Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
−2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is A0 f(x)=3x4+4x3−12x2+a. ⇒f′(x)=12x3+12x2−24x =12x(X2+x−2)=12x(x+2)(x−1)
Clearly f′(x) changes its sign from +ve to −ve in the neighbourhood of 0, hence f(x) attains maximum value at x=0. f′(x) changes its sign from −ve to +ve in the neighbourhood of x=1 & x=−2.