f ( x ) is a differentiable function satisfy the relationship f 2 ( x ) + f 2 ( y ) + 2 ( x y - 1 ) = f 2 ( x + y ) x , y R . Also f ( x ) 0 ? ? x R , and f ( 2 ) = 2 . then f ( 7 ) =

Question

$f (x)$ is a differentiable function satisfy the relationship $f^{2} (x) + f^{2} (y) + 2 (x y - 1) = f^{2} (x + y)$ $\forall x, y \in R .$ Also $f (x) > 0$ $\forall x \in R$ , and $f (\sqrt{2}) = 2$ . then $f (\sqrt{7}) =$

Smart Achievers · Accepted Answer

(1)

$Put x = 0 and y = 0 \Rightarrow f^{2} (0) = 2$

$f' (x) = \lim_{h \to 0} \frac{f (x + h) - f (x)}{h} = \lim_{h \to 0} \frac{f^{2} (x + h) - f^{2} (x)}{[f (x + h) + f (x)] h}$

$= \frac{1}{2 f (x)} \lim_{h \to 0} \frac{f^{2} (h) + 2 (x h - 1)}{h}$ $= \frac{1}{2 f (x)} \lim_{h \to 0} [\frac{2 x h}{h} + \frac{f^{2} (h) - 2}{h}]$

$= \frac{1}{2 f (x)} [2 x + \lim_{h \to 0} \frac{f^{2} (h) - f^{2} (0)}{h}]$ $= \frac{1}{2 f (x)} [2 x + \lim_{h \to 0} \frac{f (h) - f (0)}{h} (f (h) + f (0))]$

$f' (x) = \frac{1}{2 f (x)} [2 x + f' (0) \cdot 2 f (0)]$

$∴$ $f (x) \cdot f' (x) = x + f (0) \cdot f' (0)$

$\Rightarrow f (x) \cdot f' (x) = x + λ,$ $where λ = f (0) \cdot f' (0)$

$Integrating both sides$

$∴$ $\frac{f^{2} (x)}{2} = \frac{x^{2}}{2} + λ x + C$

$f^{2} (x) = x^{2} + 2 λ x + C$ $At x = 0, f^{2} (0) = 2 \Rightarrow C = 2$

$x = \sqrt{f^{2} (\sqrt{2})} \Rightarrow λ = 0$

$∴$ $f^{2} (x) = x^{2} + 2$

$\Rightarrow f (x) = \sqrt{x^{2} + 2}, (f (x) > 0)$

Solution

Q.
$f (x)$ is a differentiable function satisfy the relationship $f^{2} (x) + f^{2} (y) + 2 (x y - 1) = f^{2} (x + y)$ $\forall x, y \in R .$ Also $f (x) > 0$ $\forall x \in R$ , and $f (\sqrt{2}) = 2$ . then $f (\sqrt{7}) =$

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Solution

Q. f(x) is a differentiable function satisfy the relationship f2(x)+f2(y)+2(xy-1)=f2(x+y) ∀x,y∈R. Also f(x)>0 ∀ x∈R, and f(2)=2. then f(7)=

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4 7

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Q.
$f (x)$ is a differentiable function satisfy the relationship $f^{2} (x) + f^{2} (y) + 2 (x y - 1) = f^{2} (x + y)$ $\forall x, y \in R .$ Also $f (x) > 0$ $\forall x \in R$ , and $f (\sqrt{2}) = 2$ . then $f (\sqrt{7}) =$