Question

Example-19)
Find maximum and minimum values of following:
(1) 1+2sinx + 3cos^2x​

Answer 1

Step-by-step explanation:

range of cos²x

0≤cos²x≤1

0≤3cos²x≤3......(1)

range of sin x

-1≤sinx≤1

-2≤2sinx≤2......(2)

adding (1) and (2)

-2≤2sinx+3cos²x≤5

adding 1

-1≤1+2sinx+3cos²x≤6

max value=6

min value=-1

note: don't trust this solution fully

Answer 2

Answer:

y=1+2sinx+3cos²x

y=1+2sinx+3(1-sin²x)

y=1+2sinx+3-3sin²x

y=1-(3sin²x-2sinx-3)

taking 3 common;

y=1-3(sin²x-2sinx/3 -1)

using the completing square method

y=1-3(sin²x-sinx+1/9-1/9-1)

y=1-3[(sinx-1/3)²-10/9]

y=-3(sinx-1/3)²+13/3

so

Ymaxi.=13/3

Ymini.=-3(16/9)+13/3=-1

Ymaxi- Ymini=13/3 -(-1)

=13/3+1

=16/3

I hope it will help you