જો f(x) = {(x + 1)^2} - 1,(x ge - 1) તો ગણ S = { x:f(x) = {f^{ - 1}}(x)} એ . . . .

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1.Relation and Function

hard

જો $f(x) = {(x + 1)^2} - 1,\;\;(x \ge - 1)$ તો ગણ $S = \{ x:f(x) = {f^{ - 1}}(x)\} $ એ . . . .

A

ખાલીગણ

B

$\{0, -1\}$

C

$\{0, 1, -1\}$

D

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

(IIT-1995)

Solution

(d) Let $f(x) = {(x + 1)^2} – 1,\,\,x \ge – 1.$ Since $f(x) = {f^{ – 1}}(x)$

$\therefore \,\,{(x + 1)^2} – 1 = \sqrt {1 + x} – 1$

$ \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}.$

Standard 12

Mathematics

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