Answer

Verified

385.8k+ 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.

Recently Updated Pages

What number is 20 of 400 class 8 maths CBSE

Which one of the following numbers is completely divisible class 8 maths CBSE

What number is 78 of 50 A 32 B 35 C 36 D 39 E 41 class 8 maths CBSE

How many integers are there between 10 and 2 and how class 8 maths CBSE

The 3 is what percent of 12 class 8 maths CBSE

Find the circumference of the circle having radius class 8 maths CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Give 10 examples for herbs , shrubs , climbers , creepers

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

List some examples of Rabi and Kharif crops class 8 biology CBSE

Which are the Top 10 Largest Countries of the World?

The provincial president of the constituent assembly class 11 social science CBSE

Write the 6 fundamental rights of India and explain in detail