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Grade: 12th pass

# Question number 19: if [5sinx]+[cosx]+6=0, then range of f(x)=3^1/2 cosx + sinx corresponding to solution set of the given equation is: a) [-2,-1) b)(-(3(3^1/2)+2)/5, -1) c) [-2,-3^1/2) d)( -(3 (3^1/2) + 4)/5 , 1)

one year ago

## Answers : (4)

Vikas TU
13784 Points

The value of [5Sinx] + [cosx] must be equal to -6
5Sinx must be equal to -5 and [cosx] must be equal to -1
-1
and  -1
Find the intersection of both these solutions , and
then Put this vaue in the f(x) which you want to find the range .
one year ago
Vikas TU
13784 Points

You need to find the intersection of solution of
-1
and
-1
Take intersection of both the equation and put the value in f(x)
one year ago
Saurabh Koranglekar
askIITians Faculty
10233 Points

one year ago
Vikas TU
13784 Points

I think there is problem in software , the thing i am typing is not uploading , well
You need to find the intersection of solution of
-1 less than or equa to sinx less than or equal to -4/5
and -1 less than or equal to cosx less than or equal to 0 .
one year ago
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