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

Let a−2b+c=1, If f(x)=∣∣ ∣∣x+ax+2x+1x+bx+3x+2x+cx+4x+3∣∣ ∣∣, then:

A
f(50)=501
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
f(50)=1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
f(50)=1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
f(50)=501
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C f(50)=1
Given f(x)=∣ ∣x+ax+2x+1x+bx+3x+2x+cx+4x+3∣ ∣,
a2b+c=1
Applying R1R12R2+R3
f(x)=∣ ∣a2b+c00x+bx+3x+2x+cx+4x+3∣ ∣
Using a2b+c=1
f(x)=(x+3)2(x+2)(x+4)
f(x)=(x+3)2(x+2)(x+4)
f(x)=1
f(50)=1
f(50)=1

Alternative:
Take a=1,b=0=c and proceed

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Evaluation of Determinants
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon