The correct option is B −3
sin2x+acosx+a2>1+cosx
⇒cos2x+(1−a)cosx−a2<0
Now let cosx=t
∴t∈[−1,1], ∀x∈R
So, t2+(1−a)t−a2<0⋯(i)
For the given inequality to be true, [−1,1] should lie in between the roots of equation (i)
Hence required condition are
(1) f(−1)<0⇒a−a2<0⇒a(a−1)>0⇒a<0 or a>1(2) f(1)<0⇒2−a−a2<0⇒a2+a−2>0⇒(a+2)(a−1)>0⇒a<−2 or a>1
So from the above condtitions, we get
a∈(−∞,−2)∪(1,∞)
Hence, the largest negative integral value of a is −3