For the differentiable function f: ℝ-{0} → ℝ, let 3 f(x)+2 f(1/x)=1/x-10, then |f(3)+f′(1/4)| is equal to

Solution

Maths sir · 4 steps

Swap x with 1/x to get a second equation, then solve the pair like two linear equations.

1. Second equation. Replace x by 1/x in the given relation. This gives 3f(1/x) + 2f(x) = x - 10, so you now have two equations in f(x) and f(1/x).
2. Eliminate f(1/x). Multiply the original equation by 3 and the new one by 2, then subtract. This gives 5f(x) = 3/x - 2x - 10.

   f(x) = 1/5(3/x - 2x - 10), f'(x) = 1/5(-3/x² - 2)

3. Put in numbers. f(3) = 1/5(1 - 6 - 10) = -3. At x = 1/4, 1/x² = 16, so f'(1/4) = 1/5(-48 - 2) = -10.
4. So. |f(3) + f'(1/4)| = |-3 - 10| = 13, which is option (a).

The idea to keep. When an equation links f(x) and f(1/x), put 1/x in place of x and solve the two equations together. Find f(x) fully before you differentiate.

Answer from the NTA final answer key. Solution by PrepWiser.