Answer
Verified
428.1k+ views
Hint: Here, we are required to find the relation between the given two expressions. We will use the range of \[\cos x\] and change the limit of the integral accordingly. We will find the value of the first integral and then substitute its value in the second integral. We will solve it further to get the required relation between the two expressions.
Complete step-by-step answer:
According to the question, we are given that \[{I_1} = \int\limits_0^{n\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\]and \[{I_2} = \int\limits_0^{5\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\]
Now, as we know that \[\cos x \in \left[ {0,\pi } \right]\].
Therefore, period of \[\left| {\cos x} \right| = \pi \]
Hence, for \[{I_1} = \int\limits_0^{n\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] , we can write this as:
\[ \Rightarrow {I_1} = n\int\limits_0^\pi {f\left( {\left| {\cos x} \right|} \right)} dx\]………………………….. \[\left( 1 \right)\]
Similarly, for \[{I_2} = \int\limits_0^{5\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] ,
\[ \Rightarrow {I_2} = 5\int\limits_0^\pi {f\left( {\left| {\cos x} \right|} \right)} dx\]
Dividing both sides by 5, we get,
\[ \Rightarrow \dfrac{{{I_2}}}{5} = \int\limits_0^\pi {f\left( {\left| {\cos x} \right|} \right)} dx\] ……………………………. \[\left( 2 \right)\]
Now, substituting the value of \[\int\limits_0^\pi {f\left( {\left| {\cos x} \right|} \right)} dx\] from equation \[\left( 2 \right)\] in equation \[\left( 1 \right)\] , we get,
\[{I_1} = n\dfrac{{{I_2}}}{5}\]
Dividing both sides by \[n\], we get
\[ \Rightarrow \dfrac{{{I_1}}}{n} = \dfrac{{{I_2}}}{5}\]
Therefore, if \[{I_1} = \int\limits_0^{n\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] and \[{I_2} = \int\limits_0^{5\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] and \[\left( {n \in N} \right)\] , then \[\dfrac{{{I_1}}}{n} = \dfrac{{{I_2}}}{5}\]
Hence, option C is the correct answer.
Note: In this question, we are given two definite integrals having the lower and upper limits respectively. As we know that the range of \[\cos x\] is \[\left[ {0,\pi } \right]\] this means that the minimum value which the trigonometric function \[\cos x\] can take is 0 and the maximum value taken by it can be \[\pi \]. We have shown the closed brackets because \[\cos x\] includes both the values 0 and \[\pi \]. Whereas, in the case of \[\sec x\] , its range lies between \[\left( {0,\pi } \right)\]. Hence, these open brackets show that it can never take the values 0 and \[\pi \] and hence, it will always lie between these values. Therefore, for solving this question, it is really important to know the ranges of various trigonometric functions.
Complete step-by-step answer:
According to the question, we are given that \[{I_1} = \int\limits_0^{n\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\]and \[{I_2} = \int\limits_0^{5\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\]
Now, as we know that \[\cos x \in \left[ {0,\pi } \right]\].
Therefore, period of \[\left| {\cos x} \right| = \pi \]
Hence, for \[{I_1} = \int\limits_0^{n\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] , we can write this as:
\[ \Rightarrow {I_1} = n\int\limits_0^\pi {f\left( {\left| {\cos x} \right|} \right)} dx\]………………………….. \[\left( 1 \right)\]
Similarly, for \[{I_2} = \int\limits_0^{5\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] ,
\[ \Rightarrow {I_2} = 5\int\limits_0^\pi {f\left( {\left| {\cos x} \right|} \right)} dx\]
Dividing both sides by 5, we get,
\[ \Rightarrow \dfrac{{{I_2}}}{5} = \int\limits_0^\pi {f\left( {\left| {\cos x} \right|} \right)} dx\] ……………………………. \[\left( 2 \right)\]
Now, substituting the value of \[\int\limits_0^\pi {f\left( {\left| {\cos x} \right|} \right)} dx\] from equation \[\left( 2 \right)\] in equation \[\left( 1 \right)\] , we get,
\[{I_1} = n\dfrac{{{I_2}}}{5}\]
Dividing both sides by \[n\], we get
\[ \Rightarrow \dfrac{{{I_1}}}{n} = \dfrac{{{I_2}}}{5}\]
Therefore, if \[{I_1} = \int\limits_0^{n\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] and \[{I_2} = \int\limits_0^{5\pi } {f\left( {\left| {\cos x} \right|} \right)} dx\] and \[\left( {n \in N} \right)\] , then \[\dfrac{{{I_1}}}{n} = \dfrac{{{I_2}}}{5}\]
Hence, option C is the correct answer.
Note: In this question, we are given two definite integrals having the lower and upper limits respectively. As we know that the range of \[\cos x\] is \[\left[ {0,\pi } \right]\] this means that the minimum value which the trigonometric function \[\cos x\] can take is 0 and the maximum value taken by it can be \[\pi \]. We have shown the closed brackets because \[\cos x\] includes both the values 0 and \[\pi \]. Whereas, in the case of \[\sec x\] , its range lies between \[\left( {0,\pi } \right)\]. Hence, these open brackets show that it can never take the values 0 and \[\pi \] and hence, it will always lie between these values. Therefore, for solving this question, it is really important to know the ranges of various trigonometric functions.
Recently Updated Pages
what is the correct chronological order of the following class 10 social science CBSE
Which of the following was not the actual cause for class 10 social science CBSE
Which of the following statements is not correct A class 10 social science CBSE
Which of the following leaders was not present in the class 10 social science CBSE
Garampani Sanctuary is located at A Diphu Assam B Gangtok class 10 social science CBSE
Which one of the following places is not covered by class 10 social science CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Why is there a time difference of about 5 hours between class 10 social science CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Draw a labelled sketch of the human eye class 12 physics CBSE