If f(x)=aloge|x|+bx2+x has extremums at x=1 and x=3 , then
a=−34,b=−18
a=34,b=−18
a=−34,b=18
a=12,b=−1
at x=1,3 we have |x|=x∴f(x)=alogex+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