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

If f(x)=ex1+ex,I1=f(a)f(a)xg{x(1x)}dx and I2=f(a)f(a)g{x(1x)}dx, then the value of I2I1 is

A
2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
-1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
-3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 2
f(a)=ea1+ea=1ex+1, f(a)=ea1+eaf(a)+f(a)=1
I1=f(a)f(a)xg{x(1x)}dx=f(a)f(a)[f(a)+f(a)x]g{[f(a)+f(a)x}{1f(a)f(a)+x}]dx
=f(a)f(a)(1x)g[(1x)x]dx=f(a)f(a)g{x(1x)}dxf(a)f(a)xg[x(1x)]dx=I2I1
2I1=I2I2I1=2

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Integration of Piecewise Functions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon