Question

Solve the Question in the attachment ..

​

Answer 1

Step-by-step explanation:

$\sf f(x) = \frac{x}{\sqrt{1+{x}^{2}}} \\ \\ \sf {f}^{n}(x) = f f f f f … n times (x) \\ \\ \sf Let\:us\:find\:{f}^{2}(n) \\ \\ \sf {f}^{2}(x) = f (f(x)) \\ \\ \sf = \frac{ f(x) }{\sqrt{1 + {f(x)}^{2}}} \\ \\ \sf = \frac { \frac{x}{\sqrt{1+{x}^{2}}}}{ \sqrt{1+\frac{{x}^{2}}{1 +{x}^{2}}}} \\ \\ \sf = \frac {x}{(\sqrt{1+{x}^{2}})(\frac{\sqrt{1 +{x}^{2}+{x}^{2}}}{\sqrt{1+{x}^{2}}})} \\ \\ \sf = \frac{x}{\sqrt{1+2{x}^{2}}} \\ \\ \sf Now\:let\:us\:find \: f(f(f(x))) \\ \\ \sf {f}^{3}(x) = f(f(f(x))) \\ \\ \sf =\frac{ f(x) }{\sqrt{1 + 2{f(x)}^{2}}} \\ \\ \sf = \frac { \frac{x}{\sqrt{1+{x}^{2}}}}{ \sqrt{1+\frac{2{x}^{2}}{1 +{x}^{2}}}} \\ \\ \sf = \frac {x}{(\sqrt{1+{x}^{2}})(\frac{\sqrt{1 +{x}^{2}+2{x}^{2}}}{\sqrt{1+{x}^{2}}})} \\ \\ \sf = \frac{x}{\sqrt{1+3{x}^{2}}} \\ \\ \sf Hence\:we\:can\:generalise \: the\:result \\ \\ \sf {f}^{n}(x) = \frac{x}{\sqrt{1+n{x}^{2}}} \\ \\ \sf So\:we\:have\:to\:find\:out\:{f}^{99}(1) \\ \\ \sf {f}^{99}(x) =\frac{x}{\sqrt{1+99{x}^{2}}} \\ \\ \sf {f}^{99}(1) = \frac{1}{\sqrt{1+99({1}^{2})}} \\ \\ \sf =\frac{1}{\sqrt{1+99}} \\ \\ \sf = \frac{1}{\sqrt{100}} \\ \\ \sf = \frac{1}{10} \\ \\ \sf So\:your\:answer\:is\:\frac{1}{10}$

❣️NIRANJAN45❣️

Answer 2

Answer:

please mark me as brainlist