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∀k∈R
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.