Let $f$ be a function such that $f(x)+3 f\left(\frac{24}{x}\right)=4 x, x \neq 0$. Then $f(3)+f(8)$ is equal to
Solution
<p>$$\begin{aligned}
& f(x)+3 f\left(\frac{24}{x}\right)=4 x, x \neq 0 \quad \ldots(1)\\
& \text { replace } x \text { by } \frac{24}{x} \\
& f\left(\frac{24}{x}\right)+3 f\left(\frac{24}{24}\right)=4\left(\frac{24}{x}\right)=\frac{96}{x} \quad \ldots(2) \\
& 3 \times(2)-(1) \\
& \Rightarrow 8 f(x)=\frac{96.3}{x}-4 x \Rightarrow f(x)=\frac{36}{x}-\frac{x}{2} \\
& f(3)+f(8)=\left(12-\frac{3}{2}\right)+\left(\frac{36}{8}-4\right)
\end{aligned}$$</p>
<p>$=8+\frac{36}{8}-\frac{12}{8}=11$</p>
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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