One root of the equation \[\cos x - x + \frac{1}{2} = 0\] lies in the interval
A. \[\left[ {0,\frac{\pi }{2}} \right]\]
В. \[\left[ { - \frac{\pi }{2},0} \right]\]
C. \[\left[ {\frac{\pi }{2},\pi } \right]\]
D. \[\left[ {\pi ,\frac{{3\pi }}{2}} \right]\]
Answer
54.6k+ views
HINT
When algebraic operations are used to solve a trigonometric problem, the term "root loss" describes how some roots are lost. Usually, it happens as both sides cancel out any identical phrases. The roots that are obtained when the terms in trigonometric equations are squared are known as extraneous roots. To accurately find the answer, it is necessary to check for extraneous roots.
Complete Step-by Step Solution:
We have been provided in the question about cos function that,
Let, \[f(x) = \cos x - x + \frac{1}{2} = 0\]
And we are to find the interval where the root lies in.
Now, we have to assume that suppose,
\[{\rm{x}} = 0\]
Then on substituting\[{\rm{x}} = 0\]on the equation above, we obtain
\[f(0) = \cos 0 - 0 + \frac{1}{2}\]
Now, we have to simplify the above equation, we obtain
\[ = \frac{1}{2} + \cos \left( 0 \right)\]
We have been already known that the x-coordinate (1) is equivalent to the value of
\[cos{0^\circ }\]
Therefore, we can write it as below
\[cos{0^\circ } = 1\]
On using the above trivial identity, the equation becomes
\[ = 1 + \frac{1}{2}\]
Now, we have to simplify the above expression to less complicated form:
Thus, we obtain
\[ = \frac{3}{2} > 0\]
Now replace the equation\[f(x) = \cos x - x + \frac{1}{2} = 0\]with\[{\rm{x}} = \frac{\pi }{2}\]
On substituting the value, the equation becomes
\[f\left( {\frac{\pi }{2}} \right) = \cos \frac{\pi }{2} - \frac{\pi }{2} + \frac{1}{2} = \frac{1}{2} - \frac{\pi }{2} < 0\]
Hence, the resultant will be obtained from the previous calculation as below,
\[f(0) > 0\]and\[f\left( {\frac{\pi }{2}} \right) < 0\]
This shows that \[{\rm{f}}\] changes its sign from\[\left( {0,\frac{\pi }{2}} \right)\].
This indicates that it contains at least one root\[\left( {0,\frac{\pi }{2}} \right)\] where\[f\]changes its sign from\[\left( {0,\frac{\pi }{2}} \right)\] in the interval\[\left( {0,\frac{\pi }{2}} \right)\].
Therefore, one root of the equation \[\cos x - x + \frac{1}{2} = 0\] lies in the interval
\[\left( {0,\frac{\pi }{2}} \right)\]
Hence, option A is correct
Note
When learning trigonometry, students frequently run across mistakes, misunderstandings, and barriers. While thinking about the cases, one should carefully analyze the regions. Students found it difficult while finding intervals. In general, just solve it without any limitations. Apply the constraint after that to your collection of solutions. The interval constraint may occasionally result in a novel solution that could be useful. Students are likely to make mistakes in these types of problems as it contains intervals which are sometimes tougher to solve.
When algebraic operations are used to solve a trigonometric problem, the term "root loss" describes how some roots are lost. Usually, it happens as both sides cancel out any identical phrases. The roots that are obtained when the terms in trigonometric equations are squared are known as extraneous roots. To accurately find the answer, it is necessary to check for extraneous roots.
Complete Step-by Step Solution:
We have been provided in the question about cos function that,
Let, \[f(x) = \cos x - x + \frac{1}{2} = 0\]
And we are to find the interval where the root lies in.
Now, we have to assume that suppose,
\[{\rm{x}} = 0\]
Then on substituting\[{\rm{x}} = 0\]on the equation above, we obtain
\[f(0) = \cos 0 - 0 + \frac{1}{2}\]
Now, we have to simplify the above equation, we obtain
\[ = \frac{1}{2} + \cos \left( 0 \right)\]
We have been already known that the x-coordinate (1) is equivalent to the value of
\[cos{0^\circ }\]
Therefore, we can write it as below
\[cos{0^\circ } = 1\]
On using the above trivial identity, the equation becomes
\[ = 1 + \frac{1}{2}\]
Now, we have to simplify the above expression to less complicated form:
Thus, we obtain
\[ = \frac{3}{2} > 0\]
Now replace the equation\[f(x) = \cos x - x + \frac{1}{2} = 0\]with\[{\rm{x}} = \frac{\pi }{2}\]
On substituting the value, the equation becomes
\[f\left( {\frac{\pi }{2}} \right) = \cos \frac{\pi }{2} - \frac{\pi }{2} + \frac{1}{2} = \frac{1}{2} - \frac{\pi }{2} < 0\]
Hence, the resultant will be obtained from the previous calculation as below,
\[f(0) > 0\]and\[f\left( {\frac{\pi }{2}} \right) < 0\]
This shows that \[{\rm{f}}\] changes its sign from\[\left( {0,\frac{\pi }{2}} \right)\].
This indicates that it contains at least one root\[\left( {0,\frac{\pi }{2}} \right)\] where\[f\]changes its sign from\[\left( {0,\frac{\pi }{2}} \right)\] in the interval\[\left( {0,\frac{\pi }{2}} \right)\].
Therefore, one root of the equation \[\cos x - x + \frac{1}{2} = 0\] lies in the interval
\[\left( {0,\frac{\pi }{2}} \right)\]
Hence, option A is correct
Note
When learning trigonometry, students frequently run across mistakes, misunderstandings, and barriers. While thinking about the cases, one should carefully analyze the regions. Students found it difficult while finding intervals. In general, just solve it without any limitations. Apply the constraint after that to your collection of solutions. The interval constraint may occasionally result in a novel solution that could be useful. Students are likely to make mistakes in these types of problems as it contains intervals which are sometimes tougher to solve.
Last updated date: 05th Jun 2023
•
Total views: 54.6k
•
Views today: 0.10k
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Name the Largest and the Smallest Cell in the Human Body ?

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

A ball impinges directly on a similar ball at rest class 11 physics CBSE

Lysosomes are known as suicidal bags of cell why class 11 biology CBSE

Two balls are dropped from different heights at different class 11 physics CBSE

A 30 solution of H2O2 is marketed as 100 volume hydrogen class 11 chemistry JEE_Main

A sample of an ideal gas is expanded from 1dm3 to 3dm3 class 11 chemistry CBSE
