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

Let f(x)=1;2x<0x21; 1x2

and g(x)=|f(x)|+f(|x|). Then, in the interval (2,2), g is :

A
not continuous
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
not differentiable at one point
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
not differentiable at two points
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
differentiable at all points
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B not differentiable at one point
|f(x)|=⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪1; 2x<01x2; 0x<1x21; 1x2

f(|x|)=x21 for x[2,2]

g(x)=⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪1+x21; 2x<01x2+x21; 0x<1x21+x21; 1x2

g(x)=⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ x2 ; 2x<0 0 ; 0x<12(x21) ; 1x2

Clearly, g is continuous in (2,2)

g(x)=⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪ 2x ; 2x<0 0 ; 0x<1 4x ; 1x2

g(0)=0 and g(0+)=0
g is differentiable at x=0

g(1)=0 and g(1+)=4
g is not differentiable at x=1

flag
Suggest Corrections
thumbs-up
12
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Differentiability in an Interval
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon