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Grade 12th passAlgebra

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 image for Question number 19: if [5sinx]+[cosx]+6=0, then r

Agrim Srivastava

6 Years agoGrade 12th pass

4 Answers

Vikas TU

6 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

6 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

6 Years ago

Vikas TU

6 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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