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

If f(x)=n∣ ∣ ∣ ∣ ∣xnn!2cosxcosnπ24sinxsinnπ28∣ ∣ ∣ ∣ ∣ then find the value of (dndxnf(x))x=0for all(nZ).

Open in App
Solution

Given f(x)=n∣ ∣ ∣ ∣ ∣xnn!2cosxcosnπ24sinxsinnπ28∣ ∣ ∣ ∣ ∣
dndxn[f(x)]=n∣ ∣ ∣ ∣ ∣n!n!2cos(x+nπ2)cosnπ24sin(x+nπ2)sinnπ28∣ ∣ ∣ ∣ ∣
(dndxn[f(x)])x=0=n∣ ∣ ∣ ∣ ∣n!n!2cosnπ2cosnπ24sinnπ2sinnπ28∣ ∣ ∣ ∣ ∣
Since, C1 and C2 are identical
Δ=0

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Adjoint and Inverse of a Matrix
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon