CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

The real number k for which the equation 2x3+3x+k=0 has two dinstinct real roots in [0,1] :

A
lies between 1 and 2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
lies between 2 and 3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
lies between 1 and 0
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
does not exist
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D does not exist
Let f(x)=2x3+3x+k
f(x)=6x2+3>0 kR
Thus, f(x) is strictly increasing function
Hence, f(x)=2x3+3x+k=0 has only one real root, so two roots are not possible.

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