1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Assertion :If n is a positive integer then ∫nπ0∣∣∣sinxx∣∣∣dx≥2π(1+12+13+...+1n) Reason: In the interval (0,π2), sinxx≥2π

A
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
Assertion is correct but Reason is incorrect
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
Both Assertion and Reason are incorrect
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion∫nπ0∣∣∣sinxx∣∣∣dx=∫π0∣∣∣sinxx∣∣∣dx+∫2ππ∣∣∣sinxx∣∣∣dx+...+∫nπ(n−1)π∣∣∣sinxx∣∣∣dx =I1+I2+I3+...+In(say)Putting x=t+π,t+2π,...,t+(n−1)π respectively in I2,I3,...,In =∫π0sinxxdx+∫π0∣∣∣sin(t+π)t+π∣∣∣dt+∫π0∣∣∣sin(t+2π)t+2π∣∣∣dt+... =∫π0sinxxdx+∫π0sintt+πdt+∫π0sintt+2πdt+...so on (∵sinxx>0in(0,π)) =∫π0sinxxdx+∫π0sinxx+πdx+∫π0sinxx+2πdx+...(merely cbanging the variable) =n∑r=1∫π0sinx(n+(r−1)π)>n∑r=1∫π0sinx(π+(r−1)π) =n∑r=1∫π0sinxπrdx=n∑r=12πr=2π[1+12+13+...+1n]

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
MATHEMATICS
Watch in App
Join BYJU'S Learning Program