- Home
- Standard 12
- Mathematics
1.Relation and Function
hard
माना $f(x) = {(x + 1)^2} - 1,\;\;(x \ge - 1)$, तब समुच्चय $S = \{ x:f(x) = {f^{ - 1}}(x)\} $ है
A
रिक्त
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) माना $f(x) = {(x + 1)^2} – 1,\,\,x \ge – 1$ चूँकि $f(x) = {f^{ – 1}}(x)$
$\therefore \,\,{(x + 1)^2} – 1 = \sqrt {1 + x} – 1$
$ \Rightarrow \,\,{(x + 1)^4} = 1 + x\,\, \Rightarrow \,\,(x + 1)\,\,[{(x + 1)^3} – 1] = 0$
$ \Rightarrow \,\,x = – 1$ या ${(x + 1)^3} = 1\, \Rightarrow x + 1 = 1,\,\,\omega ,\,\,{\omega ^2}$
$ \Rightarrow \,\,x = 0,\,\, – 1,\,\frac{{ – 3 + i\sqrt 3 }}{2},\,\,\frac{{ – 3 – i\sqrt 3 }}{2}.$
Standard 12
Mathematics