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divya vikash Grade: 11
        

The value of y for which the equation 4sinx+3cosx=y^2-6y+14 has a real solution is:


3


5


-3


4

6 years ago

Answers : (2)

gOlU g3n|[0]uS
42 Points
										

ans)   4sinx + 3cosx = y2 - 6y + 14


        MAX and MIN value of 4sinx + 3cosx is 5 and -5.


        -5<= y2 - 6y + 19 <=5


         0<= y2 - 6y + 19 <=10   u get 2 eq-n


         y2 - 6y + 19<=10   and   y2 - 6y + 19>=0.


 ANS  3

6 years ago
Shivanshu Madan
16 Points
										

Using LHS, we get that


-5 <= y^2-6y+14<= 5


Case 1


y^2-6y+19>=0


This is true for all real y.


 


Case 2


y^2-6y+9<= 0


 


This eq can never be less than zero as coeff. of y is +ve.


Hence this eq is equal to 0.


i.e.


(y-3)^2=0


y=3

6 years ago
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