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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)`
5 months ago

```							The value of [5Sinx] + [cosx] must be equal to -6 5Sinx must be equal to -5 and [cosx] must be equal to -1 -1and  -1 Find the intersection of both these solutions , and then Put this vaue in the f(x) which you want to find the range .
```
5 months ago
```							You need to find the intersection of solution of -1 and -1Take intersection of both the equation and put the value in f(x)
```
5 months ago Saurabh Koranglekar
3304 Points
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5 months 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 .
```
5 months ago
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