Q. 133.5( 37 Votes )
On which of the f
Answer :
It is given that f (x) = x100 + sin x –1
Then, f’(x) = 100x99 + cosx
In interval (0,1), cos x >0 and 100x99 > 0
⇒ f’(x)>0
Therefore, function f is strictly increasing in interval (0,1).
In interval, cos x < 0 and 100x99 > 0.
Also, 100x99 > cos x
⇒ f’(x) > 0 in
Therefore, function f is strictly increasing in interval .
In interval , cos x < 0 and 100x99 > 0.
Also, 100x99 > cos x
⇒ f’(x) > 0 on
Therefore, function f is strictly increasing in interval .
Hence, function f is strictly decreasing on none of the intervals.
Rate this question :
How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
view all courses
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation


RELATED QUESTIONS :
Find the interval
RD Sharma - Volume 1Find the interval
RD Sharma - Volume 1Find the interval
RD Sharma - Volume 1Find the interval
RD Sharma - Volume 1Find the interval
RD Sharma - Volume 1Find the interval
RD Sharma - Volume 1Show that the alt
Mathematics - Board PapersFind the interval
RD Sharma - Volume 1Find the interval
RD Sharma - Volume 1Find the interval
RD Sharma - Volume 1