The correct option is D a = 0 and b = -6
Given, f(x)=x2+(a+1)x+b.
And, 2 and -3 are zeroes of f(x)
⇒f(2)=0
⇒(2)2+(a+1)×2+b=0
⇒4+2a+2+b=0
⇒6+2a+b=0…(i)
and f(−3)=0
⇒(−3)2+(a+1)×(−3)+b=0
⇒9−3a−3+b=0
⇒6−3a+b=0…(ii)
On subtracting Eq. (ii) from Eq. (i), we get
5a=0⇒a=0
⇒6+b=0 [from Eq. (i)]
⇒b=−6
Hence,a=0 and b=−6