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

Let f(x) is a polynomial of degree 4, with f(2)=1,f(2)=0,f′′(2)=2,f′′′(2)=12,f′′′′(2)=24, then the value of f′′(1) is -

A
24
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
26
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
28
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
30
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 26
f(x)=a(x2)4+b(x2)3+c(x2)2+d(x2)1
f(x)=4a(x2)3+3b(x2)2+2c(x2)+d
f(2)=0d=0
Similarly f′′(x)=12a(x2)2+6b(x2)+2c
f′′(2)=22c=2c=1
f′′′(x)=24a(x2)+6b
6b=12b=2
f′′′′(x)=24a=24a=1

Thus, f′′(x)=12(x2)212(x2)+2
f′′(1)=26

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