If f(x)=aloge⁡|x|+bx2+x has extremums at x=1 and x=3 , then

at x=1,3 we have |x|=x∴f(x)=aloge⁡x+bx2+x∴f′(x)=ax+2bx+1 According to question, f′(1)=0,f′(3)=0∴a+2b+1=0,a3+6b+1=0 On solving, we get a=−34,b=−18