wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Let f:(0,π)R be a twice differentiable function such that limtxf(x)sintf(t)sinxtx=sin2x for all x(0,π).
If f(π6)=π12, then which of the following statement(s) is (are) TRUE?

A
f(π4)=π42
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
f(x)<x46x2 for all x(0,π)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
There exists α(0,π) such that f(α)=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
f′′(π2)+f(π2)=0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are
B f(x)<x46x2 for all x(0,π)
C There exists α(0,π) such that f(α)=0
D f′′(π2)+f(π2)=0
limtxf(x)sintf(t)sinxtx=sin2x
By using L Hopital's rule on LHS, we get
limtxf(x)costf(t)sinx1=sin2x
f(x)cosxf(x)sinxsin2x=1
ddx(f(x)sinx)=1
f(x)sinx=x+C
Given, f(π6)=π12C=0
f(x)=xsinx

f(π4)=π42

We know sinx>xx36
f(x)=xsinx<x2x46

f(x)=xsinx is continuous on [0,π] and differentiable on (0,π).
Also, f(0)=f(π)=0
By Rolle's theorem, there exists α(0,π) such that f(α)=0

f(x)=sinxxcosx
f′′(x)=2cosx+xsinx
f′′(π2)+f(π2)=0



flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
L'hospitals Rule
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon