Question

Find the extreme values of 1)cos2x+cos²x
2)3sin²X+5cos²x

Answer 1

Answer:

cos2x+cos²x extreme values -1 & 2

3sin²x+5cos²x extreme values 3 & 5

Step-by-step explanation:

1)

cos2x+cos²x

= 2cos²x - 1 + cos²x

= 3cos²x - 1

- 1≤ Cosx ≤ 1

0 ≤ cos²x ≤ 1

Extreme values

= 3 * 1 - 1 = 2

& 3 * 0 - 1 =  - 1

2)3sin²x+5cos²x

= 3sin²x + 5 - 5sin²x

= 5 - 2sin²x

0 ≤ Sin²x ≤ 1

5 - 2* 0 = 5

5 - 2*1 = 3

Extreme values are  3 & 5

Answer 2

Answer:

1)    -1 and 2

2)    3 and 5

Step-by-step explanation:

1)  cos2x+cos²x

It can be written as;

= 2cos²x - 1 + cos²x

= 3cos²x - 1

Since, the range of cosx belongs to;

- 1≤ Cosx ≤ 1

So,

0 ≤ cos²x ≤ 1

Now, we have to find the Extreme values of the above function;

= 3 * 1 - 1 = 2

&

3 * 0 - 1 =  - 1

∴ The two extreme values of the above function are -1 and 2.

2)3sin²x+5cos²x

It can be written as;

= 3sin²x + 5 - 5sin²x

= 5 - 2sin²x

Since the range of sinx belongs to

0 ≤ Sin²x ≤ 1

Now, we have to find the Extreme values of the above function;

5 - 2* 0 = 5

&

5 - 2*1 = 3

Extreme values are  3 & 5

∴ The two extreme values of the above function are 3 and 5.