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The value of y for which the equation 4sinx+3cosx=y^2-6y+14 has a real solution is:
3
5
-3
4
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
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
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