The correct options are
B c=13
C d=1
D Minimum value of f(x) occurs at x=−12
f(x)=4x4+12x3+cx2+6x+d
If a bi-quadratic polynomial is a perfect square, then it must be in the form of
(ax2+px+q)2
=a2x4+2apx3+(p2+2aq)x2+2pqx+q2
Comparing the coefficient, we get
a2=4⇒a=±2
2ap=12⇒p=±3
2pq=6⇒q=±1
⇒a,p,q have same sign.
Now, d=q2⇒d=1
c=p2+2aq
=9+2(±2)(±1)
=9+4
=13
f(x)=(2x2+3x+1)2
=4x4+12x3+13x2+6x+1
The minimum value of f(x) is 0
(2x2+3x+1)2=0
⇒2x2+3x+1=0
⇒(2x+1)(x+1)=0
⇒x=−12 or −1