How-to-find-maximum-value-of-k-if-5-cos-2-sin-2k-0-pi-

Question Number 14130 by Joel577 last updated on 28/May/17

How to find maximum value of k if

\frac{5 - \cos 2 θ}{\sin θ} ⩾ 2 k 0 ⩽ θ ⩽ π

Commented by Joel577 last updated on 28/May/17

\frac{5 - \cos 2 θ}{\sin θ} ⩾ 2 k

\frac{4 + {2 \sin}^{2} θ}{\sin θ} ⩾ 2 k

\sin θ is minimum when θ = \frac{π}{2}

so k = 3

Am I right ?

Commented by prakash jain last updated on 28/May/17

\sin θ is actually maximum when

θ = \frac{π}{2}

Commented by prakash jain last updated on 28/May/17

f (θ) = \frac{4 + {2 \sin}^{2} θ}{\sin θ}

f^{'} (θ) = \cos θ (2 - {4 cosec}^{2} θ)

f^{'} (θ) = 0

a t θ = \frac{π}{2}

f^{″} (θ) = - 2 \sin θ + {8 cosec}^{3} θ - 4 cosec θ

f^{″} (\frac{π}{2}) = - 2 + 8 - 4 > 0

f (\frac{π}{2}) is local minima for f (θ)

f (\frac{π}{2}) = \frac{4 + 2}{1} = 6 \Rightarrow k = 3

Commented by Joel577 last updated on 29/May/17

t h a n k y o u v e r y m u c h

Answered by ajfour last updated on 28/May/17

4 + (1 - \cos 2 θ) ⩾ 2 k \sin θ

4 + 2 \sin^{2} θ - 2 k \sin θ ⩾ 0

\sin^{2} θ - k \sin θ + 2 ⩾ 0

\Rightarrow f o r θ = 0, π; k = a n y r e a l n u m b e r

\Rightarrow k ⩽ \frac{2 + \sin^{2} θ}{\sin θ}

l e t f (θ) = \frac{2 + \sin^{2} θ}{\sin θ}

f^{'} (θ) = \frac{2 \sin^{2} θ \cos θ - \cos θ (2 + \sin^{2} θ)}{\sin^{2} θ}

= \frac{\cos θ (\sin^{2} θ - 2)}{\sin^{2} θ}

f^{'} (θ) > 0 f o r \frac{π}{2} < θ < π

w h i l e f^{'} (θ) < 0 f o r 0 < θ < π

l o c a l m i n i m a a t x = \frac{π}{2}

\Rightarrow f (θ) i s m i n i m u m f o r f^{'} (θ) = 0

o r \cos θ = 0, θ = \frac{π}{2} .

f (\frac{π}{2}) = \frac{2 + 1}{1} = 3

k_{m a x} = m i n i m u m o f f (\frac{π}{2})

k_{m a x} = 3 .

Commented by prakash jain last updated on 28/May/17

Please see my comment I think

minimum is 6 .

Commented by ajfour last updated on 28/May/17

t h a n k y o u M r . P r a k a s h .

Commented by Joel577 last updated on 29/May/17

t h a n k y o u v e r y m u c h

Facebook Tweet Pin