JEE Main 2023 — Differentiation Question with Solution

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 $$x$$:

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

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

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

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

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

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