Given, f(x)=∣aaxax2−1aax0−1a∣
f(x)=a∣1xx2−1aax0−1a∣
=a[1(a2+ax)+1(ax+x2)]
⇒f(x)=a(x+a)2
Now differentiating the function w.r.t x
We get, f′(x)=2a(x+a)
Given, 2f′(10)−f′(5)+100=0
⇒2×2a(10+a)−2a(5+a)+100=0
⇒40a+4a2−10a−2a2+100=0
2a2+30a+100=0
⇒a2+15a+50=0
(a+10)(a+5)=0
a=−10 or a=−5
Required =(−10)2+(−5)2=125