Let f(x) = {(x + 1)^2} - 1,;;(x ge - 1). Then | vedclass

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(d) Let $f(x) = {(x + 1)^2} - 1,\,\,x \ge - 1.$ Since $f(x) = {f^{ - 1}}(x)$

$	herefore \,\,{(x + 1)^2} - 1 = \sqrt {1 + x} - 1$

$ \Rightarrow

Vedclass · Accepted Answer

$\left\{ {0,\; - 1,\;\frac{{ - 3 + i\sqrt 3 }}{2},\;\frac{{ - 3 - i\sqrt 3 }}{2}} \right\}$

Vedclass · Answer

(d) Let $f(x) = {(x + 1)^2} - 1,\,\,x \ge - 1.$ Since $f(x) = {f^{ - 1}}(x)$

$	herefore \,\,{(x + 1)^2} - 1 = \sqrt {1 + x} - 1$

$ \Rightarrow \,\,{(x + 1)^4} = 1 + x\,\, \Rightarrow \,\,(x + 1)\,\,[{(x + 1)^3} - 1] = 0$

$ \Rightarrow \,\,x = - 1$ or ${(x + 1)^3} = 1\, \Rightarrow x + 1 = 1,\,\,\omega ,\,\,{\omega ^2}$

$ \Rightarrow \,\,x = 0,\,\, - 1,\,\frac{{ - 3 + i\sqrt 3 }}{2},\,\,\frac{{ - 3 - i\sqrt 3 }}{2}.$

Let $f(x) = {(x + 1)^2} - 1,\;\;(x \ge - 1)$. Then the set $S = \{ x:f(x) = {f^{ - 1}}(x)\} $ is

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