Here, f(x)=x3+3x2+4x+bsinx+ccosx
⇒f′(x)=3x2+6x+4+bcosx−csinx
Now for f(x) to be one-one function, the only possibility is
f′(x)>0 ∀x∈R
⇒3x2+6x+4+bcosx−csinx>0∀x∈R
⇒3x2+6x+4>csinx−bcosx,∀x∈R
⇒3x2+6x+4>√b2+c2,∀x∈R
⇒√b2+c2<3(x2+2x+1)+1,∀x∈R
⇒√b2+c2<3(x+1)2+1,∀x∈R
⇒√b2+c2<1,∀x∈R
⇒b2+c2<1,∀x∈R
Also, b2+c2≥0
⇒0≤b2+c2<1,∀x∈R
∴[b2+c2]=0