First slide

Differentiability

Question

$If f : R \to R be a differentiable function such that$ $f (x + 2 y) = f (x) + f (2 y) + 4 x y \forall x, y \in R then f^{1} (1) - f^{1} (0) =$

Moderate

Solution

$f (x + 2 y) = f (x) + f (2 y) + 4 x y$
$\begin{array}{l} \Rightarrow \frac{f (x + 2 y) - f (x)}{2 y} = 2 x + \frac{f (2 y)}{2 y} \\ \Rightarrow \underset{2 y \to 0}{L t} \frac{f (x + 2 y) - f (x)}{2 y} = \underset{2 y \to 0}{L t} (2 x + \frac{f (2 y)}{2 y}) \\ \Rightarrow f^{1} (x) = 2 x + f^{1} (0) (as f (0) = 0) \\ \Rightarrow f^{1} (0) = f^{1} (1) - 2 \Rightarrow f^{1} (1) - f^{1} (0) = 2 \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App