{\text{a + b - x}}} \right){\text{dx}}} $
This is Statement II itself, hence Statement – II is true.
Let us apply this property on equation – (1), we get
$\Rightarrow {\text{X = }}\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan }}\left( {\dfrac{\pi }{2} - {\text{x}}} \right)} }}} \\
\Rightarrow {\text{X = }}\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{cot x}}} }}} \\
\Rightarrow {\text{X = }}\int_{\pi /6}^{\pi /3} {\dfrac{{\sqrt {{\text{tan x}}} }}{{1 + \sqrt {{\text{tan x}}} }}{\text{dx}}} {\text{ - - - }}\left( 2 \right) \\ $
Adding equations (1) and (2) we get
$\Rightarrow 2{\text{X = }}\int_{\pi /6}^{\pi /3} {\left[ {\dfrac{{\sqrt {{\text{tan x}}} }}{{1 + \sqrt {{\text{tan x}}} }} + \dfrac{1}{{1 + \sqrt {{\text{tan x}}} }}} \right]{\text{dx}}} \\
\Rightarrow 2{\text{X = }}\int_{\pi /6}^{\pi /3} {{\text{dx}}} \\
\Rightarrow 2{\text{X = }}\left[ {\text{x}} \right]_{\pi /6}^{\pi /3} \\
\Rightarrow 2{\text{X = }}\dfrac{\pi }{6} \\
\Rightarrow {\text{X = }}\dfrac{\pi }{{12}} \\ $
Hence we get $\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan x}}} }}} = \dfrac{\pi }{{12}}$
Therefore the statement is false.
Hence we get Statement – I is false, Statement – II is true.
Option C is the correct answer.
Note: In order to solve this type of question the key is to know the concepts of integration and definite integrals. Once we identify the statement – II is an identity, it is true. Then we use the same to compute the answer of integral in statement I without even simplifying the integral. General knowledge of how to perform integration and substitute limits later is required to perform integration problems.
Complete step-by-step solution:
Given data,
$\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan x}}} }}} $
$\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}}} = \int_{\text{a}}^{\text{b}} {{\text{f}}\left( {{\text{a + b - x}}} \right){\text{dx}}} $
Let us consider the integral function given in statement I to be ‘X’
$ \Rightarrow {\text{X = }}\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan x}}} }}} {\text{ - - - - }}\left( 1 \right)$
Now according to the property of definite integrals we know that an integral function of the form $\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}}} $can be expressed as$\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {{\text{a + b - x}}} \right){\text{dx}}} $.
I.e., $\int_{\text{a}}^{\text{b}} {{\text{f}}\left( {\text{x}} \right){\text{dx}}} = \int_{\text{a}}^{\text{b}} {{\text{f}}\left( {{\text{a + b - x}}} \right){\text{dx}}} $
This is Statement II itself, hence Statement – II is true.
Let us apply this property on equation – (1), we get
$\Rightarrow {\text{X = }}\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan }}\left( {\dfrac{\pi }{2} - {\text{x}}} \right)} }}} \\
\Rightarrow {\text{X = }}\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{cot x}}} }}} \\
\Rightarrow {\text{X = }}\int_{\pi /6}^{\pi /3} {\dfrac{{\sqrt {{\text{tan x}}} }}{{1 + \sqrt {{\text{tan x}}} }}{\text{dx}}} {\text{ - - - }}\left( 2 \right) \\ $
Adding equations (1) and (2) we get
$\Rightarrow 2{\text{X = }}\int_{\pi /6}^{\pi /3} {\left[ {\dfrac{{\sqrt {{\text{tan x}}} }}{{1 + \sqrt {{\text{tan x}}} }} + \dfrac{1}{{1 + \sqrt {{\text{tan x}}} }}} \right]{\text{dx}}} \\
\Rightarrow 2{\text{X = }}\int_{\pi /6}^{\pi /3} {{\text{dx}}} \\
\Rightarrow 2{\text{X = }}\left[ {\text{x}} \right]_{\pi /6}^{\pi /3} \\
\Rightarrow 2{\text{X = }}\dfrac{\pi }{6} \\
\Rightarrow {\text{X = }}\dfrac{\pi }{{12}} \\ $
Hence we get $\int_{\pi /6}^{\pi /3} {\dfrac{{{\text{dx}}}}{{1 + \sqrt {{\text{tan x}}} }}} = \dfrac{\pi }{{12}}$
Therefore the statement is false.
Hence we get Statement – I is false, Statement – II is true.
Option C is the correct answer.
Note: In order to solve this type of question the key is to know the concepts of integration and definite integrals. Once we identify the statement – II is an identity, it is true. Then we use the same to compute the answer of integral in statement I without even simplifying the integral. General knowledge of how to perform integration and substitute limits later is required to perform integration problems.
Recently Updated Pages
Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Why is the cell called the structural and functional class 12 biology CBSE

a Tabulate the differences in the characteristics of class 12 chemistry CBSE

Who discovered the cell and how class 12 biology CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE
