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

The number of integers a in the interval [1,2014] for which the system of equations x+y=a;x2x1+y2y1=4 has finitely many solutions is


A

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

1007

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

2013

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

2014

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

2013


x21+1x+1+y21+1y1=4x+1+1x1+y+1+1y1=4a+2+1x1+1(a1)x=4(a1)x+x+1(x1)[(a1)x]=2aa2 [for a = 2 equation have infinitely many solution]

(x - 1)[(a - 1) - x] = -1

(x - 1)[x - (a - 1)] = 1

x2 - ax + (a - 2) = 0

D > 0

equation have 2 real roots so

a can be 1, 3, 4....... 2014

Answer is 2013


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