Solution
Related Formula
For functional equations with inversion, substituting x → (1)/(x) establishes a solvable system of algebraic equations to isolate f(x) directly.
Core Logic
The given equation is:
f(x) - 6f((1)/(x)) = (35)/(3x) - (5)/(2) (1)Substitute x → (1)/(x) in equation (1):
f((1)/(x)) - 6f(x) = (35x)/(3) - (5)/(2) (2)Step 1: Eliminate f(1/x)
Multiply equation (2) by 6 and add it to equation (1):
[ f(x) - 6f((1)/(x)) ] + 6 [ f((1)/(x)) - 6f(x) ] = ( (35)/(3x) - (5)/(2) ) + 6 ( (35x)/(3) - (5)/(2) ) f(x) - 36f(x) = (35)/(3x) - (5)/(2) + 70x - 15 -35f(x) = 70x + (35)/(3x) - (35)/(2)Divide across by -35:
f(x) = -2x - (1)/(3x) + (1)/(2)Step 2: Evaluate the Limit
We are given that the following limit evaluates to a finite constant β:
x → 0 ( (1)/(α x) + f(x) ) = β x → 0 ( (1)/(α x) - 2x - (1)/(3x) + (1)/(2) ) = β x → 0 ( [ (1)/(α) - (1)/(3) ] (1)/(x) - 2x + (1)/(2) ) = βFor the limit to be a finite value, the coefficient of (1)/(x) must vanish completely:
(1)/(α) - (1)/(3) = 0 α = 3When α = 3, the limit simplifies directly to the constant term:
β = x → 0 ( -2x + (1)/(2) ) = (1)/(2)Step 3: Calculate Final Value
Substitute the determined parameters α and β:
α + 2β = 3 + 2((1)/(2)) = 3 + 1 = 4Pattern Recognition
In limit problems involving fractional components where x → 0, any term like (1)/(x) or higher negative powers must have a net coefficient of zero to guarantee existence of a finite limit value.
Chapter Mix
Class 11 Mathematics: Functions Class 11 Mathematics: Limits and Derivatives