The sum of all real values of x satisfying the equation `(x^2-5x+5)^(x^2+4x-60)=1` is: (A) 5 (B) 3 (C) -4 (D) 6

Options

A. 3
B. `-4`
C. 6
D. 5

Explanation

Given, `I,(x^(2)-5x+5)^(x^(2)+4-60)=1`
Clearly, this is possible when
`I.x^(2)+4x-60=0and x^(2)-5x+5 ne0 or`
`II. x^(2)-5x+5=1`
`III x^(2)-5x+5=-1andx^(2)+4x-60=` Even integer.
Case I When `x^(2)+4x-60=0`
`implies x^(2)+10x-6x-60=0`
`implies x(x+10)-6(x+10)=0`
`impliesx=-10 or x=6`
Note that , for these two values of `x, x^(2)-5x+ 5 ne 0`
Case II When `x^(2)-5x+5=1`
`implies x^(2) -5x+4=0`
`implies x^(2)-4x-x+4=0`
`implies x(x-4)-1(x-4)=0`
`implies (x-4)(x-1)=0impliesx=4 or x=1`
Case III When `x^(2)+5=-1`
`impliesx^(2)-5x+6=0`
`implies x^(2)-2x-3x+6=0`
`implies x(x-2)-3(x-2)=0`
`implies (x-2)(x-3) =0`
`impliesx=2or x=3`
Now, when `x=x^(2)+4x-60=4+8-60=-48,` which is an even intege.
When `x=3, x^(2)+4x-60=9+12-60=-39,` which not an even integer.
Thus in this case, we get `x=2.`
Hance, the sum of all real values of
`x=-10+6+4+1+2=3`