1.

If x = 3sin2Ɵ + 2 and y = 3cos2Ɵ + 1, then find x + y.​

Answer»

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual TWO DISTINCT SOLUTIONS, then the two solutions must be same i.e. discriminant = 0.

Hence we GET that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.


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