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

If the roots of x26kx+(22k+9k2)=0 are greater than 3, then the range of k is

A
(1,)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
(119,)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
(,1)(119,)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
(1,119)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B (119,)
Let f(x)=x26kx+(22k+9k2), whose roots are α,β
Given: α,β>3

Conditions:
(i) Δ036k24(22k+9k2)0
k1 (1)

(ii) b2a>36k2>3k>1 (2)

(iii) f(3)>09k220k+11>0
(k1)(9k11)>0
k<1 or k>119 (3)

From (1),(2) and (3), we get
k(119,)

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Nature and Location of Roots
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon