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

If α,β0, and f(n)=αn+βn and
∣ ∣ ∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣ ∣ ∣=K(1α)2(1β)2(αβ)2, then K is equal to:

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

The correct option is C 1
f(1)=α+β
f(2)=α2+β2
f(3)=α3+β3
f(4)=α4+β4
So, ∣ ∣ ∣31+f(1)1+f(2)1+f(1)1+f(2)1+f(3)1+f(2)1+f(3)1+f(4)∣ ∣ ∣=∣ ∣ ∣31+α+β1+α2+β21+α+β1+α2+β21+α3+β31+α2+β21+α3+β31+α4+β4∣ ∣ ∣
Splitting it as a product of two determinants.
=∣ ∣1111αβ1α2β2∣ ∣×∣ ∣1111αβ1α2β2∣ ∣
=∣ ∣1111αβ1α2β2∣ ∣2

C1C1C2, C2C2C3
=∣ ∣0011ααββ1α2α2β2β2∣ ∣2
=[(1α)(aβ)]2∣ ∣00111β1+αα+ββ2∣ ∣2
=[(1α)(αβ)]2(β1)2
=(1α)2(1β)2(αβ)2
Hence, K=1

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