Useful results:

• asin(x) + bcos(x) has range [-sqrt(a^{2+b}^2), sqrt(a^{2+b}^2)]
• a*$sin^2$(x) + b*sin(x)cos(x) + c$cos^2$(x): rewrite using double angle formulas, then find range

Method: For f(x) = Asin(x) + Bcos(x), write as R*sin(x + phi) where R = sqrt(A^{2+B}^2). Max = R, min = -R.

Optimization on [0, 2pi]: sin(x): max = 1 at pi/2, min = -1 at 3pi/2 cos(x): max = 1 at 0, min = -1 at pi sin(x) + cos(x) = sqrt(2)*sin(x+pi/4): max = sqrt(2), min = -sqrt(2)

JEE Tip: For expressions like sin(x)*cos(x) = sin$\frac{2x}{2}$, the max is 1/2 and min is -1/2.

For a*$sin^2$(x) + b*$cos^2$(x) = a + (b-a)*$cos^2$(x). Since $cos^2$(x) in [0,1], the range is [min(a,b), max(a,b)].