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Bunuel
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We see that the equation becomes 1 , this occurs in the following three conditions -

1) The power is 0

2) The base is 1 and the power can be anything

3) The base is -1 and the power is even

Case 1: When the power is 0

\(x^2 \)+ 8x = 0

x(x + 8) = 0

x = -8
or
x = 0

Case 2: The base is 1 and the power can be anything

\(x^2 - 5x + 5 = 1\)

\(x^2 - 5x + 4 = 0\)

\((x-4)(x-1) = 0 \)

x = 4 or x = 1

Case 3: The base is -1 and the power is even

\(x^2 - 5x + 5 = -1\)

\(x^2 - 5x + 6 = 0\)

\((x - 3)(x-2) = 0\)

x = 3
or
x = 2

But if we have x = 3, the power becomes odd, hence we have to eliminate this.

A = {-8, 0, 1, 2, 4}

Median = 1

IMO C
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If a data set A consists of all x such that \((x^2 - 5x + 5)^{(x^2 + 8x)} = 1\), then what is the median of A ?

A. -4
B. 0.5
C. 1
D. 1.5
E. 2.5

(x^2 - 5x + 5)^{(x^2 + 8x)} = 1
L.H.S will be 1 if either (x^2 - 5x + 5)=1 or x^2+8x=0
(x^2 - 5x + 5)=1 --- equation 1
x^2+8x=0 ----equation 2
Possible solution for equation 1
(x^2-5x+5)=1
=>x^2-5x+4=0
=>x^2-4x-x+4=0
=>x(x-4)-1(x-4)=0
=>(x-4)(x-1)=0
Therefore x=1,4
Again if (x^2-5x+5)=-1 and (x^2 + 8x) is even then also the value of the expression would be 1
(x^2-5x+5)=-1
=>x^2-5x+6=0
=>x^2-3x-2x+6=0
=>x(x-3)-2(x-3)=0
=>(x-3)(x-2)=0
Therefor x=2,3 For x=2 (x^2+8x) is even but for x=3 (x^2+8x) is odd
Thus the only solution here is x=2
Now x^2+8x=0
=>x(x+8)=0
There x=-8,0
All the values in set A= {-8,0,1,2,4}
Therefore the median is 1
Answer is C
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There are 3 cases where we can get RHS of 1.
- Power = 0
- Base = 1
- Base = -1 and power is an even number.

---
Harsha
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