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

The set of values of a for which the inequality (x3a)(xa3)<0 is satisfied for all x in the interval 1x3 are

A
(13,3)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(0,13)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(2,0)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
(2,3)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct options are
B (2,0)
C (0,13)
The given inequality (x3a)(xa3)<0 satisfies for all x in interval 1x3
Hence, x=1,2,3 should also satisfy the given inequality. We find solutions for x=1,2,3
x=1 (13a)(a2)<0 (13a)(a+2)>0
Solution for above inequality are a>2 and a<1/3 .......A
x=2 (23a)(a1)<0 (23a)(a+1)>0
Solution for above inequality are a>1 and a<2/3 .......B
Again,
x=3 (33a)(a)<0 (33a)(a)>0
Solution for above inequality are a>0 and a<1 .......C
A,B and C equations give solutions to be a(0,13) and a(2, 0)

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