If-a-gt-0-b-gt-0-and-the-minimum-value-of-a-sin-2-b-cosec-2-is-equal-to-maximum-value-of-a-sin-2-b-cos-2-then-a-b-is-equal-to-Answer-4-

Question Number 16179 by Tinkutara last updated on 24/Jun/17

If a > 0, b > 0 and the minimum

value of a \sin^{2} θ + b {cosec}^{2} θ is equal to

maximum value of a \sin^{2} θ + b \cos^{2} θ,

then \frac{a}{b} is equal to [Answer : 4]

Answered by ajfour last updated on 24/Jun/17

let \sin^{2} θ = t

f (t) = at + \frac{b}{t} is minimum when

f^{'} (t) = a - \frac{b}{t^{2}} = 0 \Rightarrow t_{0} = \sqrt{\frac{b}{a}}

f (t_{0}) = 2 a (\sqrt{\frac{b}{a}}) = 2 \sqrt{ab}

g (t) = at + b (1 - t) = (a - b) t + b

here i need to know if a > b or not

if a > b maximum of g (t) is

= (a - b) (1) + b = a

since \min . of f (t) = \max . of g (t)

2 a (\sqrt{\frac{b}{a}}) = a

or \frac{a}{b} = 4 (a > b)

but if a < b then

\max . of g (t) = (a - b) (0) + b = b

(remember t = \sin^{2} θ)

then from given condition

2 a (\sqrt{\frac{b}{a}}) = b

or \sqrt{\frac{b}{a}} = 2 \Rightarrow \frac{a}{b} = \frac{1}{4} (a < b) .

Commented by Tinkutara last updated on 24/Jun/17

Thanks Sir! I also got confused with

4 and \frac{1}{4} .