Question

Sum of maximum and minimum value of 2cos^2x+8sinx.cosx+2cos2x

Answer 1

Answer:

$1 + \cos(2x) + 4 \sin(2x) + 2 \cos(2x) =$

$4 \sin(2x) + 3 \cos(2x) + 1......(i)$

now,

$- \sqrt{ {4}^{2} + {3}^{2} } \leqslant 4 \sin(2x) + 3 \cos(2x) \leqslant \sqrt{ {4}^{2} + {3}^{2} }$

$- 5 \leqslant 4 \sin(2x) + 3 \cos(2x) \leqslant 5$

adding 1 both sides

$- 5 + 1\leqslant 4 \sin(2x) + 3 \cos(2x) + 1 \leqslant 5 + 1$

$- 4\leqslant 4 \sin(2x) + 3 \cos(2x) + 1 \leqslant 6$

max. value = 6, min. value = -4

required sum = 6 - 4 = 2 Ans.