Math, asked by Aniketsaini345, 1 month ago

Find the range of f(x)=2 sin⁸x-3 sin⁴ x +2.​

Answers

Answered by readygovindbhai
2

Answer:

Range = 7/8  to 2

Step-by-step explanation:

Find the range of f(x)=2 sin^(8)x-3 sin^(4)x+2​

f(x) = 2Sin⁸x - 3Sin⁴x +2

= 2Sin⁸x - 3Sin⁴x + 9/8  - 9/8 + 2

= 2Sin⁸x - (3/2)Sin⁴x - (3/2)Sin⁴x + 9/8  + 7/8

= Sin⁴x(2Sin⁴x - 3/2) - (3/4)(Sin⁴x - 3/2) + 7/8

= (2Sin⁴x - 3/2)( Sin⁴x - 3/4) + 7/8

= 2( Sin⁴x - 3/4)( Sin⁴x - 3/4) + 7/8

= 2( Sin⁴x - 3/4)² + 7/8

( Sin⁴x - 3/4)² min value = 0

f(x) = 7/8  min

Sin⁴x lies betwen 0 & 1

( Sin⁴x - 3/4)²  max when x = 0

=> 9/16

f(x) = 2*9/16  + 7/8   (max)

Range = 7/8  to 2

f(x) = 2

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