Mathematics - Functions Question with Solution | TestHub

MathematicsFunctionsFunctional EquationMedium2 minQB

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Passage / Comprehension

The function $f (x)$ is defined for all $x > 0$ and satisfies the conditions:

(i) $f (x)$ is strictly increasing on $(0, \infty)$

(ii) $f (x) > x - \frac{1}{x}$ for all $x > 0$

(iii) $f (x + \frac{1}{x}) = f (x) + f (\frac{1}{x})$ for all $x > 0$ .

The value of $f (1)$ is

Answer:

Solution:

Let $f (1) = k$ . For $x = 1$ , we have:

$k f (k + 1) = 1 \Rightarrow f (k + 1) = \frac{1}{k}$

Now, $x = k + 1$ gives:

$f (k + 1) (f (\frac{1}{k + 1} + f (k + 1))) = 1$

$\Rightarrow f (\frac{1}{k} + \frac{1}{k + 1}) = \frac{1}{f ( k + 1 )} = k = f (1)$

Since $f$ is increasing, we have:

$\frac{1}{k} + \frac{1}{k + 1} = 1 \Rightarrow k = \frac{1 \pm 5}{2}$

By contradiction, we can say that $k$ cannot be positive, so $k = \frac{1 - 5}{2}$ .

The generalized form is $f (x) = \frac{k}{x}$ .

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Functional Equation

⏱ 2mℹ️ Source: QB

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