AP EAPCET 23-Aug-2021 Shift 1 Paper

8f(x)+6f(

1

x

)=x+5 ...(i)
y=x2f(x)
Replace x by

1

x

in Eq. (i), we obtain
8f(

1

x

)+6f(x)=

1

x

+5 ...(ii)
Solve Eqs. (i) and (ii) for f(x) and f(

1

x

)
Multiply by 8 in Eq. (i) and by 6 in Eq. (ii) and subtract, we get
64f(x)−36f(x)=8x−

6

x

+10
⇒28f(x)=8x−

6

x

+10
Now, 28x2f(x)=8x3−6x+10x2
⇒28y=8x3−6x+10x2
Differentiating it,
28

dy

dx

=24x2−6+20x
⇒

dy

dx

=

24x2−6+20x

28

(

dy

dx

)x=−1 =

24(−1)2−6+20(−1)

28

=−

1

14