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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)

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)

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Grade:12th pass

4 Answers

Vikas TU
14149 Points
4 years ago
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 .
Vikas TU
14149 Points
4 years ago
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)
Saurabh Koranglekar
askIITians Faculty 10335 Points
4 years ago
576-2341_15647585461981587610160.jpg
Vikas TU
14149 Points
4 years ago
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 .

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