whats the range of 1/(|sinx| + |cosx| )i dont want steps just ans
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Hello
Sorry dear its against the rules to just give answer so i have to explain it.
let f(x) = 1/|sinx|+|cosx|
so
f(x) = 1/|sinx| + 1/|cosx|
we know that AM(Arithmetic mean) is Greater than equal to GM(Geometric mean)
use it here .
=> 1/|sinx| + 1/|cosx|
______________ >= 1/(|sinx||cosx|)^1/2
2
so now
1/|sinx|+1/|cosx| >= 2(2cosec2x)^1/2
we know
|cosec2x|>= 1
so
1/sinx +1/cosx >= 2√2
hence range or
1/|sinx|+|cosx| € [2√2 , ●●)
hence range of f(x) is [2√2,●●) .
I m again saying sorry but its against the rules.
Sorry dear its against the rules to just give answer so i have to explain it.
let f(x) = 1/|sinx|+|cosx|
so
f(x) = 1/|sinx| + 1/|cosx|
we know that AM(Arithmetic mean) is Greater than equal to GM(Geometric mean)
use it here .
=> 1/|sinx| + 1/|cosx|
______________ >= 1/(|sinx||cosx|)^1/2
2
so now
1/|sinx|+1/|cosx| >= 2(2cosec2x)^1/2
we know
|cosec2x|>= 1
so
1/sinx +1/cosx >= 2√2
hence range or
1/|sinx|+|cosx| € [2√2 , ●●)
hence range of f(x) is [2√2,●●) .
I m again saying sorry but its against the rules.
seem1512:
its right now and u have to take reciprocal of it
ya
ive solved that que in his other id
Hey there, well nice answer, keep it up!
I would just like to tell you that it's not against the rule to give short answer, but users are expected to give long explanatory answers to questions just to increase brainly's content quality.
;)
Well thanks for your help and keep posting such answers
r u moderator!
Yes
inbox me pls
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