Let f ( 1 ) = k . For x = 1 , we have: k f ( k + 1 ) = 1 ⇒ f ( k + 1 ) = k 1 Now, x = k + 1 gives: f ( k + 1 ) ( f ( k + 1 1 + f ( k + 1 ) ) ) = 1 ⇒ f ( k 1 + k + 1 1 ) = f ( k + 1 ) 1 = k = f ( 1 ) Since f is increasing, we have: k 1 + k + 1 1 = 1 ⇒ k = 2 1 ± 5 By contradiction, we can say that k cannot be positive, so k = 2 1 − 5 . The generalized form is f ( x ) = x k .

Start Free Mock LoginAlready registered? Login here

Mathematics - Functions Question with Solution | TestHub

MathematicsFunctionsFunctional EquationMedium2 minQB

MathematicsMediumsingle choice

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 $x > 0$ .

$f^{'} (x)$ is a

Options:

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

Doubts & Discussion

Loading discussions...