1.

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

Answer» if `a^x = 1`
then 3 conditions are
`a= -1,0,1`
&`x=0 or in` even no
`a= 1 ; x in R`
`a=-1 ; x in `even no
`x=0, a=1`
now, `x^2 - 5x + 5 =1`
`x^2 - 5x + 4=0 `
`x^2 - 4x - x + 4 = 0`
`(x- 4)(x-1) = 0`
`x=4,1`
now,`x^2 - 5x + 5= -1`
`x^2 - 5x + 6= 0`
`x^2 - 2x - 3x + 6= 0`
`(x-2)(x-3) =0 `
`x= 2,3`
`x^2 + 4x - 60`
for ` x=2`
`4 + 8 - 60 = 48 in ` even number
for `x=3`
`9 + 12 - 60 = 30 cancel( in) ` even no, so not possible
`x^2 + 4x - 60 = 0`
`x^2 + 10x - 6x - 60 = 0`
`(x+10)(x-6) = 0`
`x= -10,6`
`x= 4,1,2,-10,6`
sum=`4+1+2-10+6 = 3`
option 4 is correctAnswer


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