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

The smallest value of k, for which both the roots of the equation x28kx+16(k2k+1) = 0 are real, distinct and have values at least 4,is


A

1

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

2

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

0

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

3

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

The correct option is B

2


α and β 4

x28kx+16(k2k+1)=0

(i) D > 0

64 k264(k2k+1) > 0

k - 1 > 0 k > 1

(ii) 8k2 > 4 k > 1

(iii) f(4) 0

16 - 32k + 16 (k2k+1)0

k23k+20k1 or k 2

k [2,)

least value of k = 2


flag
Suggest Corrections
thumbs-up
39
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Extrema
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon