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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

7 years ago

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

7 years ago
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

7 years ago
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