Given f(x)=∣2x2+3x−2∣+sinxcosx
⇒f(x)=∣(2x−1)(x+2)∣+sinxcosx
Now, {f}^{'}(x)={\begin{matrix}4x+3+\frac{\mathrm{cos}2x}{4}, & \frac{1}{2}<x<1 \\ -(4x+3)+\frac{\mathrm{cos}2x}{4}, & 0\leq x<\frac{1}{2}\end{matrix}
For 0≤x<21⇒f′(x)<0
For 21<x≤1⇒f′(x)>0
So f(x) has minima at x=21 and maxima at x=1
Hence, f(21)+f(1)=3+21(1+2cos1)sin1