We have, ∣f(x)f′(x)f′(x)f′′(x)∣=0
⇒f(x)f′′(x)−(f′(x))2=0
⇒f′(x)f′′(x)=f(x)f′(x)
On integrating both side, we get
ln(f′(x))=lnf(x)+lnc
⇒f′(x)=cf(x)
⇒f(x)f′(x)=c
Again integrating, we get
lnf(x)=cx+k1
⇒f(x)=kecx
Since, f(0)=1=k
Therefore, f′(0)=c=2
Now, f(x)=e2x
Hence, f(1)=e2∈(6,9)