Question

What is the value of f?
Given, $f=z+\frac{2y}{(x+\frac{5}{x})z}+\frac{2}{3y}$; $x=5;y=2;z=1$

• A. $f=3$
• B. $f=1$
• C. $f=-1$
• D. $f=2$

Answer 1

The correct option is D $f=2$

On substituting the values of $x,y,z$ in the given equation
$f=z+\frac{2y}{(x+\frac{5}{x})z}+\frac{2}{3y}$

$\Rightarrow f=1+\frac{2\times 2}{(5+\frac{5}{5})\times 1}+\frac{2}{3\times 2}$

$\Rightarrow f=1+\frac{4}{6}+\frac{1}{3}$

​​​​​​$\Rightarrow f=1+\frac{2}{3}+\frac{1}{3}$

​$\Rightarrow f=2$
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