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