Q.

Let f:R{0}R be a function such that f(x)6f(1x)=353x52. If the limx0(1αx+f(x))=β; α,βR, than α+2β is equal to          [2025]

1 4  
2 3  
3 5  
4 6  

Ans.

(1)

We have, f(x)6f(1x)=353x52           ... (i)

Apply x1x, then

f(1x)6f(x)=35x352           ... (ii)

Using (i) and (ii), we get f(x)=2x13x+12

Now, β=limx0(1αx+f(x))

=limx0(1αx2x13x+12)

=[limx01x(1α13)]+12  α=3, β=12

Hence, α+2β=3+2×12=4