The correct option is C 3
CASE 1:
x2+4x−60 can be any real number and x2−5x+5=1
i.e. x2−5x+4=0
⇒x=1,4
CASE 2:
x2+4x−60=0 and x2−5x+5 can be any real number except 0 as 00 is undefined.
i.e. (x+10)(x−6)=0
⇒x=6,−10
and x2−5x+5≠0
⇒x≠5±√52
CASE 3:
(−1)even=1
i.e. x2−5x+5=−1 and x2+4x−60 is an even number
⇒x2−5x+6=0
⇒(x−2)(x−3)=0
⇒x=2,3
At x=2, the value of x2+4x−60 is
(2)2+4(2)−60=−48
Which is even
At x=3, the value of x2+4x−60 is
(3)2+4(3)−60=−39
Which is odd
∴x=3 is rejected
Sum of all the real values of x=1+4+6−10+2=3