Length-of-interval-of-range-of-function-f-cos-2-6-sin-cos-3-sin-2-2-is-1-8-2-8-3-10-4-2-10-

Question Number 20653 by Tinkutara last updated on 30/Aug/17

Length of interval of range of function

f (θ) = \cos^{2} θ - 6 \sin θ \cos θ + 3 \sin^{2} θ + 2

is

(1) 8

(2) - 8

(3) \sqrt{10}

(4) 2 \sqrt{10}

Answered by dioph last updated on 31/Aug/17

f (θ) = \cos^{2} θ - 6 \sin θ \cos θ + {3 \sin}^{2} θ + 2

\Rightarrow f (θ) = 1 + 2 \sin^{2} θ - 3 \sin 2 θ + 2

\Rightarrow f (θ) = 1 - \cos 2 θ - 3 \sin 2 θ + 3

\Rightarrow f (θ) = 4 - \cos 2 θ - 3 \sin 2 θ

f^{'} (θ) = 2 \sin 2 θ - 6 \cos 2 θ

f^{'} (θ) = 0 \Leftrightarrow \sin 2 θ = 3 \cos 2 θ

\Leftrightarrow {\begin{cases} \sin 2 θ = 3 / \sqrt{10} \\ \cos 2 θ = 1 / \sqrt{10} \end{cases} or {\begin{cases} \sin 2 θ = - 3 / \sqrt{10} \\ \cos 2 θ = - 1 / \sqrt{10} \end{cases}

f (θ) = 4 - \frac{1}{\sqrt{10}} - \frac{9}{\sqrt{10}} = 4 - \sqrt{10} (\min)

f (θ) = 4 + \frac{1}{\sqrt{10}} + \frac{9}{\sqrt{10}} = 4 + \sqrt{10} (\max)

r a n g e = 2 \sqrt{10}

Commented by dioph last updated on 31/Aug/17

I am sorry, I forgot to add 1, it is

corrected

Answered by mrW1 last updated on 31/Aug/17

f (θ) = \cos^{2} θ - 6 \sin θ \cos θ + 3 \sin^{2} θ + 2

f (θ) = 3 - 3 \sin 2 θ + 2 \sin^{2} θ

f (θ) = 4 - (3 \sin 2 θ + \cos 2 θ)

f (θ) = 4 - \sqrt{10} (\frac{3}{\sqrt{10}} \sin 2 θ + \frac{1}{\sqrt{10}} \cos 2 θ)

f (θ) = 4 - \sqrt{10} (\cos α \sin 2 θ + \sin α \cos 2 θ)

f (θ) = 4 - \sqrt{10} \sin (2 θ + α)

\max . f (θ) = 4 + \sqrt{10}

\min . f (θ) = 4 - \sqrt{10}

\max . - \min . = 2 \sqrt{10}

\Rightarrow answer (4)

Commented by Tinkutara last updated on 31/Aug/17

Thank you very much Sir!