QJEEM23Ma4131: JEE Main 2023 Mathematics - Class 12 JEEM23Ma4131 Solution: For the differentiable function , let , then is equal to

1. Given the equation: $3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$

2. Replace $x$ with $\frac{1}{x}$ in the original equation:
   $3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$

3. Now, we have two equations:

$3f(x) + 2f\left(\frac{1}{x}\right) = \frac{1}{x} - 10$

$3f\left(\frac{1}{x}\right) + 2f(x) = x - 10$

1. By adding the two equations, we can find $f(x)$ :

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

1. Now, let's differentiate both sides with respect to $x$ :

$5f'(x) = -\frac{3}{x^2} - 2$

1. Now, we can find the values for $f(3)$ and $f'\left(\frac{1}{4}\right)$ :

$f(3) = \frac{1}{5}(1 - 6 - 10) = -3$

$f'\left(\frac{1}{4}\right) = \frac{1}{5}(-48 - 2) = -10$

1. Finally, calculate the expression we are interested in :

$\left|f(3) + f'\left(\frac{1}{4}\right)\right| = \left|-3 - 10\right| = 13$