If `8f(x)+6f(1/x)=x+5` and `y=x^2(f(x),` then `(dy)/(dx)` at `x=-1` is equal to 0 (b) `1/(14)` (c) `-1/4` (d) None of these

Options

A. 0
B. `(1)/(14)`
C. `-(1)/(14)`
D. 1

Answer

C

Explanation

We have ,
`8f(x)+6f((1)/(x))=x+5`
for all x
Therefore , `8f((1)/(x))+6f(x)=(1)/(x)+5`
From Eqs.(i) and (ii) , we get `f(x)=(1)/(28)(8x-(6)/(x)+10)`
Now , `y=x^2f(x)`
`rArr y=(1)/(28)(8x^3-6x+10x^2)`
`rArr (dy)/(dx)=(1)/(28)(24x^2+20x-6)`
`therefore ((dy)/(dx))_(x=-1)=(1)/(28)(24-20-6)`
`=-(1)/(14)` .