For the differentiable function f : - { 0 } , let 3 f ( x ) + 2 f ( 1 x ) = 1 x - 10 , then | f ( 3 ) + f ( 1 4 ) | is equal to [2023]

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Solution

Q.
For the differentiable function $f : ℝ - {0} \to ℝ, let 3 f (x) + 2 f (\frac{1}{x}) = \frac{1}{x} - 10, then$ $| f (3) + f' (\frac{1}{4}) |$ is equal to [2023]

1 13

2 7

3 $\frac{33}{5}$

4 $\frac{29}{5}$

Ans.
(1)

Given, $3 f (x) + 2 f (\frac{1}{x}) = \frac{1}{x} - 10 ...(i)$

Put $x = \frac{1}{x}$ in (i), we get $3 f (\frac{1}{x}) + 2 f (x) = x - 10 ...(ii)$

Multiply (i) by 3 and (ii) by 2, then (i) − (ii):

$5 f (x) = \frac{3}{x} - 2 x - 10 \Rightarrow f (x) = \frac{1}{5} (\frac{3}{x} - 2 x - 10)$

Differentiate it w.r.t. $x,$ $f' (x) = \frac{1}{5} (\frac{- 3}{x^{2}} - 2)$

$∴$ $| f (3) + f' (\frac{1}{4}) |$ $=$ $| \frac{1}{5} (\frac{3}{3} - 2 (3) - 10) + \frac{1}{5} (- 48 - 2) |$

$=$ $| \frac{1}{5} (1 - 6 - 10 - 50) |$ $=$ $| \frac{1}{5} (- 65) |$ $= 13$

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