1.Relation and Function
hard

ધારો કે $x \ge - 1$ માટે વિધેય $f(x) = {(x + 1)^2}$ આપેલ છે. જો $g(x)$ એ વિધેય છે કે  જેનો આલેખએ વિધેય $f(x)$ ના આલેખનું રેખા $y = x$ ની સાપેક્ષ પ્રતીબિંબ હોય તો , $g(x)$ મેળવો.

A

$ - \sqrt x - 1,\;x \ge 0$

B

$\frac{1}{{{{(x + 1)}^2}}},\;x > - 1$

C

$\sqrt {x + 1} ,\;x \ge - 1$

D

$\sqrt x - 1,\;x \ge 0$

(IIT-2002)

Solution

(d) The graph of $f(x)$ has the equation $y = {(x + 1)^2},\,\,x \ge – 1$.
The reflection of the graph
of $f(x)$ is obtained by
interchanging $x,\,y$. So,
the graph of $g(x)$ has the
equation $x = {(y + 1)^2},\,\,y \ge – 1$
$\therefore$ $y = \sqrt x – 1$ because $y \ge – 1$
$\therefore$ $\phi \,(x) = \sqrt x – 1,\,\,\,x \ge 0$.

Standard 12
Mathematics

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