The maximum value of 3sin2Q + 4cos2Q is?

5 3 1 4

4

⇒ 3sin2Q+ 4cos2Q = 3sin2 Q + 4(1-sin2Q)      [ sin2Q + cos2Q = 1] = 4 - sin2Q  ⇒ The maximum value of eqn is when sin2Q is minimum i.e. When Q = 0, ⇒ sin20° = 0 ⇒ Therefore, 4 is the maximum value.