AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

Consider the function
f(x)=a⋅log|x|+bx2+x
Differentiate with respect to x
f′(x)‌‌=a‌

1

|x|

⋅‌

x

|x|

+2bx+1
‌‌=‌

ax

|x|2

+2bx+1
It is given that x=−1 is one of the extremity of f(x)
f′(−1)=0
‌

a(−1)

1

+2b(−1)+1=0
−a−2b+1=0
Also,
f′(−2)‌‌=0
a‌

(2)

4

+2b(2)+1‌‌=0
a+8b+2‌‌=0
On solving both the equations,
a=2
b=‌

−1

2

Therefore,
(a,b)=(2,−‌

1

2

)