Find-the-range-of-real-number-of-q-such-that-the-function-f-x-cos-x-q-sin-2-x-5-have-minimum-value-is-5-

Question Number 141412 by bramlexs22 last updated on 18/May/21

F i n d t h e r a n g e o f r e a l n u m b e r

o f q s u c h t h a t t h e f u n c t i o n

f (x) = \cos x (q \sin^{2} x - 5) h a v e

m i n i m u m v a l u e i s - 5 .

Commented by MJS_new last updated on 18/May/21

I get - \frac{5}{2} ⩽ q ⩽ 20 but I^{'} ve got no time to type

the path right now

Commented by bramlexs22 last updated on 18/May/21

h o w s i r ?

Answered by MJS_new last updated on 18/May/21

f (x) = (q - 5 - q \cos^{2} x) \cos x

f^{'} (x) = (5 - q + 3 q \cos^{2} x) \sin x = 0

\sin x = 0 \Rightarrow x = n π

f (n π) = \pm 5

5 - q - q \cos^{2} x = 0 \Rightarrow \cos^{2} x = \frac{q - 5}{3 q} \Rightarrow

\Rightarrow x = \pm \arccos \frac{\sqrt{q - 5}}{\sqrt{3 q}}

f (\pm \arccos \frac{\sqrt{q - 5}}{\sqrt{3 q}}) = \pm \frac{2 \sqrt{3 {(q - 5)}^{3}}}{9 \sqrt{q}}

- \frac{2 \sqrt{3 {(q - 5)}^{3}}}{9 \sqrt{q}} = - 5 \Rightarrow q = 20

+ \frac{2 \sqrt{3 {(q - 5)}^{3}}}{9 \sqrt{q}} = - 5 \Rightarrow q = - \frac{5}{2}

this might be wrong but {it}^{'} s what I did

Commented by bramlexs22 last updated on 19/May/21

f^{'} (x) = - \sin x (q - 5 - q \cos^{2} x) +

\cos x (2 q \cos x \sin x) = 0

\Rightarrow \sin x [q - 5 - q \cos^{2} x - 2 q \cos^{2} x] = 0